# How many acres is 200 feet by 200 feet?

Jerilyn Kalandek asked, updated on October 26th, 2022; Topic: 200 feet
👁 511 👍 64 ★★★★☆4.9

200 feet multiplying by 200 feet equal to 40,000 sq ft. Therefore, 40,000 sq ft divide by 43,560 sq ft equal to 0.92 acres approximately.

Either way, what is the perimeter of a 1/4 acre?

There are 43560 square feet in an acre so one quarter of an acre is square feet. If each side of the square is F feet long then the area is F^2 square feet. Thus F^2 = 10890 and F = \sqrt{10890} = 104.35 feet.

Despite that, is an acre 200x200? An area 200 ft long by 200 ft wide encompasses 40,000 sq feet. This is 40000/43560 = 0.91827acres. 200 feet x 200 feet = 0.918 acres. Or in other words, approximately 92% of an acre.

Suitably, how many acres is 100 feet by 100 feet?

We know 43,560 square feet to 1 acre. 100 ft multiplying by 100 equal to 10,000 sq ft. Therefore, 10,000 sq ft divide by 43,560 sq ft equal to 0.23 acres approximately.

How much does a quarter acre cost?

2. Average Cost of Clearing Land by Lot Size

AcreageCost Range
¼ acre$125 -$1,400
½ acre$250 -$2,800
1 acre$500 -$5,600
2 acres$1,000 -$11,200

### What is the size of an acre lot?

1 acre is approximately 208.71 feet × 208.71 feet (a square) 4,840 square yards. 43,560 square feet.

### What is linear feet?

Technically, a linear foot is a measurement that is 12 inches long (so, one foot) and that is measured in a straight line, which is why it's called linear.

### What is a linear acre?

The most standard shape for an acre is one furlong by one chain, or 660 feet by 66 feet. To find the linear measurements of other rectangular acres, just divide 43,560 by the number of feet you want on one side. A square-shaped acre would then be about 208.7 by 208.7 feet (because 208.7 x 208.7 = ~43,560).

### How many acres is 150 feet by 200?

We know 43,560 square feet to 1 acre. 200 feet multiplying by 200 feet equal to 40,000 sq ft. Therefore, 40,000 sq ft divide by 43,560 sq ft equal to 0.92 acres approximately.

### How many square feet is 200?

We know about 43,500 square feet. 20,000 sq ft is equal to 200 feet. 43,500 sq ft is equal to 0.92 sq ft.

### How big is half an acre compared to a football field?

If you calculate the entire area of a football field, including the end zones, it works out to 57,600 square feet (360 x 160). One acre equals 43,560 square feet, so a football field is about 1.32 acres in size.

### How big is a 50 by 100 plot of land?

1)For a rectangular plot, 50 by 100 refers to 50 feet by 100 feet which is equivalent to 15 meters by 30 meters and is also equal to 450 square meters. This is what people refer to as 1/8 of an acre though slightly less due to the provision for access road.

### What does 1acre look like?

As all farmers and real estate agents know, an acre is defined as an area one furlong long by 4 rods wide. ... Basically if you can picture a football field, that's pretty close to an acre in size. Officially, it is 43,560 square feet, and a football field is 48,000 square feet. Our standard acre isn't the same worldwide.

### How long does it take to walk around 4 acres?

A square acre is 208.7 feet on a side, so the perimeter of an acre is about 835 feet, or about 16 percent of a mile. If you walk a brisk pace of 3 miles an hour, you can cover a mile in 20 minutes. So you should be able to walk 835 feet in about three minutes.


# Modular Arithmetic - Are we allowed to distribute the Modularity?

Assume I have a problem such as "Prove that $\displaystyle103^{53} + 53^{103}$ is divisible by $39$." This would mean I wanted to prove that $\displaystyle103^{53} + 53^{103}\equiv0\pmod{39}$.

My starting statement would be then "$\displaystyle103^{53} + 53^{103}\pmod{39}$" and I would then equate this to "$\displaystyle103^{53}\pmod{39} + 53^{103}\pmod{39}$" and then continue.

Am I allowed to distribute the mod like that?

Thanks.

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Yes, you are... –  DonAntonio Mar 7 '14 at 16:06
@Matthew, please verify the edited version –  lab bhattacharjee Mar 7 '14 at 16:14
That should be a $+$ rather than a comma, right? In the formula right before "and then continue". –  Jack M Mar 7 '14 at 17:38

Mathematicians and programmers have two different ways of thinking about mod. To the latter, it is a binary operation in which ${\rm Mod}[a,b]$ is the remainder of $a$ when divided by $b$ (so it will return an integer in the range $0\le r<b$). To the former, it is a binary relation (a symbol that relates things in some way, like $<,=,>,\approx,\sim$ etc.) for each modulus $b$. We say that $n\color{Red}{\equiv}m$ mod $b$ if the difference $n-m$ is divisible by $b$, or equivalently if ${\rm Mod}[n,b]={\rm Mod}[m,b]$. Sometimes the equivalence symbol $\equiv$ is simply replaced by an equality symbol $=$, in which case we are understood to be equating equivalence classes. The relation $n\equiv m$ mod $b$ is in fact a congruence relation (it "respects" $+$ and $\times$), and the equivalence classes are called residue classes, or just residues.

So if $a\equiv b$ and $c\equiv d$ mod $m$ then $ac\equiv bd$ and $a+c\equiv b+d$ mod $m$. One can use this to end up proving that, in particular, $f(a)\equiv f({\rm Mod}[a,m])$ mod $m$ for integer-coefficient polynomials $f$.

Thus for example ${\rm Mod}[a,m]+{\rm Mod}[b,m]$ and $a+b$ and ${\rm Mod}[a+b,m]$ are all congruent mod $m$, however it is not strictly true that the first and last are equal as integers. Take $a,b=2$ and $m=3$, in which case ${\rm Mod}[2,3]+{\rm Mod}[2,3]=2+2=4\ne1={\rm Mod}[2+2,3]$, albeit $4\equiv1$ mod $3$.

If $n$ and $m$ are coprime, then $a\equiv b$ modulo both $n$ and $m$ if and only if $a\equiv b$ mod $nm$. In particular this means $x\equiv 0$ mod $39$ if and only if $x\equiv0$ mod $3$ and mod $13$. Compute

$$103^{53}+53^{103}\equiv 1^{53}+(-1)^{103}\equiv 1+(-1)\equiv0\mod 3$$

because $103\equiv1$ and $53\equiv-1$ mod $3$. And then compute

$$103^{53}+53^{103}\equiv(-1)^{53}+1^{103}\equiv(-1)+1\equiv0\mod 13$$

because $103\equiv-1$ and $53\equiv1$ mod $13$. Since $103^{53}+53^{103}$ is $0$ mod $3$ and $13$, it is $0$ mod $39$.

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Good, and complete. For the purposes of students going farther in mathematics, it is disastrous to keep the mindset of relying on the operation rather than the equivalence relation. When talking about $R/I$ where $R$ is a ring and $I$ is an ideal, or $G/H$ where $\supset H$ are groups, the equivalence relation outlook is essential. –  Lubin Mar 7 '14 at 18:49

If $103^{53} \equiv a \mod 39$ and $53^{103} \equiv b \mod 39$, than it is indeed true that $103^{53} + 53^{103} \equiv a+b \mod 39$. I wouldn't recommend your notation, which makes $\mod\mbox{ }$ look like an operator.

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There's nothing wrong with using $\bmod$ as an operator. –  ShreevatsaR Mar 7 '14 at 16:54
Using mod as an operator loses the properties of the equivalence class that makes it so convenient. For example, you cannot assume $a \mod M + b \mod M = (a + b) \mod M$. –  DanielV Mar 7 '14 at 17:48
@DanielV: Assuming the name "mod" refers to a proper modulus operator and not the much-less-useful "remainder" operator found in many programming languages, and one is not expecting one's numeric type to act as an algebraic ring with a modulus which is not a multiple of M, why can't one assume that (a mod M)+(b mod M) will equal ((a+b) mod M) in cases where the result of the addition is defined? –  supercat Mar 7 '14 at 18:25
Ah, if I understand, you are talking about a $\mathbb{N} \rightarrow \text{Galois}[M]$ operator, yes that would be useful indeed. –  DanielV Mar 8 '14 at 0:02

To say that $$x \equiv a \pmod{m}$$ means that there is some integer $k$ such that $$x - a = mk$$ Thus, expanding on user133281's answer (i.e. if $103^{53}\equiv a \pmod{39}$ and $53^{103}\equiv b \pmod{39}$), we have $$(103^{53}-a)+(53^{103}-b)=39k+39l$$ which is equivalent to $$(103^{53}+53^{103})-(a+b)=39(k+l)$$ so $$103^{53}+53^{103}\equiv a+b\pmod{39}$$

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Either you meant $\mod 39$ in the first line or $mk$ in the second. –  Mike Miller Mar 8 '14 at 3:45
@Mike Right you are. Thank you. –  iamnotmaynard Mar 8 '14 at 20:56
Glad to help!$\$ –  Mike Miller Mar 8 '14 at 21:37

HINT:

Please have a look into this for the properties of congruence

$\displaystyle 103\equiv-1\pmod{13}\implies 103^{53}\equiv(-1)^{53}\equiv-1$

$\displaystyle 53\equiv1\pmod{13}\implies 53^{103}\equiv(1)^{103}\equiv1$

Similarly, for $\pmod3$

Now if $13$ and $3$ both divides $a,a$ will be divisible by lcm$(13,3)$

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While this is a good hint for solving the problem "show that $103^{53} + 53^{103}$ is divisible by 39", this doesn't address the question that was asked ("am I allowed to distribute the mod") at all. –  Magdiragdag Mar 7 '14 at 16:05
@Magdiragdag, added a relevant link –  lab bhattacharjee Mar 7 '14 at 16:10

The "equivalent modulo $n$" relation is a congruence: a congruence is an equivalence relation with the additional property that for all relevant arithmetic operations (in this case, $0, 1, +, \times$, and anything derived from those), if the inputs the the arithmetic operation are congruent, then the outputs are also congruent.

In the actual ring of integers modulo $n$, though, there is no "congruence" or "mod": e.g. $39 = 0$ is a literal equality. Your question to prove

$$103^{53} + 53^{103} \equiv 0 \pmod{39}$$

in the integers is the same thing as trying to prove

$$103^{53} + 53^{103} = 0$$

in the ring of integers modulo $39$. And since we're working with actual equality, it's clear that we can simplify the two summands separately, then add the simplified results.

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# Prove that the expression is divisible by 26460

#### anemone

##### MHB POTW Director
Staff member
Problem:
Prove that $27195^8-10887^8+10152^8$ is divisible by $26460$.

Attempt:
I grouped the last two terms and manipulated them algebraically and came to the point where I suspect I might have taken the wrong path...here is the last step where I stopped and don't know how to proceed.

$\dfrac{27195^8-10887^8+10152^8}{26460}=\dfrac{3^5\cdot5^7\cdot7^{14}\cdot37^8-7013(2^6\cdot3^6\cdot47^2+3^2\cdot19^2\cdot191^2)(10152^4+10887^4)}{8}$

I'd like to ask, do you think this problem can be solved using only elementary methods?

Last edited by a moderator:

#### Opalg

##### MHB Oldtimer
Staff member
Problem:
Prove that $27195^8-10887^8+10152^8$ is divisible by $26460$.

Attempt:
I grouped the last two terms and manipulated them algebraically and came to the point where I suspect I might have taken the wrong path...here is the last step where I stopped and don't know how to proceed.

$\dfrac{27195^8-10887^8+10152^8}{26460}=\dfrac{3^5\cdot5^7\cdot7^{14}\cdot37^8-7013(2^6\cdot3^6\cdot47^2+3^2\cdot19^2\cdot191^2)(10152^4+10887^4)}{8}$

I'd like to ask, do you think this problem can be solved using only elementary methods?

The best way to make the calculation more manageable is to factorise $26460 = 2^2\cdot 3^3\cdot 5\cdot 7^2$. If you can separately show that $27195^8-10887^8+10152^8$ is divisible by each of the numbers $2^2$, $3^3$, $5$ and $7^2$, then the result will follow.

Take the factor 5, for example. Since $27195$ is a multiple of $5$, so is its eighth power. The other two terms are not multiples of $5$, but here you need to use your idea of grouping those two terms together. In fact, $a^8-b^8$ is a multiple of $a-b$. So $10887^8+10152^8$ is a multiple of $10887-10152 = 735$. That is a multiple of $5$. Putting those results together, you see that $27195^8-10887^8+10152^8$ is a multiple of $5$.

The exact same procedure shows that $27195^8-10887^8+10152^8$ is a multiple of $7^2$. You can use similar ideas to show that it is also a multiple of $3^3$ and of $2^2$. (To deal with $3^3$, notice that $a^8-b^8$ is also a multiple of $a+b$.)

