# Surface area of a curve

1. Aug 8, 2013

### stunner5000pt

1. The problem statement, all variables and given/known data
An industrial settling pond has a parabolic cross section described by the equation $y = \frac{x^2}{80}$. the pond is 40 m across and 5 m deep at the cetner. the curved bottom surface of the pond is to be covered with a layer of clay to limit seepage from the pond. determine the surface are on the clay bottom of the pond

2. Relevant equations
Length of a curve formula
$$ds = \sqrt{1+\left( \frac{dy}{dx} \right)^2 }$$

3. The attempt at a solution
I was thinking that if we simply find the length of the curve, then that would yield the surface area of the curve. This gives the following:

$$s = \int_{-20}^{20} \sqrt{1 + \frac{x^2}{1600} } dx$$
Which gives an answer of 66.9

The answer however is 1332.2 m^2. This leads me suspect that we must include the fact that this curve is to be rotated around the y axis making a bowl. I am not sure how to rotate this curve though...

2. Aug 8, 2013

### Ray Vickson

What is the formula for the surface area of the surface z = f(x,y)? (Here, my z = your y, and (x,y) are the planar coordinates on the pond's surface.)

3. Aug 8, 2013

### Zondrina

I read the question a few times over and I think I've got it. Plotting $y = \frac{x^2}{80} - 5$ on wolfram gives an accurate model of the problem ( The cross section of the pond which is 5m deep at the center ).

I believe what you would want to use here is :

$2 \pi \int_{c}^{d} x \sqrt{1 + (\frac{dx}{dy})^2} dy$ which will be the surface area of your curve when you rotate it about the y axis. The reason you want to rotate it about the y-axis is because you want the surface area of the ENTIRE cross section of the pond.

I got an answer of 1332.18.

Hint : Re-arrange your equation for y. Your limits should be from the bottom of the cross section of the pond to the top.

Last edited: Aug 8, 2013
4. Aug 8, 2013

### LCKurtz

You almost have it. The $ds$ element is at distance $x$ from the y axis, so the circumference of the circle it traces is $2\pi x$. You take that times the $ds$ length to get the surface area swept out. So you want $$\int_0^{20} 2\pi x~ds = \int_0^{20} 2\pi x\sqrt{1 + \frac{x^2}{1600} }~dx$$You only go from $0$ to $20$ in the limits because revolving it gets the "other side".

Last edited: Aug 8, 2013
5. Aug 8, 2013

### stunner5000pt

Thank you very much for all your help. I see now the integrand is the curved surface area of a cylinder where the radius is x and the height is the length of a curve formula


# Radius of a circle touching a rectangle both of which are inside a square

Given this configuration :

We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.

If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $$\sqrt{2}\times R-R$$)

I really can't understand how to solve it. Any help appreciated.

• This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this. Sep 13 '18 at 18:29
• The information you have is $R(1-\cos\alpha) = 2a$ and $R(1-\sin\alpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$. Sep 13 '18 at 18:29
• @amsmath juuuusssssttt beat me to it. Nice! Sep 13 '18 at 18:31
• If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
Sep 13 '18 at 18:31
• This question could benefit from adding the 20 cm and 10 cm to the image itself. Sep 14 '18 at 13:23

It is just using the pythagorean theorem:
$a=10$ $cm$
$b=20$ $cm$
$(r-a)^2+(r-b)^2=r^2$
$(r-10)^2+(r-20)^2=r^2$
$r^2+100-20r+r^2+400-40r=r^2$
$r^2-60r+500=0$
$r=50$ $cm$
$r=10$ $cm$
The $r=50$ $cm$ is the acceptable answer.

• It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square. Sep 14 '18 at 7:12
• @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square? Sep 14 '18 at 10:12
• A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square. Sep 14 '18 at 11:24
• @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally? Sep 14 '18 at 11:40
• @XanderHenderson, Is it better now? Sep 14 '18 at 12:19

Place the center of the cricle at $O.$

Let the radius be $R$

The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.

Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$

And the distance from this point equals the $R.$

That should put you on your way to the solution.

$(x, y) = (R-20, R-10)$ as a point on the circle $y = \sqrt{R^2 - x^2}$

$R - 10 = \sqrt{R^2 - (R-20)^2}$

$(R- 10)^2 = R^2 - (R-20)^2$

$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$

$R^2 - 60R + 500 = 0$

$(R - 50)(R-10) = 0$

$R = 50$ is the only sensible option.

• $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square. Sep 14 '18 at 8:10
• @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square... Sep 14 '18 at 8:55
• @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though. Sep 14 '18 at 9:01
• @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title. Sep 14 '18 at 13:12
• But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right? Sep 14 '18 at 13:50

The point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.

This leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_{1,2} = a+b\pm\sqrt{2ab}.$$

One solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+\sqrt{2ab}.$$

Plugging in $a=10$ and $b=20$ gives $R=50$.

You can use trig to get the same answer as those above.

Draw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.

We know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).

Find the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.

The longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.

Apply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible.


# Math Help - Completing the square

1. ## Completing the square

Hello,

I'm having problems with completing the square can anyone comment on my working?

2x^2 - 3x -1 = 0

2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

2(x - 3/4)^2 -17/16 =0

2(x - 3/4)^2 = 17/16

(x - 3/4)^2 = 17/32

x - 3/4 = +-√17/32

x = 3/4 +-√17/32

2. Originally Posted by Mouseman
Hello,

I'm having problems with completing the square can anyone comment on my working?

2x^2 - 3x -1 = 0

2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

2(x - 3/4)^2 -17/16 =0

2(x - 3/4)^2 = 17/16

(x - 3/4)^2 = 17/32

x - 3/4 = +-√17/32

x = 3/4 +-√17/32

2x^2 - 3x = 1

x^2 - (3/2)x = 1/2

(x - 3/4)^2 - 9/16 = 1/2

x - 3/4 = +-sqrt(17/16)

x = 3/4 +-sqrt(17/16)

3. Thank you very much!

4. Originally Posted by sean.1986
2x^2 - 3x = 1

x^2 - (3/2)x = 1/2

(x - 3/4)^2 - 9/16 = 1/2

x - 3/4 = +-sqrt(17/16)

x = 3/4 +-sqrt(17/16)
The answer is to be x = 3/4 +- √17/4

I can't see where I am going wrong.

5. When completing the square, just use the general formula.

$a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}$

Plug in a,b,c and you're set.

6. Second attempt

2x^2 - 3x -1 = 0

2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

2((x - 3/4)^2) -17/8 =0

2(x - 3/4)^2 = 17/8

(x - 3/4)^2 = 17/16

x - 3/4 = +-√17/16

x = 3/4 +-√17/16

Is my teacher wrong in stating that it is x = 3/4 +-√17/4?

7. Just to add my 2 cents......

$2x^2-3x-1=0$

1. First transpose the -1 to the right side of the equation.

$2x^2-3x=1$

2. Divide each term by 2

$x^2-\frac{3}{2}x=\frac{1}{2}$

3. Take half of the coefficient of x, square it and add it to both sides.

$x^2-\frac{3}{2}x+(\frac{3}{4})^2=\frac{1}{2}+\frac{9}{ 16}$

4. Noting the perfect square trinomial on the left:

$(x-\frac{3}{4})^2=\frac{17}{16}$

5. Take the square root of both sides:

$x-\frac{3}{4}=\pm\sqrt\frac{17}{16}$

6. Finally,

$x=\frac{3}{4}\pm\frac{\sqrt17}{4}$

$x=\frac{3\pm\sqrt17}{4}$

8. Hello,

Originally Posted by Mouseman
Second attempt

2x^2 - 3x -1 = 0

2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

2((x - 3/4)^2) -17/8 =0

2(x - 3/4)^2 = 17/8

(x - 3/4)^2 = 17/16

x - 3/4 = +-√17/16

x = 3/4 +-√17/16

Is my teacher wrong in stating that it is x = 3/4 +-√17/4?
Note that $16=4^2$

Therefore, $\sqrt{16}=4$, and this yields the result your teacher gave you

9. Yeah it's just a question of bracketing or simplifying.

sqrt (a/b) = sqrt(a) / sqrt(b)

so sqrt(17/16) = sqrt(17) / sqrt(16) = sqrt(17) / 4

Both answers are correct but I guess I should've simplified. Sorry mate!


# Is it necessary to assume a moment generating function exists?

Consider random variables A, B, and C. We know that A = B + C. We also know that A and C have an MGF. Is it the case that B must have a MGF?

Addition: Does this change if we know A and C both come from (different) chi-squared distributions? I am tasked with finding the distribution of B. If I can just do MGF(A) / MGF (C) = MGF (B) then it's simple... but can I even write this statement without assuming MGF (B) exists?

Generally -- You can't compute the MGF of B

In general, you can't compute the MGF of $$B$$ if you only know the MGFs of $$A$$ and $$C$$. For example, consider two possible joint distributions of $$A$$ and $$C$$:

Case 1: P( A=0 and C=0) = 1/2 and P(A=1 and C=1)=1/2. In this case, the MGFs of A and C are $$(1+\exp(t))/2$$ and the MGF of B is 1.

Case 2: P( A=0 and C=1) = 1/2 and P(A=1 and C=0)=1/2. In this case, the MGFs of A and C are $$(1+\exp(t))/2$$ and the MGF of B is $$\frac{\exp(-t)+\exp(t)}2=\cosh t$$.

Notice that in both Case 1 and Case 2 the MGFs for $$A$$ and $$C$$ were $$(1+exp(t))/2$$, but the MGF for $$B$$ changed from Case 1 to Case 2.

 Generally -- You can prove existence of an MGF for B

Although you can't compute the MGF of $$B$$, you can prove that $$M_B(t)$$ exists for $$(*)\quad t\in D=\frac12 (Dom(M_A)\cap (-Dom(M_C)).$$ Suppose $$t\in D$$. If the MGF's of $$A$$ and $$C$$ exist, then for all $$t\in Dom(M_A)$$, $$M_A(t)=||\exp(ta)||_1<\infty$$ and for all $$t\in (-Dom(M_C))$$, $$M_C(-t)=||\exp(-tc)||_1<\infty$$ where $$||g||_p=\left(\int\int |g(a,c)|^p\; f(a,c)\; da dc\right)^{1/p}$$ is the $$L_p$$-norm of $$g$$ over the joint probability space and $$f(a,c)$$ is the joint pdf of $$A$$ and $$C$$. That implies $$||\exp(ta/2)||_2 < \infty$$ and $$||\exp(-tc/2)||_2 < \infty$$. By the Hölder's inequality or the Schwarz inequality, $$||\exp(ta)\exp(-tc)||_1<\infty$$. But, $$||\exp(ta)\exp(-tc)||_1= ||\exp(t(a-c)||_1= E[\exp(tB)]=M_B(t).$$ This proves that $$M_B(t)$$ exists for $$t\in D$$.

 If A and C are independent, you can compute the MGF of B

If $$A$$ and $$C$$ are independent and $$B = A-C$$, then it must be the case that $$(**) \quad M_B(t) = M_A(t)\cdot M_C(-t)$$ whenever $$t\in Dom(M_A)\cap(-Dom(M_C))$$ (see e.g. Wikipedia). Here is a rough proof.

