# The sum of consecutive integers is $50$. How many integers are there?

I started off by calling the number of numbers in my list "$n$". Since the integers are consecutive, I had $x + (x+1) + (x+2)...$ and so on. And since there were "$n$" numbers in my list, the last integer had to be $(x+n)$. This is where I got stuck. I didn't know how to proceed because I am not given the starting point of my integers, nor an ending point.

• If there are $n$ numbers in your list and the first is $x = x + 0$, then the last is $x + n - 1$. – N. F. Taussig Jun 24 '17 at 10:37

If your $n$ is odd, then the middle number has to be $50/n$. The odd divisors of $50$ are $1$ and $5$, which gives us two solutions $50=50$ and $8+9+10+11+12=50$

If $n$ is even, then $50/n$ is the half-integer between the middle two numbers. So $n$ has to be an even divisor of $100$, but not a divisor of $50$, so $n=4$ or $20$.

If $n=4$, then $50/4 = 12.5$ and we get $11+12+13+14=50.$

If $n=20$. then $50/20 = 2.5$ and we get $-7+-6+-5+\cdots +11+12 = 50.$

So there are 4 answers: $n=1, 4, 5,$ and $20$.

Edit: As Bill points out, I missed the divisors $25$ and $100$, which give two more answers: $50 = -10+-11+\cdots+14$ and $50 = -49 +-48+\cdots +50$.

Note that each solution with negative integers is related to an all-positive solution. From the solution $11+12+13+14=50$, we just prepend the terms $-10, -9, \ldots, 10$, which add to $0$, and we have another solution.

• interesting the approach using the middle number! – G Cab Jun 24 '17 at 12:31
• Great technique! n=25 and n=100 work aswell – Mike Jun 24 '17 at 13:47
• Incomplete, e.g. the odd divisor $25\ \$ – Bill Dubuque Jun 24 '17 at 14:01
• @GCab If you think of it in terms of the "average" (mean), instead of the "middle" (median) number, then it's less surprising. The two only happen to be equal because the numbers are uniformly distributed. – Robin Saunders Jun 24 '17 at 15:24

What may be helpful is to use the formula for the sum of an arithmetic progression: if you have a sequence whose first term is $a$ and each term is $d$ more than the rest, then the sum of the first $n$ terms is $na+\frac{n(n-1)}{2}d$. In this case, since we are looking at consecutive integers, $d=1$, and so you are trying to find $a$ and $n$ such that $na+\frac{n(n-1)}2=50$, or equivalently $n(2a+n-1)=100$.

• This is surely helpful. As there are only nine divisors of 100, you can try them all easily and look at which lead to an integer $a$. The nice thing is that (unlike when using a more sophisticated solution) you won't miss the negative first terms. – maaartinus Jun 25 '17 at 13:38
• Yes, and you can make things slightly quicker using the fact that if $n$ is even then $2a+n-1$ is odd, and vice versa, so you can skip cases like $2\times 50$ where both factors are even. – Especially Lime Jun 25 '17 at 19:29

\scriptsize\begin{align} 50 &=2\times 25 &&=25\boxed{+}25 &&=24.5\boxed{+}25.5 \text{ (AP but not integer AP)}\\ &=4\times 12.5 &&=12.5+12.5\boxed{+}12.5+12.5 &&=\color{red}{11+12\boxed{+}13+14}\\ &=5\times 10 &&=10+10+\boxed{10}+10+10 &&=\color{red}{8+9+\boxed{10}+11+12}\\ &=10\times 5 &&=\underbrace{\underbrace{5+5+\cdots+5}_{5}\boxed{+}\underbrace{5+5+\cdots +5}_{5} }_{10} &&=\underbrace{\underbrace{0.5+1.5+\cdots+4.5}_{25}\boxed{+}\underbrace{5.5+\cdots+9.5}_{25}}_{50}\\ & && && \quad \text{(AP but not integer AP)}\\ &=20\times 2.5 &&=\underbrace{\underbrace{2.5+2.5+\cdots+2.5}_{10}\boxed{+}\underbrace{2.5+\cdots +2.5}_{10} }_{20} &&=\color{red}{\underbrace{\underbrace{-7+(-6)+\cdots+2}_{10}\boxed{+}\underbrace{3+4+\cdots+12}_{10}}_{20}}\\ &=25\times 2 &&=\underbrace{\underbrace{2+2+\cdots+2}_{12}+\boxed{2}+\underbrace{2+\cdots +2}_{12} }_{25} &&=\color{red}{\underbrace{\underbrace{-10+(-9)+\cdots+1}_{12}+\boxed{2}+\underbrace{3+\cdots+13+14}_{12}}_{25}}\\ &=50\times 1 &&=\underbrace{\underbrace{1+1+\cdots+1}_{25}\boxed{+}\underbrace{1+\cdots +1}_{25} }_{50} &&=\underbrace{\underbrace{-23.5+(-22.5)+\cdots+0.5}_{25}\boxed{+}\underbrace{1.5+\cdots+25.5}_{25}}_{50}\\ & && && \quad \text{(AP but not integer AP)}\\ &=100\times 0.5 &&=\underbrace{0.5+0.5+\cdots+0.5}_{50}\boxed{+}\underbrace{0.5+0.5+\cdots+0.5}_{50} &&=\color{red}{\underbrace{\underbrace{-49+(-48)+\cdots+0}_{50}\boxed{+}\underbrace{1+2+\cdots +49+50}_{50} }_{100}} \end{align}

In more detail:

$$\frac {50}n=m$$ If $n$ is even $(n=2p)$, then we want $m=a+0.5 \;\;(a\in \mathbb Z)$ $\cdots$ Condition $(1)$

• The AP would comprise $p$ consecutive integers on either side of $m$.

If $n$ is odd $(n=2p+1)$, then we want $m\in \mathbb Z$ $\cdots$ Condition $(2)$

• The AP would comprise $p$ consecutive integers on either side of $m$, as well as $m$ itself.

Try different values of $n$ (excluding the trivial case $n=1$):

• If $n=2$, then $m=25$, hence Condition ($1$) not satisfied.

• If $n=3$, then $m=16\frac 23$, hence Condition ($2$) not satisfied.

• If $n=4$, then $m=12.5$, hence Condition $(1)$ satisfied, so the AP is $\lbrace (m-1.5), (m-0.5), (m+0.5), (m+1.5)\rbrace$, i.e. $\color{red}{11,12,13,14}$ ($AP1$).
Adding negative terms and corresponding positive terms which cancel out gives: $\color{blue}{-10,-9,-8\cdots, 0\cdots, 8,9,10,}\color{red}{11,12,13,14}$ ($AP 1'$)

• If $n=5$, then $m=10$, hence Condition $(2)$ satisfied, so the AP is $\lbrace (m-2),(m-1),m,(m+1),(m+2)\rbrace$, i.e. $\color{red}{ 8,9,10,11,12}$ ($AP 2$).
Adding negative terms and corresponding positive terms which cancel out gives: $\color{blue}{-7,-6,-5\cdots, 0\cdots, 5,6,7,}\color{red}{8,9,10,11,12}$ ($AP 2'$)

• If $n=6, 7,8,9$, then $m\notin\mathbb Z$ and $m\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.

• If $n=10$, then $m=5$, hence Condition $(2)$ not satisfied - not possible.

• If $n=11,12,\cdots, 24$, then $m\notin\mathbb Z$ and $m\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.

• If $n=25$, then $m=2$, hence Condition $(2)$ satisfied, so AP is $\color{red}{\underbrace{-10,-9,\cdots 0,1}_{12\text{ terms}},2,\underbrace{3,\cdots 9,10,11,12,13,14}_{12 \text{ terms}}}$ (same as $AP 1'$)

• If $n=26,27,\cdots, 49$, then $m\notin\mathbb Z$ and $m\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.

• If $n=100$, then $m=0.5$, hence Condition $(1)$ satisfied, so AP is $\color{red}{\underbrace{-49,-48,\cdots -2,-1,0}_{50\text{ terms}},\underbrace{1,2,3,\cdots 48,49,50}_{50 \text{ terms}}}$ ($AP 3$)

• For higher values of $n$, $0<m<0.5$ - not possible.

Without reading other answers... this should tell you how an old computer programmer thinks, versus a real mathematician.

First the obvious answer is the single integer 50. However, if negative numbers are allowed, then we can scoop up the sequence from -49 to 50 for 100 consecutive numbers. This is the longest possible sequence.

If n is the starting number of a sequence and s is the number of consecutive numbers, then we end up with

50 = (n+0) + (n+1) + (n+2) +... + (n+s-1)
50 = ns + (s-1)(s)/2

This is helpful, maybe, as ns pretty much puts a box around the solution set. Since we know the nature of the rightmost term, we can tell that s is 10 or less.

Consider the transformation if we multiply both sides by 2/s:

100/s = 2n + s - 1

The right side will always be an integer. Therefore, s must always divide 100 evenly. From above we know that s is between 1 and 10, so the only possible values for s are 1, 2, 4, 5, and 10.

That reduces the problem to 5 single-degree-of-freedom equations. Let's do it.

100/1 = 2n, n = 50, seq is (50)
100/2 = 2n + 1, no solution
100/4 = 2n + 4 - 1, 25-3 = 2n, n = 11, seq is (11, 12, 13, 14)
100/5 = 2n + 5 - 1, 20 = 2n + 4, 16=2n, n = 8, seq is (8, 9, 10, 11, 12)
100/10 = 2n + 10 - 1, 10 = 2n +9, 1 = 2n, no solution

So those are all the solutions where n and s both > 0.

Using the standard summation of an arithmetic progression formula:
$S_n=\frac{n(2a_1+(n-1)d)}{2}$
here since $d=1$
$2S_n=n(2a_1+n-1)$
$2S_n=n(a_1+(a_1+n-1))$
Here $a_1+n-1$ is just the last term.
$2S_n=n(a_1+a_n)$
Rewrite as

If n is even:

### $n=\frac{2S_n}{2a_{n/2}}$

Since $a_1+a_{n/2}+a_n-a_{n/2}=2a_{n/2}$
hence

### $n=\frac{S_n}{a_{n/2}}$

and since $n\in Z$ , so $a_{n/2}$ is a divisor of $S_n$
once you get the required $n, a=a_{n/2}-n/2$
Similarly for odd $S_n$ you get

### $n=\frac{S_n}{a_{(n-1)/2}-1}$

where $a_{(n-1)/2}-1$ is a divisior of $S_n$

The sum of consecutive numbers $1\dots n$ up to $n$ starting from 1 is $\frac{n(n+1)}{2}$. Since you want only a partial sum from, say, $m+1$ to $n$, you can just subtract to get the partial sum:

$$\frac{n(n+1)}{2} - \frac{m(m+1)}{2} = \frac{n^2+n-m^2-m}{2} = 50$$

Multiplying by 2 gets you

$$100 = n^2+n-m^2-m = (n+m)(n-m) + n-m = (n+m+1)(n-m)$$

Now the trick: $n+m$ and $n-m$ are either both even or both odd. This means that in the last formula, one term must be even, the other one must be odd. Factorization of $100 = 5\cdot 5 \cdot 2 \cdot 2$ means that there are very limited solutions. For instance, one term is 25, the other is 4 (there is one other nontrivial combination, see below). Since $n+m+1$ is larger than $n-m$, in the choice shown here, you must have \begin{align} n+m+1 &= 25\\ n-m &=4 \end{align} Solving this will give you the desired numbers.

Addendum You get the combination by observing that you get all acceptable combinations of the factorisation via: \begin{align} 100 &= 1\cdot (5\cdot5\cdot2\cdot2)\\ &= 5\cdot (5\cdot2\cdot2)\\ &= (5\cdot 5)\cdot(2\cdot2) \end{align} where you have to stop as all the following factorisations will contain only even factors.

Let tere be $n\geq1$ numbers, the smallest of them being $p\in{\mathbb Z}$. We then want $$p+(p+1)+\ldots+\bigl(p+(n-1)\bigr)=50\ ,$$ which amounts to $np+{(n-1)n\over2}=50$, or $$n(n+2p-1)=100\ .$$ Going with $n$ through the $9$ divisors of $100$ we obtain the following table: $$\matrix{n:&1&2&4&5&10&20&25&50&100 \cr p:&50&&11&8&&-7&-10&&-49\cr}$$ When $n\in\{2,10,50\}$ solving $n+2p-1={100\over n}$ for $p$ does not lead to an integer $p$. It follows that there are $6$ solutions in all.

• How did you get from your first equation to the second? – Tony Ennis Jun 25 '17 at 21:26

First, welcome to Mathematics stack exchange. Second, if you start like $x+(x+1)+...+(x+n)$, you will end nowhere. $50$ is a concrete and small number so what can you try is to use that fact, sum of two consecutive numbers it is not, due to sum of two consecutive numbers can not be divided with $2$ in $\mathbb{Z}$, sum of three consecutive numbers is divisible with $3$, so it is not sum of three numbers. Four consecutive numbers- now this is on first look possible(remainder mod $4$ is $2$) but each one of them has to be around $12$, so first options that you have is $11+12+13+14=23+27= 50$, hence the solution. Maybe there are even other solution, but I feel free to interpret your question as required to see one possible solution.

