# Is there a quicker proof to show that $2^{10^k} \equiv 7 \pmod{9}$ for all positive integers $k$?

I noticed this pattern while playing with digit sums and noticed that the recursive digit sums (until you arrive at a single digit number) of numbers like, $2^{10}$, $2^{100}$, $2^{1000}$ and so on is always $7$. So, I decided to find out if it is true that for all positive integers $k$,

$$2^{10^k} \equiv 7 \pmod{9}$$

My proof is as follows:

Lemma 1. $\, 10^k \equiv 4 \pmod{6}$ for all integers $k \geq 1$

Proof. For all integers $k \geq 1$, the number $10^k + 2$ must be divisible by $6$ since it is even (implying divisibility by $2$), and its digit sum is $3$ (implying divisibility by $3$). Therefore, you can show that \begin{align*} 10^k + 2 &\equiv 0 \pmod{6} \\ 10^k &\equiv 4 \pmod{6} \,\, \text{ for all integers } k \geq 1 \end{align*}

Lemma 2. $\, 2^{4 + 6k} \equiv 7 \pmod{9}$, for all integers $k \geq 0$

Proof. If $a \equiv c \pmod{n}$ and $b \equiv d \pmod{n}$, then $a\,b \equiv c\,d \pmod{n}$. And, by extension, $a\,b^k \equiv c\,d^k \pmod{n}$ (for integers $k \geq 0$). Therefore, with \begin{align*} 2^4 \equiv 16 \equiv 7 \pmod{9} \end{align*} and \begin{align*} 2^6 \equiv 64 \equiv 1 \pmod{9} \end{align*} we can show that, \begin{align*} 2^{4 + 6k} \equiv 7 \cdot 1^k \equiv 7 \pmod{9} \,\, \text{ for all integers } k \geq 0 \end{align*}

Theorem. $\, 2^{10^k} \equiv 7 \pmod{9}$ for all integers $k \geq 1$

Lemma 1 implies that for all integers $k \geq 1$, $10^k = 4 + 6n$ where $n$ is some positive integer. Therefore, \begin{align*} 2^{10^k} &= 2^{4 + 6n} \\ 2^{10^k} \!\!\!\! \mod a &= 2^{4 + 6n} \!\!\!\! \mod a \end{align*} for any positive integer $a$. Using this result along with Lemma 2, \begin{align*} 2^{10^k} \equiv 7 \pmod{9} \,\, \text{ for all integers } k \geq 1 \end{align*} $$\tag*{\blacksquare}$$

I think the proof is correct, but I am not a fan of it. Mainly because, I think proving Lemma 2 is a much too long a way to prove this theorem - since it proves a generalization of the theorem first. Also, I discovered that lemma numerically, which feels a bit like cheating.

Either way, is there a quicker, more elegant proof which does not:

1. Use Lemma 2, or prove some generalization of the theorem first.

2. Uses Euler's Theorem. I feel that using Euler's Theorem is using a needlessly complicated theorem to prove something as simple as this.

I am sure some of the people here can come up with single line proofs. I am curious to see if there's such a proof.

• euler's theorem is one of the basic tools when working with exponents – HereToRelax Sep 3 '16 at 1:13
• Do you mean is there a shorter proof ? – Rene Schipperus Sep 3 '16 at 1:14
• Well...$2^{10}\equiv 7\pmod 9$ and $7^{10}\equiv 7\pmod 9$. Then go by induction. – lulu Sep 3 '16 at 1:14
• "I feel using Euler's theorem is using a needlessly complicated theorem to prove something as simple as this" well, that is just silly! The entire reason to have complicated theorems is because they are versitile and so you won't have to prove simple propositions like this over and over again. "Proof: it follows directly from Euler's theorem". That is a six word proof and it is enough! It's ... silly... to give a six paragraph 2 lemma prove and then complain a six word proof is too complicated. It's not like we have prove Euler th. Every time we use it. – fleablood Sep 3 '16 at 4:46
• @fleablood Moise once referred to it as something like using a tractor where a shovel would do. It's very natural to wonder if there is a more elementary solution. There is nothing inherently wrong with finding more than one way to solve a problem. How many proofs of quadratic reciprocity exists? – steven gregory Jan 2 '18 at 15:41

## 2 Answers

Induction? Base case when $k=1$ is clear, for inductive step we have: $2^{10^k}=(2^{10^{k-1}})^{10}\equiv 7^{10}\equiv 7\bmod 9$

• How did you get $7^{10} \equiv 7 \pmod{9}$ without evaluating in a computer? – XYZT Sep 3 '16 at 1:40
• we can go $7^{10}\equiv (-2)^{10}\equiv 2^{10}$. And you already know $2^{10}\equiv 9$. So we don't need no extra stuff – HereToRelax Sep 3 '16 at 1:43
• @CarryonSmiling You mean $2^{10}\equiv 7$. It simply follows from $1+0+2+4\equiv 7\pmod{9}$. – user236182 Sep 3 '16 at 5:09

Don't fear Euler's theorem. You have it. Use it. It's not like it costs a lot of gas money or you have to pay tolls.

$2^6 \equiv 1 \mod 9$ so $2^{10^k} \equiv 2^{(10^k \mod 6 = 4^k \mod 6)} \mod 9$.

One little observation. If $4^n \equiv 4 \mod 6$ then $4^{n+1} \equiv 16 \equiv 4 \mod 6$ so inductively $4^n \equiv 4 \mod 6$ for all $n$.

So $2^{10^k} \equiv 2^{4^k} \equiv 2^4 =16 \equiv 7 \mod 9$.


# Radius either integer or $\sqrt{2}\cdot$integer

Given a circle about origin with exactly $100$ integral points(points with both coordinates as integers),prove that its radius is either an integer or $\sqrt{2}$ times an integer.

What my solution is: Since circle is about origin, hence, integral points would be symmetric about the $x$-axis and $y$-axis as well as line $x=y$ and line $x+y=0$ ,i.e. if $(x,y)$ is an integral point, so are $(x,-y),(-x,-y),(-x,y),(y,x),(y,-x),(-y,x)$ and $(-y,-x)$.Therefore, we need to consider only a single octant. Since there are a total of $100$ integral points, two cases are possible:
1) radius of the circle is integer.
2) radius of the circle is not an integer.
case 1: If radius is an integer, then $4$ points on the $x$-axis and $y$-axis of the circle would be integral points and hence each octant must have $12$ points(as $100-4=96$ is a multiple of $8$). therefore, this case is consistent.
case2: If radius is not an integer, then $100$ integral points can't be divided into $8$ parts(octants),and points on $x$-axis and $y$-axis of circle are not integral points, therefore points on line $x=y$ and $x+y=0$ must be integral points so as to divide $100-4=96$ points in $8$ parts. But since point on line $x=y$ and circle is of the form $(r\cdot\cos(45^\circ),r\cdot\sin(45^\circ))$,therefore, $r/\sqrt{2}$ is an integer and hence $r=\sqrt{2}\cdot$integer. other points of circle on these lines are consistent with it.

So, i proved that either radius is an integer and if not then it has to be $\sqrt{2}\cdot$integer.

Is there any flaw in my arguments?? I couldn't find the proof to check whether mine is correct. Thanks in advance!!

-
Looks good to me. –  Gerry Myerson Jun 20 '12 at 12:52

All the ingredients are here, but the flow of the argument is not optimal. A smooth proof of the claim would begin with "Assume the set $S:=\gamma\cap{\mathbb Z}^2$ contains $100$ elements. Then $\ldots$", or it should begin with "Assume the radius of $\gamma$ is neither an integer nor $\sqrt{2}$ times an integer. Then $\ldots$".
The essential point (which does not come out clearly in your argument) is the following: The group of symmetries of $S$ is the dihedral group $D_4$, which is of order $8$. Since $100$ is not divisible by $8$ this action has nontrivial fixed points, i.e. points on the lines $x=0$, $y=0$, $y=\pm x$.
For the case of the radius being an integer, you don't have to make any argument at all. What you are trying to prove is that a circle having exactly 100 integral points implies the radius is either integral or $\sqrt 2$integral. This would still be true if none of these circles were integral. It would even be true if there were no circles with exactly 100 integral points.
That said, your argument works fine. It extends to any number of integral points equal to $4 \pmod 8$


# Thread: Density function solution check

1. ## Density function solution check

Suppose Y has density function:

f(y) = ky(1-y) when 0<=y<=1
and
f(y)= 0 elsewhere

a)Find the value of k that makes f(y) a probability density function
b)Find P(0.4<=Y<1)
c)Find P(Y<=0.4|Y<=0.8)
d)Find P(Y<0.4|Y<=0.8)

sol.

for part a I found k to be 6 (by making the integral = 1 on the interval 0,1 and solving for k

for part b also used the integral between 0.4 and 1 and got 0.648

c) d) is where I have the problem. I suppose both must have the same answer.

Since this is conditional probability can we state that:

P(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)

so I find using integral P(0.4<=Y<=0.8)=0.544
and P(y<=0.8) = P(0<=y<=0.8)=0.896

then we divide 0.544/0.896= 0.607
I feel that I am doing something wrong here, but can't see what.

2. Originally Posted by somestudent2
Suppose Y has density function:

f(y) = ky(1-y) when 0<=y<=1
and
f(y)= 0 elsewhere

[snip]
c)Find P(Y<=0.4|Y<=0.8)
d)Find P(Y<0.4|Y<=0.8)

[snip]

c) d) is where I have the problem. I suppose both must have the same answer.

Since this is conditional probability can we state that:

P(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)

so I find using integral P(0.4<=Y<=0.8)=0.544
and P(y<=0.8) = P(0<=y<=0.8)=0.896

then we divide 0.544/0.896= 0.607
I feel that I am doing something wrong here, but can't see what.
The answer to (c) and (d) is

$\frac{\Pr(Y \leq 0.4 \text \, \text{and}\, Y \leq 0.8)}{\Pr(Y \leq 0.8)} = \frac{\Pr(Y \leq 0.4)}{\Pr(Y \leq 0.8)}$ $= \frac{\int_{0}^{0.4} ky(1-y) \, dy}{\int_{0}^{0.8} ky(1-y) \, dy} = \frac{\int_{0}^{0.4} y(1-y) \, dy}{\int_{0}^{0.8} y(1-y) \, dy} = .....$


# Factorise $f(x) = x^3+4x^2 + 3x$

Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:

$$\text{Factorise}: f(x) = x^3+4x^2+3x$$

Firstly, the GCD of the above is $x$:

$$x(x^2+4x+3)$$

Now take $x^2+4x+3$ and factorise that:

$$x^2+4x+3$$

Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.

\begin{array}{|c|c|} \hline x^2 & \\ \hline & 3 \\ \hline \end{array}

Then find HCF of 3:

$$3\\ 1 | 3$$

Enter the values $1x$ and $3x$ into the other two boxes:

\begin{array}{|c|c|} \hline x^2 & 1x \\ \hline 3x& 3 \\ \hline \end{array}

Now factorise the rows and columns:

$$x^2 + 1x = x(x+1)\\ x^2 + 3x = x(x+3)\\ 1x + 3=1(x+3)\\ 3x +3=3(x+3)$$

Therefore: $$x^2+4x+3=(x+1)(x+3)$$ It follows that: $$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$

Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...