#### anemone

##### MHB POTW Director
Staff member
The best way to make the calculation more manageable is to factorise $26460 = 2^2\cdot 3^3\cdot 5\cdot 7^2$. If you can separately show that $27195^8-10887^8+10152^8$ is divisible by each of the numbers $2^2$, $3^3$, $5$ and $7^2$, then the result will follow.

Take the factor 5, for example. Since $27195$ is a multiple of $5$, so is its eighth power. The other two terms are not multiples of $5$, but here you need to use your idea of grouping those two terms together. In fact, $a^8-b^8$ is a multiple of $a-b$. So $10887^8+10152^8$ is a multiple of $10887-10152 = 735$. That is a multiple of $5$. Putting those results together, you see that $27195^8-10887^8+10152^8$ is a multiple of $5$.

The exact same procedure shows that $27195^8-10887^8+10152^8$ is a multiple of $7^2$. You can use similar ideas to show that it is also a multiple of $3^3$ and of $2^2$. (To deal with $3^3$, notice that $a^8-b^8$ is also a multiple of $a+b$.)
Awesome! I finally understand it now!

#### Bacterius

##### Well-known member
MHB Math Helper
Opalg's answer is very good! A more "heavy machinery" method could also go as follows:

$$26460 = 2^2 \cdot 3^3 \cdot 5^1 \cdot 7^2$$
And we can do the following reductions with a couple modulo operations:

$$27195^8 - 10887^8 + 10152^8 \equiv 3^8 - 3^8 + 0^8 \equiv 0 \pmod{2^2}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 6^8 - 6^8 + 0^8 \equiv 0 \pmod{3^3}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \pmod{5^1}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 9^8 + 9^8 \equiv 0 \pmod{7^2}$$
And invoking the CRT (since $2^2$, $3^3$, $5^1$, $7^2$ are pairwise coprime):

$$\mathbb{Z}_{26460} = \mathbb{Z}_{2^2} \times \mathbb{Z}_{3^3} \times \mathbb{Z}_{5^1} \times \mathbb{Z}_{7^2}$$
$$\therefore$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0 \pmod{26460}$$
This also shows that the exponent (8) is in fact irrelevant, and could be anything.

EDIT: fixed, see ILikeSerena's post below.

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Opalg's answer is very good! A more "heavy machinery" method could also go as follows:

$$26460 = 2^2 \cdot 3^3 \cdot 5^1 \cdot 7^2$$
And we can do the following reductions with a couple modulo operations:

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{2}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{3}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \pmod{5}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \pmod{7}$$
And invoking the CRT (since 2, 3, 5, 7 are pairwise coprime):

$$\mathbb{Z}_{26460} = \mathbb{Z}_{2^2} \times \mathbb{Z}_{3^3} \times \mathbb{Z}_{5^1} \times \mathbb{Z}_{7^2}$$
$$\therefore$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0 \pmod{26460}$$
This also shows that the exponent (8) is in fact irrelevant, and could be anything.
I'm afraid that the number $2 \cdot 3 \cdot 5 \cdot 7$ is also zero mod 2, mod 3, mod 5, and mod 7.
But it is not divisible by 26460.

For CRT you need to show the mod relation for $2^2$, $3^3$, $5^1$, and $7^2$.

For instance for $3^3$, we need:

$\hspace{0.5 in}27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{3^3}$

Still quite doable.

#### Bacterius

##### Well-known member
MHB Math Helper
I'm afraid that the number $2 \cdot 3 \cdot 5 \cdot 7$ is also zero mod 2, mod 3, mod 5, and mod 7.
But it is not divisible by 26460.

For CRT you need to show the mod relation for $2^2$, $3^3$, $5^1$, and $7^2$.

For instance for $3^3$, we need:

$\hspace{0.5 in}27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{3^3}$

Still quite doable.
True, my mistake. Though it's not too difficult to check, just do the same operations but using 4, 27 and 49... (which still work out, it's clear the integers in the problem were carefully chosen to cancel each other out)


# Alternative ways to find $\lim_{x\to 1}\; \frac{1-x}{1- \sqrt{x}}$

Searching on Google by a solution, I found this:

$1) \space$The numerator can be written by a diference of squares:$\space (1-\sqrt{x})(1+\sqrt{x})$

$2 \space)$Then, one can "eliminated" the commum factor beteween the numerator and the denominator.

$3) \space$The final expression looks like: $$\lim_{x\to 1}\; {1+ \sqrt{x}}=2$$

However this was not a intuitive algebraic solution. I haven't thought of this solution at the first attempts. Could you please give me other alternative algebraic solutions, if it exists. Thanks.

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Same solution, let $x=u^2$, but now instantly familiar. – André Nicolas Jan 7 '13 at 17:20
Why isn't it intuitive? If you have a $0/0$ form, the first thing one usually does is search for common factors. – David Mitra Jan 7 '13 at 17:21
The difference of two squares method was the first thing that popped into my head before I clicked the link to this question! – Clive Newstead Jan 7 '13 at 17:21
The other approach to this solution is to think - that was neat, how will I be able to spot neat solutions like that next time and the comment from @AndréNicolas was exactly what I was going to put. – Mark Bennet Jan 7 '13 at 17:22
@João What's $(1-\sqrt{x})(1+\sqrt{x})$? – WimC Jan 7 '13 at 17:35

You can compute the inverse: $$\lim_{x\to 1}\frac{1-\sqrt{x}}{1-x}=\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$$ which is, by definition, $f'(1)$ where $f(x)=\sqrt{x}$.

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As I understood, you "turn over" the fraction,by puting $\frac{1-\sqrt{x}}{1-x}$ on the denominator. Then you multiplied the denominator by $\frac{-1}{-1}$ to get the final expression, that is under the main fraction (the numerator is $1$).Finally you computed the derivative of $\sqrt{x}$ at $x=1$ to get $\frac{1}{\frac{1}{2}}$. Is that? – João Jan 7 '13 at 18:04
Yeah, that's basically it. Essentially, if $g(x)/h(x)\to A$ then $h(x)/g(x)\to 1/A$ – Thomas Andrews Jan 8 '13 at 14:12

Same solution, let $x=u^2$, but now instantly familiar.

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Perhaps we can use L'Hospital rule here ?

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By substitution:$\quad$ let $x = u^2\,.\quad$Then as $x\to 1,\;\;u^2\to 1\implies u\to 1$, giving us $$\lim_{x\to 1}\; \frac{1-x}{1- \sqrt{x}} \;\;=\;\; \lim_{u\to 1}\; \frac{1-u^2}{1- u} = \lim_{x\to 1} \frac{u^2 - 1}{u-1} \;=\;\lim_{u \to 1} \frac{(u-1)(u+1)}{u-1} = \lim_{u\to 1}(u + 1) = 2$$

Using the substitution makes the "difference of squares" route looks so much more obvious!

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If $u^2 \to 1$, then $u \to 1$ or $u \to -1$. Is this second situation if one compute the limit we get $0$.M'I rigth? – João Jan 7 '13 at 18:14
since $x \to 1$, the only root of concern is u = 1 (since $u = \sqrt{x}$ must necessarily be $\ge 0$ in this situation. – amWhy Jan 7 '13 at 18:22


# Why $1\equiv a^{p-1} \mod p$?

Let $\mathbb{Z}_p^*$, where $p$ is prime and let $a\in\mathbb{Z}_p^*$.

Consider the following equation:$$(p-1)! \equiv (p-1)! a^{p-1} \mod p$$

I've read that since $\gcd((p-1)!, p) = 1$ we can infer that $$a^{p-1} \equiv 1$$

So I have two questions:

1. Why is it true that $\gcd ((p-1)!, p)= 1$?
2. Why can we infer that $a^{p-1} \equiv 1$?
• for the frist question: if $p$ is prime then observe than $(p-1)!$ doesnt contain $p$ as a factor, hence $\gcd((p-1)!,p)=1$ – Masacroso Mar 19 '17 at 15:44
• And for the second question, you can always write the $gcd(a,b)$ as a linear combination of $a$ and $b$ with integer coefficients (this is from the Euclidean algorithm). Use this to show that if some number is relatively prime to $p$, then it has a multiplicative inverse mod $p$. – Malkoun Mar 19 '17 at 15:46
• Fermat's little theorem en.wikipedia.org/wiki/Fermat%27s_little_theorem?wprov=sfla1 – Zuo Mar 19 '17 at 15:51

1. The only divisors of $p$ are $p$ and $1$. What are the divisors of $(p-1)!$?

2. Once you know that $p, (p-1)!$ are relatively prime, consider $(p-1)! * (a^{p-1} - 1) \equiv 0 \mod p$. The relative primeness tells you that this is only possible if $(a^{p-1} - 1) \equiv 0 \mod p$

This common result is known as Fermat's Little Theorem. I hope this simple proof helps:

Consider the sequence of integers $n,2n,3n,…,(p−1)n$.

Note that none of these integers are congruent modulo $p$ to the others.

If this were the case, we would have $an≡bn \pmod p$ for some $1≤a<b≤p−1$.

Then as $gcd(n,p)=1$, and we can cancel the $n$, we get $a≡b \pmod p$ and so $a=b$.

Also, since $p∤n$ and $p∤c$, for any $1≤c≤p−1$, then by Euclid's Lemma $p∤cn$ for any such $cn$, which means $cn≢0 \pmod p$.

Thus, each integer in the sequence can be reduced $modulo \ p$ to exactly one of $1,2,3,…,p−1$.

So ${1,2,3,…,p−1}$ is the set of Reduced Residue System $modulo \ p$.

So, upon taking the product of these congruences, we see that $n×2n×3n×⋯×(p−1)n≡1×2×3×⋯×(p−1) \mod p$.

This simplifies to $n^{p−1}×(p−1)!≡(p−1)! \pmod p$.

Since $p∤(p−1)!$, we can cancel $(p−1)!$ from both sides, leaving us with $n^{p−1}≡1 \pmod p$.

• The last statement is essentially my question. Why does the fact $p\not\vert (p-1)!$ implies that you can cancel $(p-1)!$ from both sides? – OliOliver Mar 19 '17 at 16:14
• It doesn't essentially imply that $(p-1)!$ is being cancelled because of that property. It merely states that they are relatively prime to each other and, hence, we can use the property of modular arithmetic to basically remove $(p-1)!$ as if it of no use here(as we are dealing with, say, modulo $p$ and not modulo $(p-1)!$) and appears on both sides leaving it to be factored out. – HKT Mar 19 '17 at 16:25
• Carefully think about it. Suppose if $p|(p-1)!$, we would have been left with $0 \pmod p$ on the right hand side. I hope that helps. – HKT Mar 19 '17 at 16:28
• You are just over-thinking it. I assure you once you get the hang of it and review the basic modular arithmetic definitions and get a stronger grasp on the notation, there will be clarity. – HKT Mar 19 '17 at 16:30

If you write down $(p-1)!$ as $(p-1)\cdot(p-2)\cdot\cdot\cdot1$ you can easily notice that $p$ and $(p-1)!$ have no common factors.

So $gcd((p-1)!,p) = 1$.

Now, knowing that:

$$ax \equiv b \mod(m) \implies x \equiv b \cdot a^{-1} \mod(\frac{m}{gcd(m,a)})$$

• You might as well just write $ax\equiv b\bmod m\implies x=a^{-1}b\bmod m$ when $\gcd(a,m)=1$. No need to overcomplicate it. – arctic tern Mar 19 '17 at 15:55
For the first question, take, for example, $p = 7$. Then $(p-1)! = 2\cdot3\cdot4\cdot5\cdot6 = 2^4\cdot3^2\cdot5$. Notice that in the first equation all of the factors are strictly less than $p = 7$, which implies that all of the prime factors are also strictly less than $p$. Since $p$ is prime, this means that $p$ and $(p-1)!$ can have no prime factors in common.
Now this implies Fermat's little theorem because, having shown that $(p-1)!$ is relatively prime with $p$, we can apply the cancellation law.
Whenver $\gcd(a,m) = 1$, $$ax \equiv ay \mod m$$ $$\downarrow$$ $$x \equiv y \mod m$$


Rolling two dice alternatively. What is probability of winning?

Tom and Paul roll 2 dice alternatively starting with Tom. Consider they use two fair 6-faced dice. The player who rolls 6 first wins. They continue to roll until one of them wins. Find probability that Tom wins.

I have listed out the total possible outcomes below:

{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Here are all the possible outcomes of two dice rolls. And as one can see, only 5 of these comes out with a 6 first; so, I think the probability of Tom winning should be 5/36.

However, when I looked at the solution (This is a question from one midterm in one of previous semesters given at my school) says:

p is 5/36

$$P(Tom\, Wins)=\sum_{k=0}^{\infty }p(1-p)^{2k}=p\sum_{k=0}^{\infty }(1-p)^{2k}=$$ $$\frac{\frac{5}{36}}{1-\frac{31}{36}}=\frac{5}{5}$$

As one can see this answer is really rediculus! (Tom will win no matter what??)