If $$t\in Dom(M_A)\cap(-Dom(M_C))$$, then $$M_A(t)\cdot M_C(-t) = \int_{a=-\infty}^\infty \exp(t a) dF_A(a) \cdot \int_{c=-\infty}^\infty \exp(-t c) dF_C(c)$$ $$= \int_{a=-\infty}^\infty \int_{c=-\infty}^\infty \exp(t (a-c)) dF_A(a) dF_C(c)$$ $$= \int_{b=-\infty}^\infty \exp(t b) dF_B(b) = M_B(t)$$ where $$F_A, F_B$$, and $$F_C$$ are the cumulative distribution functions of $$A, B$$, and $$C$$ respectively.

 If A and C have $$\chi^2$$ distributions

In general, if $$A$$ and $$C$$ have $$\chi^2$$ distributions, you can only state that $$B$$ has an MGF and the domain of $$B$$'s MGF is \begin{align} Dom(B) &= \frac12 (Dom(M_A)\cap (-Dom(M_C))\\ &= \frac12 ((-\infty,2)\cap (-(-\infty,2))\\ &= \frac12 ((-\infty,2)\cap (-2,\infty))\\ &= \frac12 (-2,2)=(-1,1)\\ \end{align} by (*) and the fact that the domain of the MGF of a $$\chi^2$$ distribution is $$(-\infty,2)$$.

If you know that $$A$$ and $$C$$ have $$\chi^2$$ distributions, and you know that they are independent, then you can look up the MGFs for $$\chi^2$$ distributions and apply formula (**) to compute the formula for the MGF of $$B$$.

• Thank you. The question asks to prove that B is chi-squared distributed. It sounds like we know that a MGF exists, but that we can't prove it's specifically chi-squared? – purpleostrich Sep 28 at 2:06
• I learned a lot about MGFs and a bit about $\chi^2$ distributions while trying to figure out the answer to your question. :) – irchans Sep 28 at 8:24
• @purpleostrich, I suspect that if this was posed as an exercise to you then there is a missing assumption, for instance that $B$ and $C$ are independent, or as explored by irchans the assumption could be that $A$ and $C$ are independent (although I find this latter assumption somewhat less likely than the former). – pre-kidney Sep 28 at 9:01


# A stronger version of discrete “Liouville's theorem”

If a function $f : \mathbb Z\times \mathbb Z \rightarrow \mathbb{R}^{+}$ satisfies the following condition

$$\forall x, y \in \mathbb{Z}, f(x,y) = \dfrac{f(x + 1, y)+f(x, y + 1) + f(x - 1, y) +f(x, y - 1)}{4}$$

then is $f$ constant function?

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You probably wanto to add a boundedness condition. Otherwise $f(x,y)=x$ is a counterexample. –  Julián Aguirre Jul 17 '11 at 10:24
@Julian Aguirre: since $x\in\mathbb Z$, we don't have $f(x,y)\geq 0$. –  Davide Giraudo Jul 17 '11 at 11:08
@girdav You are right. The lower bound is probably enough. –  Julián Aguirre Jul 17 '11 at 13:08

You can prove this with probability.

Let $(X_n)$ be the simple symmetric random walk on $\mathbb{Z}^2$. Since $f$ is harmonic, the process $M_n:=f(X_n)$ is a martingale. Because $f\geq 0$, the process $M_n$ is a non-negative martingale and so must converge almost surely by the Martingale Convergence Theorem. That is, we have $M_n\to M_\infty$ almost surely.

But $(X_n)$ is irreducible and recurrent and so visits every state infinitely often. Thus (with probability one) $f(X_n)$ takes on every $f$ value infinitely often.

Thus $f$ is a constant function, since the sequence $M_n=f(X_n)$ can't take on distinct values infinitely often and still converge.

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I love probabilistic arguments in analysis! Very nice. –  Jonas Teuwen Jul 17 '11 at 11:46
This is indeed very nice: Question: Is it possible to change this argument in such a way that it applies for $\mathbb{Z}^n$ instead of $\mathbb{Z}^2$ only? Is it even true that a non-negative harmonic function on $\mathbb{Z}^n$ is constant for $n \geq 3$? For bounded ones this seems clear by considering the Poisson boundary. –  t.b. Jul 17 '11 at 11:53
@Byron: This paper contains the claim that it is true that "nonnegative nearest-neighbors harmonic function on $\mathbb{Z}^d$ are constant for any $d$" on page 2. –  t.b. Jul 17 '11 at 12:08
The usual Liouville theorem also holds with just a one-sided bound. –  GEdgar Jul 18 '11 at 13:56
If you know that the real part of an entire function $f(z)$ is non-negative on the complex plane, what can you say about the function $g(z)=f(z)/(1+f(z))$? –  Jyrki Lahtonen Jul 18 '11 at 14:07

I can give a proof for the d-dimensional case, if $f\colon\mathbb{Z}^d\to\mathbb{R}^+$ is harmonic then it is constant. The following based on a quick proof that I mentioned in the comments to the same (closed) question on MathOverflow, Liouville property in Zd. [Edit: I updated the proof, using a random walk, to simplify it]

First, as $f(x)$ is equal to the average of the values of $f$ over the $2d$ nearest neighbours of $x$, we have the inequality $f(x)\ge(2d)^{-1}f(y)$ whenever $x,y$ are nearest neighbours. If $\Vert x\Vert_1$ is the length of the shortest path from $x$ to 0 (the taxicab metric, or $L^1$ norm), this gives $f(x)\le(2d)^{\Vert x\Vert_1}f(0)$. Now let $X_n$ be a simple symmetric random walk in $\mathbb{Z}^d$ starting from the origin and, independently, let $T$ be a random variable with support the nonnegative integers such that $\mathbb{E}[(2d)^{2T}] < \infty$. Then, $X_T$ has support $\mathbb{Z}^d$ and $\mathbb{E}[f(X_T)]=f(0)$, $\mathbb{E}[f(X_T)^2]\le\mathbb{E}[(2d)^{2T}]f(0)^2$ for nonnegative harmonic $f$. By compactness, we can choose $f$ with $f(0)=1$ to maximize $\Vert f\Vert_2\equiv\mathbb{E}[f(X_T)^2]^{1/2}$.

Writing $e_i$ for the unit vector in direction $i$, set $f_i^\pm(x)=f(x\pm e_i)/f(\pm e_i)$. Then, $f$ is equal to a convex combination of $f^+_i$ and $f^-_i$ over $i=1,\ldots,d$. Also, by construction, $\Vert f\Vert_2\ge\Vert f^\pm_i\Vert_2$. Comparing with the triangle inequality, we must have equality here, and $f$ is proportional to $f^\pm_i$. This means that there are are constants $K_i > 0$ such that $f(x+e_i)=K_if(x)$. The average of $f$ on the $2d$ nearest neighbours of the origin is $$\frac{1}{2d}\sum_{i=1}^d(K_i+1/K_i).$$ However, for positive $K$, $K+K^{-1}\ge2$ with equality iff $K=1$. So, $K_i=1$ and $f$ is constant.

Now, if $g$ is a positive harmonic function, then $\tilde g(x)\equiv g(x)/g(0)$ satisfies $\mathbb{E}[\tilde g(X_T)]=1$. So, $${\rm Var}(\tilde g(X_T))=\mathbb{E}[\tilde g(X_T)^2]-1\le\mathbb{E}[f(X_T)^2]-1=0,$$ and $\tilde g$ is constant.

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Taxicab metric. Never heard that name (I learn maths in french at school). Funny! –  Patrick Da Silva Jul 17 '11 at 20:24
@Patrick: Also called the Manhattan metric. –  George Lowther Jul 17 '11 at 20:30
LAAAAAAAAAWL. Funnier. –  Patrick Da Silva Jul 17 '11 at 20:39
Note: A similar proof will also show that harmonic $f\colon\mathbb{R}^d\to\mathbb{R}^+$ is constant. Interestingly, in the two dimensional case, Byron's proof can be modified to show that harmonic $f\colon\mathbb{R}^2\setminus\{0\}\to\mathbb{R}^+$ is constant (as 2d Brownian motion has zero probability of hitting 0 at positive times). Neither of the proofs generalize to harmonic $f\colon\mathbb{R}^d\setminus\{0\}\to\mathbb{R}^+$ for $d\not=2$. In fact, considering $f(x)=\Vert x\Vert^{2-d}$, we see that $f$ need not be constant for $d\not=2$. –  George Lowther Jul 17 '11 at 22:54

Here is an elementary proof assuming we have bounds for $f$ on both sides.

Define a random walk on $\mathbb{Z}^2$ which, at each step, stays put with probability $1/2$ and moves to each of the four neighboring vertices with probability $1/8$. Let $p_k(u,v)$ be the probability that the walk travels from $(m,n)$ to $(m+u, n+v)$ in $k$ steps. Then, for any $(m, n)$ and $k$, we have $$f(m, n) = \sum_{(u,v) \in \mathbb{Z}^2} p_k(u,v) f(m+u,n+v).$$ So $$f(m+1, n) - f(m, n) = \sum_{(u,v) \in \mathbb{Z}^2} \left( p_k(u-1,v) - p_k(u,v) \right) f(m+u,n+v).$$ If we can show that $$\lim_{k \to \infty} \sum_{(u,v) \in \mathbb{Z}^2} \left| p_k(u-1,v) - p_k(u,v) \right| =0 \quad (\ast)$$ we deduce that $$f(m+1,n) = f(m,n)$$ and we win.

Remark: More generally, we could stay put with probability $p$ and travel to each neighbor with probability $(1-p)/4$. If we choose $p$ too small, then $p_k(u,v)$ tends to be larger for $u+v$ even then for $u+v$ odd, rather than depending "smoothly" on $(u,v)$. I believe that $(\ast)$ is true for any $p>0$, but this elementary proof only works for $p > 1/3$. For concreteness, we'll stick to $p=1/2$.

We study $p_k(u,v)$ using the generating function expression $$\left( \frac{x+x^{-1}+y+y^{-1}+4}{8} \right)^k = \sum_{u,v} p_k(u,v) x^u y^v.$$

Lemma: For fixed $v$, the quantity $p(u,v)$ increases as $u$ climbs from $-\infty$ up to $0$, and then decreases as $u$ continues climbing from $0$ to $\infty$.