• There are other solutions. The simplest is $50$. – N. F. Taussig Jun 24 '17 at 10:42


\$x_1,x_2\$ vs. \$x_1\$, \$x_2\$

Basic version of my question: Consider the fragment of some text `bla bla \$x_1,x_2,x_3\$ bla`. Is this correctly type-setted, or is `bla bla \$x_1\$, \$x_2\$, \$x_3\$ bla` better ?

Advanced version: Cf. the "Chicago Manual of Style", 16th printed ed, on pp. 589, in 12.19 it is recommended that typesetting lists of mathematical symbols, e.g. `\$x_1,x_2,x_3\$` that medium spaces between commas should be used. Does LaTeX insert these automatically ? If not, should I use `\$x_1\:,x_2,\:x_3\$` rather than `\$x_1\$, \$x_2\$, \$x_3\$` ? It appears that these strings differ slightly in length, when the document is compiled to pdf.

• It matters greatly what the three items `x_1`, `x_2`, and `x_3` -- or, say, `a`, `b`, and `c` -- denote: Are they just any three distinct elements, or do they form a sequence? In the former case, you should write `... \$x_1\$, \$x_2\$, \$x_3\$ ...`: the commas are parts of the sentence rather than parts of the formulas. In the latter case, you should definitely write `\$... x_1, x_2, x_3 ...\$`, as the commas are now a part of the overall math expression. If they form a sequence, you may also want to encase them in curly braces, i.e., write `\$\{x_1, x_2, x_3\}\$`. – Mico Sep 23 '15 at 10:45
• @Mico Sounds like an answer? – Gonzalo Medina Sep 23 '15 at 15:03
• @GonzaloMedina - Done. :-) – Mico Sep 23 '15 at 15:43

It matters greatly what the three items `x_1`, `x_2`, and `x_3` -- or `a`, `b`, and `c` -- denote. Are they just any three distinct elements, or do they form a structured entity such as a sequence?

In the former case, you should write, say,

``````The sum of \$x_1\$, \$x_2\$, \$x_3\$, etc.\ diverges because ...
``````

Don't include the commas in the math terms, because the commas are parts of the sentence rather than components of formulas. With this setup, line breaks can occur (if needed) after the commas.

In the latter case, though, you should definitely write

``````\$... x_1, x_2, x_3 ...\$
``````

because the commas are now parts of some larger math expression. And, if `x_1`, `x_2`, and `x_3` form, say, a three-element set, you may want to encase them in curly braces, i.e., write

``````\$\{x_1, x_2, x_3\}\$
``````

TeX will, in general, not insert line breaks after math-mode commas. If you must allow a line break in the formula and if the commas are sensible break-points, you'll need to tell TeX about this fact, by inserting judiciously placed `\allowbreak` directives, say,

``````\$\{a_1,a_2,\dots,a_n,\allowbreak a_{n+1}, \dots, a_{n+m}\}\$
``````
• you might also want to mention that, when entered as separate math strings, a line can break after one of the commas, but if it's a single math expression, it won't break at the end of a line. – barbara beeton Sep 23 '15 at 16:05
• @barbarabeeton - Good idea. :-) I'll add a sentence or two to this effect. – Mico Sep 23 '15 at 16:13
• First: To such a complete and carefully crafted answer, I cannot give anything but a thankful +1! Second: In the text I'm writing, `\$x_1\$` to `\$x_3\$` are something in between unrelated items and a structured entity: I'm dividing a line in three segments, called `\$A_1\$`, `\$A_2\$` and '`\$A_3\$` and I constantly have to mention things like "On the segments `\$A_1\$`, `\$A_2\$` happens X"; "in contrast, on the segment \$A_2\$`, `\$A_3\$` we have to deal with the problem Y"; "when looking at `\$A_1\$`, `\$A_2\$` and '\$A_3\$` together we can see Z". Would you agree that I should write them apart, as I just did ? – l7ll7 Sep 24 '15 at 13:54
• [...] Although, they do form an entity, as they are a partition of a line, which makes me somewhat unsure. – l7ll7 Sep 24 '15 at 13:56
• @user10324 - I'd write the segments as separate formulas, i.e., `on the segments \$A_1\$ and \$A_2\$, \$X\$ happens...`. On the other hand, I'd also write `Let \$(A_1,A_2,A_3)\$ be an exhaustive, ordered, and non-overlapping partition of the line segment \$A\$.` – Mico Sep 24 '15 at 15:01


# Question about asymptotic notation with logs of different bases

For the two functions f(n) and g(n), where $$f(n) = n^{1/2}$$ and $$g(n) = 2^{(\log_2 n)^{1/2}}$$, I am trying to determine whether $$f$$ is asymptotically bound below by $$g$$, i.e find whether there exists an N1 and a k1 > 0 such that:

$$f(n) >= k1 * g(n)$$, for all n >= N1.

By taking the log of both $$f(n)$$ and $$g(n)$$ I got:

$$f(n) = n^{1/2} = logn^{1/2} = 1/2*logn$$.

$$g(n) = 2^{(\log_2 n)^{1/2}} = log(2^{(\log_2 n)^{1/2}}) = (\log_2 n)^{1/2}log2$$.

In trying to solve this I have:

$$1/2logn >= k1 * (\log_2 n)^{1/2}log2$$

Setting k1 to 1/2:

$$1/2logn >= 1/2 * (\log_2 n)^{1/2}log2$$

Dividing both sides by 1/2:

$$logn >= (\log_2 n)^{1/2}log2$$

Subtracting $$(\log_2 n)^{1/2}log2$$ from both sides:

$$logn - (\log_2 n)^{1/2}log2 >= 0$$

I can probably figure out whether there is any N1 such that the above is true for all n >= N1 by substituting values. However, I was wondering if there is any way to further simplify the above expression. In particular is there any way to get rid of the $$(\log_2 n)^{1/2}$$ entirely? I would prefer all the logs in the inequality to have the same base. Any insights are appreciated.

• In my answer that you've accepted, I made a mistake which I've now corrected. In particular, I used $\log(ab) = \log(a) \log(b)$, which is what you've used as well, instead of the correct $\log(ab) = \log(a) + \log(b)$, when you take logarithms of both sides of the original inequality. I'm sorry for my mistake, but as you can see it does't change the overall result. – John Omielan Mar 4 at 4:33

## 1 Answer

You could finish your work using $$\log$$, I assume to base $$e$$ but, in answer to your request of

I would prefer all the logs in the inequality to have the same base.

since $$g(n)$$ uses a base of $$2$$ and has a $$\log$$ to the base of $$2$$ in the exponent, it would likely be simplest & easiest to instead take the logs of both sides to base $$2$$ so you're then comparing the exponents of $$f(n)$$ and $$g(n)$$ to that base. As such, this would change trying to prove that there exists an $$N_1$$ and a $$k_1 \gt 0$$ such that

$$f(n) \ge k_1 \, g(n) \; \forall \; n \ge N_1 \tag{1}\label{eq1}$$

to finding, by taking $$\log_2$$ of both sides, a $$k_2 = \log_2{k_1}$$ such that

$$\log_2{f(n)} \ge k_2 + \log_2{g(n)} \; \forall \; n \ge N_1 \tag{2}\label{eq2}$$

Note that

$$\log_2{f(n)} = \log_2{n^{\frac{1}{2}}} = \frac{1}{2} \, log_2 n \tag{3}\label{eq3}$$ $$\log_2{g(n)} = \log_2{2^{(\log_2 n)^{\frac{1}{2}}}} = (\log_2 n)^{\frac{1}{2}} \tag{4}\label{eq4}$$

Substituting \eqref{eq3} and \eqref{eq4} into \eqref{eq2} gives

$$\frac{1}{2} \, \log_2 n \ge k_2 + (\log_2 n)^{\frac{1}{2}} \; \forall \; n \ge N_1 \tag{5}\label{eq5}$$

In this case, consider values of $$m$$ where $$\frac{1}{2}m^2 \ge m$$. Multiplying both sides by $$2$$, moving $$2m$$ to the LHS and factoring gives $$m\left(m - 2\right) \ge 0$$. This is true for all $$m \ge 2$$. In this case, if we let $$m = (\log_2 n)^{\frac{1}{2}}$$, we can see that having $$k_2 = 0$$ and since $$(\log_2 16)^{\frac{1}{2}} = 2$$, we can use $$N_1 = 16$$. I believe you should be able to finish the rest.

Note if you had used your original idea of, I assume, the natural logarithm, it would just involve an extra factor by the change of logarithm base formula, i.e., $$\log_a{x} = \cfrac{\log_b{x}}{\log_b{a}}$$ to give in this case that $$\log_2{n} = \cfrac{\log_e{n}}{\log_e{2}}$$. Also, as I show, you would need to add the constant, not multiply by it. After that correction, this just results in an extra factor being used for $$\log_2{n}$$, plus you also have the $$\log_e{2}$$ factor, so it might change the $$k_2$$ constant you determine to use (but not necessarily the $$k_1$$ constant) and/or the value of $$N_1$$, as these values are not uniquely determined.


# What is missing in my solution of “from PDF to CDF and $P(X > 0.5)$”?

The continuous random variable $$X$$ is described with the following probability density function (pdf):

$$f_X(x) = \begin{cases} \frac{1}{9}\big(3 + 2x - x^2 \big) \; : 0 \leq x \leq 3 \\ 0 \; \;: x < 0 \; \lor \; x > 3\end{cases}$$

Find cumulative distribution function $$F_X$$ and probability $$P(X > 0.5)$$.

The task is started by verifying if the pdf is in fact correct pdf. I am checking two conditions:

1. Is the pdf nonnegative on all of its domain? Yes, hence we can write:

$$\forall_{x \in \mathbb{R}}\;f_X(x) \geq 0$$

1. The pdf has to be integrable and its total area under the curve has to be equal $$1$$:

\begin{align*} &\int_{\mathbb{R}}f_X = 1 \\ &\color{red}{\int_{-\infty}^{\infty}f_X(x)dx = 1} \\ \end{align*}

(for now assume the condition is true)

PDF plot:

Computing CDF which is defined as:

$$F_X(x) = \int_{-\infty}^{x}f_X(t)dt$$

Therefore:

If $$x < 0$$:

$$F_X(x) = \int_{-\infty}^{x} 0dt = 0$$

If $$x \geq 0 \; \land \; x \leq 3$$:

\begin{align*}F_X(x) &= \int_{-\infty}^{0}0dt + \int_{0}^{x}\frac{1}{9}\big(3 + 2t - t^2\big)dt = \\ &= 0 + \frac{1}{9}\Big(3t + t^2 - \frac{1}{3}t^3 \Big)\Bigg|^{x}_0 = \\ &= \frac{1}{9} \Big(3x + x^2 - \frac{1}{3}x^3 \Big)\end{align*}

If $$x \geq 3$$:

\begin{align*} F_X(x) &= \int_{-\infty}^{0}0dt + \int_{0}^{3}\frac{1}{9}\Big(3 + 2t - t^2 \Big)dt + \int_{3}^{x}0dt \\ &= 0 + \frac{1}{9}\Big(3t + t^2 - \frac{1}{3}t^3 \Big)\Bigg|^3_0 + 0 = \\ &= 1 \end{align*}

(this implicitly confirms the $$\color{red}{\text{red}}$$ condition)

Finally the CDF is defined as:

$$F_X(x) = \begin{cases} 0 \; \; : x < 0 \\ \frac{1}{9} \Big(3x + x^2 - \frac{1}{3}x^3 \Big) \; \; : x \geq 0 \; \land \; x \leq 3 \\ 1 \; \; : x > 3 \end{cases}$$

The CDF result agrees with:

$$\lim_{x \to \infty}F_X(x) = 1 \; \land \; \lim_{x \to -\infty}F_X(x) = 0$$

Also the function is non-decreasing and continuous.

CDF plot:

## Calculating $$P(X > 0.5)$$:

\begin{align*}P(X > 0.5) &= \int_{0.5}^{\infty}f_X(x)dx = \\ &= \int_{0.5}^{3}\frac{1}{9}(3+2x-x^2)dx + \int_{3}^{\infty}0dx = \\ &= \frac{1}{9} \Big(3x + x^2 - \frac{1}{3}x^3 \Big)\Bigg|^3_{0.5} + 0 = \\ &= \frac{175}{216} \approx 0.81\end{align*}

This probability solution does not agree with the book's solution.

The book says $$P(X > 0.5) = 1 - F_X(0.5) = \frac{41}{216} \approx 0.19$$, so it's my solution "complemented".

## My questions:

• Which final probability solution is correct?
• Is this any special kind of probability distribution, e.g. Poisson or Chi Square (well, not these)?
• Can you please point out all minor or major mistakes I have made along the way? (perhaps aside from plots that are not perfect). This is the most important for me.
• What have I forget to mention or calculate for my solution to make more sense? Especially something theoretical, perhaps e.g. definition for $$X$$.
• Looks like the book have a bug. – kludg Nov 30 '19 at 18:04

## My questions:

• Which final probability solution is correct?