I think that the key here is probably on the word "alternatively", but cannot figureout what has gone wrong here.

• You’ve computed that sum incorrectly. Observe that the exponent is $2k$, not $k$, i.e., you’ve got a geometric series in $(1-p)^2$, not $1-p$.
– amd
Apr 3 '20 at 19:50
• I didn't come out with that answer. It is answer given by someone. I looked at it and was very confused. Apr 4 '20 at 1:37

The probability that Tom wins on the first roll is $$\frac16$$. He wins on the third roll if and only if he first rolls something other than a $$6$$, then Paul rolls something other than a $$6$$, and then Tom rolls a $$6$$; the probability of this is $$\frac56\cdot\frac56\cdot\frac16$$. In general, Tom wins on the $$(2k+1)$$-st roll if and only if the first none of the first $$2k$$ rolls is a $$6$$, and Tom rolls a $$6$$ on roll $$2k+1$$; this occurs with probability $$\frac16\left(\frac56\right)^{2k}$$. Thus, the probability that Tom wins is

\begin{align*} \sum_{k\ge 0}\frac16\left(\frac56\right)^{2k}&=\frac16\sum_{k\ge 0}\left(\frac56\right)^{2k}\\ &=\frac16\sum_{k\ge 0}\left(\frac{25}{36}\right)^k\\ &=\frac16\cdot\frac1{1-\frac{25}{36}}\\ &=\frac6{11}\;. \end{align*}

This is the formula in the solution that you read, with $$p=\frac16$$.

It should not be surprising that this is slightly more than $$\frac12$$: the fact that Tom goes first gives him an advantage, but it’s a small one, since the game is otherwise very symmetric.

One can also compute the desired probability without resort to infinite series. Let $$p$$ be the probability that Tom wins, so that Paul wins with probability $$1-p$$. On the other hand, the probability that Paul when Tom first rolls something other than a $$6$$ must be $$p$$, because at that point the game is effectively starting over with Paul as the first player. Thus, Paul wins with probability $$\frac56p$$, the probability that Tom rolls something other than a $$6$$ initially and Paul then rolls a $$6$$ before Tom does. In short, $$1-p=\frac56p$$, and solving for $$p$$ again yields $$p=\frac6{11}$$.

• Thank you. I think your answer is good. I guess the solution given for this problem is probably wrong. Apr 4 '20 at 1:36
• @AmosKu: You’re welcome. Apr 4 '20 at 1:42
• I have given the question quite a bit of thought last night. I think the key is "alternatively". And whenever Tom rolls a 6, he wins. I think your answer is for the case where only one 6 face dice is rolled. I have edited my wording above, hoping to make the meaning of original question clearer. Apr 4 '20 at 17:44
• @AmosKu: Alternatively makes no sense here; I suspect that the intended word is alternately. That simply means that Tom rolls his die, then Paul rolls his, and so on. This is exactly the same problem as if they were using a single die and trading it back and forth. Apr 4 '20 at 17:52
• Possible. But I just double checked the original question. The word is " Alternatively". There are other errors in the writing of the test as well. I guess you are right. Apr 4 '20 at 20:07


# Conclusion about the dimension of C°(R)?

Tags:
1. Jul 12, 2017

### Rodrigo Schmidt

[mentor note: thread moved from Linear Algebra to here hence no homework template]

We have a linear map A:E→E, where E=C°(ℝ), the vector space of all continuous functions.
Let's suppose that Aƒ= x0 ƒ(t)dt.

By the Calculus Fundamental Theorem, d/dx(Aƒ) = ƒ, so we have a left inverse, which implies that ker(A)={0}.

Supposing that dim(E) is finite, by the rank-nullity theorem we have that im(A)=E . As a result of that:
(ƒ(x)=|x|) ∈ E ⇒ ƒ ∈ im(A)
⇒dƒ/dx ∈ E

But we know that ƒ's derivative is not continuous. So, supposing that dim(E) is finite lead us to a contradition (dƒ/dx ∈ E ∧ dƒ/dx ∉ E) therefore dim(E) must be infinite.

Is this argument valid? If not, could you guys point where does it fail? Thank you!

Last edited by a moderator: Jul 12, 2017
2. Jul 12, 2017

### Krylov

Note that the right-hand side is $(Af)(x)$, not $Af$.

The derivative is not even defined at $x = 0$.

Yes, it is valid, but it is quite convoluted. If you insist on doing it this way, I would prefer to say that you have found a linear operator $A$ that is injective but not surjective (the latter because $A$ maps into $C^1(\mathbb{R})$). This already implies infinite dimensionality of $E$, since on a finite dimensional space injectivity and surjectivity are equivalent for linear operators.

Of course, if you merely care to show that $E$ is infinite dimensional, it is more straightforward to identify an infinite linearly independent set.

3. Jul 12, 2017

### Rodrigo Schmidt

Thanks for the knowledge shared!
I hadn't tought that, that's, indeed, much simpler.
So the basis of the set of all polynomials would be enough?

4. Jul 12, 2017

### LCKurtz

The set of all polynomials is too large to be a basis for $C_0$, but $\{1,x,x^2,\dots\}$ would do.

5. Jul 12, 2017

### mathwonk

this is very nice, and shows a creative grasp of what you are learning. keep it up and you will eventually do some new research!

6. Jul 13, 2017

### Krylov

While I usually like the straightforward approach best (maybe this is one of the differences between a "pure" mathematician and an (aspiring) "applied" mathematician?), I gave it some thought and then came to the conclusion that I agree with you.

Just as an exercise, you could try to complete the direct approach as well. (You are almost there, anyway.)

7. Jul 13, 2017

### Rodrigo Schmidt

Hm, i see, but isn't that the basis of the set of all polynomials? So, by being a basis, that must be a linearly independent set, and it's also infinite, so that implies in the infinite dimensionality of C°(ℝ)?
That's where i want to get someday. Thanks for the inspiration!

8. Jul 13, 2017

### Krylov

I think that I and post #4 misunderstood you when you wrote
We (or I, at least) thought that here you asserted that the set of all polynomials itself is a basis for a linear subspace of $C^0(\mathbb{R})$, maybe because you wrote "the basis". However, from what you wrote afterwards (quoted at the top of this post), I get that you meant any (Hamel) basis for the linear space of all polyniomials. (For example, you can indeed use the canonical basis mentioned in post #4. By the way, I think there the $C_0$ was written by small mistake.)

Very good, as far as I can see, you are more than done.

Last edited: Jul 13, 2017
9. Jul 13, 2017

### Rodrigo Schmidt

Yeah, that's what i meant. Sorry if i wasn't clear enough! English is not my mother language so, mostly when writing about math and science, my texts can get a little confusing. I will try to be more specific in the next time.

Okay! Thanks for the support and knowledge shared!


# Whats the probability a subset of an $\mathbb F_2$ vector space is a spanning set?

Let $V$ be an $n$-dimensional $\mathbb F_2$ vector space. Note that $V$ has $2^n$ elements and $\mathcal P(V)$ has $2^{2^n}$.

I'm interested in the probability (under a uniform distribution) that an element of $\mathcal P(V)$ is a spanning set for $V$. Equivalently a closed form formula (or at least one who's asymptotics as $n\rightarrow \infty$ are easy to work out) for the number of spanning sets or non-spanning sets.

It's not hard to show that the probability is greater than or equal to $1/2$. Since any subset of size greater than $2^{n-1}$ must span the space. I calculated the proportion of spanning sets of size $n$ for $n$ up to $200$ which seem to be going to a number starting with $.2887$. This leads me to believe that the probability exceeds $1/2$. I couldn't nail down a formula for arbitrary sized subsets though to continue experimental calculations.

I feel like this is something that's been done before, but googling I've mostly found things concerning counting points on varieties over finite fields or counting subspaces of finite fields. Any references would be appreciated.

-
I think it would be sufficient to count how many different bases you can have in V. Then multiply that number by all possibilities of adding elements to it, since that does not perturb the spanning property of the set. –  Raskolnikov Jul 17 '12 at 15:31
@Raskolnikov The issue with that is avoiding duplicates. –  JSchlather Jul 17 '12 at 15:32
Indeed, you're right. –  Raskolnikov Jul 17 '12 at 15:33

A random subset of $V$ is owerwhelmingly likely to span $V$.

Let's look at how hard it is for a random subset not to span $V$. In order not to span $V$, there must be an $(n-1)$-dimensional subspace that contains the entire subset. There are exactly $2^n-1$ such subspaces, since they are in bijective correspondence with the nontrivial linear maps $V\to\mathbb F_2$ (each subspace is the kernel of exactly one map).

For each fixed $(n-1)$ dimensional subspace, the probability for a random subset to stay within that subspace is $2^{-2^{n-1}}$, since the $2^{n-1}$ vectors outside the subspace must all randomly decide not to be in the subset.

So the probability for a random set not to span is at most $(2^n-1)2^{-2^{n-1}} < 2^{-(2^{n-1}-n)}$ (and this is slightly too high, because there are a few non-spanning subsets that have more than one proper subspace they fit into and so are counted twice here). Everything else spans.

Even for $n$ as small as $5$ the probability for a random subset to span $V$ is above 99.9%, and the number of 9's increases exponentially with larger $n$s.

-
There are indeed $2^n - 1$ subspaces of dimension $n - 1$. There is a bijection with subspaces of dimension $1$, though not via the inner product. –  Zhen Lin Jul 17 '12 at 15:57
I think the problem is that the probability that some set $S$ is not contained in some subspace $U$ is not independent of the probability that $S$ is not contained in some other subspace $U'$. –  Zhen Lin Jul 17 '12 at 16:01
That works, and it's certainly enough to get the conclusion that the probability goes to $1$ as $n \to \infty$. –  Zhen Lin Jul 17 '12 at 16:08
(+1 to Henning) @ZhenLin: A non-degenerate bilinear form gives that correspondence (as you probably noticed) between $(2^n-1)$ non-zero vectors (=1-dim. subspaces) and $(n-1)$-dimensional subspaces. IMHO the word orthogonal might still be used - with the caveat that a vector or a subspace may be orthogonal to itself. –  Jyrki Lahtonen Jul 17 '12 at 16:23
@Jyrki: I think it is cleaner to speak of nonzero vectors in $V$ and $(n-1)$-dimensional subspaces in $V^{\ast}$ (their annihilators) as this correspondence is functorial and doesn't require the addition of extra structure. –  Qiaochu Yuan Jul 17 '12 at 20:18

It's worth mentioning that there is an explicit formula for the probability that $N$ random vectors drawn from $\mathbb{F}_q^n$ span. (Note: I am doing sampling with replacement. If you want to require the $N$ vectors to be distinct, you can do this using inclusion-exclusion, but the result will be much messier.) Of course, the answer is $0$ if $N<n$, so assume $N \geq n$.

Write down an $n \times N$ matrix whose columns are the vectors in question. Note that the following are equivalent:

1. The columns span $\mathbb{F}_q^n$.
2. The matrix has rank $n$.
3. The rows are linearly independent.

Thinking in terms of rows, we have the new question: If we draw $n$ vectors at random from $\mathbb{F}_q^N$, what is the probability that they are linearly independent?

The probability that the first vector is nonzero is $1-q^{-N}$. Given that, the probability that the second vector is not in the span of the first is $1-q^{-N+1}$. Given that, the probability that the third vector is not in the span of the first two is $1-q^{-N+2}$. Continuing in this manner, the probability that $n$ vectors in $\mathbb{F}_q^n$ are independent is $$(1-q^{-N}) (1-q^{-N+1}) \cdots (1-q^{-N+n-1}).$$ And, by the above argument, this is also the probability that $N$ random vectors will span $\mathbb{F}_q^n$.

In particular, as Henning says, once $N-n$ is of any significant size, this is extremely close to $1$.

-


By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10 001st prime number?

At first glance, a simple iteration whilst checking if index position is prime, would suffice.

A pseudocode attempt:

sum = 0
for i = 2; i <= ?; i++ {
if(isPrime(i)){
sum++
if ( sum >= 10001) {
return i
}
}
}

Using the solution in go to check if is prime from former examples:

func isPrime(x int64) bool {
var i int64 = 2
for ; i < x; i++ {
if x%i == 0 {
return false
}
}
return true
}

The solution would scale to

$$O(n^2)$$.

This is not viable. Also, there's no upper bound to set our iteration limit, meaning, we don't know up to which number we should iterate to find the 10,001 prime.

So digging further, I found about the Erathostenes Sieve:

An ancient algorithm for finding all prime numbers up to any given limit

This is exactly what we need. Let's look at the procedure:

To find all the prime numbers less than or equal to a given integer n by Erathostenes' method:

1. Create a list of consecutive integers from 2 ... n (2,3,4,...,n)
2. Initially, let p equal 2, smallest prime.
3. Enumerate multiples of p by counting in increments of p from 2p to n, and mark them in the list (these will be 2p, 3p, 4p, ...; the p itself should not be marked)
4. Find the first number greater than p in the list that is not marked, if there is not such number, stop. Otherwise, let p now equal to this number (which is next prime), and repeat step 3
5. When algorithm terminates, the numbers remain not marked in the list, are all primes below n.