Proof: We see that $\sum_u p_k(u,v) x^u$ is a positive sum of Laurent polynomials of the form $(x/8+1/2+x^{-1}/8)^j$. So it suffices to prove the same thing for the coefficients of this Laurent polynomial. In other words, writing $(x^2+8x+1)^k = \sum e_i x^i$, we want to prove that $e_i$ is unimodal with largest value in the center. Now, $e_i$ is the $i$-th elementary symmetric function in $j$ copies of $4+\sqrt{15}$ and $j$ copies of $4-\sqrt{15}$. By Newton's inequalities, $e_i^2 \geq \frac{i (2j-i)}{(i+1)(2j-i+1)} e_{i-1} e_{i+1} > e_{i-1} e_{i+1}$ so $e_i$ is unimodal; by symmetry, the largest value is in the center. (The condition $p>1/3$ in the above remark is when the quadratic has real roots.) $\square$

Corollary: $$\sum_u \left| p_k(u-1,v) - p_k(u,v) \right| = 2 p_k(0,v).$$

Proof: The above lemma tells us the signs of all the absolute values; the sum is \begin{multline*} \cdots + (p_k(-1,v) - p_{k}(-2,v)) + (p_k(0,v) - p_{k}(-1,v)) + \\ (p_k(0,v) - p_k(1,v)) + (p_k(1,v) - p_k(2,v)) + \cdots = 2 p_k(0,v). \qquad \square\end{multline*}

So, in order to prove $(\ast)$, we must show that $\lim_{k \to \infty} \sum_v p_k(0,v)=0$. In other words, we must show that the coefficient of $x^0$ in $\left( \frac{x}{8}+\frac{3}{4} + \frac{x^{-1}}{8} \right)^k$ goes to $0$.

There are probably a zillion ways to do this; here a probabilistic one. We are rolling an $8$-sided die $k$ times, and we want the probability that the numbers of ones and twos are precisely equal. The probability that we roll fewer than $k/5$ ones and twos approaches $0$ by the law of large numbers (which can be proved elementarily by, for example, Chebyshev's inequality). If we roll $2r > k/5$ ones and twos, the probability that we exactly the same number of ones and twos is $$2^{-2r} \binom{2r}{r} < \frac{1}{\sqrt{\pi r}} < \frac{1}{\sqrt{\pi k/10}}$$ which approaches $0$ as $k \to \infty$. See here for elementary proofs of the bound on $\binom{2r}{r}$.

I wrote this in two dimensions, but the same proof works in any number of dimensions

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Assuming that $f$ is bounded, you can prove that it is constant by a discrete analogue of the Borel-Caratheodory inequality, as shown in this equivalent question.

The key fact is such a function attains its maximum on a ball on the boundary, so it suffices to provide lower bounds for

$$\Gamma_N = \max_{|x|+|y|=N}f(x,y).$$

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# Getting 2 different solutions to the integral of $\frac{dx}{2x}$

I get two different answers that seem to conflict.

Is there an error in one method??

### Method 1

\begin{align} \int \frac{\mathrm{d}x}{2x}&=\frac 1 2 \int\frac{\mathrm{d}x}{x}\\ &=\frac 1 2\ln|x|+C \end{align}

### Method 2

\begin{align} \int \frac{\mathrm{d}x}{2x}&=\frac 1 2 \int \frac{2\mathrm{d}x}{2x}\\ &=\frac 1 2\int \frac{\mathrm{d}u}{u},\;\;\text{ where }u=2x,\ \mathrm{d}u=2\mathrm{d}x\\ &=\frac 1 2 \big(\ln |u| + C\big)\\ &=\frac 1 2 \ln|2x|+C \end{align}

• As the answer below details: the $C$'s are 'different', for want of a better word. Your "paradox" here is one of the reasons we are careful to always put a $C$ at the end, and also have to stay constantly aware of what it means. Oct 21, 2016 at 13:58
• Another good example is integrating $\sin(x)\cos(x)$. Try integration by parts, to get $\frac{1}{2}\sin^2(x)+C$, then try using the trig identity $\sin(x)\cos(x)=\frac{1}{2}\sin(2x)$ to get $-\frac{1}{4}\cos(2x)+D$. A hint on reconciling the answers is $\cos(2x)=1-2\sin^2(x)$. Oct 21, 2016 at 14:06
• When we get two ostensibly different solutions to an anti-differentiation problem, it is often helpful to check whether the solutions (ignoring the constants) are vertical translations of each other. If so, the two solutions must, of course, differ by a constant. In this case, $\frac 1 2\ln|x|$ translated vertically $\ln\sqrt{2}$ units yields $\frac 1 2\ln|2x|$. Oct 21, 2016 at 16:40

Hint: $\frac{1}{2}\ln{|2x|} + C = \frac{1}{2}\ln{|x|} + \frac{1}{2}\ln{2} + C = \frac{1}{2}\ln{|x|} + \left(\frac{1}{2}\ln{2} + C\right)$.


# Factoring in the derivative of a rational function

Given that $$f(x) = \frac{x}{1+x^2}$$

I have to find $$\frac{f(x) - f(a)}{x-a}$$

So some progressing shows that:

$$\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} = \frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} = \frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)}$$

Now, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?

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Can't you just use the quotient rule to find this derivative? –  user2357112 Jun 3 '14 at 21:28

Since $x-a$ is in the denominator, it makes sense to consider the possibility that $x-a$ is a factor of the numerator. (If you know the factor theorem, you can see this is the case, since the numerator equals zero when $x = a.)$ I see $x-a$ in the numerator, along with two other terms. Very little cleverness is needed at this point to write

$$x-a + xa^2 - ax^2 \; = \; x - a + ax(a-x)$$ $$= \; (x-a) - ax(x-a) \; = \; (x-a)(1-ax)$$

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Ahhhh. So much time looking at it and I didn't realise this. I don't know about the factor theorem, something to look at. Thank you for your answer. –  sidyll Jun 3 '14 at 18:08
@sidyll: For more about using the factor theorem, see my answer at Finding limit of a quotient. For a less obvious use of the factor theorem, see my comments at How to simpify this?. –  Dave L. Renfro Jun 3 '14 at 18:12
Thank you for your references –  sidyll Jun 3 '14 at 18:17
@sidyll: I just remembered that I posted a couple of old handouts of mine on the factor theorem a couple of years ago. See the files attached to this 12 January 2012 sci.math post archived at Math Forum. –  Dave L. Renfro Jun 3 '14 at 18:25
@sidyll As for derivatives, generally the arithmetic works nicely using the quotient rule - see my answer. I'll bet Dave can give links to interesting expositions in AMM, Math. Mag, etc. –  Bill Dubuque Jun 3 '14 at 19:02

\eqalign{x+xa^2-a-ax^2&= x-a+xa^2-ax^2 \\ & = x-a+x(a^2-ax) \\ &= x-a+x(a(a-x)) \\ &= x-a+x(-a(x-a)) \\ &=\color{blue}{x-a}-ax\color{blue}{(x-a)} \\ &=(x-a)(1-ax).\;\checkmark } Therefore you can conclude that: \eqalign{\require{cancel}\dfrac{f(x)-f(a)}{x-a}&=\dfrac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)} \\ &=\dfrac{\color{red}{\cancel{\color{black}{(x-a)}}}(1-ax)}{(1+x^2)(1+a^2)\color{red}{\cancel{\color{black}{(x-a)}}}} \\ &= \dfrac{1-ax}{(1+x^2)(1+a^2)}. }\tag{x\neq a}

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Thank you very much for your examples in continuing it. –  sidyll Jun 3 '14 at 18:09
+1 for nice format. I learn something new about $LaTeX$ from you. –  Tunk-Fey Jun 3 '14 at 18:10
@Tunk-Fey I've also learned a lot from your answers and your TeX style! $\overset{\cdot\cdot}\smile$ BTW you can write a more beautiful LaTeX as: $\LaTeX$ using \LaTeX, and there's also a variant for TeX namely $\TeX$ which is produced by \TeX. –  Hakim Jun 3 '14 at 18:13
@sidyll You're welcome! Glad I could help! ;-) –  Hakim Jun 3 '14 at 18:14
@حكيمالفيلسوفالضائع Thank you for the code. I didn't know that code before. $\ddot\smile$ –  Tunk-Fey Jun 3 '14 at 18:17

By the quotient rule for the difference $\ f'(x)\, :=\, \dfrac{f(x)-f(a)}{x-a}$

$$\quad \begin{eqnarray} (g/h)'(x) &=\,\ & \dfrac{\color{#c00}{g'(x)} h(a) - g(a)\color{#0a0}{h'(x)}}{h(a)h(x)}\\ \begin{array}{l}\ g(x)=x\qquad\Rightarrow\,\color{#c00}{g'(x) = 1}\\ h(x) = 1+x^2\,\Rightarrow\,\color{#0a0}{h'(x) = x+a}\\\end{array}\ \Bigg\}\!\!\!\!\!&=& \dfrac{\color{#c00}1\cdot (1+a^2)\overset{\phantom{I^I}}-a(\color{#0a0}{x+a})}{(1+a^2)(1+x^2)}\ =\ \dfrac{1-ax}{(1+a^2)(1+x^2)}\end{eqnarray}$$

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Hint: you can guess that something interesting is going to happen near $x = a$, which suggests looking to factor $x-a$.

Indeed, $x + xa^2 - a - ax^2 = (x-a) + xa(x-a)$.

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Factor: $$x+xa^2−a−ax^2=x-a-ax^2+xa^2=(x-a)-ax(x-a)=(1-ax)(x-a)$$ Thus, $$\frac{f(x)-f(a)}{x-a}=\frac{\frac{(1-ax)(x-a)}{(1+x^2)(1+a^2)}}{(x-a)}=\boxed{\frac{1-ax}{(1+x^2)(1+a^2)}}$$

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# Questions concerning smallest fraction between two given fractions.

I recently encountered this on a practice test

Find the smallest positive integer $n$ such that there exists an integer $m$ satisfying $0.33 < \frac{m}{n} < \frac{1}{3}$

\begin{align}0.33 < \frac{m}{n} < \frac{1}{3}&\implies(\frac{33}{100} < \frac{m}{n})\land(\frac{m}{n}<\frac{1}{3}) \\ &\implies(33n<100m)\land(3m<n) \\\end{align} and thus $n=3m+1$.

So \begin{align}33n<100m&\implies33(3m+1)<100m \\ &\implies 99m+33<100m\\ &\implies m>33\end{align} Taking $m\geq34$, we find that $n=34\times3+1=103$

After the test, I had the following thoughts, which I do not know how to answer.

Questions

Will the solution for $n$ always yield the smallest $m$ possible?

Does this method (find minimum value of $n$ and substitute) work for all such problems? (where $m,n\in\mathbb{Z}$)

Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}<\frac{m}{n}<\frac{c}{d}$).

note: sorry for asking multiple questions in one post (which I know some people frown on) but I feel posting multiple questions with the same 'introduction' (problem+proof) would clutter and be somewhat cumbersome.

• In my opinion the smallest is $n=103$ because the smallest $m$ is $34$. Indeed if you solve for positive $n$ $$\frac{34}{n}<\frac{1}{3}$$ you get $n>102$ – Raffaele Oct 29 '17 at 11:01
• @mathlove typo by me, thanks. I've also noticed that in our example $\frac{33}{100}<\frac{34}{103}<\frac{1}{3}$, the numerator of the middle fraction is the sum of the other numerators, and the denominator for the middle fraction is the sum of the other denominators. I wonder if this is a coincidence or not, though I won't put that into the question (three questions in one post is already plenty) – user472341 Oct 29 '17 at 11:03
• The continued fraction for $\frac13$ is $(0;3)=(0;3,\infty)$ and the continued fraction for $0.33$ is $(0;3,33)$. The simplest between them is $(0;3,34)=\frac{34}{103}$. – robjohn Oct 29 '17 at 11:04

Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}<\frac{m}{n}<\frac{c}{d}$).