Yours answer is right and the book's isn't. They presumably have mistakenly computed $$\mathbb P(X < 0.5)$$ instead of $$\mathbb P(X > 0.5)$$.

• Is this any special kind of probability distribution, e.g. Poisson or Chi Square (well, not these)?

Not a common one, no. I found this page on "U-quadratic distributions" (a term I've never heard before), and this would be the vertical inverse of one of these described in the "related distributions" section, but I don't think this is a particularly common term or distribution.

EDIT: Whoops, this isn't even quite the vertical inverse of a U-quadratic distribution, is it? Such a distribution would apparently not truncate the left side of the parabola as this one does. The better answer to your question is: "No, this distribution is neither named nor important."

• Can you please point out all minor or major mistakes I have made along the way? (perhaps aside from plots that are not perfect). This is the most important for me.

I'd love to, but I didn't find any!

• What have I forget to mention or calculate for my solution to make more sense? Especially something theoretical, perhaps e.g. definition for $$X$$.

I didn't spot any holes or anything that needs to be improved.

EDIT: One thing you could do to clean this up a bit: when you compute $$\mathbb P(X > 0.5)$$, you're redoing the integration you already did in your CDF. Instead, you could just use that result that you already obtained: $$\mathbb P(X > 0.5) = 1 - \mathbb P(X \leq 0.5) = 1 - F_X(0.5) = 3(0.5) + (0.5)^2 - \frac{1}{3}(0.5)^3 = \dots$$ That said, your answer isn't wrong, it's just a bit inefficient.

• So the efficient way is to compute the CDF at $x=0.5$ and subtract the result from $1.$ It seems likely that whoever wrote the answer key forgot to subtract. I likewise did not find any mistake in the calculations in the question. – David K Nov 30 '19 at 18:30
• Can I write:$$f_X: \; \mathbb{R} \longrightarrow \left[0, \frac{4}{9} \right]$$ where $\frac{4}{9}$ denotes mode of PDF? And also, $$F_X : \; \mathbb{R} \longrightarrow \left[0,1\right]$$ for the CDF? Thanks for help. I will accept today. – weno Dec 1 '19 at 4:56
• @weno Sure - both those statements are true, though the first is questionably useful and the second is trivially true for all CDFs. – Aaron Montgomery Dec 1 '19 at 4:58


# Area of the part of the cylinder $x^2+y^2=2ay$ outside the cone $z^2=x^2+y^2$

Problem: Find the area of the part of the cylinder $$x^2+y^2=2ay$$ that lies outside the cone $$z^2=x^2+y^2$$.

My attempt: So I thought we could do this by projecting the surface onto the $$yz$$-plane and taking the surface integral of the function $$x=g(y,z)=\sqrt{z^2-y^2}$$. I.e letting $$S$$ be the surface and $$E$$ be the projection onto the $$yz$$-plane where we have a $$2$$ before the integral over $$E$$ since we have both $$x<0$$ and $$0\leq x$$: \begin{align*}\iint_{\mathcal{S}}x \ \mathrm{d}S &=2\iint_{E}x\underbrace{\sqrt{1+\left(\frac{\partial x}{\partial y}\right)^2+\left(\frac{\partial x}{\partial z}\right)^2} \ \mathrm{d}z\mathrm{d}y}_{\mathrm{d}S} \\ &=2\iint_{E}x\sqrt{1+\frac{z^2}{x^2}+\frac{y^2}{x^2}} \ \mathrm{d}z\mathrm{d}y\\ &=2\iint_{E}\sqrt{x^2+z^2+y^2}\ \mathrm{d}z\mathrm{d}y\\ &=2\iint_{E}\sqrt{2}z\ \mathrm{d}z\mathrm{d}y \end{align*} Now in the projection it seems to me that we have the following bounds on $$z$$ and $$y$$ since the cylinder has radius $$a$$ and the cone and the surface intersect at $$z=\sqrt{2ay}$$ $$0\leq z \leq \sqrt{2ay} \quad \text{and} \quad 0\leq y \leq 2a$$ so: \begin{align*}2\iint_{E}\sqrt{2}z\ \mathrm{d}z\mathrm{d}y &= \sqrt{2}\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}2z \ \mathrm{d}z\mathrm{d}y \\ &=\sqrt{2}\int_{0}^{2a} 2ay \ \mathrm{d}y\\ &=4\sqrt{2}a^{2}\end{align*} However my book says its $$16a^2$$ so what is my mistake(s)?

PS. I think this is also possible with polar coordinates but I would like to use the surface integral with projection onto the $$yz$$-plane.

PSDS. Picture is not totally acurate as $$a=4$$

Edit:

As Ninad Munshi pointed out I was projecting the wrong surface and I used the wrong formula for the surface area. My thoughts are

Would it be correct to say that $$\iint\mathrm{d}S$$ is the surface area, and would $$\mathrm{d}S$$ be $$\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2} dzdy$$? If so I still seem to be off by a factor of $$2$$ as \begin{align*}\iint_{\mathcal{S}} \mathrm{d}S &= 2 \iint_{E}\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2}dzdy \\ &=2\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2}dzdy=8a^{2}\end{align*}

• You used the wrong surface in the beginning. It wants the area of the cylinder, not the cone, so you have to project the cylinder down. – Ninad Munshi May 12 '20 at 21:33
• Also, $\iint x\:dS$ does not give you surface area. – Ninad Munshi May 12 '20 at 21:35
• @NinadMunshi I solved it, Thank you for your help! – André Armatowski May 13 '20 at 2:42

The correct way to solve this question is to start with the cylinder $$x^2+y^2=2ay$$ that we wish to project onto the $$yz$$ plane. This is done by first calculating $$\mathrm{d}S$$ in $$\iint_{\mathcal{S}}dS$$ which gives the surface area
We have that $$dS = \sqrt{1+\left(\frac{\partial x}{\partial y}\right)^{2}+\left(\frac{\partial x}{\partial z}\right)^{2}} \ \mathrm{d}z\mathrm{d}y=\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$
Now the problem I had is that when I projected the cylinder: I only considered the symmetric areas of $$x<0$$ and $$0\leq x$$ while we infact have two more symmetries: namely $$z<0$$ and $$0\leq z$$.
In summary: We have four areas that are equal (and not two) so letting $$E$$ represent the area of the projection of the cylinder onto the $$yz$$-plane in the first octant we get: $$\iint_{\mathcal{S}}\mathrm{d}S=4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$ The particular limits of $$z$$ and $$y$$ are still correct that is $$0\leq z \leq \sqrt{2ay} \quad \text{and} \quad 0\leq y \leq 2a$$ So: \begin{align*}4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y & = 4\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y \\ &= 4 \int_{0}^{2a}\sqrt{2ay}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}y\\ &= 4(4a^2)=16a^2\end{align*} Which is the correct answer


# Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls

I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.

### My thoughts

Let $A = \{\text {Exactly one 6 in first ten rolls of a die} \}$ and $B = \{\text {Exactly two 6s in twenty rolls of a die} \}.$
Then I want to find $$P[A\mid B] = \frac{P[A \cap B]}{P[B]}.$$

By the binomial distribution formula, we get that $$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$

Furthermore I think that $P[A \cap B]$ is equal to the probability of rolling exactly one $6$ in ten rolls and then rolling exactly one $6$ in another set of ten rolls. That is,
$$P[A \cap B] = \left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2.$$

Am I correct in thinking this?
If so, then it follows that the required probability is $$P[A \mid B] = \frac{\left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2}{{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}},$$ which, I know, can be simplified further!

• Shouldn't $P[B]$ be $$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$ ? – Mohamad Misto Jun 4 '15 at 14:45
• A curiosity is that in this case $P(B \mid A)=P(A)$ so $P(A \cap B)=P(A^2)$ and $P(A \mid B)=\dfrac{P(A)^2}{P(B)}$ – Henry Jun 4 '15 at 14:59
• Well laid out. Apart from a minor slip, all correct. You will notice near total collapse when you simplify. Now can you simplify the argument? – André Nicolas Jun 4 '15 at 14:59
• You should get $10/19$. – André Nicolas Jun 4 '15 at 15:26
• There are $\binom{20}{2}$ equally likely ways to choose the locations of the two $6$'s. There are $(10)^2$ ways to choose one in each half. So the probability is $100/\binom{20}{2}$. – André Nicolas Jun 4 '15 at 15:29

I took a different approach to the question. Suppose B. There are three ways to get two 6's in twenty rolls:

• $B_1$: Both 6's come in the first 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen.
• $B_2$: One 6 comes in the first 10 rolls, and the second comes in the next 10 rolls. There are $\begin{pmatrix} 10 \\ 1\end{pmatrix} \begin{pmatrix} 10 \\ 1\end{pmatrix} = 100$ ways for this to happen.
• $B_3$: Both 6's come in the second lot of 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen.

Now $$P(A|B) = P(B_2|B_1 \cup B_2 \cup B_3) = \frac{100}{45+100+45} = \frac{10}{19}.$$

• Your reasoning seems sound to me. It doesn't agree with all of the other answers, though. I am not yet sure why... – Caleb Owusu-Yianoma Jun 4 '15 at 15:24

I think you are over thinking this. We know we get exatly two sixes in twenty rolls how many ways can that happen? Consider a roll to be 6 or not 6 we don't care what number it is otherwise.

One of the sixes arived in any of the twenty rolls and the other in an of the nineteen remaining rolls and since a 6 is a 6 we divide by two because the order does not matter.

There are thus $\dfrac{20 \cdot 19}{2} = 190$ ways we can get exactly 2 6's in 20 rolls.

In how many ways can we have exactly one 6 in 10 rolls? Well it can be any of the 10 rolls, and it must be not 6 in the other 9 so there are 10 ways to get exactly one 6 in ten rolls, and 10 ways to get the second 6 in the last 10 rolls making

The answer is simply $\dfrac{10 \cdot 10}{190} = \dfrac{100}{190} = \dfrac{10}{19}$.

• Surely, when considering the number of ways in which we can obtain exactly one $6$ in the first ten rolls, we must account for the fact that we obtain a $6$ in the second set of ten rolls, too. In that case, the number of ways of rolling exactly one $6$ in the first ten rolls AND exactly one $6$ in the second ten rolls is $10^2 = 100.$ Is my reasoning sound? – Caleb Owusu-Yianoma Jun 4 '15 at 15:20
• @CKKOY there are 100 ways you can roll exacly one 6 in the first 10 rolls and exacly 1 six in the next 10 rolls but that's not what the question is asking. The question is asking how many ways you get exactly 1 six in the first 10 rolls given all the ways you can get 2 6's in 20 rolls including one where both 6's occur in the first 10 rolls or both 6's occur in the last 10 rolls. – Warren Hill Jun 4 '15 at 15:26
• Is your interpretation of the question equivalent to the following? 'The question is asking how many ways you can roll exactly one $6$ in the first ten rolls (out of twenty), given that you rolled two $6$s in the total twenty rolls.' Please see the other answers and comments, which imply that the correct answer is $10/19$. – Caleb Owusu-Yianoma Jun 4 '15 at 15:30
• @CKKOY sorry you are correct. I've corrected my answer. – Warren Hill Jun 4 '15 at 15:35
• Apology accepted. – Caleb Owusu-Yianoma Jun 4 '15 at 15:40

Since all drawings are independent, is confusing to use the conditional probability formula and it is not necessary at all. Think as you are rolling simultaneously two dices 10 times each. Having in both groups exactly one 6 has the same probability p. Because these two are independent the simultaneous event (one 6 in each group) has the p^2 squared probability. The answer is what you thought A and B was: [(10C1)⋅(1/6)⋅(5/6)^9]^2 = 95,367,431,640,625/914,039,610,015,744 = 0.104336. (far to be 10/19)

This is what Adelafif said too, but he made an error on the coefficient n-k=9 (not 5).

What Paul Wright said is very funny but totally wrong. Having one 6 in ("the first") 10 rollings has the same probability independently how many time you are rolling the dice after!. This probability is (10C1)⋅(1/6)^2⋅(5/6)^8 (much less than Paul calculated). With Paul's theory, we should win a lottery pot considerably easier.

• Your answeris wrong, and $10\over 19$ is the correct one. According to mathematics. – user228113 Jun 4 '15 at 16:43
• @Laszlo: Your calculation of $0.104336$ is the rounded value of $P[A \cap B].$ However, the final answer, $10/19$, is roughly equal to $0.104336/{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$ – Caleb Owusu-Yianoma Jun 4 '15 at 16:48
• @G.Sassatelli Mathematics is not a matter of enouncing something. You must argument your opinion, as I did. What is wrong with "my mathematics"? – Laszlo Jun 4 '15 at 18:01
• @CKKOY Exactly, as I said. You DO NOT need the denominator. – Laszlo Jun 4 '15 at 18:05
• To avoid any other useless dispute: I made a simulation for one million of 20 times rolling sets (it is only two lines in Mathematica!). The number of cases with one six in the first 10 and the second 10 rolls was: 104,524. Repeated this 100 times, got a normal distribution with the mean value exactly at 104336 (it took 20 minutes). Still sustain that my solution is wrong? – Laszlo Jun 4 '15 at 18:34

((10C1)(1/6)(5/6)^5)(10C1)(1/6)(5/6)^5

• It would be helpful if you wrote a brief explanation of the answer that you have written. – Caleb Owusu-Yianoma Jun 4 '15 at 15:13


# For which $N$ is it possible to arrange all whole numbers from $1$ to $N$ in such a way that every adjacent pair sums up to a Fibonacci number?