Here's the code I came up:

package main

import (
"fmt"
"math"
"time"
)

func primeSieve(n, limit int) int {
// Create and populate array of values
// lim := c
a := make([]bool, limit+1)
for i := 0; i < limit+1; i++ {
a[i] = true
}

// p * p <= limit === p <= int(math.Sqrt(float64(limit)))
for p := 2; p <= int(math.Sqrt(float64(limit))); p++ {
// At first all will be true
if a[p] == true {
// i = 2 * 2, i = 3 * 2, i = 4 * 2, ..., i = i * i
// Calculate multiples of 2, then 3, then 5, then 7
for i := p * 2; i <= limit; i += p {
a[i] = false
}
}
}

// Count primes up to n
// if primes == 10001, return sievePrime
var primes, sievePrime int
for p := 2; p <= limit; p++ {
if a[p] == true {
primes++
// Sum up primes and print 10,001th prime
if primes <= n {
sievePrime = p
}
}
}
return sievePrime
}

func main() {
start := time.Now()
// Nth prime to find
var n, limit int = 10001, 105000 // Arbitrary limit find  by  testing

fmt.Println("Execution time: ", time.Since(start))
fmt.Println("10,001th Prime: ", primeSieve(n, limit))
}


# Limit of $2^{1/n}$ as $n\to\infty$ is 1

How do I prove that:

$\lim \limits_{n\to \infty}2^{1/n}=1$

Thank you very much.

-

Using the binomial theorem for integer exponents:

Can you see that $(1+\frac 1 n)^n > 2>1$

Take the nth root, to give:

$(1+\frac 1 n) > 2^{\frac 1 n} >1$

-
oh, that's really nice. how to you prove that $(1+\frac{1}{n})^n>2$? Thanks! –  Anonymous Mar 28 '12 at 20:18
$(1+x)^n = 1+nx+$ other positive terms –  Mark Bennet Mar 28 '12 at 20:20
And $n$ has to be greater than 1. –  Mark Bennet Mar 28 '12 at 20:21
sweet, thank you. –  Anonymous Mar 28 '12 at 20:25
Now look at Sivaram Ambikasaram's answer and catch hold of how the two are related - then you'll have learned some maths. –  Mark Bennet Mar 28 '12 at 20:33

Note that for $a\gt 0$, $$a^b = e^{b\ln a}.$$ So $$\lim_{n\to\infty}2^{1/n} = \lim_{n\to\infty}e^{\frac{1}{n}\ln 2}.$$ Since the exponential is continuous, we have $$\lim_{n\to\infty}e^{\frac{1}{n}\ln 2} = e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln 2}.$$

Can you compute $\displaystyle\lim_{n\to\infty}\frac{\ln 2}{n}$ ?

-
Thank you for the help, however I don't want to prove this using ln and e. –  Anonymous Mar 28 '12 at 20:09
@Anonymous: Thank you for your comment; perhaps next time you can state, in your post, what it is you do and do not want, so that other people may not waste their time giving you information you don't care about. –  Arturo Magidin Mar 28 '12 at 20:11
I really appreciate your time and sorry for the wasted time for me, however there are probably people who would find this informative and want to know how to prove it this way. I didn't state that I don't want lan and e because I wasn't aware that it can be proven this way. Thank you very much again. –  Anonymous Mar 28 '12 at 20:16
This answer deserves more than one upvote, as it not only answers the question, but also deals neatly with the ambiguity in the way the question is asked. –  Mark Bennet Mar 28 '12 at 21:37

HINT: Use squeeze theorem.

Since $1 < 2$, we have $1 = 1^{1/n} < 2^{1/n}$ for all $n \in \mathbb{N}$.

To bound the limit from above, note that $1 + n \epsilon < \left( 1 + \epsilon \right)^n$.

Hence, given any $\epsilon >0$, $\forall n > \displaystyle 1/{\epsilon}$, we have $2 < 1 + n \epsilon < \left(1 + \epsilon \right)^n$ and hence $2^{1/n} < 1 + \epsilon$.

-

Here are a few different proofs which don't use $e$ or $\log$ and can be regarded completely elementary.

Proof 1)

We use the following theorem:

If $a_n \gt 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim_{n\to\infty} a_n^{1/n} = L$

This is a standard theorem, and a proof can be found in almost any textbook. You can also find a proof in my answer here: Show that this limit is equal to $\liminf a_{n}^{1/n}$ for positive terms.

Apply the theorem to the sequence $a_n = 2$.

Proof 2)

We use the $\text{AM} \ge \text{GM}$ inequality on $n-1$ ones and one $2$.

$$\frac{1 + 1 + \dots + 1 + 2}{n} \ge 2^{1/n}$$

$$\frac{n+1}{n} \ge 2^{1/n}$$

Thus we have

$$1 + \frac{1}{n} \ge 2^{1/n} \ge 1$$

so by Squeeze theorem, $\lim_{n \to \infty} 2^{1/n} = 1$.

Proof 3)

We can use Bernouli's inequality (essentially similar to Sivaram's answer) to show that

$$\left(1 + \frac{1}{n}\right)^n \ge 1 + \frac{1}{n} \times n = 2$$

and we get inequalities similar to the proof in 2).

Proof 4)

The sequence $a_n = 2^{1/n}$ is bounded below (by $1$) and montonically decreasing.

Thus it is convergent, to say $L$. Since $a_{2n}$ also converges to $L$, we have that $L = \sqrt{L}$, as $2^{1/2n} = \sqrt{2^{1/n}}$. So $L = 0$ or $L = 1$. Since the limit is not less than $1$ ($2^{1/n} \ge 1$), the limit is $1$.

Proof 5)

For $n \gt 2$, we have that $1 \le 2^{1/n} \le n^{1/n}$.

Now use the fact that $\lim_{n \to \infty} n^{1/n} = 1$.

An elementary proof of that can be found here: http://math.stackexchange.com/a/115825/1102. Any proof for $n^{1/n}$ now becomes a proof for $2^{1/n}$. Proof 1) above can also be used for $n^{1/n}$.

Proof 6)

Using combinatorics.

The number of $n$ digit numbers in base-$n+1$ is $(n+1)^n$ (allowing for leading zeroes). The number of $n$ digits numbers in base-$n$ is $n^n$. We can show that $(n+1)^n \ge 2 \times n^n$: consider the base-$n$ numbers. Replace the last digit with $n$. You get a base-$n+1$ $n$ digit number. Counting the base-$n$ numbers (which are also base-$n+1$ numbers) and the "last digit modified" numbers, gives us the inequality.

This inequality implies that $1 + \frac{1}{n} \ge 2^{1/n}$ and can be used to give a proof using the squeeze theorem, similar to proofs 2 and 3.

-
Sorry, had some free time :-) I am pretty sure there are more proofs... –  Aryabhata Mar 28 '12 at 21:31
Very nice list! –  Pedro Tamaroff Mar 29 '12 at 23:50
@PeterT.off: Thanks! –  Aryabhata Mar 29 '12 at 23:59


# Sum of series

## Homework Statement

Find the sum of the series:

$$S = 1~+~\frac{1 + 2^3}{1 + 2}~+~\frac{1 + 2^3 + 3^3}{1 + 2 + 3}~+~\frac{1 + 2^3 + 3^3 + 4^3}{1 + 2 + 3 + 4}~+~...$$

upto 11 terms

## Homework Equations

Sum of first 'n' natural numbers: $$S = \frac{n(n + 1)}{2}$$
Sum of the squares of the first 'n' natural numbers: $$S = \frac{n(n + 1)(2n + 1)}{6}$$
Sum of the cubes of the first 'n' natural numbers: $$S = \left(\frac{n(n+1)}{2}\right)^2$$

## The Attempt at a Solution

In the series, the $n^{th}$ term is given by:

$$T_n = \frac{1 + 2^3 + 3^3 + ... + n^3}{1 + 2 + 3 + ... + n}$$

$$T_n = \frac{\left(\frac{n(n+1)}{2}\right)^2}{\left(\frac{n(n + 1)}{2}\right)}$$

$$T_n = \frac{1}{2}(n^2 + n)$$

Hence,

$$S_n = \frac{1}{2}\left(\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\right)$$

On substituting n = 11, I get:

$$S_n = 286$$

But this isn't one of the options I got on the test. All the options were between 300-400 and all had their last digit '9' [that's all i remember.. they took the question paper away].

Last edited:


# What is $\Bbb E(X)$?

Let there be $$100$$ balls in a box out of which $$50$$ are red and $$50$$ are green. Pick $$75$$ balls at random from the box and throw them away. Now pick one ball at random from the remaining balls in the box. Let $$X$$ be the random variable which takes the value $$100$$ when the ball drawn is red in colour and takes the value $$25$$ if the ball drawn is green in colour. Find the expectation $$\Bbb E(X).$$

My attempt $$:$$ Let $$Y$$ denote the number of red balls thrown away. Then the number of green balls thrown away is $$75-Y.$$ Clearly $$Y \geq 25.$$ So \begin{align*} \Bbb P(X=100) & = \sum\limits_{n=25}^{49} \Bbb P(X=100 \mid Y=n)\ \Bbb P(Y=n) \\ & = \sum\limits_{n=25}^{49} \frac {\binom {50-n} {1}} {\binom {25} {1}} \times \frac {\binom {50} {n}} {\binom {100} {n}} \end{align*} Similarly \begin{align*} \Bbb P(X=25) & = \sum\limits_{n=25}^{49} \Bbb P(X=25 \mid (75-Y) = n)\ \Bbb P((75-Y) = n) \\ & = \sum\limits_{n=25}^{49} \Bbb P(X=25 \mid Y = 75-n)\ \Bbb P(Y = 75-n) \\ & = \sum\limits_{n=25}^{49} \frac {\binom {50-n} {1}} {\binom {25} {1}} \times \frac {\binom {50} {75-n}} {\binom {100} {75-n}} \end{align*}

Then the required expectation would be $$100\ \Bbb P(X=100) + 25\ \Bbb P(X=25).$$ But the computation is very tough. Is there any simpler way to approach the problem?

Any help will be highly appreciated. Thank you very much.

• It seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So P(X=100) = P(X=25) = 1/2 – Ant Jun 13 '20 at 8:51
• But @Ant after throwing $75$ balls away it might not be the case. Obviously the green and red balls in the box are present with different proportions after throwing $75$ balls. Right? Then the red and green balls are not equally likely to be drawn. – Phi beta kappa Jun 13 '20 at 8:55
• I have posted an answer, so we can keep discussing it there. To answer your question, they will be present in a different proportion after being drawn, but that's just a single observation. Their distribution is the same; the distribution of how many red vs green balls end up in the urn after you throw some away is the same. Therefore the distribution of which ball you pick after is also the same. So probabilities are the same. Right? – Ant Jun 13 '20 at 9:02
• It is clear by symmetry that $$\mathbb P(X=25)=\mathbb P(X=100)=\frac12.$$ There is no need for elaborate computation here. – Math1000 Jun 13 '20 at 10:09
• @Math1000 can you show it mathematically without making some vague assertion? I know that you are a genius; but it's better not to think others as genius as you. – Phi beta kappa Jun 13 '20 at 10:22

It seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So $$P(X=100) = P(X=25) = 1/2$$

Which implies

$$E(X) = 100\cdot P(X=100) + 25 \cdot P(X=25) = 62.5$$

A simple script confirms that the expectation is indeed $$62.5$$. You can also double check by computing the probability. Using your terminology,

$$P(Y=n) = \frac{{75 \choose n} \cdot {25 \choose 50-n}} {100\choose 50}$$

So that

$$P(X=100) = \sum_{n=25}^{50} \frac{50-n}{25} \cdot \frac{{75 \choose n} \cdot {25 \choose 50-n}} {100\choose 50}$$

And you can check numerically that this is equal to $$1/2$$

• How have you computed $\Bbb P(Y=n)$? – Phi beta kappa Jun 13 '20 at 9:43
• The probability of removing n red balls can be computed as follows: Line up all the 100 balls in a row. Count how many of the permutations end up with n balls in the first 75 spots. Divide by total number of permutations, and you're done. To compute the first number, you have 75 balls of which n are red. So the permutations are 75!/n!/(75-n)!, i.e. 75 choose n. Then you have to multiply with the number of permutations of the remaining 25 balls; it's the same, except that you now have 50-n red balls. The second number is the total number of permutations, therefore 100 choose 50. – Ant Jun 13 '20 at 9:48
• I don't think so. If according to your assumption the balls are indistinguishable then we have just $26$ ways to throw $75$ balls from the urn and each way is equally likely. If the balls are indistinguishable then it doesn't matter in which order you throw them away or which red/green balls are thrown away. In this case which really matters that how many red/green balls you throw away out of $75$ balls. Right? If I present them as ordered pairs as (number of red balls thrown away, number of green balls thrown away) then there are $26$ such pairs $(25,50),(26,50), \cdots, (50,25).$ – Phi beta kappa Jun 13 '20 at 10:15
• Each pair is equally likely to occur. So for each $25 \leq n \leq 50$ we have $$\Bbb P(Y=n) = \frac {1} {26}.$$ – Phi beta kappa Jun 13 '20 at 10:17
• No. I know it for the case of distinguishable but not for the case of indistinguishable. – Phi beta kappa Jun 13 '20 at 10:28


# The Dynamics of Contrapositive Proofs

The Wikipedia Link for contrapositive proofs states that proving if p then q is the same as proving if not q then not p. I don't completely follow why.