In the following, $a,b,c,d,m,n$ are positive integers.

$\frac ab\lt \frac mn\lt \frac cd$ is equivalent to $$na\lt mb\qquad\text{and}\qquad \frac{md}{c}\lt n$$

From the second inequality, we can set $n=\lfloor\frac{md}{c}\rfloor+1$ where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.

So, from the first inequality, we get $$\left(\left\lfloor\frac{md}{c}\right\rfloor+1\right)a\lt mb\tag1$$

As a result, we can say that the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\frac ab\lt\frac mn\lt\frac cd$ is given by$$\left\lfloor\frac{Md}{c}\right\rfloor+1$$ where $M$ is the smallest integer $m$ satisfying $(1)$.

If $c=1$, then the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\frac ab\lt\frac mn\lt\frac 1d$ is given by$$\left(\left\lfloor\frac{a}{b-da}\right\rfloor+1\right)d+1$$

Yes to both of your questions. This is essentially the mediant for two terms in a Farey sequence. There is some deep theory here. Read up on Farey sequences for more information.

• I had three questions - i'm assuming you're talking about the first two? either way, thanks for the links. – user472341 Oct 29 '17 at 11:09
• The mediant is not always the smallest. Take for instance $\frac13$ and $\frac34$: the mediant is $\frac47$, which is between the two, but $\frac12$ is the fraction with the smallest denominator. – robjohn Oct 29 '17 at 11:12
• I think the mediant is only relevant when the fractions are consecutive terms in some Farey sequence. In the example @robjohn gives, $1/3$ and $3/4$ are not consecutive terms in any Farey sequence. – Gerry Myerson Oct 29 '17 at 11:17
• You can iterate the algorithm using the mediant. For 1/3, 3/4 we get 4/7 and then with 1/3, 4/7 we get 1/2. – i. m. soloveichik Oct 29 '17 at 11:27


# Function notation terminology

Given the function $f:X\longrightarrow Y$, $X$ is called the domain while $Y$ is called the codomain. But what do you call $f(x)=x^2$ in this context, where $x\in X$? That is to say - what is the name for the $f(x)$ notation?

And while I'm here, what is the proper way to write a function like this? Would it be $f:\mathbb{R}\to\mathbb{R},\;f(x)=x^2$?

Edit:

I figured I'd add this to add a bit of context into why I'm asking. I'm writing a set of notes in LaTeX, and I'd like to use the correct terminology for the definition of a function.

A function from set $A$ to set $B$, denoted by $$f:A\to B;x\mapsto f(x)$$ is a mapping of elements from set $A$, (the $\textit{domain}$) to elements in set $B$ (the $\textit{codomain}$) using the $\color{blue}{\sf function}$ $f(x)$. The domain of a function is the set of all valid elements for a function to map from. The codomain of a function is the set of all possible values that an element from the domain can be mapped to. The $\textit{range}$ (sometimes called the $\textit{image}$) of a function is a subset of the codomain, and is the set of all elements that actually get mapped to by the function $f$.

Here I'm pretty sure the highlighted word "function" is not right.

• I suggest \to or \rightarrow instead of \longrightarrow. The last one is there for if you need the arrow to be longer because you're writing something over it. As in $\overset{\text{text}}{\longrightarrow}$ instead of $\overset{\text{text}}{\to}$.
– Jim
Feb 22, 2013 at 23:16
• why do you think that function is not right = Feb 22, 2013 at 23:36
• Some would call "$f(x)=x^2$" the rule of the function $f$. Feb 22, 2013 at 23:44
• @DominicMichaelis Well, function is used in two places to mean two slightly different things; I don't like using function to describe part of a function. Feb 22, 2013 at 23:47

I can remember to read this text, and being puzzled with the exact same question. From what I've learned from my teacher, you're right, writing down something as "the function $f(x)$..." is sloppy notation. However, many books/people will use it this way.

If you're are very precise, $f(x)$ is not a function or an map. I don't know of a standard way to refer to $f(x)$, but here is some usage I found on the internet:

• The output of a function $f$ corresponding to an input $x$ is denoted by $f(x)$.
• Some would call "$f(x)=x^2$" the rule of the function $f$.
• For each argument $x$, the corresponding unique $y$ in the codomain is called the function value at $x$ or the image of $x$ under $f$. It is written as $f(x)$.
• If there is some relation specifying $f(x)$ in terms of $x$, then $f(x)$ is known as a dependent variable (and $x$ is an independent variable).

A correct way to notate your function $f$ is: $$f:\Bbb{R}\to\Bbb{R}:x\mapsto f(x)=x^2$$

Note that $f(x)\in\Bbb{R}$ and $f\not\in\Bbb{R}$. But the function $f$ is an element of the set of continuous functions, and $f(x)$ isn't.

In some areas of math it is very important to notate a function/map specifying it's domain, codomain and function rule. However in for example calculus/physics, you'll see that many times only the function rule $f(x)$ is specified, as the reader is supposed to understand domain/codmain from the context.

You can also check those questions:

• Nice, Kasper +1 Feb 23, 2013 at 1:50

normally you say $f:X\rightarrow Y$; $x\mapsto f(x)$, as a functions takes an element from $X$ and give you one from $Y$. $$y=f(x)$$ Is an equation, and not a definition of a function in a strict sense.

The proper way would be $$f:\mathbb{R}\rightarrow \mathbb{R}; \ x\mapsto x^2$$

The Image of a function is definied as $$\operatorname{im}f:=\{f(x)|x\in X\}$$ It is often written as $$f(X):=\{f(x)|x\in X\}$$ Notice that $X$ is a set, not an element. so $f(\{x\})=\{f(x)\}\subseteq Y$ while $f(x)\in Y$

• I was wondering - is $f(x)$ called the image? I've seen that word used but only know it to be synonymous with Range. Feb 22, 2013 at 23:15
• not $f(x)$ but $f(X)$ note the different as $X$ is a set but $x$ is an element Feb 22, 2013 at 23:19
• @agent154 You can say "$f(x)$ is the image of the element $x$ under the function $f$". Perhaps this is what you were thinking of (here of course "image" means something different than the image of the function). Feb 22, 2013 at 23:32
• @davidMitra i would be careful with that, as the image is a set but $f(x)$ is an element of $Y$ Feb 22, 2013 at 23:34
• The notation $f(X)$ is also written (in set theory) as $f[X]$ and $f''X$ sometimes, often because the elements of the set are sets themselves, and sometimes $f(x)\neq f''x$. Feb 22, 2013 at 23:35

From what I understand, $f$ is the function, so saying "a function $f(x)$" would be wrong. Instead, you could say "a function $f$", or if you didn't want to assign it a name, "a function $x\mapsto x^2$", or if you want to specify the domain and codomain, "a function $f:X\to Y$".

If you just want to define the function's input-output relationship, $f(x)=x^2$ suffices, maybe with a $\forall x\in X$ at the beginning if you want to do it properly. That doesn't necessarily define the domain though, that set could just be a subset of the domain. $f(x)$ is really no different from $f(1)$, except that $x$ is a (qualified) variable. It's basically a rule defining the input and output for an improper subset of its domain.

The addition of "$:X\to Y$" is useful if you want to specify both the domain and codomain when defining the function. As far as I can tell, you can add it after a function like $f$ and the whole expression still refers to the function. You can't really use it with the $f(x)$ style, so I'd do something like

$f:\mathbb R\to\mathbb R=x\mapsto x^2$


# What is the variance of a binomial distribution with -1 and 1?

I am struggling to see how to solve the following problem:

I have n i.i.d bernoulli trials. The result can be -1 or 1. I can figure out the expected value of this = n(p-(1-p) but how do I know what the variance is if p is known?

I know that: $${\rm Var}(X) = E[X^2] - E[X]^2$$ I don't know where to go from here.

• In general, for a Bernoulli Random Variable E(X) is $P(X=x_1)*x_1+P(X=x_2)*x_2$. In the "default" case $x_1 = 1, x_2 = 0$, which reduces this to E(X) = p. In your case, $x_1 = 1$ and $x_2 = -1$. You can then plug in what you get for E(X) into the formula $Var(X) = E(X^2) - E(X)^2$. The variance of the default case is $p*(1-p)$ – Pugl Jul 27 '17 at 13:57
• If B is a 0-1 Bernoulli variable with probability of a 1 at $p$, then you're describing a variable, ($Z$ say) where $Z=2B-1$. Just apply basic properties of variance to that linear transformation. – Glen_b Jul 28 '17 at 2:06

When you work with variances, know these facts (in addition to the definition of variance):

1. The variance of a sum of independent (or just uncorrelated) variables is the sum of their variances.

2. The expectation of a sum of variables (independent or not) is the sum of their expectations.

3. The expectation (of any discrete variable) is the sum of each possible value multiplied by its probability.

As an example, let's apply them to your case. You may model the sum of your Bernoulli trials with a variable $X$ expressed as the sum of $n$ independent variables $X_i$. Each $X_i$ takes on the value $1$ with probability $p$ and the value $-1$ with the probability $1-p$.

Fact $(3)$ asserts $$\mathbb{E}(X_i) = p(1) + (1-p)(-1) = 2p-1$$ and $$\mathbb{E}(X_i^2) = p(1)^2 + (1-p)(-1)^2 = p + (1-p) = 1.$$

Fact $(2)$ asserts \eqalign{\mathbb{E}(X) &= \mathbb{E}(X_1+\cdots+X_n) = \mathbb{E}(X_1) + \cdots + \mathbb{E}(X_n) = n\mathbb{E}(X_1) \\&=n(2p-1).}

The definition of variance now tells you$$\operatorname{Var}(X_i) = \mathbb{E}(X_i^2) - \mathbb{E}(X_i)^2 = 1 - (2p-1)^2 = 4p(1-p).$$

Consequently, fact $(1)$ yields \eqalign{\operatorname{Var}(X) &= \operatorname{Var}(X_1+\cdots+X_n) = \operatorname{Var}(X_1) + \cdots + \operatorname{Var}(X_n)=n \operatorname{Var}(X_1) \\&= n(4p(1-p)).}

The appearance of the factor of $4$ might seem somewhat mysterious. There's another extremely useful fact you might consider using to shortcut these considerations and identify the origin of that factor:

(4) The variance of a shifted, rescaled variable is the square of the scale factor times the variance. In mathematical symbols, $$\operatorname{Var}(\sigma X + \mu) = \sigma^2 \operatorname{Var}(X)$$ no matter what values the numbers $\sigma$ and $\mu$ might have.

This applies by noting that your $X_i$ can all be expressed as $2Y_i-1$ where $Y_i$ is a true Bernoulli (that is, $0-1$) variable. It's simple to show this: since $2\times 1-1=1$ and $2\times 0 - 1=-1$, the values of $1$ and $0$ taken on by $Y_i$ become $1$ and $-1$, respectively, for $X_i$. The probabilities are unchanged.