Recently I came up with a problem regarding Fibonacci numbers:

For which $$N$$ is it possible to arrange all whole numbers from $$1$$ to $$N$$ in such a way that every adjacent pair sums up to a Fibonacci number?

I have manually tested a bunch of cases and I was also able to prove almost every case. The results that I was able to prove are the following:

• If $$N$$ is a Fibonacci number or exactly one less than a Fibonacci number, I was able to prove that an arrangement exists. I used an argument by induction to achieve this.
• If $$F_k+2 \leq N \leq F_{k+1} -3$$ , it is completely impossible, because the numbers $$F_k$$, $$F_k + 1$$ and $$F_k + 2$$ have only one possible pair and therefore have to be at the end of the arragement. This obviously gives a contradiction, because arrangements only have two ends (the first number and the last number).

The cases I was not able to solve are $$N=F_k+1$$ and $$N=F_k-2$$. My theory is that $$N=9$$ is the only working case of the form $$N=F_k+1$$, and $$N=11$$ the only working case of the form $$F_k-2$$. I expect every other $$N$$ of these two forms to be impossible.

Does anybody know a full proof to this problem or maybe the name of the official theorem (if this exists)?

• I ran some code to check if there are solutions for small numbers. everything you said checks out so far - 1-5, 7,8,9, 11 solvable, and also 12,13, 20,21, 33,34, 54,55. none of the others are solvable. I checked up to 55 so far. I'm gonna run the code at night, but I don't think it's gonna get much farther than that (the code isn't too smart). Nov 11 '20 at 0:07
• The only sequence in the OEIS matching these conditions is oeis.org/A259624, which it sounds like is not your sequence if $N=6$ cannot be done. If you have confirmed values for $N$ up to $34$ or so, you may wish to submit it as a new sesquence. Nov 11 '20 at 3:45
• @RavenclawPrefect problem is OEIS search engine finds $2,3,4,5,7,8,9,11,12,13,20,21,33,34,54,55$ but no result for $1,2,3,4,5,7,8,9,11,12,13,20,21,33,34,54,55$. Nov 28 '20 at 15:20

Thanks to @Rei Henigman data, I could find OEIS sequence A079734 and from there the "Fibonacci Plays Billiards" paper by Elwyn Berlekamp and Richard Guy, where they state with theorem 1 on page 3, that the only solutions are $$N = 9$$, $$N = 11$$, $$N = F_k$$, $$N = F_k − 1$$ for $$k \ge 4$$. Note that $$1$$ is excluded from the OEIS sequence.


# Help with Implicit Differentiation: Finding an equation for a tangent to a given point on a curve

When working through a problem set containing Implicit Differentiation problems, I've found that I keep getting the wrong answer compared to the one listed at the back of my book.

The problem is given as such: Use implicit differentiation to find the equation of the tangent line to the curve at a given point

x^2 + xy + y^2 = 3

With given point (1, 1). I also am told that it is an ellipse.

To solve this, I evidently must differentiate both sides of the problem:

1: dy/dx ( x^2 + xy + Y^2 ) = dy/dx(3)

2: dy/dx (2x + 1y'+ 2yy') = 0

3: 1y' + 2yy' = 0 - 2x

4: y'(1+2y) = -2x

5: y' = -2x/(1+2y)

Hurray, so now since I have the first derivative of Y. I can use it to find the slope at the point.

Slope at Point (1,1)= -2(1) / (1+2(1)

Slope at Point (1,1)= -2/3

So now that I've got my slope, I know the equation of the tangent will be in the form:

y=mx+b

So, given I now know the slope:

y=-2/3x + b

Substitute in the known point:

1 = -2/3(1) + b

b = 5/3

So the final answer I get is: y = -2/3x + 5/3

But according to the answer, it is supposed to be: -x + 2, I don't know where I went wrong, and I've done it twice to make sure I'm getting the same answer. Could someone please help me?

• The differentiation of $xy$ requires the product rule: $\frac{d}{dx}(xy)=y+xy'$. Oct 20, 2014 at 14:49
• Consider the product rule on the term $xy$: It should read: $\frac{d}{dx}(xy) = \frac{d}{dx}x \cdot y + x \cdot \frac{d}{dx}y.$ Oct 20, 2014 at 14:49
• @Clayton: AH! Fantastic find! I never saw that Oct 20, 2014 at 14:50

You made an error when you differentiated implicitly. You did not apply the Product Rule $(fg)' = f'g + fg'$ to the term $xy$. Keeping in mind that $y$ is a function of $x$, you should obtain

$$(xy)' = 1y + xy' = y + xy'$$

Therefore, when you differentiate

$$x^2 + xy + y^2 = 3$$

implicitly with respect to $x$, you should obtain

$$2x + y + xy' + 2yy' = 0$$

Solving for $y'$ yields

\begin{align*} xy' + 2yy' & = -2x - y\\ (x + 2y)y' & = -2x - y\\ y' & = -\frac{2x + y}{x + 2y} \end{align*}

As you can check, evaluating $y'$ at the point $(1, 1)$ yields $y' = -1$. Therefore, the tangent line equation is $y = -x + 2$.


# How come $\left(\frac{n+1}{n-1}\right)^n = \left(1+\frac{2}{n-1}\right)^n$?

I'm looking at one of my professor's calculus slides and in one of his proofs he uses the identity:

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

Except I don't see why that's the case. I tried different algebraic tricks and couldn't get it to that form.

What am I missing?

Thanks.

Edit: Thanks to everyone who answered. Is there an "I feel stupid" badge? I really should have seen this a mile a way.

• Because $n+1=(n-1)+2$. – vadim123 Feb 23 '15 at 14:39

Just write $$\left(\frac{n+1}{n-1}\right)^n = \left(\frac{n-1+2}{n-1}\right)^n =\left(1+\frac{2}{n-1}\right)^n$$
$1+\frac{2}{n-1}=\frac{n-1}{n-1}+\frac{2}{n-1}=\frac{n+1}{n-1}$
Note that $$\frac{n+1}{n-1} = \frac{n-1+2}{n-1} = \frac{n-1}{n-1} + \frac{2}{n-1} = 1 + \frac{2}{n-1}.$$
$(1 + \frac {2}{n-1})^n = (\frac {n-1 +2}{n-1})^n = (\frac{n+1}{n-1} )^n$


# Inverse of the Pascal Matrix

Let $P_n$ be the $(n+1) \times (n+1)$ matrix that contains the numbers of Pascal's triangle in the upper triangle. For example in the case of $n=3$ $$P_3 = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$ or in general $$(P_n)_{ij} = \binom{j}{i} \lfloor i \leq j \rceil ~~~\text{for}~~ i,j \in \{0,...,n \}$$ using the definition $$\lfloor A \rceil := \begin{cases} 1 & \text{A is true} \\ 0 & \text{A is not true} \end{cases}$$ This matrix is invertible since $\det P_n = 1$. For smaller cases like $n=3$, I calulated the inverse of the matrix by hand and found $$P_3^{-1} = \begin{pmatrix} 1 & -1 & 1 & -1 \\ 0 & 1 & -2 & 3 \\ 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$ $n=2,4$ led to similar results, so I'm guessing that the inverse should be $$(P_n^{-1})_{ij} = (-1)^{j+i}(P_n)_{ij} = (-1)^{j+i} \binom{j}{i} \lfloor i \leq j \rceil$$ But I have not been able to prove or disprove this yet. So far I tried multiplying the two matirces which gives $$\sum^j_{k=i} (-1)^{j+k} \binom{j}{k} \binom{k}{i} = \delta_{ij}$$ if one asumes that the result is the unit matrix. For $i=0$ and $j>0$ this gives $$\delta_{0j} = 0 = \sum^j_{k=0} (-1)^{j+k} \binom{j}{k} \binom{k}{0} = \sum^j_{k=0} (-1)^{k} \binom{j}{k}$$ which is an identity I know to be true, so it reasures me a little bit that the above should also be true.

• Instead of proving the above mentioned formula for all $(j,i)$ you can also use induction and prove it only for $(n,i)$, maybe it is easier. The $(j,n)$ entries are easy to handle Jul 8, 2016 at 6:23
• Use $\binom jk \binom ki = \binom ji \binom {j-i}{k-i}$. Jul 8, 2016 at 6:36
• Binomial transform Jul 27, 2016 at 11:10

In my opinion this is a bit simpler to prove, if we interpret the matrix $$P_n$$ as a linear transformation.

Consider the space $$V_n$$ of polynomials of degree $$\le n$$ (over, say $$\Bbb{Q}$$, but you are welcome to use reals or complex numbers, or any other field actually).

The mapping $$T: f(x)\mapsto f(x+1)$$ for all $$f(x)\in V_n$$ is obviously linear. My key point is to observe that $$P_n$$ is the matrix $$M_{\mathcal B}(T)$$ of $$T$$ with respect to the obvious basis $$\mathcal{B}=\{1,x,x^2,\ldots,x^n\}$$ of $$V_n$$. This is because for any $$k, 0\le k\le n$$ the binomial formula says that $$T(x^k)=\sum_{\ell=0}^k\binom k\ell x^\ell.$$ From this we can read the coordinates of $$T(x^k)$$ with respect to $$\mathcal{B}$$, and see that those coincide the $$k$$th column of $$P_n$$ (numbered from $$0$$ to $$n$$). That's exactly what the claim was.

It is obvious that the inverse of $$T$$ is given by the recipe $$T^{-1}:f(x)\mapsto f(x-1).$$ A similar application of the binomial formula shows that the matrix $$M_{\mathcal B}(T^{-1})$$ then is exactly your prescribed inverse of $$P_n$$ containing binomial coefficients - this time with alternating signs.

But, by basic linear algebra $$M_{\mathcal B}(T^{-1})=M_{\mathcal B}(T)^{-1}.$$

• Nice solution! I especially like that it does not require to guess the form of the inverse. It is also very close to the reason I got confronted with the problem in the first place. Because I was looking at the change of base from $\{x^j\}$ to $\{(x+1)^j\}$. Jul 22, 2016 at 18:39
• @JyrkiLahtonen This is brilliant, really, a pity that I can only give (+1). I just wanted to know the formula for the inverse of the Pascal matrix and searched the web for it. Now I found not only the formula, but this higher level view on it which makes everything so clear. It feels similar to the moment when I was told that the addition theorems for sin and cos are just real and imaginary parts of the power rule for the complex exponential function. May 30, 2020 at 10:10

By the suggestion of Sungjin Kim one can use $$$$\binom{j}{k}\binom{k}{i} = \binom{j}{i}\binom{j-i}{k-i}$$$$ which is useful since it leads to the sum only going over one of the binomial coefficients. It leads to \begin{align} \sum^j_{k=i} (-1)^{j+k} \binom{j}{k} \binom{k}{i} &= \sum^j_{k=i} (-1)^{j+k} \binom{j}{i}\binom{j-i}{k-i} \\ &= (-1)^j \binom{j}{i} \sum^j_{k=i} (-1)^{k} \binom{j-i}{k-i} \\ &= (-1)^{j} \binom{j}{i} \sum^{j-i}_{k=0} (-1)^{k+i} \binom{j-i}{k} \\ &= (-1)^{j+i} \binom{j}{i} (1 - 1)^{j-i} = \delta_{ij} \end{align} The identity used is easily proven by $$\binom{j}{k}\binom{k}{i} = \frac{j!}{k!(j-k)!} \frac{k!}{i!(k-i)!} = \frac{j!}{i!(j-i)!} \frac{(j-i)!}{(j-k)!(k-i)!} = \binom{j}{i}\binom{j-i}{k-i}$$ Another way of showing the identity involving $$\delta_{ij}$$ is to start with $$x^j$$ and using the binomial thoerem twice $$x^j = (x+0)^j = ((x+1) - 1 )^j = ~...~ = \sum^j_{i=0} \sum^j_{k=i} (-1)^{j+k} \binom{j}{k} \binom{k}{i} x^i$$ comparing coefficients of the two sides of the equation, then also leads to the desired result. You can find the whole calcultion here, along whith other examples of when adding 0 really counts.

In the following it's somewhat more convenient to equivalently consider the transpose $P_n^T$ of the matrix $P_n$ and its inverse.