Is there any way to understand what's happening without involving logic tables ? If logic tables are inevitable, can anyone give me a basic explanation?

I can't seem to follow how if p then q gets simplified to not p or q I understand that the truth tables of the two are same. But, that's that.

-
It could help to think the about contrapositive of some simple theorems, like the Pythagorean theorem. – Ragib Zaman Jul 8 '12 at 16:43
Semiformally speaking: Suppose $P\implies Q$. Further suppose $Q$ is false; if $P$ were true then $Q$ would also be true and we'd have a contradiction - impossible - so $P\implies Q$ and $\lnot Q$ implies $\lnot P$, i.e. $P\implies Q$ implies $\lnot Q\implies\lnot P$, ie an implication entails its contrapositive. Of course this means the contrapositive entails the contrapositive's contrapositive, ie the original statement, hence we have an equivalence, $(P\implies Q)\iff (\lnot Q\implies \lnot P)$. – anon Jul 8 '12 at 16:54

The sentence "if $p$ then $q$" asserts that in every single situation (model) in which $p$ is true, $q$ must be true. The only way this can be incorrect is if we have a situation in which $p$ is true but $q$ is false. We can show this cannot be the case by showing that whenever $q$ fails, $p$ must fail, that is, by showing that if "not $q$" then "not $p$."

Conversely, suppose that in every situation in which $p$ is true, then $q$ must be true. Then there cannot be a situation in which $q$ fails and $p$ is true. So if "not $q$" holds, then "not $p$" must hold.

Thus the two assertions "if $p$ then $q$" and "if not $q$ then not $p$" hold in precisely the same cases.

Remark: The answer above is much too abstract. To understand what's going on, it is useful to examine a number of concrete cases. Suppose that we want to show that if $x^2=2$ then $x$ is not an ordinary fraction (integer divided by integer). So $p$ is the assertion $x^2=2$, and $q$ is the assertion that $x$ is not a fraction.

We prove the result by showing that if $x$ is a fraction, then $x^2$ cannot be equal to $2$, that is, by showing that "not $q$" implies "not $p$." That shows that if $x^2=2$, then $x$ could not possibly be a fraction, which is exactly what we wanted.

-
Much Clearer now. Thanks André. – Inquest Jul 8 '12 at 17:23

"If $P$ then $Q$" tells you that whenever you know that $P$ is true, you may automatically conclude that $Q$ is true.

"If not $Q$, then not $P$" tells you that whenever you know that $Q$ is not present, you may conclude that $P$ is not present.

If ($P$ implies $Q$) is true, then knowledge of the presence of $P$ always gives us that $Q$ is present. If $Q$ isn't there, then there must be no $P$ to have put it there. Likewise, if the absence of $Q$ automatically gives us the absence of $P$, and if $P$ is present, there can be no not-$Q$ to have given us a not-$P$.

Similarly, either we have $P$ or we don't have $P$. In the case that we have $P$, we are guaranteed to have $Q$. The other case is that we don't have $P$. So we either don't have $P$, or we have $Q$ (which we got by having $P$): not-$P$ or $Q$.

On the other hand, let's say all we know is that either $P$ isn't true or $Q$ is true. If $P$ isn't true, then $P\Rightarrow Q$ is trivially true. Let's deal with the case where $P$ is true. So we know that $P$ is true and also that either $P$ isn't true or $Q$ is true. That is, since $P$ and not-$P$ can't both be true, if $P$ is true, then $Q$ is true.

-
+1. Took me a few reads to understand. The last paragraph's overall flow and structure is why most of my non-math friends don't speak to me when I'm doing math. – Inquest Jul 8 '12 at 17:22
Sorry. I'll take a look to revise it. – Neal Jul 8 '12 at 17:26


# Is every correlation matrix positive definite?

I'm talking here about matrices of Pearson correlations.

I've often heard it said that all correlation matrices must be positive semidefinite. My understanding is that positive definite matrices must have eigenvalues $> 0$, while positive semidefinite matrices must have eigenvalues $\ge 0$. This makes me think that my question can be rephrased as "Is it possible for correlation matrices to have an eigenvalue $= 0$?"

Is it possible for a correlation matrix (generated from empirical data, with no missing data) to have an eigenvalue $= 0$, or an eigenvalue $< 0$? What if it was a population correlation matrix instead?

Consider three variables, $X$, $Y$ and $Z = X+Y$. Their covariance matrix, $M$, is not positive definite, since there's a vector $z$ ($= (1, 1, -1)'$) for which $z'Mz$ is not positive.

However, if instead of a covariance matrix I do those calculations on a correlation matrix then $z'Mz$ comes out as positive. Thus I think that maybe the situation is different for correlation and covariance matrices.

My reason for asking is that I got asked over on stackoverflow, in relation to a question I asked there.

• If, for example, two attributes are one thing, only having different names, the matrix is singular. If two attributes add to a constant, it is again singular, et cetera. – ttnphns Nov 21 '15 at 16:54
• If a covariance matrix is singular correlation matrix is singular as well. – ttnphns Nov 21 '15 at 17:05
• Near-duplicates: Is every correlation matrix positive semi-definite? which has less focus on the definite versus semi-definite angle, and Is every covariance matrix positive definite? which is relevant because a covariance is essentially a rescaled correlation. – Silverfish Nov 21 '15 at 21:20

Correlation matrices need not be positive definite.

Consider a scalar random variable X having non-zero variance. Then the correlation matrix of X with itself is the matrix of all ones, which is positive semi-definite, but not positive definite.

As for sample correlation, consider sample data for the above, having first observation 1 and 1, and second observation 2 and 2. This results in sample correlation being the matrix of all ones, so not positive definite.

A sample correlation matrix, if computed in exact arithmetic (i.e., with no roundoff error) can not have negative eigenvalues.

• May be worth mentioning the possible effects of missing values on the sample correlation matrix. Numerical fuzz isn't the only reason to get a negative eigenvalue in a sample correlation/covariance matrix. – Silverfish Nov 21 '15 at 21:22
• Yes, I didn't make it explicit, but I was assuming, per the question statement, "with no missing data". Once you get into the wild, wacky world of missing data and adjustments therefor, anything goes. – Mark L. Stone Nov 21 '15 at 21:35
• Yes, sorry, you're quite right the question said "no missing data" - just thought it worth mentioning somewhere since future searchers might be interested even if the OP's appetite is sated! – Silverfish Nov 21 '15 at 21:40

The answers by @yoki and @MarkLStone (+1 to both) both point out that a population correlation matrix can have zero eigenvalues if variables are linearly related (such as e.g. $X_1 = X_2$ in the example of @MarkLStone and $X_1 = 2X_2$ in the example of @yoki).

In addition to that, a sample correlation matrix will necessarily have zero eigenvalues if $n<p$, i.e. if the sample size is smaller than the number of variables. In this case covariance and correlation matrices will both be at most of rank $n-1$, so there will be at least $p-n+1$ zero eigenvalues. See Why is a sample covariance matrix singular when sample size is less than number of variables? and Why is the rank of covariance matrix at most $n-1$?

• True 'dat. I suppose I could have and should have provided this info as well, but my goal was to produce a counterexample to refute the OP's hypothesis, thereby showing its invalidity Nevertheless, you should adjust your second sentence to be "In this case covariance and correlation matrices will be at most rank n−1, so there will be at least (p−n+1) zero eigenvalues." – Mark L. Stone Nov 21 '15 at 15:21

Consider $X$ to be an r.v. with mean 0 and variance of 1. Let $Y=2X$, and calculate the covariance matrix of $(X,Y)$. Since $2X=Y$, $E[Y^2]=4E[X^2]=\sigma_Y^2$, and $E[XY]=2E[X^2]$. Due to the zero mean configuration, the second moments are equal to the suitable covariances, for instance: $\mbox{Cov}(X,Y)=E[XY]-EXEY=E[XY]$.

So the covariance matrix will be: $$\Lambda = \left(\array{1 & 2 \\ 2 & 4 }\right),$$ having a zero eigenvalue. The correlation matrix will be: $$\Lambda = \left(\array{1 & 1 \\ 1 & 1 }\right),$$ having a zero eigenvalue as well. Due to the linear correspondence between $X$ and $Y$ it is easy to see why we get this correlation matrix - the diagonal will always be 1, and the off-diagonal is 1 because of the linear relationship.

• Just for math-challenged readers like myself, let me point out that the 2 in $\Lambda$ is the $cov(X,Y)= \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)= 2\mathbb{E}[X^2]=2\left(\sigma_X^2+[E(X)]^2\right)$ this last equality resulting from: $E(X^2)=\text{Var}(X)+[E(X)]^2$. – Antoni Parellada Nov 21 '15 at 19:45
• +1 for your post. I wanted to make it easy to follow for everyone by including the $diag \Lambda^{-1/2}\,\Lambda\, diag \Lambda^{1/2}$ formula, but the comments format won't allow it. Do you think there is any valid point in including it in your post? – Antoni Parellada Nov 21 '15 at 20:44
• @AntoniParellada , I'm not exactly sure what you mean - the covariance here is a direct calculation. But I will edit and make that clearer. Thanks. – yoki Nov 22 '15 at 14:59


Toronto Math Forum

APM346-2022S => APM346--Lectures & Home Assignments => Chapter 2 => Topic started by: Xiangmin.Z on January 17, 2022, 04:40:37 PM

Title: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 04:40:37 PM
Hello, I have a question about example 4 from W2 L1:
We are given :$u_{t}+xu_{x}=xt$
after calculation we get:
$x=Ce^t$

$du=xt dt=Cte^t dt$, so $u=C(t-1)e^t+D=x(t-1)+D$,
but how did we get $D=\phi({xe^{-t}} )$? we know it is a constant, but why is D depended on $xe^{-t}$?

Also, why would the initial condition $u|_{t=0} =0$ implies that $\phi({x}) =x$ ?

Thanks.

Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 05:27:38 PM
I checked the calculation and I think the answer for x is correct, and does anyone know why D is depended on $xe^{-t}$?
Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Victor Ivrii on January 17, 2022, 07:48:38 PM
Now it is correct $x=Ce^{t}$ and then $C=?$
Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 08:19:53 PM
$C=xe^{-t}$, so D is a function of $\phi$, therefore $D=\phi({xe^{-t}} )$ ? Yes, but "$D$  is a function of it"


# Prove that $f(x) = x^3 -x$ is NOT Injective

Okay, so I know there are plenty of injective/surjective (and thus, bijective) questions out there but I'm still not happy with the rigor of what I have done. So I'd really appreciate some help!

So the question actually asks me to do two things:

(a) give an example of a cubic function that is bijective. Explain why it is bijective.

(b) give an example of a cubic function that is not bijective. Explain why it is not bijective.

So for (a) I'm fairly happy with what I've done (I think):

$$f: \mathbb R \rightarrow \mathbb R , f(x) = x^3$$

So we know that to prove if a function is bijective, we must prove it is both injective and surjective.

Proof: $f$ is injective

Let: $$x,y \in \mathbb R : f(x) = f(y)$$ $$x^3 = y^3$$ (take cube root of both sides) $$x=y$$

Proof: $f$ is surjective

Let: $$y \in \mathbb R$$

$$x = \sqrt[3]{y}$$

$$f(x) = (\sqrt[3]{y})^3 = y$$

So I believe that is enough to prove bijectivity for $f(x) = x^3$. Keep in mind I have cut out some of the formalities i.e. invoking definitions and sentences explaining steps to save readers time. This is just 'bare essentials'.

So for (b)

$$f: \mathbb R \rightarrow \mathbb R , f(x) = x^3 - x$$

Now I'm just going to try and prove it is NOT injective, as that should be sufficient to prove it is NOT bijective.

Let:

$$x,y \in \mathbb R : f(x) = f(y)$$ $$x^3 - x = y^3 - y$$

This is about as far as I get. Send help. Thanks everyone.