• If you want to check your work, multiply out $x(x+1)(x+3)$, and you will find it equals $x^3+4x^2+3x$. Nov 18, 2016 at 18:40

Had never heard of the box method before I saw you use it!

When you got to the part of factoring $x^2 + 4x + 3$ I would go and find the roots of it, because with the roots one can also factor the polynomial.

Upon finding that $-1$ and $-3$ are the roots, I would know $x^2 + 4x + 3 = (x - (-3))(x - (-1)) = (x + 1)(x + 3)$.

This works for a general polynomial of degree $n$. If a $n$-degree polynomial has roots $\lambda_1, \cdots, \lambda_n$, then the polynomial is equal to $(x - \lambda_1)\cdots(x - \lambda_n)$

Of course we not always have access to all the polynomial's roots at once, but we can partially factorise and work our way through that. Let us go through the polynomial $p = x^4 + 3x^3 - x^2 - 3x$.

First obvious thing is that $\lambda_1 = 0$ and thus $p = x^4 + 3x^3 - x^2 - 3x = x(x^3 + 3x^2 - x - 3)$.

Now comes the tricky part. Finding the roots of $p_1 = x^3 + 3x^2 - x - 3$. What I always start by doing is trying some small numbers. Guessing $\lambda_2 = 1$ turns out to be fine and thus we can factor $p_1$. Now there is some polynomial $p_2$ of degree 2 that multiplied by $(x - \lambda_2) = (x - 1)$ gives $x^3 + 3x^2 - x - 3$. To calculate such polynomial I refer you to Ruffini's rule, which is just a faster way to factorize a polynomial when you know one root. Then we get $p_2 = x^2 + 4x + 3$ which was what you factorized above. Since this is a 2nd degree polynomial, we could keep trying to guess roots, use your box method, or using the quadratic formula to find its roots.

• The box method is really cool, particularly when in particular multiplying trinomials with trinomials. Nov 18, 2016 at 18:44
• First time for me too ! Nov 18, 2016 at 18:45
• @imranfat had never heard of it. should I google it and learn it?
– RGS
Nov 18, 2016 at 18:47
• Bear in mind guys, I'm a total novice! I like systematic approaches to problems, and this one helps me remember the steps required. Video I used is here: youtube.com/watch?v=_Wb_CT-1VN8&app=desktop
– Dan
Nov 18, 2016 at 18:49
• @DanB your approach is fine. Also note that the method I talked about always works. If you are able to find the roots, then you are able to factorise it! I will add something to my answer.
– RGS
Nov 18, 2016 at 18:53

I wouldn't know if this really qualifies as an answer, but what you're asking for is essentially an opinion of methodology.

First of all, your factorisation is correct: you can check the result simply by doing the multiplication.

That said, it looks like you could have saved a lot of effort by using a couple of different techniques.

To begin with, the polynomial $x^3+4x^2+3x$ is clearly divisible by $x$ (as you noted), so we factor it out to get $x(x^2+4x+3)$.

Here is where it gets interesting: to factor the degree 2 polynomial, I usually take one of these two roads.

First: observe that $(x+a)(x+b)=x^2+(a+b)x+ab$, so if we can guess two numbers $a$ and $b$ such that, in our case, we have $ab=3$ and $a+b=4$, we are done.

In this case you might have been able to spot that $3\cdot 1=3$ and $3+1=4$, so $(x+1)(x+3)$ is the factorization we're looking for. Trust me, it gets easier with practice and it's a huge time- and effort-saver for simple degree 2 polynomials with integer coefficients.

Else we could try and guess just one root, and then use polynomial long division. This has the advantage of working with any degree of polynomials, as long as you can guess one root. I usually try $1$, $-1$ and maybe $2$, but if those don't work I'm better off using some other technique.

And finally, for degree 2 polynomials you have the general formula to find the roots, which has the advantage of always working (in the sense that it will always find all real roots there are to find), but can get messy and is, in my honest opinion, rather boring.

Now, this is not to say that any of what you did is incorrect or even strictly less efficient: it is ultimately a matter of personal taste, but I feel that having as many tools as possible in your pocket can only do you good.

• Thanks! It was indeed really a question of methodology, although I wasn't completely sure of the answer! The guessing method sounds like something that with practice would indeed save a lot of time.
– Dan
Nov 18, 2016 at 19:00

Everything is more or less correct about the way you approach the problem. You could have also opted for the middle term factorisation method or the method of vanishing method. However you have to correct one thing ...

Now factorise the rows and columns: $$x^2 + 1x = x(x+1)\\ x^2 + 3x = x(x+3)\\ 1x + 3=1(x+3)\\ 3x +3=1(x+3)$$

The last line ought to be $3x +3=3(x+1)$.

• Thank you! I've edited my post. Will look up the methods you've suggested, can never hurt to have more tools in the arsenal!
– Dan
Nov 18, 2016 at 19:01
• @DanB You're welcome. Nov 18, 2016 at 19:02

To factor $x^3+4x^2+3x$, we notice that we can factor $x$ out. Therefore, we get$$x^3+4x^2+3x=x(x^2+4x+3)\tag1$$ Now, we need to see if $x^2+4x+3$ can be factored as a product of two linear terms. An easy way to factor a monic polynomial is to find two numbers $r,s$ that sum to the negated value of $b$ and have a product of $c$ in $x^2+bx+c$.

In other words, we have \begin{align*} & r+s=-b\\ & rs=c\end{align*}\tag2 for $x^2+bx+c$. In your example, we see that $-b=-4$ and $c=3$. Messing around, we see that when $r=-1,s=-3$, the requirements are met. Thus,$$x^3+4x^2+3x=x(x-1)(x-3)$$

X(X+1)(X+3)

Therefore (X+1)(X+3) = X^2+4X+3.

Therefore multiplying this by X you get X^3+4X^2+3X.

So yes well done that is correct.

Well done. And for those of us unaware of the box method, the following would have also worked: $$x^2+4x+3=x^2+4x+4-1=(x+2)^2-1=(x+2-1)(x+2+1)=(x+1)(x+3)$$

Avoid the 'box method' which will only work for quadratics with nice integer roots and doesn't really tell you anything about what's going on.

Suppose we have a quadratic $p(x)$ that factorizes as:

$$p(x) = (x + s)(x + t)$$

where we don't know what $s$ and $t$ are yet. If we multiply out the brackets, then we get:

$$p(x) = x^2 + sx + tx + st = x^2 + (s+t)x + st$$

Now suppose that we are given $p(x)$ in the form

$$p(x) = x^2 + bx + c$$

In order to factorize $p(x)$, all we need to do is find two numbers $s$ and $t$ such that $$s+t=b$$ and $$st=c$$

In your case, you want to find $s$ and $t$ such that $s+t=4$ and $st=3$. It shouldn't take you very long to realize that $s=3$ and $t=1$ will do. Then you can immediately factorize the polynomial as $(x+3)(x+1)$.

Slogan: To factorize a quadratic of the form $x^2+bx+c$, just find two numbers that add to give $b$ and multiply to give $c$.

Just for interest

This method should work for all the quadratics you see in Math 1, and I hope that it's a bit clearer what's going on compared to the 'Box Method'. You might be interested in how we solve quadratic equations that don't have nice solutions. For example, suppose we were trying to factorize $$x^2+x-1$$ We want to find two numbers that add together to give $1$ and multiply together to give $-1$. It turns out that the right two numbers are $$\frac{1+\sqrt{5}}{2}$$ and $$\frac{1-\sqrt{5}}{2}$$ How did I work those out? Well, I'll show you a method that was invented by the ancient Babylonians. Suppose we have two numbers $b$ and $c$ and we're trying to find two numbers $s$ and $t$ such that $s+t=b$ and $st=c$.

If we square the first equation, we get $$b^2 = (s+t)^2 = s^2 + 2st + t^2$$ Now since $st=c$, we can subtract $4c$ from the left hand side and $4st$ from the right hand side to get: $$b^2-4c = s^2 - 2st + t^2$$ But $s^2 - 2st + t^2=(s-t)^2$, so we may write $$s-t = \sqrt{b^2-4c}$$ (Here, we are free to assume that $s\ge t$, so this is the positive square root.) Now we can recover $s$ and $t$: $$s = \frac{(s+t) + (s-t)}{2} = \frac{b+\sqrt{b^2-4c}}{2}$$ $$t = \frac{(s+t) - (s-t)}{2} = \frac{b-\sqrt{b^2-4c}}{2}$$


# Proving recurrence using induction where there is an upper limit on the number of integers to prove it for

Given $$x_0=1$$ and $$x_j=x_{j-1}\frac{N-(j-1)}{N}+x_{j+1}\frac{j+1}{N}$$ for $$j=1,...,N-1$$, the formula $$x_j={N\choose j}$$ can be proven by induction. I do not see why we are able prove it by induction, seeing that using induction, we prove the base case, we assume the formula holds for n, then show it holds for n+1, then we claim it holds for every integer. In this case it only holds up to N-1. So why does the induction proof work? I think that the induction proof should fail.

The inductive proof: $$x_0=1$$, Suppose the result is true for $$k \le j$$

\begin{align}x_{j+1} &=\frac{N}{j+1}\left(x_j-\frac{N-j+1}{N}x_{j-1}\right)\\&=\frac{N}{j+1}\left(\frac{N!}{j!(N-j)!}-\frac{N-j+1}{N}\frac{N!}{(j-1)!(N-j+1)!}\right)\\ &\text{after some simplification}\\&={N\choose{j+1}} \end{align}

See it works but I think that it should fail.