We obtain for $n=3$

\begin{align*} P_3^T&= \begin{pmatrix} 1 & & & \\ 1 & 1 & & \\ 1 & 2 & 1 & \\ 1 & 3 & 3 & 1 \\ \end{pmatrix}=\left(\binom{i}{j}\right)_{0\leq i,j\leq 3} \\ \left(P_3^T\right)^{-1}&= \begin{pmatrix} \begin{array}{rrrr} 1 & & & \\ -1 & 1 & & \\ 1 & -2 & 1 & \\ -1 & 3 & 3 & 1 \\ \end{array} \end{pmatrix}=\left((-1)^{i+j}\binom{i}{j}\right)_{0\leq i,j\leq 3} \end{align*}

The elements of the Pascal matrix can be considered as coefficients of binomial inverse pairs. These are sequences $(a_i)_{0\leq i \leq n},(b_i)_{0\leq i\leq n}$ which are related for $n\geq 0$ via \begin{align*} a_n=\sum_{i=0}^n\binom{n}{i}b_i\qquad\text{resp.}\qquad b_n=\sum_{i=0}^n(-1)^{i+n}\binom{n}{i}a_i\tag{1} \end{align*}

One relation implies the other in (1). This can be seen via exponential generating functions.

Let $A(x)$ and $B(x)$ be two exponential generating functions

$$A(x)=\sum_{n=0}^\infty a_{n}\frac{x^n}{n!} \qquad\qquad\text{ and }\qquad\qquad B(x)=\sum_{n=0}^\infty b_{n}\frac{x^n}{n!}$$

The expressions in (1) are the coefficients of

The most important characterization of the pair of inverse relation in (1) is the orthogonal relation the pair implies. This follows by substituting one of the pair into the other. We obtain for $n\geq 0$
\begin{align*} a_n&=\sum_{j=0}^n\binom{n}{j}b_j\\ &=\sum_{j=0}^n\binom{n}{j}\sum_{i=0}^j(-1)^{i+j}\binom{j}{i}a_i\\ &=\sum_{i=0}^n a_i\sum_{j=i}^n(-1)^{i+j}\binom{n}{j}\binom{j}{i} \end{align*} Hence the orthogonal relation is \begin{align*} \delta_{ni}=\sum_{j=i}^n(-1)^{i+j}\binom{n}{j}\binom{j}{i} \end{align*} with $\delta_{ni}$ the Kronecker Delta.
The inverse of $$P_n$$ also follows from a familiar representation of $$GL(2,\mathbb{R})$$. Let $$\{u,v\}$$ be a basis of $$\mathbb{R}^2$$ and $$S^n(\mathbb{R}^2)$$ denote the vector space of homogeneous polynomials of degree $$n$$ in the variables $$u$$ and $$v$$. Then, any $$A\in GL(2,\mathbb{R})$$ acts linearly on $$p(u,v)\in S^n(\mathbb{R}^2)$$ by $$A\cdot p(u,v)=p(Au,Av).$$ Choose $$\{u^{n-j}v^j\}_{j=0}^n$$ as a basis of $$S^n(\mathbb{R}^2)$$. Then, the given group action induces a homomorphism $$\varphi_n: GL(2,\mathbb{C})\to GL(n+1,\mathbb{C})$$ defined by $$\begin{pmatrix}a & b\\ c & d\end{pmatrix}\mapsto \left([t^i](a+ct)^{n-j}(b+dt)^j\right)_{i,j=0}^n$$ where $$[t^i]f(t)$$ denotes the coefficient of $$t^i$$ in $$f(t)$$. Let $$M=\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}.$$ Then, we see that $$P_n=\varphi_n(M).$$ Since $$\varphi_n$$ is a homomorphism, we get $$P_n^{-1}=\varphi_n(M^{-1})$$. But $$M^{-1}=\begin{pmatrix}1 & -1\\ 0 & 1\end{pmatrix}$$. Hence, by definition of $$\varphi_n$$, we obtain explicit indices as you described.


# Munkres's Strong induction principle vs. “traditional” mathematical induction

Question:

In Munkres's Topology (2nd ed.), he gives a proof for the following:

Theorem 4.2 (Strong induction principle). Let $A$ be a set of positive integers. Suppose that for each positive integer $n$, the statement $S_n\subset A$ implies the statement $n\in A$. Then $A=\mathbb{Z}_+$.

This appears to be slightly different from the "traditional" (strong) induction principle which can be stated as: let $P_n$ be a sequence of statements. Suppose $P_1$ is true and (($P_m$ is true $\forall\, m < n$) implies $P_{n}$ is true) then $P_k$ is true $\forall\, k\in \mathbb{Z}_+$.

How can one reconcile the two?

Attempt:

I'm thinking of something along the lines of:

Let $A$ be a set of positive integers and $P_n$ be a sequence of propositions. Suppose that for each positive integer $n$, the statement $S_n\subset A$ implies the statement ($n\in A$ and $P_n$ is true). Then $A=\mathbb{Z}_+$ and $P_k$ is true $\forall\, k \in \mathbb{Z}_+$.

This can proved to be equivalent to the "traditional" form in the following way:

1. Proceed with Munkres's proof and obtain $A=\mathbb{Z}_+$.
2. $\because S_1=\varnothing\subset A$ is true, our supposition for $n=1$ is equivalent to assuming that $P_1$ is true.
3. ??? (how do we show that the supposition "$S_n\subset A$ implies the statement ($n\in A$ and $P_n$ is true)" is equivalent to "($P_m$ is true $\forall\, m < n$) implies $P_{n}$ is true"?)

Actually I just noticed that this point is discussed in Chapter 1 $\S$ 7 after Lemma 7.2 (pg 45 in 2nd ed.)

To use the principle to prove a theorem "by induction", one begins the proof with the statement "Let $A$ be the set of all positive integers $n$ for which the theorem is true," and then one goes to prove that $A$ is inductive, so that $A$ must be all of $\mathbb{Z}_+$

Using this one can write:

Let $P_n$ be a sequence of propositions and $A$ be the set of all positive integers $n$ for which $P_n$ is true. Suppose that for each positive integer $n$, the statement $S_n\subset A$ implies the statement $n\in A$. Then $A=\mathbb{Z}_+$.

The equivalence with the "traditional" induction principle is easily demonstrated:

1. For $n=1$, the assumption ($S_1 = \varnothing \subset A \Rightarrow 1 \in A$) is equivalent to ($P_1$ is true).
2. For $n>1$, the assumption ($S_n \subset A \Rightarrow n \in A$) is equivalent to (($P_m$ is true $\forall\, m<n$) $\Rightarrow$ $P_n$ is true)
3. The conclusion ($A=\mathbb{Z}_+$) is equivalent to ($P_n$ is true $\forall\, n\in \mathbb{Z}_+$).
• Yes, this is the intended argument. – Brian M. Scott Aug 31 '16 at 22:10


Double Antiderivation problem

I have to find $f(x)$ given $f''(x)$ and certain initial conditions.

$$f"(x) = 8x^3 + 5$$ and $f(1) = 0$ and $f'(1) = 8$

$$f'(x) = 8 \cdot \frac{x^4}{4} + 5x + C = 2x^4 + 5x + C$$

Since $f'(1) = 8 \Rightarrow 2 + 5 + C = 8$ so $C = 1$

$$f'(x) = 2x^4 + 5x + 1$$

$$f(x) = 2 \cdot \frac{x^5}{5} + 5 \cdot \frac{x^2}{2} + X = \frac{2}{5} x^5 + \frac{5}{2} x^2 + X + D$$

Since $f(1) = 0$ then:

$$\frac{2}{5} + \frac{5}{2} + 1 + D = 0$$ $$\frac{29}{10} + 1 + D = 0$$

So $D = \dfrac{-39}{10}$

So $$f(x) = \frac{2}{5} \cdot x^5 + \frac{5}{2} \cdot x^2 + x - \frac{39}{10}$$

Does that look right?

• Yes it's correct. An alternative way to do it is to work with definite integrals and write $f'(x) - f'(1) = \int_1^x (8t^3+5)\,{\rm d}t$ and $f(x)-f(1) = \int_1^x f'(t)\,{\rm d}t$. This avoids solving for the integration constants, but it's pretty much the same thing. Mar 12, 2018 at 17:32
• Yes, you can also check it by yourself Mar 12, 2018 at 17:33

$$f(x) = \frac{2}{5} \cdot x^5 + \frac{5}{2} \cdot x^2 + x - \frac{39}{10}\implies f'(x)= 2x^4+5x+1\implies f''(x)=8x^3+5$$
$$f(1)=\frac{2}{5}+ \frac{5}{2} + 1 - \frac{39}{10}=\frac{4+25+10-39}{10}=0$$
$$f'(1)=2+5+1=8$$


+0

# Here's a mind-boggler

0
347
4
+40

A positive whole number that is the same when reading from right to left and left to right is called a palindromic number. If all of them were put into a sequence that is from small to big:1,2...11,22...,101,111...,10001,10101,.....

Then which term is the number 78987 in this sequence?  For example : 1 is the 1st term,2 is the second....etc

yomyhomies  Oct 22, 2017
#2
+89840
+1

Starting with 0,  78987  is the 889th palindrome

[BTW...889 is known as the palindrome's rank ]

Starting with 1, it is the 888th palindrome

This can be verified with the calculator here :

http://rhyscitlema.com/algorithms/generating-palindromic-numbers/

[ The calculator is about half-way down the page.....there are also other intems of interest related to palindromes on this page ]

A procedure is desribed for finding the nth palindrome, but I don't see one that tells us how to determine the rank of any particular palindrome  [ although I did not look at the page with a fine tooth comb]

Maybe heureka knows of a proceedure to produce this???

CPhill  Oct 22, 2017
#3
+20024
+3

which term is the number 78987 in this sequence?

Here 0 is the 1st term:

The position of a palindrome within the sequence can be determined almost without calculation:

If the palindrome has an even number of digits,

prepend a 1 to the front half of the palindrome's digits.

Examples: 98766789=a(19876)

If the number of digits is odd, prepend the value of front digit + 1 to the digits from position 2 ... central digit.

Examples: 515=a(61), 8206028=a(9206), 9230329=a(10230).

which term is the number 78987 in this sequence?

The number of digits is 5 is odd, prepend the value of front digit + 1 to the digits from position 2 ... central digit.

$$\begin{array}{|rrrrll|} \hline & & & \Rsh & & \text{ until central digit} \\ &7 & 8 & 9 & 8 & 7 \\ &| & | & | \\ &+1 & | & | \\ &\downarrow & \downarrow & \downarrow \\ \text{term is } & \color{red}8 &\color{red}8 &\color{red} 9 \\ \hline \end{array}$$

Starting with 1, it is the 888th palindrome

heureka  Oct 23, 2017
#4
+89840
+1

Wow!!!!....thanks, heureka.....that's almost too easy....!!!

CPhill  Oct 23, 2017


+0

# Suppose $f(x)$ is a rational function such that $3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$ for all $x \neq 0$. Find $f(-2)$.

0
258
3

Suppose f(x) is a rational function such that $$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$$ for all x =\=0. Find f(-2).

Guest May 22, 2018
#1
+93629
+1

Suppose f(x) is a rational function such that

$$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$$

for all $$x\ne0$$. Find f(-2).

We have

$$3 f \left( \frac{1}{-2} \right) + \frac{2f(-2)}{-2} = (-2)^2\\ 3 f \left( \frac{1}{-2} \right) - f(-2) =4\qquad \qquad(1)\\ and\\ 3 f \left( \frac{1}{-1/2} \right) + \frac{2f(-1/2)}{-1/2} = (-1/2)^2\\ 3 f \left( -2 \right) -4f(\frac{-1}{2}) =\frac{1}{4}\\ -4f(\frac{-1}{2}) +3 f \left( -2 \right)=\frac{1}{4}\qquad\qquad (2)\\--------------\\~\\ \quad 3 f \left( \frac{1}{-2} \right) - f(-2) =4\qquad \qquad(1)\\ \quad 12f \left( \frac{1}{-2} \right) - 4f(-2) =16\qquad \qquad(1b)\\ -4f(\frac{-1}{2}) +3 f \left( -2 \right)=\frac{1}{4}\qquad\qquad (2)\\ -12f(\frac{-1}{2}) +9 f \left( -2 \right)=\frac{3}{4}\qquad\qquad (2b)\\~\\ (1b)+(2b)\\ 5f(-2)=16\frac{3}{4}\\ f(-2)=3\frac{7}{20}\\$$

The method is correct but you need to check for careless errors.