You might already know that $Y=Y_1+\cdots + Y_n$, the sum of $n$ independent Bernoulli variables with common probability $p$, is called a Binomial variable. It has a Binomial distribution. You can remember or look up its variance, which is $np(1-p)$. Since $$X = X_1+\cdots+X_n = (2Y_1-1) + \cdots + (2Y_n-1) = 2(Y_1 + \cdots + Y_n) - n=2Y-n,$$ you can take $\sigma=2$ in applying fact $(4)$, which immediately tells you $$\operatorname{Var}(X) = 2^2 \operatorname{Var}(Y) = 4np(1-p).$$

• I personally find the observation that $4=\frac{1}{0.5(1-0.5)}$ soothing as it shows that the variance is the ratio of $pq(X_i)$ to $pq(0.5)$ (forgive the abuse of notation). – Jared Goguen Jul 27 '17 at 16:09
• In what sense is the second part of the answer $X_i=2Y_i-1$ ... different from my answer ? – user83346 Jul 28 '17 at 5:41
• @fcop It's exactly the same approach--but the explanation is more extensive and thorough. If you take a look over the entirety of my answer, I hope you will see the pattern: it focuses on stating and illustrating useful principles. These are often overlooked or unknown to people who come to our site with questions. My purpose is to collect, unify, and illustrate some (obvious, well-known) techniques that can be found by future visitors and referenced in future answers. In that sense, my answer differs from yours, which focuses on solving the immediate problem. – whuber Jul 28 '17 at 13:37
• Well you could have referenced it no? – user83346 Jul 28 '17 at 14:08
• @fcop One ordinarily does not reference obvious or well-known things. (If that were the practice, then most posts would be 90% references.) The point of this answer is to explain the reasoning--not just the results--at the same level of understanding as the question. Its contribution is not to make the observation that $X_i=2Y_i-1$ (you're welcome to take credit for that), but rather to show as clearly and explicitly as possible, with all reasons supplied, how that produces the same answer derived earlier in the question. – whuber Aug 13 '17 at 13:10

if you take $x=2b-1$ where $b$ is bernoulli, then then $x$ is either 1 or -1, and mean and variance follow from Bernoulli $\mu_b=p$ and $\sigma^2_b=p(1-p)$, the extension to 'Binomial' is then just a sum (assuming independence),

The mean of $x$ is then $\mu_x=2\mu_b-1=2p-1$, variance $\sigma_x^2=4\sigma_b^2=4p(1-p)$,

The sum of $n$ independent such $x$'s has mean $n\cdot(2p-1)$ (your result) and the variance is $4\cdot n \cdot p(1-p)$

EDIT after your question in the comment

It is as @Pegah says: $b$ is the ''usual'' Bernoulli with a success probability $p$ and outcomes 0 and 1. It's expected value is $1 \times p + 0 \times (1-p)=p$ and the variance $p(1-p)$ (also from the definition.

So $b$ is the known Bernoulli with outcome 0 and 1 and $x$ is just a linear function of it. The linear function is such that $x$ has outcomes -1 and 1, and the mean and variance can be obtained from the mean and variance of $b$.

• In fcop's derivation, $\mu_b$ is the expected value of the random variable b, which is, as I understand a "default" binomial random variable (i.e. it can take on values 1 or 0, see also my comment above). Here the expected value is just p. – Pugl Jul 27 '17 at 15:31


# Simplifying to a desired expression structure

My book has this expression:

\begin{align} ((n(n+1)(2n+7))/6)+(n+1)(n+3) \end{align}

And then the book simplified it, and ended up with the desired expression:

\begin{align} ((n+1)(n+2)(2n+9))/6 \end{align}

I tried to do such simplification. But I ended up with this: \begin{align} (2n^3+15n^2+31n+18)/6 \end{align}

Even though the "value" in my expression is the same as the desired result's, I need the structure of my expression to be the same as well. But I don't understand how to do such.

Here are the full steps my book shows. Of course, I understand that the steps make sense, but I don't understand why did my book take that approach? Is that the only approach possible to reach the desired expression? How was I supposed to know that? I mean, I know how to simplify, but clearly my book is using a different "style" or "path"

\begin{align} ((n(n+1)(2n+7))/6)+(n+1)(n+3)\\ (n(n+1)(2n+7)+6(n+1)(n+3))/6\\ ((n+1)[n(2n+7)+6(n+3)])/6\\ ((n+1)(2n^2 +7n + 6n + 18))/6\\ ((n+1)(n+2)(2n+9))/6 \end{align}

So yes, how should I simplify to get a desired expression structure?

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Ah! I... didn't see that. Somehow. I'm eager to see how to extract such though. –  Zol Tun Kul Jul 3 '12 at 1:49
I wrote out a little something about extracting. –  André Nicolas Jul 3 '12 at 1:59

Below I explain in detail the solution given in your book.

$\begin{eqnarray} &&\rm\ \ \,n(n\!+\!1)(2n\!+\!7)/\color{#C00}6+(n\!+\!1)(n\!+\!3)\\ \rm put\ all\ over\ a\ common\ denominator = \color{#C00}6:\quad &= &\rm\ (n(\color{#0A0}{n\!+\!1})(2n\!+\!7)\,+\,\color{#C00}6\,(\color{#0A0}{n\!+\!1})(n\!+\!3))/\color{#C00}6\\ \rm pull\ out\ the\ common\ factor\ \color{#0A0}{n\!+\!1}:\quad &=&\rm\ (\color{#0A0}{n\!+\!1})\,(n(2n\!+\!7)+6(n\!+\!3))/6\\ \rm apply\ the\ distributive\ law:\quad &=&\rm\ (n\!+\!1)\,(2n^2 + 13n + 18)/6\\ \rm factor\ the\ quadratic,\ see\ below:\quad &=&\rm\ (n\!+\!1)\,(n\!+\!2)\,(2n\!+\!9)/6 \end{eqnarray}$

In order to factor the quadratic $\rm\:f = 2n^2+13n+18\:$ one can apply the AC-method as follows:

$$\begin{eqnarray}\rm 2f\, &=&\rm\ \ 4n^2\ +\ 13\cdot 2n\, +\, 2\cdot 18 \\ &=&\rm\ (2n)^2 + 13\,(2n) + 2\cdot 18 \\ &=&\rm\ \ N^2\ +\ 13\, N\ +\ 36\quad for\quad N = 2n \\ &=&\rm\ (N\ +\ 4)\,(N\ +\ 9) \\ &=&\rm\ (2n\, +\, 4)\,(2n\, +\, 9) \\ \rm f\, &=&\rm\ (\ n\ +\ 2)\,(2n\ +\ 9) \end{eqnarray}$$

-
I'm enlightened. Thank you. –  Zol Tun Kul Jul 3 '12 at 3:58

Your expression is $$\frac{2n^3+15n^2+31n+18}{6}.$$ Let's forget about the $6$. Also, I would like to change the $n$ to $x$, for no good reason except familiarity. So we want to factor $$2x^3+15x^2+31x+18.$$ To do this, we look for rational roots of the polynomial $2x^3+15x^2+31x+18$. By the Rational root Theorem, such roots must have shape $a/b$, where $a$ divides $18$ and $b$ divides $2$. (There may be no rational roots, though in this case there are three.)

It is easy to spot the root $x=-1$. So $x-(-1)$, that is, $x+1$, divides our polynomial. Do the division, using the ordinary division process for polynomials, which is much like "long division." We get $$2x^3+15x^2+31x+18=(x+1)(2x^2+13x+18).$$ Now factor the quadratic as one did in school. Or else note that $x=-2$ is a root of $2x^2+13x+18$.

Remark: The book grabbed the obvious common factor of $n+1$. The rest turned out to be a quadratic that factors nicely. You multiplied out, meaning you buried the $n+1$ term. It can be extracted from your expression, and the above calculation shows how, but why bury and then extract? A factored or partly factored expression is often more useful. Anyway, "taking out" common factors usually simplifies calculations.

-


# Show that set of all $2 \times 2$ matrices forms a vector space of dimension $4$

I have this question:

Show that the set of all $2 \times 2$ matrices with real coefficients forms a linear space over $\Bbb R$ of dimension $4$.

I know that the set of the matrices will basically form a linear combination which will define the vector space and they satisfy the axioms defined for the vector space.

I do not know how to show that this is possible. Do the vectors have to be linearly independent?

Any sort of help is appreciated.

Thanks.

Begin by listing the axiom satisfied by a general vector space over $\mathbb{R}$. Now consider your set of matrices. What does you addition look like? What is your zero element? What does your scalar multiplication look like? Does your scalar multiplication and your addition behave as they should? Finally, show that all 2 by 2 matrices can be written as a linear combination of 4 special matrices. A good choice is to pick matrices who each have precisely 3 zero entries and 1 non zero entry. Can you show that these 4 matrices are linearly independent?

• Can those four matrices be [a 0 0 0], [0 b 0 0], [0 0 c 0], [0 0 0 d]? I can form a linear combination of them and show they are linearly independent. Will that work? – Suvrat Jan 31 '16 at 0:05
• That is exactly what I hoped that you would do! – Carl Christian Jan 31 '16 at 0:05
• oh very well then. Thanks a lot! – Suvrat Jan 31 '16 at 0:06
• Can you tell me how these 4 matrices show the formation of linear space for generalized situation, i.e. all 2x2 matrices? – Suvrat Jan 31 '16 at 20:14
• For the sake of simplicity pick a=b=c=d=1. Then write the matrix [e f g h] as e*[1 0 0 0] + f*[0 1 0 0] + g [0 0 1 0] + h*[0 0 0 1]. – Carl Christian Jan 31 '16 at 20:23

One can easily see that $\mathbb{R}^{2\times 2}$ is in fact exactly the same as the vectorspace $\mathbb{R}^4$, only the notation of the elements differs. And of course whether or not something is a vector space does not depend on the way some unimportant species of primates decides to write its elements.

Hint Can you find a basis of the set of $2 \times 2$ matrices consisting of four elements? (There is a natural choice of basis here that includes the matrix $\pmatrix{1&0\\0&0}$.)

Alternatively, can you find a vectorspace isomorphism from the space of $2 \times 2$ matrices to some vector space you know to be $4$-dimensional, e.g., $\Bbb R^4$?

• I don't know how to find a vectorspace isomorphism. But i can work with basis, which is similar to Carl's solution. Thanks! – Suvrat Jan 31 '16 at 0:11
• You're welcome. – Travis Jan 31 '16 at 0:42


How can I prove this statement about subsets?

Let $$A$$, $$B$$, $$C$$ be sets. Prove that if $$A \subseteq C$$ and $$B \subseteq C$$ then $$A \cup B \subseteq C$$.

This is an exercise in mathematical logic.

My attempt to progress forward: This statement can be written as$$(A \subseteq C) \land (B \subseteq C) → A \cup B \subseteq C\\ (x \in A → x \in C) \land (x \in B → x \in C) → A \cup B \subseteq C$$ But I am not even sure that is how I am supposed to do it so I am a bit stuck. Can anyone explain to me how to get through this proof? Thanks in advance.