• You need to use "Strong" induction in this case. First show it works for $j=0$ and $j=1$, then show if it works for $j-1$ and $j$ then it works for $j+1$. Jan 15 '20 at 21:04
• @DonaldSplutterwit Yes. But, it works only up to N-1 and not all the integers. So induction should not prove this formula. Jan 15 '20 at 21:10
• I think you are misunderstanding what the statement is claiming. It is not claiming $x_j={N\choose j}$ for all $N$ and doing induction on $N$. Is is claiming that $x_j={N\choose j}$ for all $j$ (up to $N-1$) and doing induction on $j$. Part of what is confusing you is that we are doing induction on a variable $j$ that has an upper limit, $N-1$, and so our result is not for all natural numbers but just for natural numbers from $1$ to $N$. This is okay. We can do induction for finite values. Jan 15 '20 at 21:48
• Both $x_j$ and ${N \choose j}$ are only defined for $j\le N$ so that's our upper bound. Our statement is $P(j)=$ if [$j \le N$] then [$x_j = {N\choose j}$. We can prove $P(j)$ for all natural $j$ because if $j> N$ then [$j \le N]$ is false and "if [FALSE] then $anything$" is a true statement so if $j > N$ then $P(j)=$ if [$j \le N$] then [$x_j = {N\choose j}$; is a true statement. Jan 15 '20 at 22:18
• "Nowhere do we specify an upper bound" Uh.... yes we do! ... it says "for j=1,...,N−1" in black and white!.... So induction will be successful up to $j+1 \le N$. For $j > N$ it will fail but we don't give a flying funky if it fails for values we don't care about. We only care that it is successful on the values we DO care about. ... "In this case it only holds up to N-1" For $j+1$ where $j \le N-1$. So $j+1 \le N$. And we don't CARE if it holds for any higher values? Why on earth would we? Jan 15 '20 at 22:52

The statement is not for every positive integer but only for positive integers up to $$N$$. It is not trying to claim it is true for any positive integer greater than $$N$$.

Consider: $$P(j )=$$: If $$j \le N$$ then something, call it $$Q(j)$$ is true.

Let's say we can show that if $$k< N$$ that $$Q(k)\implies Q(k+1)$$ but only if $$k < N$$. I claim we can still prove $$P(k)$$ is true for all natural $$k$$.

Base case: $$P(1)$$. We show that $$Q(1)$$ is true and as $$1 < N$$ then \$P(1) is true.

Induction step: $$P(k)\implies P(k+1)$$.

Assume if $$k\le N$$ then $$Q(k)$$ is true.

Case 1: $$k \ge N$$.

Then $$k+1 > N$$ and [$$k+1 \le N$$] is false: $$FALSE \implies Q(k+1)$$ is vacuously true whether $$Q(k+1)$$ is true or not. So $$P(k+1)$$ is true.

Case 2: $$k < N$$.

The $$k+1 \le N$$. We showed that $$Q(k)\implies Q(k+1)$$. So if [$$k+1 \le N]\implies Q(k+1)$$ is true. So $$P(k+1)$$ is true.

SO our induction step works.

We have proven:

For any natural $$j$$, $$P(j)$$ is true.... or in other words,

if $$j \le N$$ then $$Q(j)$$ is true... or in other words,

$$Q(j)$$ is true for every natural $$j \le N$$.

That's all the induction is trying to claim.

The induction of $$Q(j)$$ works..... up to $$j \le N$$. There is nothing invalid about this.

Given the recurrence you have written, you will need two initial values to get going. Solving the recurrence for $$x_{j+1}$$ gives $$x_{j+1} = \frac{n x_j+ (j-n-1) x_{j-1}}{j+1}.$$ Note that if $$j=n$$, this gives $$x_{n+1} = \frac{n\cdot 1 + (n-n-1)\cdot n}{n+1} = 0,$$ just as you would expect. So in fact the inductive proof holds for all $$j$$ provided you define $$\binom{n}{j}=0$$ for $$j<0$$ or $$j>n$$.


How to Inscribe a Circle in a Triangle using just a compass and a straightedge. Example 5. Home List of all formulas of the site; Geometry. A circle can be drawn inside a triangle and the largest circle that lies in the triangle is one which touches (or is tangent) to three sides, is known as incircle or inscribed. Find AD,BE and CF ( these 3 are altitudes of triangle ABC ) . The radius of the circle inscribed in the triangle is. Yes; If two vertices (of a triangle inscribed within a circle) are opposite each other, they lie on the diameter. The area within the triangle varies with respect to … We want to find area of circle inscribed in this triangle. I have solved for the diameter and I got 2. 2.A movie company surveyed 1000 people. 640×482. If AB=5 cm, BC=12 cm and < B=90*, then find the value of r. Now draw a diameter to it. For the right triangle in the above example, the circumscribed circle is simple to draw; its center can be found by measuring a distance of 2.5 units from A along ¯ AB. Radius of a circle inscribed in a right triangle . Solve for the third side C. The side opposite the right angle is called the hypotenuse (side c in the figure). Small. Answers. Examples: For each inscribed quadrilaterals find the value of each variable. In this situation, the circle is called an inscribed circle, and its center is called the inner center, or incenter. This triangle, this side over here also has this distance right here is also a radius of the circle. Inscribe: To draw on the inside of, just touching but never crossing the sides (in this case the sides of the triangle). May 2015 13 0 Canada May 14, 2015 #1 Hi everyone, I have a question. A circle is inscribed in the triangle if the triangle's three sides are all tangents to a circle. So let's say that this is an inscribed angle right here. Calculator Technique. abc is a right angle triangle right angled at a a circle is inscribed in it the length of two sides containing angle a is 12 cm and 5 cm find the radi - Mathematics - TopperLearning.com | 42jq3mpp Original. Every non-equilateral triangle has an infinitude of inscribed ellipses. It's going to be 90 degrees. So, Area A: = (base * height)/2 = (2r * r)/2 = r^2 A right triangle (American English) or right-angled triangle (British English) is a triangle in which one angle is a right angle (that is, a 90-degree angle). Illustration showing the diameter of a circle inscribed in a right triangle is equal to the difference between the sum of the legs and the hypotenuse. Because the larger triangle with sides 15, x, and 25 has a base as the diameter of the circle, it is a right triangle and the angle opposite the diameter must be 90. Since the triangle's three sides are all tangents to the inscribed circle, the distances from the circle's center to the three sides are all equal to the circle's radius. This is a central angle right … Let me draw another triangle right here, another line right there. Right angles are typically denoted by a square drawn at the vertex of the angle that is a right angle. We bisect the two angles and then draw a circle that just touches the triangles's sides. asked Oct 1, 2018 in Mathematics by Tannu ( 53.0k points) circles BEOD is thus a kite, and we can use the kite properties to show that ΔBOD is a 30-60-90 triangle. D. 18, 24, 30 . Question 188171: 1.A circle with a radius of 1 in. The sheet of Circle Theorems may help you. What is the length of $BD?$ What is the length of $DC?$. You know the area of a circle is πr², so you’re on the lookout for π in the answers. But I just don't understand how to get the largest and smallest. In the given figure, a cradle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. A right triangle is a triangle in which one angle has a measurement of 90° (a right angle), such as the triangle shown below.. A circle is inscribed in an equilateral triangle with side length x. The center of the incircle is a … One of them is a circle, and one of them is the Steiner inellipse which is tangent to the triangle at the midpoints of the sides. The inverse would also be useful but not so simple, e.g., what size triangle do I need for a given incircle area. Find the area of the black region. Thus, the Pythagorean theorem can be used to find the length of x. x 2 + 15 2 = 25 2 Rather than do the calculations, notice that the triangle is a 3-4-5 triangle (multiplied by 5). For the 3,4,5 triangle case, the radius can be found algebraically or by construction. A circle circumscribing a triangle passes through the vertices of the triangle while a circle inscribed in a triangle is tangent to the three sides of the triangle. Right triangle. Δ ABC is a right angled triangle with ∠A = 90°, AB = b cm, AC = a cm, and BC = c cm A circle is inscribed in this triangle. If AB=5 cm, BC=12 cm and < B=90*, then find the value of r. If the two sides of the inscribed triangle are 8 centimeters and 10 centimeters respectively, find the 3rd side. Right Triangle: One angle is equal to 90 degrees. For the right triangle in the above example, the circumscribed circle is simple to draw; its center can be found by measuring a distance of $$2.5$$ units from $$A$$ along $$\overline{AB}$$. Show Step-by-step Solutions. Let's call this theta. Alex drew a circle with right triangle prq inscribed in it, as shown below: the figure shows a circle with points p, q, and r on it forming an inscribed triangle. Find the radius of its incircle. Forums. 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# Infinite Sum of Sines With Increasing Period

A while ago, I was thinking about the Weierstrass function, which is a sum of sines with increasing frequencies in such a way that the curve is a fractal. However, I wondered what would happen if one took the sum where the frequencies decreased; in particular, noting that $|\sin(x)|\leq x$, it is clear that the function $$f(x)=\sum_{n=1}^{\infty}\sin\left(\frac{x}{s_n}\right)$$ converges pointwise for any sequence $s_n$ such that the sum of $\frac{1}{s_n}$ converges absolutely - and, in fact, yields an $f$ which is analytic. Of particular interest to me is the sequence of square numbers - that is, the function $$f(x)=\sum_{n=1}^{\infty}\sin\left(\frac{x}{n^2}\right).$$ I created the following plot of the function from the first 10,000 terms in the series:

What I find interesting here is that, for some reason I can't determine, it looks like $f(x)$ might be asymptotic to $\sqrt{x}$. I've checked numerically for higher arguments and this seems to continue to be the case. This strikes me as odd, since I had expected it to appear more or less periodic, with long-term variation in amplitude and frequency.

1. Is $f(0)=0$ the only (real) zero of $f$?

2. Does $f$ grow without bound? What is it asymptotic to?

• that's 3 questions. There are three types of people in the world: Those who can count to three, and those who can't. (LOL) – Mark Fischler Oct 16 '14 at 23:42
• It often helps to analyze these kinds of functions as real or imaginary parts of the same series with the exponential function in place of $\sin$ or $\cos$. In this case, you have $\Im\sum_{i=1}^\infty\exp\left(i\frac{x}{n^2}\right)$. – alex.jordan Oct 16 '14 at 23:59
• Here's something funky about this function: If you take the Fourier transform of $f(x)$ and move the sum outside the integral (which I'm not sure is justified) you get zero. Yet $f(x)$ certainly looks to have a quasi-periodic structure. – Mark Fischler Oct 17 '14 at 0:18
• Well, I plugged $\int_{1}^{\infty}\sin(\frac{x}{n^2})$ into Mathematica and it returned $\sqrt{2 \pi } \sqrt{x} C\left(\sqrt{\frac{2}{\pi }} \sqrt{x}\right)-\sin (x)$ where $C$ is the Fresnel integral - $\int_{0}^{x}\cos(\pi x^2/2)$, which converges to $\frac{1}2$ - so the expression $\int_{1}^{\infty}\sin(x/n^2)$ is indeed asymptotic to $\sqrt{x}$. I wonder if there's some way to get the analogous discrete result from there. – Milo Brandt Oct 19 '14 at 14:32

Here's a proof that $f$ only vanishes at $x = 0$ (you can use a similar method to get some asymptotic results as well).