Melody  May 22, 2018
#2
+20024
+1

Suppose

$$f(x)$$

is a rational function such that  $$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$$

for all $$x \neq 0$$. Find $$f(-2)$$.

$$\begin{array}{|lrclcl|} \hline & 3f(\frac1x) + \frac2x f(x) &=& x^2 \\ & 3f(\frac1x) &=& x^2 - \frac2x f(x) \\ & f(\frac1x) &=& \frac{x^2}{3} - \frac{2}{3x} f(x) \qquad (1) \\ \\ \text{Set }x=\frac1x: & 3f(x) + 2x f(\frac1x) &=& \frac{1}{x^2} \\ & 2x f(\frac1x) &=& \frac{1}{x^2}-3f(x) \\ & f(\frac1x) &=& \frac{1}{2x^3}-\frac{3}{2x}f(x) \qquad (2) \\\\ \hline (1) = (2): & \frac{x^2}{3} - \frac{2}{3x} f(x) &=& \frac{1}{2x^3}-\frac{3}{2x}f(x) \\ & \frac{3}{2x}f(x) - \frac{2}{3x} f(x) &=& \frac{1}{2x^3} -\frac{x^2}{3} \\ & f(x)\left( \frac{3}{2x} - \frac{2}{3x} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{9x-4x}{6x^2} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{5x}{6x^2} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{5x}{1} \right) &=& \frac{3-2x^5}{x} \\\\ & \mathbf{f(x)} & \mathbf{=} & \mathbf{\dfrac{3-2x^5}{5x^2}} \\ \hline \end{array}$$

The rational function is  $$f(x) = \dfrac{3-2x^5}{5x^2}$$

$$f(-2)=\ ?$$

$$\begin{array}{|rclcl|} \hline f(-2) &=& \frac{3-2(-2)^5}{5(-2)^2} \\ &=& \frac{3+2^6}{5\cdot 4} \\ &=& \frac{3+64}{20} \\ &=& \frac{67}{20} \\ \mathbf{f(-2)} & \mathbf{=} & \mathbf{3.35} \\ \hline \end{array}$$

$$\text{ f(-2) is 3.35 }$$

graph:

heureka  May 22, 2018
edited by heureka  May 22, 2018
#3
+93629
+1

That is interesing Heureka, I had not realised that there was enough infromation to draw the graph. :)

Melody  May 22, 2018

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# How should I write vectors like this?

If I'm trying to write basic vectors, just as simple as the magnitude being 5 and the direction being zero, how would I do this? Would it be a row vector with parenthesis:$$\overrightarrow{v} = (5, 0)$$, a row vector with brackets: $$\overrightarrow{v} = [5, 0]$$, a column vector with parenthesis: $$\overrightarrow{v} = \begin{pmatrix} 5\\ 0\\ \end{pmatrix}$$, or a column vector with brackets: $$\overrightarrow{v} = \begin{bmatrix} 5\\ 0\\ \end{bmatrix}$$? Thank you if you can tell me what the correct notation for this simple vector is, everywhere I go seems to write them differently and the inconsistency makes me want to rip my hair out.

• You can write it however you want, as long as you are consistent. – Morgan Rodgers Oct 14 '18 at 21:19
• You might also want to consider what you are doing with these vectors. If you are frequently left-multiplying them by a matrix (e.g., $Av$), make them (thin) column vectors. If you are frequently right-multiplying them by a matrix (e.g., $vA$), make them (wide) row vectors. If you are performing both of these operations just as frequently, try not to mix row and column vectors too much (just pick one convention and apply transposes when necessary). As for the bracketing, just be consistent within one document (see Morgan Rodgers' comment). – parsiad Oct 14 '18 at 21:22
• @MorganRodgers and parsia thank you, I'll take this into consideration! – jstowell Oct 14 '18 at 21:23
• One thing I want to point out: the coordinates of a vector are not their magnitude and direction. The magnitude of the vector (3, 4) is 5, and the direction is about 53.13 degrees above the x-axis. – Deusovi Oct 16 '18 at 16:15

So long as you make sure that you are consistent throughout the entire text, it's completely up to you. There are many different ways to represent vectors. In Linear Algebra, people like to use column notation, with either parentheses or square brackets, as this is the most convenient in linear algebra. Some people also write something like $$\vec{v}=\begin{bmatrix}5&0\end{bmatrix}^T$$ to have a column vector, but be able to write it nicely inline. In high school (perhaps up to freshman) level classes, the notation $$\vec{v}=\langle 5,0\rangle$$ is also used, but this ignores the fact that vectors are matrices with 1 row or 1 column, so is less widely used at a higher level. Ultimately it's your choice, whatever is most convenient to you, you should use it.
Usually in linear algebra context vectors $$\vec v$$ are considered colummn vector and transponsed vectors $$\vec v^T$$ are row vectors that is
$$\overrightarrow{v} = \begin{pmatrix} 5\\ 0\\ \end{pmatrix} \quad \overrightarrow{v^T} = \begin{pmatrix} 5 &0\\ \end{pmatrix}$$


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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # Jennifer can buy watches at a price of B dollars per watch, which she  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 50544 Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 03:18 2 7 00:00 Difficulty: 45% (medium) Question Stats: 70% (01:48) correct 30% (02:08) wrong based on 141 sessions ### HideShow timer Statistics Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) – B)/(100B) E. 100(T – NB)/N Kudos for a correct solution. _________________ Manager Joined: 15 Aug 2013 Posts: 54 Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 05:11 2 1 Cost price of N watches = BN (cost of each watch, B * total nuber of watches, N) Total Profit = T But this profilt is also equal to BN (x/100), where x is the profit percentage Hence, BN (x/100) =T or x = 100T/ (NB) Hence (A). CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2689 Location: India GMAT: INSIGHT WE: Education (Education) Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 05:19 Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) – B)/(100B) E. 100(T – NB)/N Let's say Jennifer marks up by x% Total Profit = Marked up price of N watches - Cost of N watches i.e. T = NB[1+(x/100)] - NB i.e. (T) / NB = [(x/100)] i.e. x = 100(T)/NB Answer: Option _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2689 Location: India GMAT: INSIGHT WE: Education (Education) Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 05:32 Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) – B)/(100B) E. 100(T – NB)/N Let's say The price of each watch, i.e. B = 7, Let's say Mark up = 400%==> 400% of 7 is 28, and the$7 watches are marked up $28, then the selling price per watch =$35.

The profit per watch is $28, and so if she sells N = 11 watches Then Total profit, T = 28*11 =$308.

OK, now Substitute T = 308 = 11*28, N = 11, and B = 7, in options and hope to get 400 as our answer.

(A) 100T/(NB) = 100*11*28/(11*4) = 100*28/7 = 110*4 = 400 = works!

(B) TB/(100N) = 11*28*7/(100*11) = 28*7/100 = doesn’t work!

(C) 100TN/B = 100*11*28*11/7 = 100*11*4*11 = doesn’t work!

(D) ((T/N) – B)/(100B) = [(11*28/11) – 7]/(7*100) = (28 – 7)/(7*100) = 21/(7*100) = 3/100 = doesn’t work

(E) 100(T – NB)/N = 100(11*28 – 11*7)/7 = 100*11*21/7 = 100*11*3 = doesn’t work!

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Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

### Show Tags

18 Feb 2015, 07:51
1
Using smart numbers:

Jennifer Purchases watches for $5: B=5 Markup of 100%; now selling for$10: x=10
She sells 5 watches: N=5

Her total profit will be (N*x)-(B*N)
(5*10)-(5*5)= 25
T=25

Now using the variables we must look at the answer choices that will solve for the % markup of the watches, which we chose to be 100.

B=5
T=25
N=5

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Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

### Show Tags

18 Feb 2015, 21:51

Cost price per watch = b

Selling price per watch = $$\frac{t}{n}$$

Let x = percent of the markup from her buy price to her sell price

$$x = 100 *\frac{t}{n} * \frac{1}{b} = \frac{100t}{nb}$$
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Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

### Show Tags

22 Feb 2015, 11:17
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION

Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)

That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.

T = total profit = $10t N = total number of items sold $$\frac{10t}{3b*n}$$ Hence, A pushpitkc is my solution correct ? Manager Joined: 14 Jun 2018 Posts: 179 Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 23 Oct 2018, 09:47 Cost Price = B No of items sold = N Profit after selling N items = T ; Profit per item = $$\frac{T}{N}$$ Let "k" be the markup percentage , therefore selling price = $$\frac{(100+K)}{100} * B$$ We know , Selling Price - Cost Price = Profit => $$\frac{(100+K)}{100} * B - B = \frac{T}{N}$$ => $$K = 100 (\frac{T}{NB} + \frac{B}{B}) - 100$$ => $$K = 100(\frac{T}{NB})$$ Senior PS Moderator Joined: 26 Feb 2016 Posts: 3301 Location: India GPA: 3.12 Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 23 Oct 2018, 11:28 1 dave13 wrote: Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) – B)/(100B) E. 100(T – NB)/N Kudos for a correct solution. let B be the cost +mark up price 2+1 =$3
T = total profit = $10t N = total number of items sold $$\frac{10t}{3b*n}$$ Hence, A pushpitkc is my solution correct ? Hey dave13 Unfortunately, this is wrong. I am not quite clear what you trying to do here If you are assigning values B =$5(cost per watch).
She marks up the watches and sells for a total profit of T = $1 Assuming there is a N = 1 watch and she keeps a profit of 20%(which is 0.2*$5 = \$1)

Now, working backward, trying to calculate the percentage by substituting the values
of B,N, and T, we will get

A. 100T/(NB) = 100/5 = 20 - Our solution is Option A($$\frac{100T}{NB}$$)

We don't need to substitute the values in the other 4 answer options as we have a match!

Hope that helps.
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Re: Jennifer can buy watches at a price of B dollars per watch, which she  [#permalink]

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25 Oct 2018, 08:19
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?

A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N

The profit (or markup) per watch is T/N. So the markup is:

(T/N)/B x 100 = 100T/(NB)

percent of the purchase price.

Alternate Solution:

Jennifer’s profit per watch is T/N. Using the formula for profit, we can compute Jennifer’s sell price:

profit = sell price – buy price

T/N = sell price – B

T/N + B = sell price

We now use the percent markup formula:

% markup = (Sell price – Buy price) / Buy price x 100

% markup = (T/N + B – B) / B x 100

% markup = T/N / B x 100

% markup = T/BN x 100

% markup = 100T / (BN)

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Re: Jennifer can buy watches at a price of B dollars per watch, which she &nbs [#permalink] 25 Oct 2018, 08:19
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# Derivative of $(a\,x)^{b\,x}$

is there any rule to differentiate the function $(a\,x)^{b\,x}$?

I've got to find the derivative of $(x^2+1)^{\arctan x}$ and Wolfram|Alpha says the answer is $$\tan^{-1}(x) (x^2+1)^{\tan^{-1}(x)-1} \left(\frac{d}{dx}(x^2+1)\right)+\log(x^2+1) (x^2+1)^{\tan^{-1}(x)} \left(\frac{d}{dx}(\tan^{-1}(x))\right)$$ Is there any general rule to do that? Thanks.

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Your derivative of $(x^2+1)^{\arctan x}$ is the particular case for $u(x)=x^2+1$ and $v(x)=\arctan x$ of

$$\frac{d}{dx}\left(\left[ u(x)\right] ^{v(x)}\right)=v(x)\left[ u(x)\right] ^{v(x)-1}u^{\prime }(x)+\left( \ln u(x)\right) \left[ u(x)\right] ^{v(x)}v^{\prime }(x),$$

or omitting de variable $x$:

$$\left( u^{v}\right)^{\prime }=vu^{v-1}u^{\prime }+\left( \ln u\right) u^{v}v^{\prime }.$$

This can be derived observing that, since $u=e^{\ln u}$, we have $u^v=e^{v\ln u}$:

$$\begin{eqnarray*} \frac{d}{dx}\left( u^{v}\right) &=&\frac{d}{dx}\left( e^{v\ln u}\right) \\ &=&e^{v\ln u}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right) \\ &=&u^{v}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right) \\ &=&u^{v}\ln u\frac{dv}{dx}+u^{v}\frac{v}{u}\frac{du}{dx} \\ &=&u^{v}(\ln u)v'+u^{v-1}vu'. \end{eqnarray*}$$

Another possibility is to consider the variables $u$ and $v$ (both depending on $x$) in the function

$$z=f(u(x),v(x))=\left[ u(x)\right] ^{v(x)}$$

and determine its total derivative with respect to $x$:

$$\begin{eqnarray*} \frac{dz}{dx} &=&\frac{d}{dx}\left( \left[ u(x)\right] ^{v(x)}\right) \\ &=&\frac{% \partial z}{\partial u}\frac{du}{dx}+\frac{\partial z}{\partial v}\frac{dv}{% dx} \\ &=&v(x)\left[ u(x)\right] ^{v(x)-1}u^{\prime }(x)+\left[ u(x)\right] ^{v(x)}\left( \ln u(x)\right) v^{\prime }(x) \end{eqnarray*}$$

because

$$\frac{\partial z}{\partial u}=\frac{\partial }{\partial u}\left( u^{v}\right) =vu^{v-1}$$

and

$$\frac{\partial z}{\partial v}=\frac{\partial }{\partial v}\left( u^{v}\right) =\frac{\partial }{\partial v}\left( e^{\ln u\cdot v}\right) =e^{\ln u\cdot v}\ln u=u^{v}\ln u.$$

For $u(x)=ax,v(x)=bx$, we get

$$\frac{d}{dx}\left( \left( ax\right) ^{bx}\right) =bx\left( ax\right) ^{bx-1}a+\left( ax\right) ^{bx}\left( \ln (ax)\right) b.$$

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Assuming you mean $(ax)^{bx}$, I'd just write it as $(e^{\ln(ax)})^{bx}$ and use the chain rule (ie $e^{\ln(ax)bx} = e^{u(x)}$ and go from there).