• ... intuitively it's obvious, right? ... Sep 26 '18 at 15:32
• @user202729 Intuitively many things are obvious, but proving them is another matter... ;) ex. A sphere bounds a given volume with minimal area... Sep 27 '18 at 1:14

4 Answers

Take any $$x\in A\cup B$$. That means either $$x\in A$$ or $$x\in B$$. Anyway you get $$x\in C$$.

This answer is expanding a bit on the other answers to give a widely applicable outline on how to prove mathematical statements in general.
The final proof will be equal to Mark's answer, but hopefully this answer sheds a bit of light on what is actually going on in the thought process, at least I like to think in the terms defined below while I am proving.

In general you need a "proof calculus" which states how proofs can be constructed and which ones are valid. Unless you are doing logic and proof theory itself, in most settings in maths you are well off using the natural deduction proof calculus for first-order logic. In short — cut in light of this question:

• To prove $$\varphi \land \psi$$, prove $$\varphi$$ and $$\psi$$ separately.
• To prove $$\varphi \lor \psi$$, give a proof for $$\varphi$$ or alternatively for $$\psi$$. One proof for either $$\varphi$$ or $$\psi$$ is already sufficient.
• To prove $$\varphi → \psi$$, assume $$\varphi$$ and prove $$\psi$$ under this assumption.

The implication captures the notion of "if we have $$\varphi$$ and $$\varphi → \psi$$ established somewhere, then we can conclude $$\psi$$." Most, if not all, theorems can be seen as implications: if you fulfill the requirements, then you can use its consequences.
Hence it makes sense to be allowed to assume $$\varphi$$ in the proof of $$\varphi → \psi$$.
• To prove $$\forall x. \varphi(x)$$ ($$\varphi$$ may depend on $$x$$), introduce a fresh variable not used before (e.g. $$c$$) and prove $$\varphi(c)$$ (i.e. every occurrence of $$x$$ is filled with $$c$$).

Introduction of a fresh variable is required to guarantee that your proof of $$\varphi(c)$$ does not depend on any previous assumptions on $$x$$. Hence, having proved $$\varphi(c)$$ really amounts to having proved the proposition $$\varphi(x)$$ for every possible $$x$$.

Now consider your statement:

$$\left((x \in A → x \in C) \land (x \in B → x \in C)\right) → A \cup B \subseteq C$$

You already did a good job translating it to more formal language! But we can even go a step further. Actually what the above line means is the following: $$\left((\forall x. x \in A → x \in C) \land (\forall x. x \in B → x \in C)\right) → (\forall x. x \in A \cup B → x \in C)$$

On the top level, we have a $$\varphi → \psi$$-kinded expression. Applying the corresponding rule, we begin with

(1) Assume $$(\forall x. x \in A → x \in C) \land (\forall x. x \in B → x \in C)$$.

Our proof goal, as stateted in the rule, is now the right side $$(\forall x. x \in A \cup B → x \in C).$$ This is of kind $$\forall x. \varphi(x)$$ with $$\varphi(x) := x \in A \cup B → x \in C$$. So,

(2) Let $$y$$ be a fresh variable.

I called it $$y$$ to avoid confusion with the set $$C$$. When one defines freshness more formally, one sees that $$x$$ would have worked as well as a fresh variable.
Now we have to show $$\varphi(y)$$, which is $$y \in A \cup B → y \in C$$. Again, apply the $$→$$ rule from above:

(3) Assume $$y \in A \cup B$$.

Our final proof goal is now: $$y \in C$$. This can be proved by case-by-case analysis:

(4) We know that $$y$$ is in $$A$$ or $$B$$. In the first case, $$y \in A$$ and per line (1) we especially know $$y \in A → y \in C$$. Since we have $$y \in A$$ just established, we can conclude $$y \in C$$. Case finished. The next and last case is $$y \in B$$ which works analogously.

If $$A \subset C$$, and $$B \subset C$$;

$$A \cap C =A$$; and $$B \cap C = B$$;

$$A\cup B = (A \cap C)\cup (B \cap C) =$$

$$C \cap (A \cup B) \subset C$$

I haven't noticed one in the in the answers, so here's a proof by contradiction.

The statement we wish to prove is: $$(\forall x)(x \in A \cup B \implies x \in C) \tag{1}$$

It's negation is:

$$(\exists x)(x \in {A} \cup {B} \wedge \neg(x \in C) ) \tag{2}$$ $$x \in A \cup B \implies (x \in A \vee x \in B)$$ $$x \in A \wedge A \subseteq C \implies x \in C \tag{3}$$ $$or$$ $$x \in B \wedge B \subseteq C \implies x \in C \tag{4}$$ Since both $$(3)$$ and $$(4)$$ are in contradiction with $$(2)$$, it must be false. Therefore our starting premise $$(1)$$ is correct.


# Fourier series, pointwise convergence, series computation

#### Markov

##### Member
Let $f(x)=-x$ for $-l\le x\le l$ and $f(l)=l.$

a) Study the pointwise convergence of the Fourier series for $f.$
b) Compute the series $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}.$
c) Does the Fourier series of $f$ converge uniformly on $\mathbb R$ ?

-------------

First I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\displaystyle\sum_{n=1}^\infty b_n\sin\frac{n\pi x}l$ where $b_n=\dfrac 2l\displaystyle\int_0^{l} f(x)\sin\frac{n\pi x}l\,dx$ so I'm getting $\displaystyle-\frac{x}{{{l}^{2}}}=\frac{1}{\pi }\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{n}\sin \frac{n\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?

Thanks for the help!

#### Sudharaka

##### Well-known member
MHB Math Helper
Let $f(x)=-x$ for $-l\le x\le l$ and $f(l)=l.$

a) Study the pointwise convergence of the Fourier series for $f.$
b) Compute the series $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}.$
c) Does the Fourier series of $f$ converge uniformly on $\mathbb R$ ?

-------------

First I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\displaystyle\sum_{n=1}^\infty b_n\sin\frac{n\pi x}l$ where $b_n=\dfrac 2l\displaystyle\int_0^{l} f(x)\sin\frac{n\pi x}l\,dx$ so I'm getting $\displaystyle-\frac{x}{{{l}^{2}}}=\frac{1}{\pi }\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{n}\sin \frac{n\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?

Thanks for the help!
Hi Markov,

Firstly the definition of the function $$f$$ seem to be erroneous, both $$f(x)=-x\mbox{ for }-l\leq x\leq l$$ and $$f(l)=l$$ cannot be true. So I shall neglect the latter part: $$f(l)=l$$.

I think the Fourier series that you have obtained is also incorrect. It should be,

$-x=\frac{2l}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi x}{l}\right)$

Since both $$f(x)=-x$$ and $$f'(x)=-1$$ are continuous on $$[-l,\,l]$$ the Fourier series converges point-wise on the interval $$(-l,\,l)$$ (Refer Theorem 5.5 here).

Substitute $$x=1\mbox{ and }l=2$$ and we obtain,

\begin{eqnarray}

-\frac{\pi}{4}&=&\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi}{2} \right)\\

&=&\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{2n+1}\sin\left(\frac{(2n+1)\pi}{2} \right)\\

&=&\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{2n+1}(-1)^n\\

&=&\sum_{n=0}^{\infty}\frac{(-1)^{3n+1}}{2n+1}\\

\therefore\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}&=&\frac{\pi}{4}

\end{eqnarray}

When $$x=l$$ we have,

$\left|f(l)-\sum_{n=1}^{N}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi l}{l}\right)\right|=l$

Therefore by the definition of uniform convergence (Refer this) it is clear that the Fourier series of $$f$$ is not uniformly convergent on $$\Re$$.

Kind Regards,
Sudharaka.

#### CaptainBlack

##### Well-known member
Hi Markov,

Firstly the definition of the function $$f$$ seem to be erroneous, both $$f(x)=-x\mbox{ for }-l\leq x\leq l$$ and $$f(l)=l$$ cannot be true. So I shall neglect the latter part: $$f(l)=l$$.
This is very likely to be intended to indicate the periodic extension: $$f(x)=f(x+2l)$$ (or there is a mistake with the value at either $$+l$$ or $$-l$$, both end points would not normally be included in the domain for a Fourier Series).

Without the periodic extension the question of uniform convergence is moot, since the Fourier Series is periodic and so does not converge to the function outside of the interval.

CB


# What can I use to solve this?

1. Jan 10, 2005

### recon

I've run into a calculation hurdle that I cannot cross in a problem I'm trying to solve.

May I know what values of z satisfy the following equation:

$$z^4 + 2z^3 - 5z + 1 = 0$$

I've never encountered this in mathematics class before.

2. Jan 10, 2005

### dextercioby

Would u please post the problem??Maybe you wouldn't get a quartic after all...

Daniel.

3. Jan 10, 2005

### Nguyen Thanh Nam

Neither do I, I can solve those that miss ax^3, never before for ax^2

4. Jan 10, 2005

### recon

A ladder 3m long rests against a wall with one end a short distance from the base of the wall. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m wide. What is the maximum distance up the wall that the ladder can reach?

5. Jan 10, 2005

### dextercioby

The maximum height on the wall is attained when the ladder touches the storage box.
The equations i have found read
$$(a+1)^{2}+(b+1)^{2} =9$$(1)
$$ab=1$$ (2)

Then u find the eq.for "a"
$$a^{2}+\frac{1}{a^{2}}+2(a+\frac{1}{a})=7$$ (3)
,which can be put in the form
$$u^{2}+2u-9 =0$$ (3)

,where
$$u=:a+\frac{1}{a}$$ (4)

Solve for "u" and then for "a".Then add 1 to the "a" and find the final result.

Daniel.

6. Jan 10, 2005

### recon

Yes, that's how I solved it. I never thought of substituting $$a+\frac{1}{a}$$ for u. Very smart. Thanks a lot.

7. Jan 10, 2005

### Hurkyl

Staff Emeritus
Use the symmetry of the problem . With dexter's notation, notice that whenever s is a solution, so is 1/s. So, we can take his quartic:

z^4 + 2z^3 - 7z^2 + 2z + 1 = 0

And write down its four factors:

(z - s) (z - 1/s) (z - t) (z - 1/t) = 0

If we group them into their symmetric pairs and multiply:

(z^2 + az + 1) (z^2 + bz + 1)

So one quadratic corresponds to the two solutions s and 1/s and the other to t and 1/t. In particular, notice, now, that our unknowns only have two possible values, instead of four, which means we should be able to determine them from a quadratic equation:

Since the polynomials should be equal, we should expand and equate them:

z^4 + (a+b)z^3 + (2+ab)z^2 + (a+b)z + 1
=
z^4 + 2z^3 - 7z^2 + 2z + 1 = 0

Equating coefficients gives us a system of equations to solve.

Once we have a and b, we can then solve for s and t and get the four possible solutions. (Two should be complex, I imagine)

(Incidentally, I don't see how you got your equation, Recon)

8. Jan 10, 2005

### dextercioby

I multiple checked it.Can u prove your statement???I frankly doubt it...

Daniel.

9. Jan 10, 2005

### Hurkyl

Staff Emeritus
Another (similar) approach:

We can arrange the factors into these pairs:

(z - s)(z - t) and (z - 1/s)(z - 1/t)

So each quadratic contains exactly one solution from each of the symmetric solutions (instead of both).