• What happens if $x=0$ or $x=1$? – Cameron Williams Mar 28 '16 at 2:59
• To prove that a function $f$ is not injective, you just need to find two different numbers $x$ and $y$ such that $f(x)\not=f(y)$. – Christopher Carl Heckman Mar 28 '16 at 3:02
• @Carl you mean two numbers $x,y$ such that $x\neq y$ but $f(x)=f(y)$ – JMoravitz Mar 28 '16 at 3:03
• @CarlHeckman as JMoravitz said, you may confuse op... – YoTengoUnLCD Mar 28 '16 at 3:07
• Thanks for the help guys, +1's given. – Rubicon Mar 28 '16 at 3:53

Injectivity:

A function $f$ from $X\to Y$ is said to be injective iff the following statement holds true:

for every $x_1,x_2\in X$ if $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$

The negation of this then yields:

A function $f$ from $X\to Y$ is not injective iff there exists $x_1,x_2\in X$ such that $x_1\neq x_2$ but $f(x_1)=f(x_2)$

In the case of the cubic in question, it is an easily factorable polynomial and we can find multiple distinct roots. $f(x)=x^3-x=x(x^2-1)=x(x+1)(x-1)$

We know that a root of a polynomial is a number $\alpha$ such that $f(\alpha)=0$

So, $f(1)=f(0)=f(-1)=0$ despite $1,0,-1$ all being distinct unequal numbers in the domain. This shows that it is not injective, and thus not bijective.

As an aside, one can prove that any odd degree polynomial from $\Bbb R\to \Bbb R$ must be surjective by the fact that polynomials are continuous and the intermediate value theorem.

• This is great. I think it has helped my understanding significantly. So what you're sort of saying is because f(1) and f(-1) equal 0, there is more than one-to-one mapping to the elements in the co-domain (hence not injective)? So in this case the set might even look like this: {(1,0) , (-1,0)} (just as a way to visualise it)? – Rubicon Mar 28 '16 at 3:53

Regarding (a), when you say "take cube root of both sides" you are (at least implicitly) assuming that the function is injective -- if it were not, the cube root would not be a well-defined operation. Depending on what you are / are not allowed to assume, that could be regarded as a circular argument.

However, there is another approach. Beginning with the assumption that $x^3=y^3$, you can manipulate the expression as follows: $$x^3-y^3=0$$ $$(x-y)(x^2+xy+y^2)=0$$ Now if you can find a convincing argument that the second factor, $x^2+xy+y^2$, is always positive for any choice of $x$ and $y$, then it follows that the only way $x^3=y^3$ can be true is if $x=y$.

The factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$, by the way, could also give you some insight into how to handle (b).

• $$x^2+xy+y^2 = {x^2+y^2\over 2} + {1\over 2}(x^2 + 2xy+y^2) = {x^2+y^2\over 2} + {1\over2}(x+y)^2 \ge 0.$$ The only way to get equality is to have $x=0$ and $y=0$, but then $x=y$ in that case as well. – Christopher Carl Heckman Mar 29 '16 at 6:30
• @CarlHeckman Yes, I know. Also $x^2+xy+y^2 = (x + \frac{1}{2}y)^2 + \frac{3}{4}y^2$. I was intentionally leaving that out to leave the OP something to finish. – mweiss Mar 29 '16 at 14:16


# Can an eigenvalue $\lambda_i=0$?

I was doing some work with diagonalization of a matrix $A$ in order to find a matrix $P$ such that $\,P^{-1}AP\,$ was diagonal. In order to that I set $\;\lambda I_{n}=0\;$ and found the characteristic polynomial and its roots.

When I factored my characteristic polynomial I obtained $\;\lambda^2(\lambda-2),\,$ so $\,\lambda=0,\,2$.

I was taught that the eignenvalues$\,\lambda_{i}\,$ I found become the entries of the diagonal matrix $\,P^{-1}AP.\,$ If this is indeed true, then two of the diagonal entries would be $\,0.\,$ Is this allowed, or must a diagonal matrix strictly have non-zero diagonal entries?

• Zero is allowed. You may be thinking of eigenvectors --- the zero vector can't be an eigenvector. But for eigenvalues, no problem. – Gerry Myerson Jun 21 '13 at 23:44

Absolutely yes: It is very possible $\lambda = 0$. Zero is allowed.

You may be mixing up what you know about eigenvectors --- the zero vector cannot be an eigenvector.

But for an eigenvalue $\lambda$, it is certainly possible and admissible that $\lambda = 0$.

With respect to your last question:

"($\lambda = 0$): Is this allowed, or does a diagonal matrix strictly have to have the diagonal entries as non-zero?"

Yes, it is allowed for zero's to be on the diagonal. No, the diagonal entries need not be non-zero.

• Dang! Beat me to the punch with the eigenvalue vs. eigenvector observation. +1 – Cameron Buie Jun 21 '13 at 23:58
• +1 what that Dang! means Amy? Dang Dang... Is that like a sound of a big Bell? – mrs Jun 22 '13 at 0:09
• +1 it's a non-profane way of saying something like "Damn": meaning, if I hadn't posted what I posted, Cameron intended to. – amWhy Jun 22 '13 at 0:12
• @amWhy Thanks; makes perfect sense now. I think I was confusing eigenvalues with eigenvectors. – Sujaan Kunalan Jun 22 '13 at 0:36
• @amWhy: Spot on +1 – Amzoti Jun 22 '13 at 0:38

A matrix is called diagonal if its off-diagonal entries ($a_{ij}$ for $i \not= j$) are all zero. This does not require the diagonal entries ($a_{jj}$) to be nonzero. For instance, the zero matrix is diagonal.


# Solving problem using related rates yields incorrect result

I've been trying for a while to figure out what I did wrong on this problem, help would be appreciated.

"A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially 20 ft away from the wall, how fast does the ladder move up the wall 5 sec after we start pushing?"

$$x$$ = distance from wall, $$y$$ = height of ladder on wall, $$h$$ = length of ladder $$\frac {dx}{dt} = -1, t = 5$$ $$x^2 + y^2 = h^2$$ $$x^2 +y^2 = 25^2$$ $$2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0$$ $$\frac{dy}{dt} = \frac{-x\frac{dx}{dt}}{y}$$ Since the ladder is moving toward the wall at 1 ft/sec for 5 sec, $$x(t = 5) =20-5 = 15$$

$$y(t=5) = \sqrt{25^2 - 15^2}=20$$ Substitute variables in equation: $$\frac{dy}{dt} = \frac{-(15)(-1)}{20}$$ $$\frac{dy}{dt} = \frac{3}{4}$$ The issue is that if $$\frac{dy}{dt} = \frac{3}{4},$$ the ladder would have risen 3.75 ft. in 5 seconds instead of 5 feet it SHOULD have risen, which is given by: $$\Delta y = y(t=5) - y(t=0)$$ $$(\sqrt{25^2 - 15^2}) - (\sqrt{25^2 - 20^2})$$ $$15-20$$ $$5$$ Thank you for reading and any answers.

At the start you have $$y(0)=\sqrt {25^2-20^2}=15$$. You are correct that the top has risen $$5$$ feet in $$5$$ seconds, but that does not mean that it has risen steadily at $$1$$ ft/sec for the whole time. Because $$\frac {dx}{dt}=-1$$ you found $$\frac {dy}{dt}=\frac xy$$ At $$t=0$$ that means that $$\frac {dy}{dt}=\frac {20}{15}=\frac 43$$ and at $$t=5$$ that means $$\frac {dy}{dt}=\frac {15}{20}=\frac 34$$. There is no contradiction here.

• Oh, I assumed that the rate of change had to be constant for some reason. Thank you for your help! – Kebab Boy Mar 14 at 20:07

Keeping your notation, you have that $$x(t) = 20 - t, \quad y(t)=\sqrt{25^2-x(t)^2},$$

and so, $$y'(t) = \frac{-2 x'(t)x(t)}{2 \sqrt{25^2-x(t)^2}},$$

which means that $$y'(5)=\frac{15}{\sqrt{25^2-15^2}}=\frac{3}{4}$$

The vertical displacement does not have to match the horizontal one. They are just connected through the expression $$x(t)^2 + y(t)^2=25^2$$.


Solving uartics

BloodyFrozen
Is there a method to finding the roots of quartics besides Ferrari's formula?

I have the equation

$$x^{4}+5x^{2}+4x+5=0$$

I know one of the factor is something like $$x^{2}+x+1$$ and the other one can be found using sythetic division, but how can I find the factors without knowing one of them in the first place?

Thanks.

Homework Helper
If the quartic has "nice" quadratic factors, then they will be of the form:

$$x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)$$

Can you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the $\pm$ operator.

BloodyFrozen
If the quartic has "nice" quadratic factors, then they will be of the form:

$$x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)$$

Can you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the $\pm$ operator.

Oh, I see, but how do we know when the factors are "nice"?

Homework Helper
Oh, I see, but how do we know when the factors are "nice"?

We don't. We can always check to see if they are, just like we check to see if there are any rational roots.

With the quartic $x^4+5x^2+4x+5$ we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, $x^4+6x^2+4x+5$ then we find no solutions for a,b in the system of equations.

BloodyFrozen
Ok, thanks!

BloodyFrozen
Ah, I believe I've seen this before. The method is nice, but it seems a little lengthy. Thanks for the link

BloodyFrozen
We don't. We can always check to see if they are, just like we check to see if there are any rational roots.

With the quartic $x^4+5x^2+4x+5$ we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, $x^4+6x^2+4x+5$ then we find no solutions for a,b in the system of equations.

I end up getting two systems but one of them has no solution. Did I proceed correctly?

Homework Helper
I end up getting two systems but one of them has no solution. Did I proceed correctly?

I don't know, what did you get?

BloodyFrozen
Sorry for not having LATEX since I'm on my phone, but I got:

System 1:

a+b=0
5+1+ab=5
5a+b=4

System 2:

a+b=0
-5-1+ab=5
-5a-b=4

Homework Helper
Sorry for not having LATEX since I'm on my phone, but I got:

System 1:

a+b=0
5+1+ab=5
5a+b=4
Right, so simplifying this system, we have

a+b=0
ab=-1
5a+b=4

Which we can then deduce,
a=-b
therefore,
a2=1 -> a=$\pm1$
4a=4 -> a=1, b=-1

System 2:

a+b=0
-5-1+ab=5
-5a-b=4
For this system we have no real solution, so what does that tell you?

BloodyFrozen
For the second system, I don't get any solutions. Therefore, that can't be the right system.

Homework Helper
For the second system, I don't get any solutions. Therefore, that can't be the right system.

Exactly, so what must your factors be?

BloodyFrozen
Ah, that's what I thought. Thanks!


# The idea behind the sum of powers of 2

I know that the sum of power of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place.

For example, sum of n numbers is $\frac{n(n+1)}{2}$. The idea is that we replicate the set and put it in a rectangle, hence we can do the trick. What is the logic behind the sum of power of $2$ formula?

• This addition of powers of two, and many other examples, are all geometric series. There's a lot of visualisations of this out there. – Parcly Taxel Oct 29 '16 at 10:58
• Write the numbers in base 2: The powers of $2$ starting from $1=2^0$ will be in binary, $1+10+100+1000$ will always be a number that will be a n with all binary digits 1. This is the largest number having that many digits. SO it is of the form $2^{n+1}-1$ – P Vanchinathan Oct 29 '16 at 11:01
• I think there's a counting/probabilistic way to approach this something about having $n$ binary choices – BCLC Oct 31 '16 at 11:17

## 8 Answers

The binary expansion of $$\sum_{k=0}^n2^k$$ is a string of $$n+1$$ 1's: $$\underbrace{111\dots111}_{n+1}$$ If I add a 1 to this number, what do I get? $$1\underbrace{000\dots000}_{n+1}$$ 1 followed by $$n+1$$ 0's, hence $$2^{n+1}$$. Therefore $$\sum_{k=0}^n2^k=2^{n+1}-1$$

• Very clever proof! – user3932000 Nov 28 '20 at 4:16

This works for any partial sum of geometric series.

Let $$S = 1 + x + x^2+\ldots +x^n$$. Then $$xS = x + x^2 + \ldots +x^n + x^{n+1} = S - 1 + x^{n+1}$$.

All you have to do now is solve for $$S$$ (assuming $$x\neq 1$$).

• Yes, but the OP said that he already knew this. What he is asking for is a visual representation of the proof. – Alex M. Oct 29 '16 at 11:01
• Actually, they said that they knew proof by induction, but didn't know how one would come up with the formula in the first place. At least that is how I understood the question. @AlexM. – Ennar Oct 29 '16 at 11:02

There is a geometrical explanation.

Take a box long enough to put twice the amount of items equal to the last term of the sum, and try to pack it.

Let's take a box of length $$2 * 2^n$$.

Let's put the items from the first term ($$2^n$$) into the box. Now exactly one half of the space is left for the other terms, from $$2^{n-1}$$ down to $$1$$, so we repeat this process, starting from the next largest term.

As we put the items from each term, we notice that they take exactly one half of the empty space left in the box, because both the largest term left and the space left decreases in half on each step.

At some point we reach the first term, which is equal to 1, and there are two empty places left in the box for just one item, so after we put the last item into the box, there is still space for one more item.