Write $f(x)/x$ as \begin{align*} {f(x)\over x} &= \sum_{n\geq 1} {\operatorname{sinc}{(x/n^2)}\over n^2} \end{align*} Since $f(x)/x$ is even, we need only treat the case $x\geq 0$. Split the sum into the regions where $x/n^2$ is smaller or greater than $\pi$. Since $\operatorname{sinc}{\lambda}>0$ for $|\lambda|\leq \pi$, and since $\operatorname{sinc}{\lambda}\geq 2/\pi$ for $|\lambda|\leq \pi/2$, we have \begin{align*} \sum_{x/n^2\leq \pi} {\operatorname{sinc}{(x/n^2)}\over n^2} &\geq {2\over \pi}\sum_{x/n^2\leq \pi/2} {1\over n^2} = {2\over \pi} \sum_{n\geq (2x/ \pi)^{1/2}}{1\over n^2} > {2\over \pi} \int_{\lceil(2x/\pi)^{1/2}\rceil}^\infty {dt\over t^2} = {2\over \pi}{1\over \lceil(2x/\pi)^{1/2}\rceil}. \end{align*} On the other hand, since $\operatorname{sinc}{\lambda}\leq 1/\lambda$ for all $\lambda>0$, we have \begin{align*} \left|\sum_{x/n^2> \pi} {\operatorname{sinc}{(x/n^2)}\over n^2}\right| & \leq \sum_{x/n^2>\pi} {1\over x} \leq {1\over x}\left\lfloor\left({x\over \pi}\right)^{1/2}\right\rfloor\leq \left({1\over \pi x}\right)^{1/2}. \end{align*} So \begin{align*} f(x)/x> {2\over \pi}{1\over \lceil(2x/\pi)^{1/2}\rceil} - {1\over (\pi x)^{1/2}}, \end{align*} which is positive when $x\geq \pi$. Since all terms in the sum are positive if $0\leq x < \pi$, it follows that $f(x)/x$ is always positive.

By the way, here's another heuristic (which can be made precise without too much trouble I think). We have \begin{align*} {f(x)\over x} & = {1\over 2}\hat g(x) = {1\over 2}\int_{-1}^1 g(t)e^{-ixt}\,dt = {1\over 2x}\int_{-x}^x g(t/x)e^{-it}\,dt, \end{align*} where $g = \sum_{n\geq 1} \chi_n$ is the sum of the characteristic functions of the intervals $[-n^{-2},n^{-2}]$. (Since $g$ is an $L^1$ function, this tells us at once that $f(x)/x\to 0$ as $x\to\infty$.) Note that $\{y:g(y)>n\} = [-n^{-2},n^{-2}] = \{y: y^{-1/2}>n\}$ (or something similar), so we should roughly expect $g$ to look like $y^{-1/2}$, and so we should expect $\hat g(x)$ to be approximately $x^{-1/2}$ (as can be seen from the last integral). You can use the same idea to get a sense of what the function would look like if you replace $\sin{(x/n^2)}$ with $\sin{(x/n^\alpha)}$ for $\alpha > 1$ (it should I think look like $x^{1/\alpha}$).

• Sorry if this approach is similar to Kirill's. I was working on it and didn't notice his answer until it was too late. – Nick Strehlke Oct 30 '14 at 1:30
• One more comment: The function $g$ I mentioned toward the end should actually be $\lfloor y^{-1/2}\rfloor$; that is, you should be able to write (for $x>0$) $$f(x) = \int_0^\infty \lfloor (x/y)^{-1/2}\rfloor \cos{y}\,dy.$$ (There may be another factor of two or something.) – Nick Strehlke Oct 30 '14 at 7:27

These are my thoughts on the problem. They do not answer any of the three questions, but do not fit on a comment.

Let $S_N$ be the $N$-th partial sum of the series and $L_n$ the lowest common multiple of $\{1,\dots,n\}$. Then $S_N$ is periodic of period $2\,L_n^2\,\pi$. $L_n$ is known to be of the order $e^{n(1+o(1))}$, so that the period is very large. Computations show that $S_N$ changes sign on $[0,2\,L_n^2\,\pi]$. Here is the grapf of $S_4$ on $[0,288\,\pi]$.

On the other hand, for any $\delta\in(0,1)$ let $\alpha\in(0,\pi/2)$ be such that $(\sin\alpha/\alpha)=\delta$. Then we have the lower bound $$f(x)\ge S_N(x)+\delta\,\Bigl(\sum_{n=N+1}^\infty\frac{1}{n^2}\Bigr)\,x,\quad0\le x\le\alpha\,N^2.$$ This is the graph of $S_{10000}$ (in blue) and the above lower bound with $n=10$, $\delta=3/\pi=0.95493$ and $\alpha=\pi/6$.

This shows that $f(x)>0$ on $(0,50\,\pi/3]$. An strategy to prove that $f(x)>0$ for all $x>0$ is to show that $$S_N(x)+\delta\,\Bigl(\sum_{n=N+1}^\infty\frac{1}{n^2}\Bigr)\,x>0,\quad0\le x\le\alpha\,N^2.$$ I have checked it up to $n=100$.

As for upper bounds, we have $$|f(x)-S_N(x)|\le\sum_{n=N+1}^\infty\min\Bigl(1,\frac{x}{n^2}\Bigr)\le \Bigl(\sum_{n=N+1}^\infty\frac{1}{n^2}\Bigr)\,x.$$

Here's a slightly informal way to get the asymptotic expansion of this function.

Split the region of summation into intervals of the form $$\pi k < \frac{x}{n^2} < \pi(k+1),$$ for $k\in\mathbb{Z}_{\geq0}$, and write the sum as $$\sum_{n\geq 1}\sin \frac{x}{n^2} = \sum_{k\geq0} \sum_{\frac{x}{\pi(k+1)} < n^2 < \frac{x}{\pi k}} \sin \frac{x}{n^2}.$$ The first term $k=0$ can be approximated with an integral, giving $$\sum_{n^2 > x/\pi}\sin\frac{x}{n^2} \sim \int_{\sqrt{x/\pi}}^{\infty} \sin\frac{x}{n^2}\,dn = \sqrt{2\pi x}C(\sqrt{2}),$$ where $C$ is the Fresnel integral.

On the rest of the sum we can approximate $\sin\frac{x}{n^2}$ with its average value $\frac2\pi$. However, the rest of the sum should be handled with some care, because the regions on which $\sin$ is positive have different lengths from the regions on which it's negative, which will produce a net contribution to the asymptotic term $O(\sqrt{x})$.

Approximating the sum $$\sum_{\frac{x}{\pi(k+1)} < n^2 < \frac{x}{\pi k}} \sin \frac{x}{n^2} \sim \frac{2(-1)^k}{\pi}\left(\sqrt{\frac{x}{\pi k}} - \sqrt{\frac{x}{\pi(k+1)}}\right),$$ and summing over $k\geq1$ gives the value $$\frac{\sqrt{x}}{\pi^{3/2}}\left((3\sqrt{2}-2)\zeta(\tfrac12) - \sqrt{2}\zeta(\tfrac12,\tfrac32) \right) = B\sqrt{x},$$ where $\zeta(z,a)$ is the generalized zeta function.

Putting everything together gives the asymptotic form $$\sum_{n\geq 1}\sin\frac{x}{n^2} \sim \left(\sqrt{2\pi}C(\sqrt{2}) + B\right)\sqrt{x} \approx 1.25038\sqrt{x}.$$

Another comment that is too long to be a comment:

The heuristic reason that the function is asymptotically proportional to $\sqrt{x}$ is that for very large $x$,

$\cdot$ The contributions of the terms in $S_n(x)$ for $n$ much less than $\sqrt{x}$ behave as pseudo-random numbers, restricted to $[-1,1]$. Thus $S_n(x)$ for $n << \sqrt{x}$ can be thought of as roughly having a mean value of zero, and a $\sigma$ on the order of $\frac{1}{2}\sqrt{x}$.

$\cdot$ The contributions of the terms in $S_n(x)$ for $n^2$ greater than about $2x$ can be well-estimated by approximating $\sin \frac{x}{n^2} \approx \frac{x}{n^2}$. Adding these from $n=\sqrt{x}$ to infinity looks like $x \int \frac{1}{n^2}$ which will be about $\frac{x}{\sqrt{2x}} = \frac{\sqrt{x}}{\sqrt{2}}$. And these contributions are all positive.

$\cdot$ The contributions of the terms in $S_n(x)$ for $n^2 \approx \frac{2}{\pi} x$ are all roughly $1$ since we are near the top of the sine curve, and there are about $\frac{8}{3\pi} \frac{1}{2n} = \frac{4\sqrt{2}}{3\sqrt{\pi}}\sqrt{x}$ of them between $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ on the sine curve, for a contribution of pretty nearly $\sqrt{x}$.

$\cdot$ The contributions of the terms in $S_n(x)$ lying on the falling side of that first part of the sine curve almost exactly cancel with the contributions from the negative part of the first period of the sine curve (because $n^2$ is greater in that part of the curve, at at any rate, the contribution is a lot smaller than that of the flat section of the positive arch).

So all in all, you would expect $f(x)$ to behave about like $\frac{3}{2}\sqrt{x}$ for large $x$.

The last 3 bullets can be made somewhat more rigorous.

But by this reasoning, you would also expect larger fluctuations than we see in the original graph. So the "effectively ergodic" argument made in the first bullet is an over-estimate of the fluctuations, and I don't have a plausible reason why.

• I think that first bullet you have might be key; I'm not sure how to rule out fluctuations, but I'll bet a statement to the tune of, "for any $\varepsilon_1,\varepsilon_2>0$, we can find arbitrarily large $x$ and $n$ such that that $S_{n-1}(x)<\varepsilon_1\sqrt{x}$ and $(1-\varepsilon_2)\frac{x}{n^2}<\sin(\frac{x}{n^2})<(1+\varepsilon_2)\frac{x}{n^2}$" might be a good way to formalize it (that particular statement might be false, but there could be a similar one of use) - it would prove that $f$ has a sequence that's asymptotic to the square root function. – Milo Brandt Oct 18 '14 at 2:59
• You can start a line with an asterisk (*) to make it an item in a bulleted list. – user856 Oct 25 '14 at 9:47


# Covering a Cube with a Square

Suppose you are given a single unit square, and you would like to completely cover the surface of a cube by cutting up the square and pasting it onto the cube's surface.

Q1. What is the largest cube that can be covered by a $1 \times 1$ square when cut into at most $k$ pieces?

The case $k=1$ has been studied, probably earlier than this reference: "Problem 10716: A cubical gift," American Mathematical Monthly, 108(1):81-82, January 2001, solution by Catalano-Johnson, Loeb, Beebee.