Again, we break the quartic into quadratic factors:

(z^2 + az + b) (z^2 + cz + d)

But, in this case, we know that we can convert from one to the other by applying the symmetry. In particular,

if (z^2 + az + b) = 0 then (1/z)^2 + c(1/z) + d = 0
Rewriting the second gives:

z^2 + (c/d)z + (1/d) = 0
which must be the same polynomial as
z^2 + az + b = 0

Since they have the same solutions.

So we can equate coefficients here, then multiply them to equate coefficients to the quartic polynomial. Again, we can solve for our unknowns, and then we know how to solve quadratics to get the answers to our problem.

Lesson learned: use symmetry to break the problem into parts. We've seen two different ways to do it:

(1) Each part corresponds to one of the different "orbits" under the symmetry. (e.g. {s, 1/s})

(2) Each part consists of exactly one element from each "orbit". (e.g. {s, t})

Then, we can solve the problem in two steps:

(i) Determine the two subproblems.
(ii) For each subproblem, get the corresponding answers.

10. Jan 10, 2005

### Hurkyl

Staff Emeritus
Dexter gave a solution using a third approach. What precisely did he do, and why did it work? Is it really a different approach?

The variable he defined, u, has the property that it is invariant under the symmetry. Notice that if you replace z with 1/z, u remains unchanged.

Because of this invariance under symmetry, there are fewer solutions for u than there are for z, making them easier to find.

Actually, in some sense, both of my approaches make use of invariants under symmetry. For instance, (z - s)(z - 1/s) is invariant under swapping s and 1/s. My second approach found two quadratics that are each invariant under swapping s and t.

P.S. for a polynomial which is the same as its reverse, making dexter's solution is the standard way of getting a lower degree polynomial.

(The reverse of a polynomial is the one with the coefficients in the opposite order. Equivalently, the multiplicative inverse of the roots of the polynomial are precisely the roots of the reverse. To make sense, of course, this assumes that 0 is not a root)

Last edited: Jan 10, 2005
11. Jan 10, 2005

### Hurkyl

Staff Emeritus
And just for fun, here's a trig approach:

Let theta be the angle the ladder makes with the base. A is the distance from the box to the base, B is the distance from the box tot he top of the ladder.

So, the length of the ladder is 1/sin theta + 1/cos theta = 3.

With trig identities, and a change of variable, I can get a quadratic in sin phi.

12. Jan 10, 2005

### Zaimeen

I was a bit sleepy at that time Dex. I did checked it just now. You are right. Sorry Dex!!

________________________________________________________________________

"The beautiful mind goes faster than the hand"

13. Jan 10, 2005

### recon

I got it the same way as dextercioby. I just made a mistake multiplying the right side of the equation with z^3 instead of z^2. I multiplied everything on the left side with z^2.

EDIT: This is the equation I'm talking about.

$$z^2 + 2z + \frac{2}{z} + \frac{1}{z^2} = 7$$

Last edited: Jan 10, 2005


# Expected value with piecewise probability density function (PDF)

I am continuing the prepare for an exam by reviewing handouts from an old statistics course I took. The handout came with a set of solutions prepared by the instructor, but I suspect that one of the answers is wrong. If the answer the instructor provided isn't wrong, then I must be missing something, so I'd like to ask the CV community to help where I've gone wrong. Here is the question:

Suppose that $X$ is a continuous random variable with pdf given by:

$f(x) = \left\{ \begin{array}{lr} x & 0 < x < 1\\ (2x-1)/12 & 1< x < 3 \end{array} \right.$

if $g(x) = x$ for $x>2$ and $g(x)=0$ for $x \le 2$, find the expected value $Eg(x)$.

The instructor put the solution as $10/9$. However, I've reworked my solution several times and continue to arrive at the answer $Eg(x)=61/72$. Here is how I approached this problem:

By definition, we know:

$$Eg(X) = \int_{-\infty}^{\infty}g(x)f(x)dx\tag{1}$$ Since $f(x)$ and $g(x)$ are $0$ everywhere, except from $2<x<3$, I only need to integrate (1) over these values of $x$. Doing so yields: $$Eg(X) = \int_{2}^{3}g(x)f(x)dx=\int_{2}^{3}{x(2x-1)\over{12}}dx=\int_{2}^{3}{2x^2-x\over{12}}dx$$ $$={1\over{12}}\left({{2x^3}\over{3}}-{x^2\over{2}}\right)\bigg\rvert_{x=2}^{x=3}={1\over{12}}\left[\left({{54}\over{3}}-{9\over{2}}\right)-\left({{16}\over{3}}-{4\over{2}}\right)\right]\bigg\rvert_{x=2}^{x=3}=61/72$$

I think the instructor accidentally integrated over $1<x<2$ instead of over $2<x<3$, but as I said earlier, I'm just hoping someone can verify if I'm correct, or perhaps I'm just not understanding something here. Thanks.

• I think you're right. If you can get correct answers for similar problems then you should have no reason to worry. – mark999 Apr 22 '17 at 6:53
• I agree with your answer. – Glen_b -Reinstate Monica Apr 22 '17 at 9:54
• Much appreciated, to you both! This makes me feel a lot better! – StatsStudent Apr 22 '17 at 15:49
• could you please add a tag-wiki for your new tag piecewise-pdf ? – kjetil b halvorsen Apr 23 '17 at 11:20


This note has been used to help create the Mental Math Tricks wiki

Hey guys, I saw a faster way to find cube roots.

We already know some basic cube numbers

$$0^{3}$$=0

$$1^{3}$$=1

$$2^{3}$$=8

$$3^{3}$$=27

$$4^{3}$$=64

$$5^{3}$$=125

$$6^{3}$$=216

$$7^{3}$$=343

$$8^{3}$$=512

$$9^{3}$$=729

Now, the common thing here is that each ones digit of the cube numbers is the same number that is getting cubed , except for 2 ,8 ,3 ,7 .

now let us take a cube no like 226981 .

to see which is the cube root of that number , first check the last 3 digits that is 981 . Its last digit is 1 so therefore the last digit of the cube root of 226981 is 1 .

Now for the remaining digits that is 226

Now 226 is the nearer & bigger number compared to the cube of 6 (216)

So the cube root of 226981 is 61

Let us take another example - 148877

Here 7 is in the last digit but the cube of seven's last digit is not seven. But the cube of three has the last digit as 7.

So the last digit of the cube root of 148877 is 3.

Now for the remaining digits 148.

It is the nearer and bigger than the cube of 5 (125).

Therefore the cube root of 148877 is 53.

Let us take another example 54872.

Here the last three digit's (872) last digit is 2 but the cube of 2's last digit is not 2 but the last of the cube of 8 is 2.

So the last digit of the cube root of 54872 is 8.

Now of the remaining numbers (54). It is nearer and bigger to the cube of 3 (27). So therefore the cube root of 54872 is 38.

Note by Kartik Kulkarni
3 years, 5 months ago

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How about to find cube roots of a number which answer is three-digit number ?? For example 111^3, 267^3, etc

- 3 years, 5 months ago

After doing the last three digits , try to find which is the nearest cube number to it for the remaining digits E.g -

$$\sqrt[3]{1860867}$$

Done with the last three digits and the last digit , & you get 3 as the last digit of $$\sqrt[3]{1860867}$$

Now find the nearest cube number of1860 & it is 12 (1728)

So therefore $$\sqrt[3]{1860867}$$ = 123

- 3 years, 5 months ago

So we just do the same ways... Thank you so much..

- 3 years, 5 months ago

waitwaitwaitwaitwait..whaaaaaaaaaaat? Where did that 3 even come from? The last digit of 1860867 is 7.....

- 3 years, 4 months ago

Read the note properly , it says 7 is in the last digit but the cube of seven's last digit is not seven. But the cube of three has the last digit as 7.

- 3 years, 4 months ago

So basically the cube of the number you are looking for must have the same last digit as the number in the problem?

- 3 years, 4 months ago

No, dude. It occasionally happens, but it ain't no rule. It happens for 1 (1³ = 1), 4 (4³ = 64), 5 (5³ = 125), 6 (6³ = 216), 9 (9³ = 729) and 0 (0³ = 0). But, here we see, it doesn't happen for 2 (2³ = 8), neither 3 (3³ = 27), nor 7 (7³ = 343) and 8 (8³ = 512). I'll always have to check this before find cube roots by this method.

- 3 years, 4 months ago

2, 3, and 7, 8 has at their unit place have their 10's compliments. Rest have the same number as said earlier.

- 3 years, 4 months ago

Thank you, a great method to solve the cube roots, so bad it doesn't work with every cubic root, it would save a lot of time in tests. Anyway, thanks!

- 3 years, 4 months ago

Another Interesting fact:: (A) cube of 2= unit digit 8 .....cube of 8=unit digit 2 (B) cube of 3=unit digit 7...... cube of 7= unit digit 3

- 3 years, 4 months ago

1, 4, 5, 6, 9 have the unit place of their cubes as the number themselves. But cubes of 2,3 and 8,7 has there unit place as their compliment of 10.

- 3 years, 4 months ago

It works for groups of threes. How adorable.

- 3 years, 4 months ago

Cub of 2, 3, 7, 8 we have its compliment of 10.
For 2: it is 10-2=8...........for 8, it is 10-8=2............................ 3, it it 10-3=7, ......for 7, it is 10-7=3

- 2 months, 1 week ago

Data handling

- 6 months, 2 weeks ago

But it is not apllicable everywhere..

- 7 months, 4 weeks ago

Comment deleted 5 months ago

We should take the number which has highest cube less than 117 in your case but not nearest.

- 6 months, 3 weeks ago

I gave the first person who introduced this to me a very good comment. Features of natural numbers can occasionally been found. Important thing is never let this concept to mislead ourselves when the situation is not whole numbers. I recalled and remember ed again but not really memorized properly. Understand why could make me a better memory perhaps. Hope I can memorize from today onwards!

- 1 year, 1 month ago

But it is not useful for non perfect cubes

- 1 year, 1 month ago

Write a comment or ask a question...if m=29 and e=13, then m=m+e e=m-e m=m-e then find the new value of m and e??

- 2 years, 10 months ago

Very nice & thanks.

- 3 years, 2 months ago

Good Method ....!!! Amazing...!!

- 3 years, 3 months ago

thats just for a sure perfect cube

- 3 years, 3 months ago

Cool.......

- 3 years, 4 months ago

Really very useful trick Thanks:)

- 3 years, 4 months ago

It's really coolest method ever.but can any1 suggests me methods for square root of a decimal number.for eg:square root of 0.56

- 3 years, 4 months ago

how to work out cube root of 216216. The answer on face is 66 but that is not the cube root.

- 3 years, 4 months ago

216216 is not perfect cube ! This method is only for perfect cube

- 4 months ago

you have to it by long division method

- 3 years, 4 months ago

good one

- 3 years, 4 months ago

do you want to know the exact long division method of finding cube roots though it tedious... :)

- 3 years, 4 months ago

sure.