The length of the box is $$2*2^n = 2^{n+1}$$, but it could be shorter by one, which is $$2^{n+1} - 1$$, and this is our formula.

For example, let's pack: $$\sum_{i=0}^3 2^i$$

Box length: $$2 * 2^3 = 16$$

2^3    |● ● ● ● ● ● ● ●|               |
2^2    |○ ○ ○ ○ ○ ○ ○ ○|● ● ● ●|       |
2^1    |○ ○ ○ ○ ○ ○ ○ ○|○ ○ ○ ○|● ●|   |
2^0    |○ ○ ○ ○ ○ ○ ○ ○|○ ○ ○ ○|○ ○|●| |

It could be shorter by one: $$2^{3+1}-1 = 15$$

By the way, similar geometrical explanation can be used for the ever decreasing geometric progression $$\sum_{i=0}^\infty 2^{-i}$$ with the only difference that we do not need to account for the last piece of empty space, because it tends to 0. Instead, we take some continuous medium as an example, such as a ribbon or a thread, which can be divided virtually infinitely so it sums to exactly a length of 2 units.

There is a combinatorial interpretation. Consider the collection of all binary sequences of length $$n+1$$ with at least one $$1$$ (call this set $$E$$). There are $$2^{n+1}-1$$ such sequences because only $$0...0$$ isn't. Now let $$E_j$$ be the set of binary sequences of length $$n+1$$ such that the first $$1$$ is in the $$j$$ th component for $$j=1,\dotsc, n+1$$. Then $$|E_j|=2^{n+1-j}$$. Then the $$(E_j)$$ partition $$E$$ and we have that $$1+2+2^2+\dotsb+2^n=\sum_{j=1}^{n+1}2^{n+1-j}=2^{n+1}-1.$$ Alternatively consider the telescoping sum: $$\sum_{k=0}^n 2^{k}=\sum_{k=0}^n (2^{k+1}-2^k)=2^{n+1}-1.$$

Another famous approach is to count the $$2$$-player games to determine a winner among $$2^{n+1}$$ people. It's $$2^{n+1}-1$$, the number of people to eliminate. It's also $$2^n$$ people eliminated in the first round, halving each round until we reach the final.

• Very beautiful and efficient proof ! – projetmbc Oct 23 '20 at 9:57

Another pictorial proof, similar to the box packing reply, is to count the number of nodes in a complete binary tree with n+1 levels.

$$\sum_{i = 0}^{n} 2^i$$

is the number of nodes in the complete binary tree for the levels 0 (the root) through n.

If you draw an example and number the nodes like this

       1
/   \
2     3
/ \   / \
4   5 6   7
/ \
8   ...

etc. you will notice that the left-most node of each level is a power of 2 (1, 2, 4, 8, 16, ...).

so the number of nodes in the tree from root to level n is 2^{n+1} - 1.

$$\sum_{i = 0}^{n} 2^i = 2^{n+1} - 1 \,.$$

Another geometric interpretation.

Start with a line segment, AB. Extend this into a line. Use compassto mark off a point C by setting center at B and passing radius through A. Then lengths, AB=BC. Create point D in a similar way. We have thus constructed a length double that of the original line segment.

This doubling can continue AS MUCH S you like.

Now ignore the first line segment, AB.

The series of segments now starts with BD which has double the length of AB.

In general the nth segment in the first case is the n-1th segment upon removal of AB. Further, each is double the length of its corresponding section.

So By subtracting a part we've doubled what remains, but with fewer sections to account for.

This gives the formula.

Not sure if my answer has been posted yet, but here's how I understand it.

You are trying to understand why

$$2^{n+1} - 1 = 2^n + 2^{n-1} + 2^{n-2} . . . + 2^1 + 2^0$$

Suppose we take 2^n in the sum. We know since these are powers of two, that the previous term will be half of 2^n, and the term before that a quarter of 2^n.

Let n in 2^n be 1, or 2^1 = 2. The term before in the sum will be half of 2, so we can also write the entire sum as:

$$2^1 + \frac{1}{2}(2^1)$$

If you do this but for different values of n for 2^n you will find you can rewrite the sums as:

$$2^n + \frac{ 2^n - 1}{2^n} ( 2^n)$$

You can simplify this because you're dividing (2^n) - 1 by 2^n, and then multiplying it by 2^n which cancel each other out. The new simplified version with the cancellation will look like:

$$2^n + 2^n - 1$$

This can also be written as: $$2( 2^n) - 1$$

Further simplified into: $$2^{n + 1} - 1$$

And that is why that formula shows up! Hope this helps ( I'm not sure if this was the mathematical induction proof you already knew, I'm young and discovered this on my own so I don't know what this is).


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# Elementary Techniques used in the IMO (International Mathematical Olympiad) - Cauchy Schwarz Inequality

If people asked you, what is the most elementary inequality you know? I bet your answer would be AM-GM. But in this series of posts I will try to show you the power that Cauchy Schwarz has over that of AM-GM. But as an introduction, let us first state and prove the theorem.

Cauchy Schwarz Inequality: Let $$(a_1, a_2, \ldots , a_n)$$ and $$(b_1, b_2, \ldots, b_n)$$ be two sequences of real numbers, then we have:

$$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \geq (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$$

In particular, equality holds iff there exists $$k \in \mathbb{R}$$ for which $$a_i = k b_i$$ for $$i = {1, \ldots, n}$$.

Proof: We will present $$2$$ proofs, one originating from analysis on the equality case, the other by wishful thinking on small cases of $$n = 2,3$$.

(i) Consider defining the following function $$f$$:

$$f(x) = (a_1x - b_1)^2 + (a_2x - b_2)^2 + \ldots + (a_nx - b_n)^2$$.

We will expand this to get:

$$f(x) = (a_1^2 + a_2^2 + \ldots + a_n^2)x^2 - 2(a_1b_1 + a_2b_2 + \ldots a_nb_n)x + (b_1^2 + b_2^2 + \ldots + b_n^2)$$

From our first way of representing $$f(x)$$, we can conclude that $$f(j) ≥ 0 \leftrightarrow ∆_f \leq 0$$ or

$$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \geq (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$$

Equality holds if the equation $$f(x) = 0$$ has one root.■

(ii) Just remark that:

$$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) - (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2 = \sum_{i,j=1}^n (a_ib_j + a_jb_i)^2 ≥ 0$$■

Let us note the following positive things regarding Cauchy-Schwarz:

• it is effective in proving symmetric inequalities

• Try to form squares

• Helps to clear up square roots

Look forward to the next few posts to see applications of this extremely elegant inequality!

Note by Anqi Li
3 years, 3 months ago

Sort by:

Hi everyone, I'm part of #TorqueGroup too. My assignment was to post olympiad level problems for the community that I found interesting. Since Anqi is posting these great topics about math things useful in such math, it seems like we're a perfect fit. So, here are some examples:

Show that for real numbers $$a_1, a_2, \cdots$$ and $$b_1, b_2, \cdots$$

$$\displaystyle \sum_{i = 1}^{k} \frac {a_i^2}{b_i} \geq \frac{(\sum_{i = 1}^k a_i)^2}{\sum_{i = 1}^k b_i}$$

IMO 1995

Let $$a, b, c \in \mathbb{R}^+$$ such that $$abc = 1$$. Prove that

$$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}$$.

..

Hint: It may help to make a symmetric substitution for $$a = \alpha, b = \beta, c = \gamma$$ such that the condition $$\alpha \beta \gamma = 1$$ still holds.

USAMO 2009

For $$n \geq 2$$ let $$a_1, a_2, \cdots, a_n$$ be positive reals such that

$$(a_1 + a_2 + \cdots + a_n)(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}) \leq (n + \frac{1}{2})^2$$

Prove that $$max (a_1, a_2, \cdots, a_n) \leq 4 min (a_1, a_2, \cdots, a_n)$$.

..

Hint: The LHS should scream cauchy. Choose which $$a_i$$ multiplies which $$b_k$$ to isolate a desired max and min, and then do some algebra. · 3 years, 3 months ago

I will be happy to post my solutions if requested (as long as you try them first!) · 3 years, 3 months ago

Could you add these examples to the Applications of Cauchy Schwarz Inequality Wiki page? Staff · 2 years, 5 months ago

For the second one, following your hint, let $$a = \frac{1}{x}, b = \frac{1}{y}, c = \frac{1}{z}$$. The expression is then $$\sum_{cyc} \frac{x^3yz}{y+z}$$. Since $$xyz = 1$$, this is $$\sum_{cyc} \frac{x^2}{y+z}$$. Using Cauchy, this is greater than $$\frac{(x+y+z)^2}{2(x+y+z)}$$. So now it is left to prove that $$x+y+z \geq 3$$. This is simple using AM-GM, so $$\frac{x+y+z}{3} \geq \sqrt[3]{xyz}$$, and we are done. $$\blacksquare$$

For the second one, WLOG assume $$a_1$$ is the maximum and $$a_n$$ is the minimum. Using Cauchy,

$$(a_1 + a_2 + \cdots + a_n)(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}) \geq (\sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2)$$

$$(n + \frac{1}{2})^2 \geq (\sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2)^2$$

Taking square roots of both sides, we get

$$n + \frac{1}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2$$

$$\frac{5}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}}$$

Now, squaring, we get

$$\frac{25}{4} \geq \frac{a_1}{a_n} + 2 + \frac{a_n}{a_1}$$

Multiplying both sides by $$a_1 a_n$$ we get

$$a_1^2 - \frac{17}{4} a_1 a_n + a_n^2 \leq 0$$

Factoring...

$$(a_1 - 4a_n)(a_1 - \frac{1}{4}a_n) \leq 0$$

Since $$a_1$$ and $$a_n$$ are positive reals, $$(a_1 - 4a_n)$$ must be nonpositive. Thus,

$$(a_1 - 4a_n \leq 0 \rightarrow a_1 \leq 4a_n$$, and we are done. · 3 years, 3 months ago

Could you explain the second one as in how did you use Cauchy Schwartz in it. · 2 years, 1 month ago

What are symmetric inequalities? · 2 years, 8 months ago

Hope you share many more such techniques!!thnx a lot for this. · 3 years, 3 months ago

Yup I will. In fact, I posted some on invariants and I also wrote some worked examples in my newest post. :) · 3 years, 3 months ago

@Anqi Li Can you add this to the Cauchy Schwarz Inequality Wiki page? Staff · 2 years, 5 months ago

One can easily prove Cauchy-Schwarz using vectors too. It is similar to the proof above, but is much cleaner. The interesting this is that almost every other inequality can be derived from this inequality. · 3 years, 3 months ago


# Calculate limit without L'Hopital's rule

I have to calculate this limit whitout using L'Hopital's rule or Taylor polynomials:

$$\lim_{ x\to \pi/4 } \frac{1 - \tan(x)}{x-\frac{\pi}{4}}$$

I know how to make it using L'Hopital and that the result is $-2$ ,but I'm getting nowhere when I try without it. Any advice?

Hint:

Let $t=x-\frac{\pi}{4}$, then $$\frac{1-\tan x}{x-\frac{\pi}{4}}=\frac{1-\tan\left(t+\frac{\pi}{4}\right)}{t}=\frac{1}{t}\cdot\left(1-\frac{\tan t+\tan\frac{\pi}{4}}{1-\tan \frac{\pi}{4}\tan t}\right)=\frac{1}{t}\left(1-\frac{\tan t+1}{1-\tan t}\right)=\frac{-2\tan t}{t(1-\tan t)}$$ Now, take the limit as $t\to 0$: \begin{align} \lim_{x\to \frac{\pi}{4}}\frac{1-\tan x}{x-\frac{\pi}{4}}&=\lim_{t\to 0}\frac{-2\tan t}{t(1-\tan t)}\\ &=\lim_{t\to 0}\frac{-2\tan t\cos t}{t(1-\tan t)\cos t}\\ &=-2\lim_{t\to 0}\frac{\sin t}{t(\cos t-\sin t)}\\ &=-2\left(\lim_{t\to 0}\frac{\sin t}{t}\right)\left(\lim_{t\to 0}\frac{1}{\cos t-\sin t}\right)\\ &=-2(1)(1)\\ &=\color{blue}{-2} \end{align}

• While all the answers are valid and helped me,I chose this because it doesn't imply the use of more advanced concepts like derivatives.Thanks! – Der Rosenkavalier Oct 4 '15 at 20:48

Hint: What is the definition of the derivative of $\tan(x)$ at $x=\pi/4$?