(This was discussed in an MSE Question.) The depicted solution results in a cube edge length of $1/(2\sqrt{2}) \approx 0.35$.

As $k \to \infty$, there should be no wasted overlaps in the covering of the 6 faces, and so the largest cube covered will have edge length $1/\sqrt{6} \approx 0.41$. What partition of the square leads to this optimal cover?

Q2. For which value of $k$ is this optimal reached?

I have not found literature on this problem for $k>1$, but it seems likely it has been explored. Thanks for any pointers!

• I wonder if anyone in the packaging industry has the answer. May 4 '12 at 0:37
• I believe there was a "Mathematical Games" column on dissections which had a Greek cross rearranged into a square. Perhaps that article also mentioned this problem? Gerhard "Testing Your Martin Gardner Fu" Paseman, 2012.05.04 May 4 '12 at 15:51
• If memory serves, Martin Gardner is also a source for the puzzle of covering a unit cube with a properly folded 1x7 strip. Apr 1 '17 at 2:52
• Cool! The solution to that 1x7 puzzle informs the answer to the following: over all rectangles which can be folded to cover the unit cube, what is the lim inf of their areas? Gerhard "Hint: The Answer Is Six" Paseman, 2017.04.03. Apr 3 '17 at 17:43
• When I read the title I thought it was about a 2D version $\varphi:[0,1]^2\to[0,1]^3$ of Peano area-filling (or space-filling) curves $[0,1]\to[0,1]^2$... Mar 29 '19 at 12:04

Four pieces, using the tessellation technique I learned from Harry Lindgren's Geometric Dissections (1964):

• I've been looking for that! Gerhard "Was It Behind The Headboard?" Paseman, 2017.04.01. Apr 1 '17 at 10:57
• Also, if you shift the (smaller) black squares a little to the right, the one piece that almost looks like two pieces looks more like one piece. Gerhard "Tilting His Head Over This" Paseman, 2017.04.01. Apr 1 '17 at 11:06
• Beautiful! I posted a colorized version to make the dissection of the net more self-evident. Apr 1 '17 at 11:54
• Thanks! GerPas: Yes, the hexomino tessellation cqn be moved a bit to the right (and independently also a bit up, though that doesn't help your cause). JO'R: I see that you were also the one who made a three-dimensional model of a cube covered by 5 pieces of a square, but evidently it will take more time (if you bother to do it at all) to re-do for the new 4-piece record. Apr 1 '17 at 14:37
• @NoamD.Elkies: I'll make the model eventually, as there will be a certain pleasure in seeing your dissection fully deployed. Apr 1 '17 at 15:05

You can cut a $\sqrt{6}\times\sqrt{6}$ square into 24 pieces that then cover the $1\times1\times1$ cube. Two triangles from the figure below plus one parallelogram make up one $1\times1$square. parts of pieces sticking out to the left can obviously fit back in the right, so 18 pieces, plus 6 parts sticking out equals 24. You can improve on this by stitching pieces across the cube edge to make one bent piece and by stitching some of the parallelograms back to the triangles.

![cube.png][1]

[Added by O'Rourke:] Just to make Yoav's construction more explicit, here is how two triangles and a parallelogram fit together to form a $1 \times 1$ square:

[Added by Kallus:] Here's an illustration of a construction similar to Fedja's construction but with only five pieces. The first figure is the $\sqrt{6}\times\sqrt{6}$ square. The second is the $2\times3$ rectangle, which we fold into a cube by taking away the two yellow squares, folding the remainder, and adding the squares as the two missing faces.

• Brilliant!! :-) May 3 '12 at 23:51
• Actually, any two polygons of the same area are equidecomposable and the surface of the cube can be unfolded into a polygon, so the result is nice but not terribly surprising. Of course, the question about the minimal number of pieces remains. May 4 '12 at 0:15
• Assuming that "pieces" mean "connected polygonal pieces", we can take a 3 by 2 rectangle, cut it into a T-shape and two unit squares and then use the standard "sliding cut" to turn it into a square, giving the total of 6 pieces to cover the unit cube. Can we do better? May 4 '12 at 0:34
• OK. Posted to soon earlier. The change from the T tetromino to the S tetromino actually allows going down to five pieces instead of six. May 4 '12 at 4:21
• Wow! $\mbox{}$ May 4 '12 at 10:18

Just illustrating Noam Elkies' 4-piece solution:

Bottom face is mostly yellow (except for a little green); two hidden back faces are mauve.

• Lol, I can't believe you actually made a model of it. I haven't used glue and paper and scissors to cut out shapes since arts and crafts in elementary school. Though this is significantly more complicated. The wide breadth of tasks that comes with the profession of mathematics, lol.
– user78249
Apr 1 '17 at 17:51

Sorry, this is an answer to an other question. (I did not read the question carefully.)

Question: For which $$k$$, $$k$$ squares can tile the surface of cube.

Answer: $$k=6\cdot(n^2+m^2)$$.

Here is a tiling with $$k=30$$, $$n=1$$ and $$m=2$$.

(source: psu.edu)

It is obvious if the tiling is vertex-to-vertex.

If the tiling is not vertex-to-vertex, you get a closed geodesic formed by overlaping sides. Then you can shift squares on one side of the geodesic to make the tiling "more vertex-to-vertex". Repeating this operation you can make the tiling to be vertex-to-vertex.

• @Anton: Did you intend $n \ge 1 , m \ge 1$, or, say, $n \ge 1 , m \ge 0$ ? May 3 '12 at 18:03
• Yes, $k$ has to be positive; so $n\ge 1$ and $m\ge 0$. May 3 '12 at 18:33
• @Anton: Sorry to be slow :-/, but could you describe a partition of the square into 6 pieces that exactly cover the cube? May 3 '12 at 19:21
• @Joseph: the partition into faces (the cube has 6 faces). May 3 '12 at 19:35
• @Anton: I apologize for being so dense, but it may be that you are answering a different question than I asked...? I asked for how to cut up one square to cover a cube, not how to cover a cube with many squares. If I've diagnosed this correctly (unsure), it may explain why we seem to be talking past one another? May 3 '12 at 23:17

No promises that these are optimal, but here are some lower bounds:

With $k=2$, side length $3/8=0.375$ (with one piece flipped over), and with $k=3$, side length $2/5=0.4$:


# Are the closed intervals of $\mathbb{R}$ precisely the compact connected sets?

Equip $\mathbb{R}$ with the topology generated by open intervals $(a, b)$. A subset of $\mathbb{R}$ is compact iff it's closed and bounded.

Is every closed bounded connected subset of $\mathbb{R}$ a closed interval $[a, b]$ (and conversely)?

Is every open bounded connected subset of $\mathbb{R}$ an open interval $(a, b)$ (and conversely)?

Is this somehow related to the fact that removing one point from $\mathbb{R}$ splits it into 2 disconnected pieces (how is this property called anyway)?

• All true if you agree to refer to a point as a closed interval. – Mikhail Katz Apr 24 '17 at 17:03
• Do you consider $[b,b]= \{b\}$ to be a closed interval? – fleablood Apr 24 '17 at 17:04
• No, I'm considering only non-trivial intervals. – étale-cohomology Apr 24 '17 at 17:04
• beware also of the title, $[0,+\infty[$ is a closed interval but not bounded. – zwim Apr 24 '17 at 17:06
• Do you want an informal or formal answer? It's intuitively obvious that only singletons and intervals are connected. And intuitively obvious that among finite intervals that only [a,b] are closed (and always closed) and (a,b) are open (and always open), and thus, yes, your statements are all true. But showing these formally via definitions is ... not hard, but tedious.... but good practice. – fleablood Apr 24 '17 at 17:30

Clearly a closed and bounded interval is compact and connected.

Conversely, if a set is connected, then it is an interval, meaning it is a set $I$ with the following property: for all $x,y \in I$, if $x < z<y$, then $z \in I$. All the sets with this property must have one of the forms $$\{a\},\,[a,b],\, ]a,b[, \,[a,b[, \,]a,b], \,]a,+\infty[,\, [a,+\infty[,\, ]-\infty,b[ \mbox{ or } ]-\infty,b].$$ Among these, only $\{a\}$ and $[a,b]$ are compact, hence the answer to your question is yes. Notice that $\{a\}$ is an interval, by definition.

• Thank you. What about the 2nd question, about open sets? – étale-cohomology Apr 24 '17 at 17:14
• Also yes, with the same reasoning: use connectedness to narrow the list of "suspects" and then see that the only open bounded set in that list is $]a,b[$. – Ivo Terek Apr 24 '17 at 17:23

Lemma 1: Only intervals and singletons and the empty set are connected and all intervals and singletons and the empty set are are connected.

Lemma Z: $K\subset \mathbb R$ is a interval if and only if for all $x,y \in K$ then for all $k; x < k < y; k \in K$.

Proof: Should be self-evident. If $K$ is an interval than $K = [(a,b)]$ (for sake of notation $a$ can be $-\infty$ and $b$ can be $\infty$). and $a \le x < y \le b$ and for all $k: x < k < y$ then $a < k < y$ so $k \in K$.

If there exists a $k$ so that $x < k < y$ with $k \not \in K$ and $x,y \in K$ then there is no $a,b$ (not even $\pm \infty$) so that $a \le x; b \ge y$ and for all $r \in \mathbb R$ $a < r < b$; $r \in K$. (as $a < k < b$ but $k \not \in K$). So $K$ would not be an interval.

Proof of Lemma 1: If $K$ = $\emptyset$ or $K = \{x\}$, some singleton then $K$ can not be partitioned into two partitions so $K$ is connected.

Let $E \subset \mathbb R$. And let $E$ be such that there exist $x,y\in E; x< y$ and there exists a $k \not \in \mathbb R$ so that $x < k < y$. Let $A = E \cap (\infty, k)$ and $B= E \cap (k, \infty)$ then $A, B$ are non-empty partitions of $E$ and $\overline A \subset (-\infty,k]$ is disjoint from $B \subset (k,\infty)$ and $\overline B \subset [k,\infty)$ is disjoint from $A \subset (-\infty, k)$ so $E$ is not connected.

So the only connected subsets of $E$ are intervals, singletons, and the empty set.

If $K$ is an interval and $K = A\cup B$ and $A,B$ non empty and $A \cap B = \emptyset$. Let $a \in A$ and $b \in B$ and wolog $a < b$. Let $K = \{x| x \ge a; \forall k; a\le k \le x: k \in A\}$ and let $L = \{y \in K| y > a; y \in B\}$. It's easy to prove $K$ is non-empty $(a \in K)$ and bounded above (by $b$) and $L$ is non-empty ($b$ is in it) and bounded below (by $a$) and that $\sup L = \inf L$. As $K$ is an interval and $a \le j = \sup K = \inf L\le b$ then $j \in K$. And $j \in \overline A$ and $k \in \overline B$. So either $j \in \overline A \cap B$ or $j \in \overline B \cap A$. So $K$ is connected.