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HINT $\rm\ \ (F^G)'\ =\ (e^{G\:ln\ F})'\: =\ F^G\ (GF'/F + G'\ ln\ F)$

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I think there is a typo in the sign. –  Américo Tavares Mar 6 '11 at 16:26
@Americo: Fixed it, thanks! –  Bill Dubuque Mar 6 '11 at 16:32

Not answering the math part, since Stijn has already done that, but if you click on the "show steps" button, Wolfram|Alpha shows you one possible path for the derivation. I've included the image for the derivation of $\frac{\partial}{\partial x} ((a x)^{b x})$

A similar set of steps is supplied for the other derivative that you want to take: http://www.wolframalpha.com/input/?i=d/dx((x^2%2B1)^(tan^(-1)(x)))

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The link for the first example is wolframalpha.com/input/?i=D[(ax)^(bx),x]. Does anyone know how to get Wolfram|Alpha links working properly in the markup? –  Simon Mar 6 '11 at 11:41
The parentheses and the caret confuse Markdown. Try escaping ^ with %5E, ( with %28, ) with %29, [ with %5B and ] with %5D like this: WA Link (you obviously have to use the [title](URL) syntax) –  kahen Mar 6 '11 at 13:21


# Solution gives wrong answer to probability problem

Great Northern Airlines flies small planes in northern Canada and Alaska. Their largest plane can seat 16 passengers seated in 8 rows of 2. On a certain flight flown on this plane, they have 12 passengers and one large piece of equipment to transport. The equipment is so large that it requires two seats in the same row. The computer randomly assigns seats to the 12 passengers (no 2 passengers will have the same seat). What is the probability that there are two seats in the same row available for the equipment?

This problem was posed on Brilliant last week, and now that the official solution is posted, I would like to know why my solution gives wrong result. Here is my solution:

The number of ways we can seat 12 (identical) people on 16 seats is $16\choose 12$. Now the equipment can occupy any of the rows (8 possibilities), and the 12 people must be seated on the remaining 14 seats ($14 \choose12$ ways). Thus the desired probability is $$\frac{8 {14\choose12}}{16\choose12}=\frac25$$ The official solution gives $\frac5{13}$, but I don't understand what is wrong with mine.

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Note that the link you provided is unique to you. You need to use the "Share this problem" link for others to view the problem properly. – Calvin Lin May 10 '13 at 0:55
@CalvinLin Oh, I'll keep that in mind. – Dave May 10 '13 at 12:02

You haven't factored in the case in which two rows are devoid of people - that setup is being double-counted in your count.

Think of it this way - you can place four seats, with the restriction that at least one pair must be together in a row. So subtract off the number of ways you can have no pairs from the total number of combinations $${16\choose4}-{8\choose4}\cdot2^4 = 700$$ as each of the four rows containing an empty seat can have it in either of the two seats. So the probability is $$\frac{700}{1820}=\frac{5}{13}$$ (note that this means that you were double-counting 28 combinations... which makes sense, as $28 = {8\choose2}$, and that's the number of ways you can have two pairs of empty seats)

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Your answer is too big, because the cases where the people are packed into 6 rows, leaving 2 empty rows for the equipment, have been counted twice.

In general, when you are getting the wrong answer to this kind of problem, and you can't see what you're doing wrong, it may help to try doing the same problem with smaller numbers. In this case, consider the same problem with 2 people and 3 rows of seats. Now the correct probability is one; there will always be room for the equipment. What do you get for an answer if you do it your way?

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# Collecting proofs for $\sum_{n=2}^{\infty} \, \frac{n-1}{2^n} = 1$ [duplicate]

Update: The summation I came across has the form shown in title, and that exact question appears to be new. I could ask for proofs that take on this summation directly (without reducing it to summations starting at $$n = 0$$ or $$n =1$$), and that would be the preferred answer and the only one I will accept. But might as well let this fly as is, collecting all proofs of the identified equivalent variants.

I just stumbled across the fact that

$$\tag 1 \sum_{n=2}^{\infty} \, \frac{n-1}{2^n} = 1$$

This is equivalent to

$$\tag 2 \sum_{n=1}^{\infty} \, \frac{n}{2^n} = 2$$

I discovered $$\text{(1)}$$ using a 'matrix/combinatorial' argument, but it would need work to turn it into a formal proof.

I googled and found this Quora link, explaining how to show $$\text{(2)}$$.

I didn't find the question on this site, prompting this 'collecting proofs post':

Please supply a proof demonstrating either $$\text{(1)}$$ or $$\text{(2)}$$. If you use any theory or technique, mention that at the start of your answer.

One approach is to note that for $$S_k:=\sum_{n\geq k}\frac{1}{2^n}$$ you have $$S_0=2$$ and $$S_{k+1}=\frac{1}{2}S_k$$. Now rearranging terms in the summation yields $$\sum_{n\geq1}\frac{n}{2^n} =\sum_{n\geq1}\sum_{k=1}^n\frac{1}{2^n} =\sum_{k\geq1}\sum_{n\geq k}\frac{1}{2^k},$$ corresponding to the following picture: The inner sums equal the $$S_k$$, so we can simplify this to $$\sum_{k\geq1}S_k =\sum_{k\geq1}\frac{1}{2^k}S_0 =S_0\sum_{k\geq1}\frac{1}{2^k}=S_0S_1=2.$$

• Anyone care to explain the downvote? – Servaes Dec 12 '18 at 15:05
• Well not me! In fact, I can use your approach to get a direct solution to the $n \ge 2$ summation. I wind up with $S_2 \times S_0 = 1$. So don't feel bad - accepting and upvoting your answer! – CopyPasteIt Dec 12 '18 at 15:14
• Glad to hear that! And indeed the argument can be generalized (or repeated?) to evaluate $\sum_{n\geq 0}\frac{f(n)}{2^n}$ for any polynomial $f$. – Servaes Dec 12 '18 at 15:25
• This is a nice answer, but I think it is important to keep the answers to this question in one place. Users should search the site to avoid duplicating material already present. This applies with extra force to experienced users who should have a fairly good idea of what type of questions have already been covered. – Jyrki Lahtonen Dec 13 '18 at 6:03

A probability theory flavored approach: The expression $$\sum_{n\geq1} \frac{n}{2^n}$$ is also the expected number of IID fair coin flips it takes to gets a heads. Assuming you know this number is finite (by some root/ratio test business), let $$H$$ be the expected time. Then from conditioning on seeing heads or tails on the first flip, respectively, $$H$$ satisfies the recursion $$H = \frac{1}{2}(1) + \frac{1}{2}(H+1),$$ so $$H = 2$$.

A brute-force approach: by induction or otherwise, $$\sum_{k=1}^n \frac{k}{2^k} = 2 - \frac{n+2}{2^n}.$$ Sending $$n \to \infty$$ recovers the desired result.

The standard technique is $$\sum_{n\ge 1}r^n=\frac{1}{1-r}-1\implies\sum_{n\ge 1}nr^{n-1}=\frac{1}{(1-r)^2}$$.

We have that for $$|x|<1$$ by

$$f(x)=\sum_{k\ge1} x^n=\frac x{1-x} \implies f'(x)=\sum_{k\ge1} nx^{n-1}=\frac1{(1-x)^2}$$

therefore

$$\sum_{k\ge1} nx^{n}=x\cdot \sum_{k\ge1} nx^{n-1}=\frac x{(1-x)^2}$$

A solution by examining the differences between successive terms.

Let us define

$$S = \sum_{n=1}^{\infty} \, \frac{n}{2^n} = \sum_{n=1}^{\infty} u_n$$

Note (Edit): the definition $$u_n = \frac{n}{2^n}$$ will be used for $$n=0$$, i.e. for a term not involved in the series.

We have:

$$u_n - u_{n+1} = \frac{n-1}{2^{n+1}} = \frac{u_{n-1}}{4}$$

It follows immediately:

$$0 = S - S = \sum_{n=1}^{\infty} u_n - \sum_{n=1}^{\infty} u_{n+1} -u_1 = -u_1 + \frac{u_0}{4} + \frac{S}{4}$$

And therefore, noting that $$u_0=0$$ and $$u_1 = \frac{1}{2}$$ $$S = \sum_{n=1}^{\infty} \, \frac{n}{2^n} = 2$$

Note: the same procedure can be used to show that $$\sum_{k=1}^n \frac{k}{2^k} = 2 - \frac{n+2}{2^n}$$

• Your answer needs some work. One problem, $u_n - u_{n-1}$ is never a positive number. Also, $u_0$ is not defined. – CopyPasteIt Dec 12 '18 at 15:52
• Sorry, I just discovered a typo: it was $u_n - u_ {n+1}$. I will correct and detail my answer a little bit more – Damien Dec 12 '18 at 16:49
• and $u_n - u_{n-1} = \frac{n}{2^{n}} - \frac{n-1}{2^{n-1}} = \frac{2-n}{2^{n}} = \frac{-(n-2)}{2^{n-2}\times 2^2} = \frac{-u_{n-2}}{4}$ - three subscripts involved – CopyPasteIt Dec 12 '18 at 16:52
• I have corrected the typo. It should be more clear now – Damien Dec 12 '18 at 16:55
• For me, removing all mention of $u_0$ is the clearest exposition, but OK... – CopyPasteIt Dec 12 '18 at 17:02

I stumbled on this by realizing that since

$$\tag 1 \sum_{n=1}^{\infty} \, \frac{1}{2^n} = 1$$

it must be true that

$$\tag 2 (\sum_{n=1}^{\infty} \, \frac{1}{2^n}) \times (\sum_{n=1}^{\infty} \, \frac{1}{2^n}) = 1$$

Using the rearrangement approach (and the identity $$\sum_1^n \, 1 = n$$) found in Servaes' answer,

$$\tag 3 (\sum_{n=1}^{\infty} \, \frac{1}{2^n}) \times (\sum_{m=1}^{\infty} \, \frac{1}{2^m}) = \sum_{n+m =k}^{\infty} \, \frac{1}{2^k} = \sum_{k=2}^{\infty} \, \frac{k-1}{2^k} = 1$$

demonstrating the identity equation.


# Elevator probability calculation

I'm learning probability, specifically techniques of counting, and need help with the following problem :

There are $5$ people in an elevator, $4$ floors in the building and each person exits at random. Find the probability that :

$(1)$ no one exit on the first floor;

$(2)$ at least one person exit on the first floor and at least one person on the second floor.

Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.

$(1)$ The number of ways to assign the $3$ remaining floors to the $5$ people is $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^5$ because for each person we can choose one of the $3$ remaining floors. By the same argument, there are $4^5$ ways to assign $4$ floors to $5$ people. Therefore, the requested probability is $$\frac{3^5}{4^5}.$$

Is my work correct for $(1)$? Any help for $(2)$ will be greatly appreciated.

• The statement "each person exits at random" is kind of vague. At each floor, does each person have a certain probability of leaving, or does each person, at the beginning, pick a random floor to get off at? – Frpzzd Jun 16 '17 at 20:55
• @Nilknarf I'm translating this exercise from my notes which are in greek so this is possibly not the most accurate translation. I think what is meant here is that all floors are equally likely. – user347616 Jun 16 '17 at 21:04
• Okay, I'll leave you an answer. :) – Frpzzd Jun 16 '17 at 21:05
• Your work is correct on 1). – Doug M Jun 16 '17 at 21:14

Nobody gets off on the First floor: $243$

Nobody gets off on the Second floor: $243$

Nobody gets off on at either floor: $2^5 = 32$

Note that nobody gets off at either floor is a subset of both nobody gets of at the first floor it is also a subset of nobody gets off at the second floor.

Nobody gets off on the First floor or Nobody gets off on the second floor: $243 + 243 - 32 = 454$

We have to subtract 32 to avoid double counting.

Somebody gets off at both the first floor and the second floor $= 4^5 - 454 = 570$


# Significance of starting the Fibonacci sequence with 0, 1…

DISCLAIMER: I do not deal with in-depth mathematics on a daily basis as some of you may, so please pardon my ignorance or lack of coherence on this topic.

QUESTION: What is the significance of starting the Fibonacci sequence with $0,1$ ?

For instance, if I picked any two random integers, say 2 and 7, to start a sequence would I actually be creating some multiple or derivation of the Fibonacci sequence?

Is there a general mathematical explanation for the relationship between any sequence represented by $a[0] = x, a[1] = y, a[n] = a[n-1] + a[n-2]$ and the Fibonacci sequence?

Or, back to my example sequence, is there a general mathematical relationship between:

$2,7,9,16,25,41,66,107,173,280...$

and

$0,1,1,2,3,5,8,13,21,34...$

Perhaps the Golden Ratio explains it somehow? Any help would be appreciated.