• $\frac{1}{\cos^2(\frac{\pi}{4})}$ But what's the point? I can't use L'Hopital's rule because the exercise ask it, not because I don't how to use it.I don't understand how can I use the derivative of $\tan(x)$ at $\frac{\pi}{4}$ in this exercise. – Der Rosenkavalier Sep 30 '15 at 21:47
• @DerRosenkavalier By definition: $$\tan'(\pi/4)=\displaystyle\lim_{x \to \pi/4}\frac{\tan(x)-1}{x-\pi/4}$$. How does this compare to your limit? – Reveillark Sep 30 '15 at 21:49
• L'Hopital isn't being used here; just the definition of the derivative. It's important to know the difference. – zhw. Sep 30 '15 at 22:00
• @zhw I agree with your comment. This type of question sometimes rasises debate as to "allowed ways forward. For example, is the use of asymptotic analysis permitted here? IMHO, yes absolutely. Yet others would argue that that approach is tantamount to the use of LHR. ;-) – Mark Viola Sep 30 '15 at 22:21
• @DerRosenkavalier : My answer posted here explicitly explains how you can use the value of the derivative of $\tan x$ at $\pi/4$. ${}\qquad{}$ – Michael Hardy Oct 1 '15 at 4:32

Recall that $$f'(a) = \lim_{x\to a}\frac{f(x) - f(a)}{x-a}.$$ Apply it to the case where $f(x) = \tan x$ and $a=\pi/4$. Then $f(a) = 1$ and $f'(a) = \sec^2 a = \sec^2(\pi/4) = 2$. Therefore $$2 = \lim_{x\to\pi/4}\frac{\tan x - \tan(\pi/4)}{x-\pi/4}.$$

So $-2$ is the answer to the question as you've posed it.

We have the identities

\begin{align}\frac{1-\tan x}{x-\pi/4}&=-\frac{\tan(x-\pi/4)}{x-\pi/4}(1+\tan x)\\\\ &=-\frac{\sin(x-\pi/4)}{x-\pi/4}\frac{1+\tan x}{\cos(x-\pi/4)} \end{align}

Can you finish from here?

• Yes,I could use that $\lim_{ x\to 0 }\frac{\sin(x)}{x}=1$ . Thanks for the help! – Der Rosenkavalier Oct 4 '15 at 20:58
• You're welcome. My pleasure. This is a very efficient approach that circumvents use of L'Hospital's Rule, as you requested. – Mark Viola Oct 4 '15 at 21:07

you can use the following hint ,

try to write $\tan(x)$ in terms of $\sin(x)$ and $\cos(x)$. A little bit of rearranging and then use the following

$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$ A little bit of work yields the result .


How to formalize the fact that $f(i)=\lceil i/2\rceil$ is surjective but not one-to-one from $\Bbb N$ to itself?

If the set $S$ is countably infinite, prove or disprove that if $f$ maps $S$ onto $S$ (i.e. $f\colon S\to S$ is a surjective function), then $f$ is one-to-one mapping.

Please give a formal mathematical proof for this statement.

I have a counter-example: Suppose the mapping is from $\Bbb N$ (the natural numbers, a countably infinite set) to $\Bbb N$. And $f(i) = \lceil i/2\rceil$. It is onto function but not one-one.

But I am not getting that how this mapping is onto. Intuitively, since $1$ and $2$ will be mapped to $1$; $3$ and $4$ will be mapped to $2$; $5$ and $6$ will be mapped to $3$; similarly, $n-1$ and $n$ will be mapped to $n/2$. So in the codomain, there will be some elements that have no pre-image for this countably infinite set. So, please correct me where I am wrong.

• Welcome to MSE. Please use MathJax. – José Carlos Santos May 15 '18 at 16:35
• "So in the codomain, there will be some elements that have no pre-image for this countably infinite set." Why do you say that? You just seemed to give an argument that it would. Anyway for any $n$ then both $2n$ and $2n+1$ will map to $n$. So $f$ is onto. – fleablood May 15 '18 at 16:47
• sir , since , cardinality of both sets are same and every 2 elements of domain are mapped to one element of codomain . so intuitively , it looks that some elements of codomain will not have any pre-image. please correct me – ankit1729 May 15 '18 at 16:51
• You just should that $1,2,3$ and $\frac n2$ (assuming $n$ is even) all have pre-images. What elements don't have pre-images? I honestly don't understand why you said that. – fleablood May 15 '18 at 16:51
• Are you thinking because you will run out of elements? But the set is infinite. You will never run out. So it's perfectly fine to have 2, 3 or even an infinite number of elements all map to the same element and still be onto. In fact that is exactly what you are proving. It is possible to be surjective and not 1-1. – fleablood May 15 '18 at 16:54

2 Answers

To prove that a function $f\colon A\to B$ is surjective, you need to be able to prove that given $b\in B$, there is some $a\in A$ such that $f(a)=b$.

In your case, $A=B=\Bbb N$, and $f(x)=\lceil \frac x2\rceil$.

So you need to prove that given any $m\in\Bbb N$, there is some $n\in\Bbb N$ such that $\lceil\frac n2\rceil=m$. Even though this $n$ is not unique, I am sure that you can come up with a way to find such $n$, from a given $m$.

Maybe I'm missing something. For every n in codomain, 2n and 2n-1 are in preimage, so the mapping is onto but not 1-1

• sir , it may or may not be one-one – ankit1729 May 15 '18 at 17:11


# what is the magnitude of the difference vector?

I read this question in Sears and Zemansky's University Physics book:

"Two displacement vectors, S and T, have magnitudes S = 3 m and T = 4 m. Which of the following could be the magnitude of the difference vector S - T ?(There may be more than one answer.) i. 9 m; ii. 7 m; iii. 5 m; iv. 1 m; v. 0m; vi. -1m "

My question is: shouldn't the direction of both vectors be specified in order for me to solve it? Or do I just assume that S and T are positive and negative respectively as per the equation S - T

• The question is asking you what the possible values of $\mathbf S - \mathbf T$ could be for all possible directions. So for example if one of the options was $666$m you need consider if there is any possible arrangement of the two vectors that could give $|\mathbf S - \mathbf T| = 666$. This freedom to orient the vectors in any direction means more than one of the answers can be correct. – John Rennie Jan 12 '17 at 16:20
• That's what I thought initially, but I had doubts. Thanks for confirming! – DigiNin Gravy Jan 12 '17 at 16:23
• Re Answer (vi): can the magnitude of a vector be negative? – DJohnM Jan 12 '17 at 17:18
• @DJohnM I know that the magnitude can never be negative, but the vector can. – DigiNin Gravy Jan 12 '17 at 18:20
• @DigiNinGravy A vector which is anti-parallel to, say, the $x$-axis of a coordinate system is sometimes called "a vector in the negative-$x$ direction." But it is an incorrect oversimplification to call such an object "a negative vector." – rob Jan 13 '17 at 7:26

The magnitude of the difference vectors depends on the orientation of $$\bf\vec{S}$$ and $$\bf \vec{T}$$. If they are parallel then $$|\bf \vec{S}-\bf \vec{T}|=|\,|\bf \vec{S}|-|\bf \vec{T}|\,|$$ and if they are anti-parallel then $$|\bf \vec{S}-\bf \vec{T}|=|\bf \vec{S}|+|\bf \vec{T}|$$.

Thus the possible values of $$|\bf \vec{S}-\bf \vec{T}\|$$ lie in the range:

$$|\,|\bf \vec{S}|-|\bf \vec{T}|\,| \leq |\bf \vec{S}-\bf \vec{T}| \leq |\bf \vec{S}|+|\bf \vec{T}|$$

I'll let you work out which answers comply with this.

• |S - T| = 5 or | |S| - |T| | = -1 or |S| + |T| = 7. I just don't understand how |S - T| = |S| + |T|. I mean to have |S| + |T| both vectors must be parallel - having the same direction - so | |S| - |T| | seems off, unless we consider T as -T? – DigiNin Gravy Jan 12 '17 at 17:09
• @DigiNinGravy I think you gave the aswer yourself both $\bf T and$-\bf T$have se same magnitude, and the magnitude does not give the direction of the vector. – Mikael Fremling Jan 12 '17 at 18:10 In above Figure move the end point$\:\mathrm{B}\:$of the vector$\:\mathbf{S}\:$around a circle of radius$\:|\mathbf{S }|=3\:$. Try to find the length$\:|\mathbf{S}-\mathbf{T}|\:$of the vector$\:\mathbf{S}-\mathbf{T}\:$when$\:\mathrm{B}\:$is on points$\:\mathrm{P_{ii}},\mathrm{P_{iii}},\mathrm{P_{iv}}\:\$.

• Thanks, but you reversed the magnitudes of S and T. Can you tell me the name of the program you used? – DigiNin Gravy Jan 12 '17 at 18:17
• @DigiNin Gravy : GeoGebra – Frobenius Jan 12 '17 at 18:35


# Finding $n$ permutations $r$ with repetitions

I have given $n$ items and I have to find ${}^nP_r$ and $\sum_{r=0}^n{}^nP_r$ with repetitions allowed. Is there any closed formula for this?

For $n=3$ and $r=1$, possible permutations are:
$\{1\},\{2\},\{3\}$
Total: 3

For $n=3$ and $r=2$, possible permutations are:
$\{1,1\},\{2,2\},\{3,3\},\{1,2\},\{2,1\},\{1,3\},\{3,1\},\{2,3\},\{3,2\}$
Total: 9

For $n=3$ and $r=3$, possible permutations are:
$\{1,1,1\},\{2,2,2\},\{3,3,3\},$
$\{1,1,2\},\{1,2,1\},\{2,1,1\},$
$\{1,2,2\},\{2,1,2\},\{2,2,1\},$
$\{1,1,3\},\{1,3,1\},\{3,1,1\},$
$\{1,3,3\},\{3,1,3\},\{3,3,1\},$
$\{2,2,3\},\{2,3,2\},\{3,2,2\},$
$\{2,3,3\},\{3,2,3\},\{3,3,2\}$,
$\{1,2,3\},\{1,3,2\}$,
$\{2,1,3\},\{2,3,1\}$,
$\{3,1,2\},\{3,2,1\}$,
Total: 27

Can we come up with any closed formula for individual ${}^nP_r$ with repetitions and also for their sum i.e. here $3+9+27=39$

I understand that I cannot call this exactly the permutation, since ${}^3P_3$ is strictly $6$, while above its $27$, since I allow repetitions, but then whatever it is, how do I get the count?

Note that permutations with repetition is usually the well known case corresponding to $\frac{n!}{n_1!n_2!...n_i!}$, which is not what I am asking here. Is what am asking also some well know case, and I am stupidly not able to guess it? My primary guess is that, there cannot be any closed formula. Is it right?

• What exactly is the condition you're looking for? I notice that the permutation $1,2,3$ is excluded - is this intentional? – platty Aug 20 '17 at 15:31
• In fact, it looks like you left out all the permutations of $1,2,3$ - is this intentional? – platty Aug 20 '17 at 15:41
• Nope, I missed it. Sorry. – anir Aug 20 '17 at 15:43
• I dont know it feels fuzzy. Is it just ${}^nP_r \text{with repetition} = n^r$?. Feels like so. Let me read question and answers again. – anir Aug 20 '17 at 15:47
• Your formula is not for permutations with replacements. It's for combinations with replacements. Are you actually looking for the expression for multinominal coefficients? – Vim Aug 20 '17 at 15:48

If you just want unrestricted strings consisting of $r$ letters, chosen with replacement from $n$, you can just use the multiplication rule to get $n^r$; there's $n$ choices for the first one, $n$ choices for the second, etc.

Extending this, we can use this to find the number of strings with length up to $r$ by summing the intermediate results: $$1+n+n^2+\dots +n^r$$ This is the sum of a geometric series, which means we can apply the formula to get $\frac{n^{r+1}-1}{n-1}$.

If you mean "permutations with replacements" then the answer is just $n^r$. But this doesn't match your result for $n=r=3$ which should be $27$? If you actually mean combinations with replacements, then see the below hint.

Hint: you are basically putting $r$ identical balls into $n$ different jars, or equivalently, finding the number of non-negative integer solutions to $$x_1+\cdots+x_n=r$$ To solve this, first let $y_i:=x_i+1$ (which are bijections) then try finding the number of positive integer solutions to the equation $$y_1+\cdots+y_n=r+n.$$ Try Stars & Bars technique.

• I don't think this is correct - OP has order mattering in the listed examples. – platty Aug 20 '17 at 15:39
• @platty this is really confusing, because in this case the third result should be 27. Clearly there is some misphrasing in the question. – Vim Aug 20 '17 at 15:41
Note that in your last paragraph, you say that $\frac{n!}{n_1!n_2! \dots n_i!}$ is the formula for permutations with repetition, but perhaps a better way to word it would be that this is a generalized formula for partitioning a set of $n$ objects into $i$ cells with each "cell" being distinguishable, but elements within their respective cells are not.
As noted before, permutations with repetition allowed can be represented with $n^r$, which @platty already explained. This is the formula you would use to solve the second example I gave, as well as what you seem to be looking for in your example.
the number of words, from alphabet $\{1,\cdots,n\}$, of length $r$ and max repetition $r$,
i.e. simply the number of words, from alphabet $\{1,\cdots,n\}$, of length $r$
which are $n^r$.