• My edit was solely to add a missing dollar-sign that disrupted some of the formatting. – DanielWainfleet Apr 25 '17 at 13:07


# Is $0$ a removable discontinuity of $\frac{|x-2|-2}{x}$?

Given the function $f(x) = \large\frac{|x-2|-2}{x}$ ,

Is it true to say that the function isn't defined at $x=0$ (because of the denominator!)? Thus it is a removable discontinuity ?

The problem is, that if I try to remove the absolute value, I get that in the region $x<2$ : $f(x) = -1$

What is the correct logical definition I need to use?

Thanks

-
What's wrong having with $f(x)=-1$ in the region $x<2$ ? – Ted Dec 21 '12 at 17:34
I changed the title because it made no sense and seemed to have no relevance to the question. Hope you like it! – rschwieb Dec 21 '12 at 17:51

As you said, $f$ isn't defined at $0$. However, $$\lim_{x\to 0}\frac{\left|x-2\right|-2}{x}=\lim_{x\to 0}\frac{2-x-2}{x}=-1$$ and so if we define $f(0)=-1$, $f$ becomes continuous at $0$.

-
Thanks. That's indeed what I thought. – joshua Dec 21 '12 at 17:34
@joshua: Note that when $x\to 0$, we always consider $x\neq 0$. – Babak S. Dec 21 '12 at 17:34
@BabakSorouh : great, thanks! – joshua Dec 21 '12 at 17:36

Given the function $$f(x) = \frac{|x-2|-2}{x},$$ Is it true to say that the function isn't defined at $x=0$ (because of the denominator!)?

Yes, that's correct. As currently defined, the function is undefined at $x = 0$.

But that doesn't mean that the limit of $f(x)$ as $x \to 0$ is undefined. Recall, we are interested in what is happening as $x$ gets very very close to $0$ (not what is happening AT zero).

As $x \to 0, |x - 2| = 2 - x$, so $$\lim_{x \to 0}\frac{\left|x-2\right|-2}{x}=\lim_{x \to 0}\frac{2-x-2}{x}=-1$$

Thus it is a removable discontinuity ?

Yes, indeed, by simply defining $$f(x) = \begin{cases} \frac{\left|x-2\right|-2}{x} & x\neq 0\\ \\ -1 & x = 0\\ \end{cases}$$

$f(x)$ is then continuous at $x = 0$, hence it is a removable discontinuity.

-
@Sami Me too: via googling some question I had, three of the first hits turned up math.stackexchange.com! – amWhy Jul 15 '14 at 11:45


# General square form of Taylor expansion polynomial

Suppose we have a Taylor expansion for a function $$f$$ with respect to t up to $$M$$-th order.

$$$$T_M = \sum^M_{k=0}\frac{1}{k!}f^k(x)\Delta t^k = f(x) + f'(x)\Delta t + \frac{1}{2}f''(x)\Delta t^2 + \cdots$$$$

What would be the general form of $$T_M^2$$ with respect to $$\Delta t$$?

$$$$T_M\times T_M = (\sum^M_{k=0}\frac{1}{k!}f^k(x)\Delta t^k)^2 = (???) + (???)\Delta t + (???)\Delta t^2 + \cdots$$$$

Perhaps Multinomial theorem can help?

• It's just the general form for squaring a polynomial, but you can drop higher order terms if all you want is the $M$th order approximation. Note: the $1/k!$ is inside the summation. May 27 '19 at 13:51

Since we are squaring a sum we can utilize $$\left( \sum_{k=1}^{n} a_k \right)^2 = \sum_{k=1}^{n} \sum_{j=1}^{n} a_ka_j$$ Applying this formula onto our problem yields
$$\left(\sum_{k=0}^{M} \frac{1}{k!}f^k(x)\Delta t^k\right)^2 = \sum_{k=0}^{M}\sum_{j=0}^{M} \frac{1}{k!}\frac{1}{j!}f^k(x)f^j(x)\Delta t^{k+j}$$ Taking a closer look at the coefficients of $$\Delta t^{k+j}$$, we see that there are $$k+j+1$$ different combinations of $$k$$ and $$j$$. For example ($$k+j=3$$ cf. Jose Brox' answer), we have the combinations for: $$(k,j) \in \{(0,1),(1,2),(2,1),(3,0)\}$$
Thus we can rewrite the above sum $$\sum_{k=0}^{M}\sum_{j=0}^{M} \frac{1}{k!}\frac{1}{j!}f^k(x)f^j(x)\Delta t^{k+j} = \sum_{k=0}^{M}\left[ \sum_{j=0}^{k}\frac{1}{j!}\frac{1}{(k-j)!}f^j(x)f^{k-j}(x)\right]\Delta t^k$$ where the coefficient corresponding to the $$k$$-th exponent is $$\sum_{j=0}^{k}\frac{1}{j!(k-j)!}f^j(x)f^{k-j}(x)$$
• In the second last equation, is that truncated to the order $M$? Jun 5 '19 at 10:32
• My fault, I think it should be $k$ instead of $M$ in the second sum term. Nov 5 '19 at 15:58
Like with usual polynomial multiplication, the term of degree $$k$$ comes from adding up all products of two monomials whose degrees sum to $$k$$. For example, for degree $$3$$ you have $$(0,3)+(1,2)+(2,1)+(3,0)$$.


# Lecture 005

## Summing a Random Number of i.i.d. Random Variable

So you have $X_1, X_2, ...$ random variables with the same distribution. And $N$ discrete, positive, integer random variable with $X_i \perp N$

Let us compute $S = \sum_{i = 1}^N X_i$

Theorem: Let $X_1, X_2, ...$ be i.i.d. (Independent and identically distributed) random variables, where $X_i \sim X$. Let $S = \sum_{i = 1}^N X_i$ where $N \perp X_i$, then:

\begin{align} E[S] =& E[N]E[X]\\ E[S^2] =& E[N]Var(X) + E[N^2]E[X]^2\\ Var(S) =& E[N]Var(X) + E[X]^2Var(N)\\ \end{align}

### Expectation

\begin{align*} E[S] =& E[\sum_{i = 1}^N X_i]\\ =& \sum_{n = 1}^\infty E[\sum_{i = 1}^N X_i | N = n] \cdot Pr\{N = n\}\\ =& \sum_{n = 1}^\infty E[\sum_{i = 1}^n X_i | N = n] \cdot Pr\{N = n\}\\ =& \sum_{n = 1}^\infty E[\sum_{i = 1}^n X_i] \cdot Pr\{N = n\} \tag{by $X_i, N$ independent}\\ =& \sum_{n = 1}^\infty nE[X_i] \cdot Pr\{N = n\}\\ =& E[X_i]\sum_{n = 1}^\infty n \cdot Pr\{N = n\}\\ =& E[X_i]E[N]\\ \end{align*}

### Variance

You want to compute $Var(S) = E[S^2] - E[S]^2$. We now compute $E[S^2]$

\begin{align*} &E[S^2]\\ =& E[(\sum_{i = 1}^N X_i)^2]\\ =& \sum_n E[(\sum_{i = 1}^N X_i)^2 | N = n] Pr\{N = n\}\\ =& \sum_n E[(\sum_{i = 1}^n X_i)^2] Pr\{N = n\}\\ =& \sum_n E[(X_1 + X_2 + ... + X_n)^2] Pr\{N = n\}\\ =& \sum_n nE[X_1^2] + (n^2 - n)E[X_1X_2] Pr\{N = n\}\tag{Note that $X_1 \not\perp X_1$, $X_1 \perp X_2$}\\ =& \sum_n nE[X^2] Pr\{N = n\} + \sum_n (n^2 - n)E[X]^2 Pr\{N = n\}\\ =& E[N]E[X^2] + E[N^2]E[X]^2 - E[N]E[X]^2\\ =& E[N](E[X^2] - E[X]^2) + E[N^2]E[X]^2\\ =& E[N]Var(X) + E[N^2]E[X]^2\\ \end{align*}

With above $E[S^2] = E[N]Var(X) + E[N^2]E[X]^2$:

\begin{align*} Var(S) =& E[S^2] - E[X]^2\\ =& E[N]Var(X) + E[N^2]E[X]^2 - E[X]^2\\ =& E[N]Var(X) + E[N^2]E[X]^2 - E[X]^2E[N]^2 \tag{by $E[S] = E[X]E[N]$ above}\\ =& E[N]Var(X) + E[X]^2(E[N^2] - E[N]^2)\\ =& E[N]Var(X) + E[X]^2Var(N)\\ \end{align*}

### Example: Tree Growing

We have time $t = 0, 1, 2, 3...$, and start from a node. On each step, with probability $\frac{1}{2}$, a leaf will stay inert, with probability $\frac{1}{2}$, a leaf will split to 2.

Define $X_t = \text{number of leaves in time} = t$, then we see:

Define $Y_i = \begin{cases} 1 & \text{with half probability}\\ 2 & \text{with half probability}\\ \end{cases}$

\begin{align*} X_t =& \sum_{i = 1}^{X_{t - 1}} Y_i\\ =& \sum_{i = 1}^{X_{t - 1}} 1 \cdot Pr\{\text{this leaf stay inert}\} + 2 \cdot Pr\{\text{this leaf split to 2}\}\\ \end{align*}

We can easily calculate:

1. $E[Y] = \frac{3}{2}$
2. $E[Y^2] = 1^2 \frac{1}{2} + 2^2 \frac{1}{2} = \frac{5}{2}$
3. $Var(Y) = \frac{10}{4} - \frac{9}{4} = \frac{1}{4}$

Now using the formula we calculate:

\begin{align*} &E[X_t]\\ =& E[X_{t - 1}] \cdot E[Y]\\ =& E[X_{t - 1}](\frac{3}{2})\\ =& (\frac{3}{2})^{y - 1}(\frac{3}{2}) \tag{by unfolding $E[X_{t - 1}]$}\\ =& (\frac{3}{2})^y\\ \end{align*}
\begin{align*} &Var(X_t)\\ =& E[X_{t - 1}]Var(Y) + E[Y]^2Var(X_{t - 1})\\ =& (\frac{3}{2})^{t - 1} \cdot \frac{1}{4} + (\frac{3}{2})^2 Var(X_{t - 1})\\ \end{align*}

## Stochasticall Domination

Stochastically Dominate: $X$ stochastically dominates $Y$ (write as $X \geq_{st} Y$) if $(\forall i)(Pr\{X > i\} \geq Pr\{Y > i\})$

### Jensen's Inequality

E[X^2] \geq E[X]^2
E[X^s] \geq E[X]^s (\text{for integer } x \geq 2)

Definition of convex function: (all point on convex function sit above a line that pass through it) A real-valued function $g(x)$ on interval $S \subseteq \mathbb{R}$ is convex on $S$ if:

1. $g(t) = at + b$
2. $(\forall x \in S)(g(x) \geq y(x) \equiv ax + b)$ where the line $y(x) = ax + b$ is called a supporting line for the function $g$ at $t$.