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I think it's just because 0 and 1 are the 2 lowest numbers that you can start from. –  Cruncher Oct 30 '13 at 20:16
As in Brian Scott's answer, note that any sequence with that additive property is a "linear combination" of the Fibonacci numbers and the Lucas numbers. That is, if you make such a sequence and call it $\mbox{frogpelt}_n,$ then there will be constants $A,B$ such that $\mbox{frogpelt}_n = A F_n + B L_n.$ –  Will Jagy Oct 30 '13 at 20:29
Fibonacci sequence. A series is an infinite sum. –  Jack M Oct 30 '13 at 23:00

Yes, such sequences are closely related, and the relationship does involve the golden ratio.

Let $\varphi=\frac12(1+\sqrt5)$ and $\widehat\varphi=\frac12(1-\sqrt5)$; $\varphi$ is of course the golden ratio, and $\widehat\varphi$ is its negative reciprocal. Let $a_0$ and $a_1$ be arbitrary, and define a Fibonacci-like sequence by the recurrence $a_n=a_{n-1}+a_{n-2}$ for $n\ge 2$. Then there are constants $\alpha$ and $\beta$ such that

$$a_n=\alpha\varphi^n+\beta\widehat\varphi^n\tag{1}$$

for each $n\ge 0$. Indeed, you can find them by substituting $n=0$ and $n=1$ into $(1)$ and solving the system

\left\{\begin{align*} a_0&=\alpha+\beta\\ a_1&=\alpha\varphi+\beta\widehat\varphi \end{align*}\right.

for $\alpha$ and $\beta$. In the case of the Fibonacci numbers themselves, $\alpha=\frac1{\sqrt5}$ and $\beta=-\frac1{\sqrt5}$; in the case of the Lucas numbers $L_n$, for which the initial values are $L_0=2$ and $L_1=1$, $\alpha=\beta=1$.

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Thank you very much, good sir! –  frogpelt Oct 30 '13 at 20:28
@frogpelt: You’re very welcome. –  Brian M. Scott Oct 30 '13 at 20:29

Let's call a sequence "fibonacci-like" if it satisfies the recurrence $$s_{n+2} = s_n + s_{n+1}$$ for all $n$. It's easy to see (or to show) that if $\{s_i\}$ and $\{t_i\}$ are two fibonacci-like sequences, then so is $\{s_i+t_i\}$, and so is $\{cs_i\}$ where $c$ is any constant. So the collection of fibonacci-like sequences forms a vector space. The dimension of the vector space is 2, since specifying two elements of the sequence (say $s_0$ and $s_1$) are enough to determine it completely.

So let's abbreviate such a sequence as $[s_0, s_1]$. The standard Fibonacci sequence $0,1,1,2,3,\ldots$ is written as $[0,1]$ in this notation. The Lucas sequence $\mathcal L_i = 1,3,4,7,11,\ldots$ is written $[1,3]$.

Since the space of all fibonacci-like sequences is a two-dimensional vector space, any two elements will form a basis for it, unless one is a multiple of the other. For example, a simple and standard basis for this vector space is the two vectors $[0,1]$ and $[1,0]$. The former is simply the standard Fibonacci sequence. The latter is the sequence $1,0,1,1,2,3,5,\ldots$, which is just the standard Fibonacci sequence shifted right by one position; its $i$th element is $f_{i-1}$, the $(i-1)$th Fibonacci number.

Now consider the general fibonacci-like sequence $[p,q]$:

$$[p,q] = p[1,0] + q[0,1]$$

So the $i$th element of the sequence $[p,q]$ is exactly $$pf_{i-1} + qf_i.$$ For example, the Lucas sequence has $$\mathcal L_i = f_{i-1} + 3f_i.$$ Simiarly your example sequence is $[2,7]$ and is therefore related to the Fibonacci sequence by $$[2,7] = 2[1,0] + 7[0,1] = 2f_{i-1} + 7f_i.$$

Any two sequences form a basis of the space as long as they are not multiples of one another. For example, any fibonacci-like sequence can be expressed in the form $s_i = af_i + b\mathcal L_i$ for some constants $a$ and $b$. For your $[2,7]$ sequence, we want $[2,7] = a[0,1] + b[1,3] = [b,a+3b]$. So $b=2$ and $a=1$, and we get $[2,7] = f_i + 2\mathcal L_i$.

Now consider the Fibonacci sequence, but shifted left by $k$ places, whose $i$th element is $f_{i+k}$ for each $i$. Then the first two terms of the fibonacci-like sequence $\{f_{i+k}\}$ are $f_k$ and $f_{k+1}$, we get $$\{f_{i+k}\} = [f_k, f_{k+1}] = f_k[1,0] + f_{k+1}[0,1] = f_kf_{i-1} + f_{k+1}f_i$$ and we have just proved the sum-of-indices formula for fibonacci numbers. Take $i=k$ in this and we get $$f_{2k} = f_kf_{k-1} + f_{k+1}f_k$$ which is useful for calculating extremely large Fibonacci numbers quickly.

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For any sequence $a_0, a_1, a_2, a_3, \dots$ that satisfies the Fibonacci recurrence, we have $$a_n=(a_1-a_0)f_n+a_0f_{n+1},$$ where $f_0,f_1, f_2, \dots$ is the Fibonacci sequence.

To prove this, let $b_n=(a_1-a_0) f_n+a_0f_{n+1}$. Note that $b_0=a_0$ and $b_1=a_1$. The two sequences $(a_n)$ and $(b_n)$ "begin" in the same way, and satisfy the same recurrence, so they are the same.

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I think this is slightly off. If $n=1$, you get $b_1=a_1+a_0$. –  Mike Oct 30 '13 at 21:18
@Mike: Thanks, cleaned up. –  André Nicolas Oct 30 '13 at 21:32


# Thread: Eve/odd numbers question

1. ## Eve/odd numbers question

Hey guys,

Let set A contain the integers 1-48 inclusive.

The sum of all the integers is 48-1+1=48 and so we use n(n+1)/2 to get 1176.

How do we sum just the odd numbers OR just the even numbers?

2. Think like Gauss. Gauss invented the formula for summing the first n integers by pairing up numbers. So, you paired up 1 with 48, 2 with 47, 3 with 46, and so on. Those sums are all the same: 1 + 48 = 49, 2 + 47 = 49, etc. How many of those pairs are there? n/2. Hence, (n/2)(n+1) is the sum. I bet you could reproduce this thinking for just the evens or just the odds. What do you think?

3. Hmm. Would we find an n that satisfies (2n-1) and then take n^2 for the odd ones?

Not sure for the even ones

4. Try this on for size: for the evens from 1 to 48, inclusive:

2 + 48 = 50
4 + 46 = 50
...
How many of these pairs are there? Well, there are 24 even numbers from 1 to 48, inclusive, so I'd say there are 12 pairs.

That is, the sum is equal to ... what do you think?

5. Even number is 2n. So, 2(24)=48. then the answer would be 24^2?

6. If you're not finding my hints helpful, please just say so, and I can try a different track. You're not following my reasoning at all. This forum's purpose is not to just give you the answer, but to help you own the answer for yourself. That won't happen unless you do most of the work. We'll help you get unstuck, but that's as far as we go.

7. Another way to go is:

2 + 4 + ... + 48 =

2 (1 + 2 + ...+ 24)

8. True, although it doesn't work quite so easily for the odds. You can jury-rig it, I guess.

9. Originally Posted by Ackbeet
True, although it doesn't work quite so easily for the odds. You can jury-rig it, I guess.
Jury-rigging MacGyver style! Here's how I would approach the odd numbers (nothing against Ackbeet's approach)

LaTeX:

Code:
1+3+5+...+47=\sum_{n=1}^{24}(2n-1)=2\sum_{n=1}^{24}n-\sum_{n=1}^{24}1
Renderer

10. Without disrepsect to the many clever solutions so far:

either
1,3,5,7,9,....,47 is a linear progression and there are standard methods for summing those. It can easily be transformed into a hypergeometric progression (with ratio 1) and there are also standard methods for summing those.

or
You have a formula for the sum of all integers in the range, and undefined gave you a clever formula for the sum of all even numbers in the range. The sum of all odd numbers is the difference between the two

ie
1+2+3+4+5+....+47+48 = 1176 (in question)

2+4+6+8+...48 = 600 (from undefined's clever post)

So
1+3+5+7+9+...47 = 1176-600

11. What about this trick...I think it works for almost everything:

Use formula a+(n-1)+d=x

where a= first number in progression, d=difference in progression and x=last number in progression. Solve for n.

Then, take n(a+x)/2=Sum of numbers.

So, if we want to sum the odd numbers, we see that we havev 1+3+5+...+47

a=1 and d=2 and x=47.

1+2(n-1)=47
n=24.
(24(48))/2=576

I think you can use this to find all even, odd, numbers divisible by 7 etc etc

12. Hello, sfspitfire23!

Here's a kooky method . . .

Set $A$ contain the integers 1 - 48 inclusive.

The sum of all the integers is: . $\frac{(48)(49)}{2} = 1176$

How do we sum just the odd numbers OR just the even numbers?

$\text{We have: }\;\begin{array}{ccccccc}
\text{Even} &=& \overbrace{2 + 4 + 6 + \hdots + 48}^{\text{24 terms}} & [1] \\ \\ [-3mm]
\text{Odd} &=& \underbrace{1 + 3 + 5 + \hdots + 47}_{\text{24 terms}} & [2] \end{array}$

$\text{Subtract [1] - [2]: }\;\text{Even} - \text{Odd} \:=\:\underbrace{1 + 1 + 1 + \hdots + 1}_{\text{24 terms}}$

$\text{Hence: }\;\text{Even} - \text{Odd} \;=\;24 \quad\Rightarrow\quad \text{Even} \;=\;\text{Odd} + 24$

$\text{Since }\,\text{Even} + \text{Odd} \:=\:1176,\,\text{ we have: }\:(\text{Odd} + 24) + \text{Odd} \;=\;1176$

. . $\text{Hence: }\;2\text{(Odd)} \:=\:1152$

$\text{Therefore: }\;\begin{Bmatrix}\text{Odd} &=&576 \\ \text{Even} &=& 600 \end{Bmatrix}$


# Sitting $n$ men and $n$ woman around a table having the wedding couple seated together

The problems says

At the wedding of John and Mary there are $n$ men and $n$ women. In how many ways they can sit at a round table, so that no two men is next to each other and John and Mary sit together?

I have some doubts because of the answer to his problem, which doesn't match the one I got.

My attempt:

Let suppose that John and Mary are already seated and that people begin sitting from the side of Mary. We have two cases. One where the person next to Mary is a man and the other when it is a woman.

1st Case:

We can set the men en $n!/n = (n-1)!$ different ways.

The are $n-1$ gaps to put the women (since next to a man are Mary and John together. So there are $(n-1)!$ different ways to sit the women.

By the rule of product we have that there are $[(n-1)!]^2$ number of ways of sitting n men and n women beginning with a man next to Mary.

2nd Case (A woman sits next to Mary):

By the same reasoning as above we find that there are $[(n-1)!]^2$ number of ways of sitting them.

Finally by the rule of sum we have that there is a total of $2[(n-1)!]^2$ number of ways of sitting n men and n women having John and Mary always seated together.

What I got matches the answer of the textbook but I feel the reasoning I followed wasn't that right. The fact of having assumed that ''The are $n-1$ gaps to put the women..'' doesn't convince me. Certainly next to the first man is Mary (that is, the gap of that side is occupied) but nothing is telling that after putting the last men there will be no gap between him and John, so there may be anyway $n$ gaps to put the women

• Surely you mean "wedding" couple rather than "weeding"? :) – hypergeometric Nov 1 '16 at 17:56
• @hypergeometric Ups. Yep, I mean that (: – Jazz Nov 1 '16 at 18:03

First of all the question is a little bit ambigous. We don't know whether the table is a circle of a row, but from your reasoning it can be concluded that it's a circle, although not for certain. Also are there $n$ men and $n$ women total or $n$ men and $n$ women plus John and Mary? Considering the textbook answer I would go with the first option. Anyway from now on I will accept the mentioned assumptions as true.

Note that there have to be an empty space between each men. Or to make it more simple if there are $2n$ numbered seats man can sit in the even or odd seats. So we have to put $n$ men in $n$ seats around a round table. The number of possibilities is $(n-1)!$

Now as Mary has to sit next to John we have two options for her, left of John or right of him. Next the remaining $n-1$ women can be seated in $n-1$ seats in $(n-1)!$. Eventually as the events are not related we have that the total number of combinations is $2\left((n-1)!\right)^2$

I guess this explanation will fill the "holes" in your solution.

• It's probably a circle since the question says "around". – hypergeometric Nov 1 '16 at 18:24

The rule that men and women alternate should also apply to John and Mary. Hence only your 1st case applies, i.e. $[(n-1)!]^2$, and not your 2nd case.

However, you should also consider the "mirror configuration" of the 1st case, i.e. where John is seated on Mary's left/right.

Hence the number of configurations should be multiplied by $2$, giving $2[(n-1)!]^2$.

NB - this is not the same as beads on a ring where you can flip over the ring and consider the configuration to be the same - here you can't flip over the dinner table!