Theorem: Let $X$ be a random variable that takes on values in an interval $S$ and let $g : S \rightarrow \mathbb{R}$ be a convex function on $S$, then $E[g(X)] \geq g(E[X])$

Table of Content


# Tagged: row space

## Problem 709

Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
$\mathbf{v}_{1}= \begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{2}= \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix} ,\;\mathbf{v}_{3}= \begin{bmatrix} 1 \\ 5 \\ -1 \\ 5 \end{bmatrix} ,\;\mathbf{v}_{4}= \begin{bmatrix} 1 \\ 1 \\ 4 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{5}= \begin{bmatrix} 2 \\ 7 \\ 0 \\ 2 \end{bmatrix} .$ Find a basis for the span $\Span(S)$.

## Problem 708

Let $A=\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 &3 & 0 & 5 \\ 1 & 1 & 6 & 3 \end{bmatrix}$.

(a) Find a basis for the nullspace of $A$.

(b) Find a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

## Problem 704

Let $A=\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 &3 & 0 & 5 \\ 1 & 1 & 6 & 3 \end{bmatrix}$.
(a) Find a basis for the nullspace of $A$.

(b) Find a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

## Problem 604

Let
$A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.$

(a) Find a basis for the null space $\calN(A)$.

(b) Find a basis of the range $\calR(A)$.

(c) Find a basis of the row space for $A$.

(The Ohio State University, Linear Algebra Midterm)

## Problem 366

Let $A=\begin{bmatrix} 1 & 0 & 1 \\ 0 &1 &0 \end{bmatrix}$.

(a) Find an orthonormal basis of the null space of $A$.

(b) Find the rank of $A$.

(c) Find an orthonormal basis of the row space of $A$.

(The Ohio State University, Linear Algebra Exam Problem)

Let $A$ be an $m\times n$ matrix. Prove that the rank of $A$ is the same as the rank of the transpose matrix $A^{\trans}$.


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At that point, they'll want you to differentiate to find the maximums and minimums; at this point, you'll find the vertex, since the vertex will be the maximum or minimum of the related graphed. Learning Calculus limits is a breeze with this problem and introduction. 5, which should be followed by the multivariable max-min problems in section 6. These are sometimes called quadratic word problems. aphing Sine and Cosine functions: amplitude, phase shift, and vertical slide Cofunctions (Notes pp. Find the global maximum value and the global minimum value of the function f (x) = x 4-2 x 2 + 3 on [-1, 2]. For success with such word problems, Professor Edwards stresses the importance of first framing the problem with precalculus, reducing the equation to one independent variable, and then using calculus to find and verify the answer. Shormann provides 21st Century math and science homeschool curriculum with a Christian foundation featuring expert video instruction, automated grading, and email support. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Pre -Calculus 11 Extra Practice - Word Problems & Max/Min Name _____ Block _____ 1. What are the values of t at which the speed obtains its (local) maximum values? Answer: x = a, c, and e. To find this value, we set dA/dx = 0. This new title. Check that this value is a minimum or maximum and read exactly what form the answer should be. The constraints may be equalities or inequalities. Minimum And Maximum Some of the worksheets for this concept are Assignment date period, For each problem find all points of absolute minima and, Name, Work 17 maxima and minima, Quadratic work name maximums and minimums, Bwc payroll reporting guidelines, Ira required minimum distribution work, Calculus work problems. Maximum And Minumum. MAXIMUM AND MINIMUM IN APPLICATIONS To find a maximum or minimum, solve f'(x) = 0. Calculus: Derivatives Maximum/Minimum Word Problems Topics include cost function, ellipse, distance, volume, surface area, and more. Question: How to solve if instead, the problem asked for the minimum profit? In some problems, the minimum is the value of x (example: the 0. Now you have an equation of one variable. Optimization problems (also called maximum-minimum problems) occur in many fields and contexts in which it is necessary to find the maximum or minimum of a function to solve a problem. We don't go higher than that in many problems, but the second derivative is an important--the derivative of the derivative is an important thing to know, especially in problems with maximum and minimum, which is the big application of derivatives, to locate a maximum or a minimum, and to decide which one it is. Max & Min applications. A complete set of Class Notes, Handouts, Worksheets, PowerPoint Presentations, and Practice Tests. MAXIMUM AND MINIMUM VALUES. Such problems, involving the determination of the form of a curve having a certain maximum or minimum property, were quite different from the ordinary maximum and minimum problems of the differential calculus. There is another way to measure an angle, which involves arc length. Justify your answer. This worksheet generates AB Calculus Topics/Questions: To keep server load down, there is a maximum of 100 questions per worksheet. Behind every calculus problem, neatly packaged and ready to solve, is a word problem. Find two positive numbers such that their product is 192 and the sum of the first plus three times the second is a minimum. Area and perimeter worksheets. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. asked by Marissa on August 16, 2007; Algebra2. One that is very useful is to use the derivative of a function (and set it to 0) to find a minimum or maximum to find either the smallest something can be, or the largest it can be. Hairy inflection point problem. Maximum and Minimum of. Find the dimensions of the rectangle with the maximum area that can be in-scribed in a circle of radius 10. Improve your math knowledge with free questions in "Find the maximum or minimum value of a quadratic function" and thousands of other math skills. Solution of exercise 3. Maximum And Minumum - Displaying top 8 worksheets found for this concept. exponential functions 25. Think about the English meaning of the word constraint, and remember that the constraint equation will have an equals sign. 2 The chapter also has maximum-minimum and optimization word problemsin 2. com Description: Considered to be the hardest mathematical problems to solve, word problems continue to terrify students across all math disciplines. How to Find Maximum and Minimum Points Using Differentiation ? In this section, we will see some example problems of finding maximum and minimum values of the function. For example, in any manufacturing business it is usually possible to express profit as function of the number of units sold. Topic: Applications of Derivatives • Find the maximum or minimum value of a function in an optimization problem by finding its critical points and applying the second derivative test. The tip of the corner is no more than 4 inches above the bottom edge of the paper. Basic calculus with applications to problems in the life and social sciences. Three major examples are geometry, number theory, and functional equations. ©a 62C0z1 b3e DKGuStwa S rS9odf 6tVwVaHr Ve f 7LBLhC U. Check that this value is a minimum or maximum and read exactly what form the answer should be. Find the Local Maxima and Minima. As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. feet and then drops the same distance. More information. The largest is the absolute maximum whereas the smallest is the absolute minimum. 512 The only critical point of A is r =1. Pre-Calculus Online. The best ticket prices to maximize the revenue is then: $10−0. 75 cm and at the sides 0. This can be regarded as the special case of mathematical optimization where the objective value is the same for every solution, and thus any solution is optimal. Linear Approximation and Differentials- Finding approximate values of a function using linearizations, definition of the differential and how it relates to linear approximations. Some have short videos. Determine a feasible domain • 4. These are often referred to as word problems. For example, companies often want to minimize production costs or maximize revenue. In the optimization sections (2. word problem involving the maximum or minimum of a quadratic function. Furthermore, a global maximum (or minimum) either must be a local maximum (or minimum) in the interior of the domain, or must lie on the boundary of the. 512 is a global maximum. 371 in the problem above) after differentiating the given equation and equating it to 0. Absolute maximum. Such areas must be learned outside class. By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, −4). Piecewise Functions Worksheet. Maximum Minimum. Mathematics has its own “grammar”. Listed below are word lessons that focus on giving students instruction on how to solve most types of word problems commonly found in algebra, geometry, and trigonometry. 18 for each thing-a-ma-bob. Rather, the purpose is to show them real world situations in which calculus is helpful. 3 Increasing & Decreasing Functions and the 1st Derivative Test F: intervals where f is incr, decr relative extrema stepbystep: 1st Deriv Test Calculus Home Page Problems for 3. If θ is the angle between the ground and her line of sight to the balloon, at what rate is this angle changing at the instant the balloon hits the ground?. ) > I'm really keen to understand if it is possible to calculate maximum and > minimum points of cubic graphs WITHOUT the use of calculus. To get started a sketch may help you get a sense of what the curve looks like. We’ll break these two big Stages into smaller steps below. All students take calculus. Louis University Brody Dylan Johnson (St. Maximum and Minimum Value Word Problems - Quadratic Equations This algebra video tutorial explains how to solve word problems that asks you to calculate the maximum value of a function or How to solve word problems with quadratic equations Geometry Teachers Never Spend Time Trying to Find. You'll find a variety of solved word problems on this site, with step by step examples. 10) Give an example function f (x) where f '' (0) = 0 and there is a relative maximum at x = 0. Students must know how to perform the first and second derivative tests and how to identify maximum, minimum and inflection points from those tests. Chapter 6: Word Problems - Rate And Distance; Chapter 7: Word Problems - Percentages; Chapter 8: Word Problems - Investments; Chapter 9: Word Problems - Ratios; Chapter 10: Word Problems - Age; Chapter 11: Word Problems - Rates Of Performing Work; Chapter 12: Word Problems - Slope-Intercept Equations; Chapter 13: Word Problems With 2 Variables. Reduce the primary equation to one variable by substitutions from other equations/information given • 3. Page Navigation. Maximum And Minumum. Because the derivative provides information about the gradient or slope of the graph of a function we can use it to locate points on a graph where the gradient is zero. Pre-Calculus 11- Course Outline: File Size: Maximum and Minimum Problems. 79 to make each thing-a-ma-bob and the company charges$ 5. 1 decade ago. In the optimization sections (2. Here is a set of practice problems to accompany the Finding Absolute Extrema section of the Applications of Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University. By using this website, you agree to our Cookie Policy. Linear Approximation and Differentials- Finding approximate values of a function using linearizations, definition of the differential and how it relates to linear approximations. 1 through 3. Worksheet # 17: Maxima and Minima 1. use differentiation to solve related rate problems in a variety of pure and applied contexts. Proof for all n is delayed until we discuss numerical sequences and series later on. Examples train understanding, translation into the mathematical language (eg, equations), solve it, check the accuracy and solution discussion. Nov 24, 2009 Equation Calculator - Quadratic Solver. Word problems involving integrals usually fall into one of two general categories: alien related and non-alien related. Complete each question on lined paper. They illustrate one of the most important applications of the first derivative. Sample Minimum/Maximum Problem (05:27) Steps for finding the minimum of a function are demonstrated. 265 Clove Road, New Rochelle, New York 10801 Pre-Calculus. " That x-coordinate of a relative maximum or minimum value of the function. We we've seen, there are many useful applications of differential calculus. Finding a maximum for this function represents a straightforward way of maximizing profits. She is watching the balloon as it travels at a steady rate of 20 feet per second towards the ground. Calculus Optimization Problems/Related Rates Problems Solutions 1) A farmer has 400 yards of fencing and wishes to fence three sides of a rectangular field (the fourth side is along an existing stone wall, and needs no additional fencing). l'hopital's rule 26. A local extrema is the point where the function takes on the largest or smallest value in a small region around the point. The best ticket prices to maximize the revenue is then: $10−0. Write a function for each problem, and justify your answers. Reduce the primary equation to one variable by substitutions from other equations/information given • 3. cost, strength, amount of material used in a building, profit, loss, etc. Know how to compute absolute maxima and minima on closed regions. •• This might not be the case in applied “word problems. This is what Wolfram|Alpha does with elementary word problems: it not only gives you the answers, but it also helpfully translates the information in the problems into mathematical symbols, showing you the first (and most important) steps toward finding a solution. In the pdf version of the full text, clicking. However after t months the value of the investment, in dollars, is. The expert examines calculus and inequalities in a word problem. It is through word problems that we find value in math. Folsom Lake College's mathematics program provides students with the ability to think logically and abstractly and develop the problem-solving and computational skills necessary for success in any field of study. ©a 62C0z1 b3e DKGuStwa S rS9odf 6tVwVaHr Ve f 7LBLhC U. There are three problems, each of which has a background discussion, an illustrative example, and an exercise for you to do. Get an answer for 'Maximum profit, given revenue and cost equations. (1 pt) A company that makes thing-a-ma-bobs has a start up cost of$ 47060. If each rented room costs $10. 3 - Radians So far we have measured angles in degrees. What is a Limit? Local Maximum and Minimum Values/ Function of Two Variables The Simplex Method – Finding a Maximum / Word Problem Example, Part 5. 01 Minimum length of cables linking to one point;. MAT 111 - Pre-Calculus Chapter 6 – Trigonometric Functions 6 6. Trakimas Math WHS. Determine a feasible domain • 4. Be able to solve word problems involving maxima and minima. Read the problem carefully. Using Derivatives to Find the Absolute Maximum and Minimum Values of f(x)OBJECTIVES• Find absolute extrema using Maximum-Minimum Principle I. Optimization Problems; Introduction to Optimization. 3 Increasing & Decreasing Functions and the 1st Derivative Test F: intervals where f is incr, decr relative extrema stepbystep: 1st Deriv Test Calculus Home Page Problems for 3. (b) Find the velocity and acceleration vectors of the particle. The functions that maximize or minimize the functionals are can be found using the Euler - Lagrange of the calculus of variations. For each$1. Word problems with max/min Example: Optimization 1 A rancher wants to build a rectangular pen, using one side of her barn for one side of the pen, and using 100m of fencing for the other three sides. Can anyone help me to solve these derivative word problems pls. Find the global maximum value and the global minimum value of the function f (x) = x 4-2 x 2 + 3 on [-1, 2]. Rational Equations Word Problems. Max Min Word Problems Our approach to max min word problems is modeled after our approach to related rates word problems. Now look at the same places and think about what the slope is at those two locations. Pre-Calculus Assignment Sheet Unit 4 - Graphing & Writing Sine & Cosine Functions; Application Problems October 21 to November 5th, 2013 Date Topic Assignment Monday Gr 10/21 changes. Pre-Calculus 11- Course Outline: File Size: Maximum and Minimum Problems. We don't go higher than that in many problems, but the second derivative is an important--the derivative of the derivative is an important thing to know, especially in problems with maximum and minimum, which is the big application of derivatives, to locate a maximum or a minimum, and to decide which one it is. Can You Show Me Examples Similar to My Problem? Optimization is a tool with applications across many industries and functional areas. Applications of Extrema of Functions of Two Variables. 2017 AP® CALCULUS AB FREE-RESPONSE QUESTIONS 2. Linear Programming A linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to system of linear constraints. Graphing with Calculus. In the process of solving these problems we will develop a procedure in which all “word problems” can be effectively solved. The analytical tutorials may be used to further develop your skills in solving problems in calculus. The course concentrates on the various functions that are important to the study of the calculus. Proof for all n is delayed until we discuss numerical sequences and series later on. A high point is called a maximum (plural maxima). OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS - BC CALCULUS. The unit cost C (the cost in dollars to make each copy machine) depends on the number of machines made. Maxima and minima mc-TY-maxmin-2009-1 In this unit we show how diﬀerentiation can be used to ﬁnd the maximum and minimum values of a function. 3 FINDING THE MAXIMUM AND MINIMUM OF WORD PROBLEMS. Optimization Problems (Calculus Fun) Many application problems in calculus involve functions for which you want to find maximum or minimum values. Some worksheets contain more problems than can be done during one discussion section. There is some vital stuff in this unit. While they might not actually work out the quadratic function to come up with a precise number, managers at movie theaters,. Read the problem- write the knowns, unknowns and draw a diagram if applicable L y 8 3 x-3 x 2. 2) There are three levels of information. Below is the graph of some function, $$f\left( x \right)$$. There was not a good enough understanding of how the Earth, stars and planets moved with respect to each other. Listed below are word lessons that focus on giving students instruction on how to solve most types of word problems commonly found in algebra, geometry, and trigonometry. Identify all of the relative extrema and absolute extrema of the function. Get an answer for 'Maximum profit, given revenue and cost equations. Give all decimal answers correct to three decimal places. One-dimensional and two-dimensional gravity problems, range, vector components of velocity, etc. Clearly, negative values are not allowed by our problem, so we are left with only two cut points and the following line graph: Therefore the minimum occurs for x 3. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. If a function is continuous on a closed interval, then by the extreme value theorem global maxima and minima exist. In this maximum and minimum worksheet, students solve and complete 10 various types of word problems. Max and Min's. We shall see that such. We don't go higher than that in many problems, but the second derivative is an important--the derivative of the derivative is an important thing to know, especially in problems with maximum and minimum, which is the big application of derivatives, to locate a maximum or a minimum, and to decide which one it is. Topic: Applications of Derivatives • Find the maximum or minimum value of a function in an optimization problem by finding its critical points and applying the second derivative test. (4 Credits) Differential calculus for engineers and scientists. Worksheets are Assignment date period, For each problem find all points of absolute minima and, Work 17 maxima and minima, Work 5, Lessonunit plan name key features of graphs swbat, Fha maximum mortgage work, Name, Quadratic work name maximums and minimums. If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height is a function of time, given by s(t) = -16t^2 + 54t. opens downward and thus has a maximum point. A ball is thrown vertically upward. Review: Review the definitions and first derivative test from previous lesson. for which the stated problem makes sense. The concavity of functions is discussed. mapped lines to other lines. Hey thanks for all your help but you kinda confused me on a few: 1)Determine whether f(x)=-5x^2-10x+6 has a maximum or minimum value and find that value A. MAT 111 - Pre-Calculus Chapter 6 – Trigonometric Functions 6 6. Constrained Optimization: Step by Step Most (if not all) economic decisions are the result of an optimization problem subject to one or a series of constraints: • Consumers make decisions on what to buy constrained by the fact that their choice must be affordable. And the absolute minimum point for the interval happens at the other endpoint. After the ball has hit the floor for the first time it rises 10. 5 applied maximum and minimum problems We have used derivatives to help find the maximums and minimums of some functions given by equations, but it is very unlikely that someone will simply hand you a function and ask you to find its extreme values. 2 show that a square has the maximum area inscribed in a circle. The following problems range in difficulty from average to challenging. We de ned a critical number of f(x) on domain D to be any number x = c on the domain D such that. New Calculus 5. Some of rst year calculus repeats senior school calculus: including \Word problems" which have always been di cult for students. Lesson 10: Max and Min Problems 3. (1 pt) A company that makes thing-a-ma-bobs has a start up cost of $47060. Each question is accompanied by a table containing the main learning objective(s), essential knowledge statement(s), and Mathematical Practices for AP Calculus that the question addresses. Such problems, involving the determination of the form of a curve having a certain maximum or minimum property, were quite different from the ordinary maximum and minimum problems of the differential calculus. How to Solve World Problems in Calculus reviews important concepts in calculus and provides solved problems and step-by-step solutions. K Worksheet by Kuta Software LLC. Optimization problems (also called maximum-minimum problems) occur in many fields and contexts in which it is necessary to find the maximum or minimum of a function to solve a problem. (solution by Calculus). What are the dimensions if the printed area is to be a maximum? 2) A cylindrical container with circular base is to hold 64 cubic centimeters. I plan on working through them in class. AP Calculus Sec 3. 265 Clove Road, New Rochelle, New York 10801 Pre-Calculus. Maximum Minimum. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume. Login to reply the answers Post; Ray. So the function has a relative maximum at x=-5. So let's look. One example is the path of an airplane. 50$ , with 27,000+300(5) = 28,500 spectators and a revenue of \$ R(5) = 270,750. While they might not actually work out the quadratic function to come up with a precise number, managers at movie theaters,. Maximizing/Minimizing word problem: Minimum and Maximum Values. Area and perimeter worksheets. Maximum and Minimum of. To get started a sketch may help you get a sense of what the curve looks like. Displaying all worksheets related to - Maximum Minimum. Quadratic applications are very helpful in solving several types of word problems (other than the bouquet throwing problem), especially where optimization is involved. Read the problem carefully. To find the local maximum and minimum values of the. Next, the first- and second-derivative tests for local extrema are studied. What calculus is useful for is science, economics, engineering, industrial operations, finance, and. But we will not always be able to look at the graph. However once you know these 6 steps, then you should be able to solve any Calculus related rates problems you like. Nov 24, 2009 Equation Calculator - Quadratic Solver. superprof resources. of cardboard, what are the dimensions of the biggest box that can be made? 2. OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS - BC CALCULUS. Maximum and Minimum Value Word Problems - Quadratic Equations This algebra video tutorial explains how to solve word problems that asks you to calculate the maximum value of a function or How to solve word problems with quadratic equations Geometry Teachers Never Spend Time Trying to Find. Limit of Sequence Problems. Rates of change: the derivative, velocity, and acceleration. More Lessons for Calculus Math Worksheets A series of free Calculus Video Lessons. Precalculus review and Calculus preview - Shows Precalculus math in the exact way you'll use it for Calculus - Also gives a preview to many Calculus concepts. Using calculus you can calculate its average cruising altitude, velocity and acceleration.