- 3 years, 4 months ago

- 3 years, 4 months ago

I HAVE SOME CONFUSION THAT WHEN HAM LOG SAME NO. KO LIKHEGE OR KAB NHI........AS 1ST SUM MEN.......226 KA 6 LIKHE AND 981 KA 1 SO ANS. IS 61 BUT 148877 MEN 148 KA 5 KYU LAST NO TO 8 HA SO COMPLEMENTRY IS 2 BUT HERE IS 5..

- 3 years, 4 months ago

I would prefer not to use Hindi cause it is confusing me that you have mixed up English & Hindi

- 3 years, 4 months ago

Only works for whole numbers. It's interesting however that you have found this method. How did you come across it?

- 3 years, 4 months ago

excellent method!!! upvoted young mind :)

- 3 years, 4 months ago

I like it

- 3 years, 4 months ago

Who discovered this method? It's really awesome

- 3 years, 4 months ago

i like that method

- 3 years, 4 months ago

Real nice method. I liked it.

- 3 years, 4 months ago

I like this method .

- 3 years, 4 months ago

nice

- 3 years, 4 months ago

you mean x^3 of 226981 , 226971 , 226961 , 226981 , 226881 , all is 61 only by your way. which is incorrect

- 3 years, 4 months ago

you have to know that it it works only for a perfect cube

- 3 years, 4 months ago

I'm sorry I did not understand

- 3 years, 4 months ago

By this trick cube root for last 3 digit is depends on unit place digit only? if we consider these numbers which all have 1 as unit place digit , 226981 , 226971 , 226961 , 226221 , 226881 so by the rule cube root should be 61 for all these numbers. which is actually incorrect because numbers are different.

- 3 years, 4 months ago

This method is only applicable for cube numbers that have the cube root with no numbers after the decimal point

- 3 years, 4 months ago

As I have mentioned in another comment, if the number is not a perfect cube, we at least know the floor and the ceiling of this number.

- 3 years, 4 months ago

excellent

- 3 years, 4 months ago

Thank U Very Much.I like Ur Way To solve The Problem.

- 3 years, 4 months ago

Great! Interesting!

- 3 years, 4 months ago

- 3 years, 4 months ago

Really good method... I like it!

- 3 years, 4 months ago

ecellent method .its working

- 3 years, 4 months ago

Good solution

- 3 years, 4 months ago

Write a comment or ask a question... Super

- 3 years, 4 months ago

Thanks

- 3 years, 4 months ago

Excellent method

- 3 years, 4 months ago

Just noticed. It actually isn't applicable to numbers other than perfect cubes. For example, if you calculate the cube root of 1,216 using this method, you get 16; actual root is 10.67. They're almost 5.5 numbers apart. If you have any better ways, please post it.

- 3 years, 4 months ago

In that case we know between which two integers the actual cube root lays.

- 3 years, 4 months ago

I had answered to a similar question , & this method is only applicable for numbers which have their cube roots with no numbers after the decimal point

- 3 years, 4 months ago

Awesome and unique way to do it!! Thanks!!

- 3 years, 4 months ago

Nice note

- 3 years, 4 months ago

Would largely help me for finding Karl Pearson's coefficient. Thanks.

- 3 years, 5 months ago

Fantastic method Thanks

- 3 years, 5 months ago

Brilliant! Good to learn this from you. Thanks.

- 3 years, 5 months ago

maths is not about approximation and estimation!!!!

- 3 years, 5 months ago

can someone prove it mathematically?

- 3 years, 5 months ago

- 3 years, 4 months ago

a + 10 b + 100 c + 1000 d + 10000 e + 100000 f could roughly prove it I guess.

- 3 years, 5 months ago

@Kartik Kulkarni .... really a nice one ... but i hav a doubt ... take 1331 ..... u get 11 by the method stated above .... if u take 1441 ..... 11 isnt correct ..... in that case .... u cant find whether a no. is a cube no. or not using this method.... rite???

- 3 years, 5 months ago

also 1441's cube root is somewhat 11 And many more numbers after the decimal points

- 3 years, 5 months ago

Well actually,this method is only applicable for actual cube numbers

- 3 years, 5 months ago

Great buddy. Here is the actual method https://brilliant.org/discussions/thread/long-divison-method-of-cube-root/

- 3 years, 4 months ago

@Kartik Kulkarni ... as soon as i saw ..... i found this interesting and also concluded this is applicable for perfect cubes ... but ur inference of 1441's cube root is around 11 is wrong ..... eg: take 1721 ..... if u infer by the same method as u did above ... it is around 11 ... but actually it can be estimated to 12 ..... (Note: cube root of 1721 = 11.98)

- 3 years, 5 months ago

well , I didn't think about the estimation part

- 3 years, 5 months ago

cool

- 3 years, 5 months ago

Really cool way...I m looking forward to u to post some cool ways of finding the sum of series....

- 3 years, 5 months ago

I love this mathed

- 3 months, 2 weeks ago

3√79510

- 4 months, 2 weeks ago

according to this cube root of 125486 should be 56 but actually it is not

- 3 years, 3 months ago

I just found out when this method works,

for eg 125486. last 3 digits = 6, first 3 digist =5,

here 125 is perfect cube , hence it doesnt work.

MY findings = This method only works when neither of the components( 1st 3 digits & last 3 digits) are perfect cube but the number that is comprised of the components is a perfect cube.

In ur case 125486 aint a perfect cube cum 125 which i call a component is.

Hows my Theorem? Thumbs up!!

- 3 years, 3 months ago

- 3 years, 2 months ago

An adition to my findings: Either both the components aint perfect cube or both are.

125000, fr last 3 digits =0, fr first 3 digits =5

cube root of 125000 is 50.

(Notice that 000 is nothing but 0 and not 1000, 0^3 is 0)

- 3 years, 2 months ago

216216

- 2 years, 6 months ago

Comment deleted Apr 05, 2015

Comment deleted Apr 08, 2015

Sorry i didnt get u.

- 3 years, 2 months ago

125486 is not a perfect cube and so this method is not applicable for that number

- 3 years, 1 month ago

- 3 years, 4 months ago

Very nice and interesting solution

- 3 years, 4 months ago

whats's wrong with these four numbers(2,3,7 and 8)? i mean these are the number which you will never find at the end of any "squared number"( at ones place i mean). and here too the same four number have different digits at ones place. by the way nice trick. thanks!

- 3 years, 4 months ago

what if we have 7 digited number could u explane me how to do it please

- 3 years, 4 months ago

I just explained it to Jonathan Christianto above

- 3 years, 4 months ago

kk:":":":":":":thanku

- 3 years, 4 months ago

then we have to go by long divison actual method

- 3 years, 4 months ago


Simplify $2(\cos^2 x - \sin^2 x)^2 \tan 2x$

Simplify: $$2(\cos^2 x - \sin^2 x)^2 \tan 2x$$ After some sketching, I arrive at: $$2 \cos 2x \sin2x$$ Now according to the answer sheet, I should simplify this further, to arrive at $\sin 4x$. But how do I derive the latter from the former? Where do I start? Hoe do I use my double-angle formulas to arrive there?

• Hint: $\sin(2u) = 2\cos(u)\sin(u)$. What is $u$ in this case? – MathematicsStudent1122 Dec 27 '15 at 9:46

Notice, using double angle identity $\cos^2A-\sin^2A=\cos 2A$, one should get

$$2(\cos^2x-\sin^2x)^2\tan 2x$$ $$= 2\cos^2 2x\tan 2x$$ $$= 2\cos^2 2x\left(\frac{\sin 2x}{\cos 2x}\right)$$ $$= 2\sin 2x\cos 2x$$ using double angle identity $2\sin A\cos A=\sin 2A$, $$=\sin 2(2x)$$$$=\color{red}{\sin 4x}$$

• Sorry, there was a typo in the initial post. – Apeiron Dec 27 '15 at 9:45
• Alright, so I have made correction accordingly – Harish Chandra Rajpoot Dec 27 '15 at 9:48

$2\cos(2x)\sin(2x)=\sin(2(2x))=\sin(4x)$

• Yeah, this just doesn't explain why I can do this. I know $2 \cos x \sin x = \sin 2x$, but how do I get from your first step to the second. This is not obvious for me. Would $2 \cos 3x \sin 3x = \sin 6x$? – Apeiron Dec 27 '15 at 9:40
• First step: use $2x$ instead of $x$ in the sine double angle formula Second step: $2*2x=4x$ Yes, $2 \cos 3x \sin 3x = \sin 6x$ – GNUSupporter 8964民主女神 地下教會 Dec 27 '15 at 9:42

Use $$\cos^2x-\sin^2x=\cos2x$$ and $$\tan A=\dfrac{\sin A}{\cos A}$$


# Calculus

Use the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm confused about how to solve this one.

1. Let V = ∫sin^2x dx
Since cos^2x is a mirror image of sin^2x, ∫cos^2x dx = V

Now, since sin^2x+cos^2x = 1,

2V = ∫[0,π/2] 1 dx = π/2
V = π/4

check:

sin^2x = (1-cos2x)/2
∫sin^2x dx = ∫(1-cos2x)/2 dx
= 1/2 (x - 1/2 sin2x) [0,π/2]
= 1/2[(π/2)-(0)]
= π/4

posted by Steve

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# How to find the derivative of a definite integral that has unusual lower and upper bounds?

I'm not sure how to deal with upper and lower bounds in integrals when using the first part of the fundamental theorem of calculus to work with them.

The question I'm looking at asks me to find the derivative of the function, where the function is a definite integral. The question is explicitly telling me to use the fact that $$\frac{d}{dx} \int f(x)dx = f(x)$$, i.e. the first part of the fundamental theorem of calculus, to answer the question.

The function is:

$$\int_{\sqrt{x}}^{\pi/4} \theta \cdot tan\theta \cdot d\theta = g(x)$$

In words: if a function corresponds to an integral where the upper bound on the integral is $$\pi/4$$, the lower bound is $$\sqrt{x}$$, and the function being integrated is $$\theta \cdot tan\theta$$ with respect to $$\theta$$, then what is the derivative of the function?

I've tried setting $$u = \pi/4$$ and applying the chain rule, that's worked in the past but it doesn't give me the right answer here. I'm guessing I also have to incorporate the lower bound, that $$\sqrt{x}$$, in my solution somehow, but I have no idea how.

Any help would be greatly appreciated.

• Hint: chain rule. – GEdgar Nov 24 '18 at 22:34
• As a suggested reading, check out Leibniz Rule for differentiation of integrals. – Shubham Johri Nov 24 '18 at 22:44
• Hint: $\int_{-\infty}^x f(x) dx = f(x)$, so $\int_a^bf(x) dx = \int_{-\infty}^b f(x) dx - \int_{-infty}^a f(x) dx$ – eSurfsnake Nov 25 '18 at 7:19

So,$$g(x)=\int_{\sqrt x}^{\frac\pi4}\theta\tan\theta\,\mathrm d\theta=-\int_{\frac\pi4}^{\sqrt x}\theta\tan\theta\,\mathrm d\theta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule?

• Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC? – James Ronald Nov 24 '18 at 22:42
• No, the lower bound does not matter. – José Carlos Santos Nov 24 '18 at 22:45

After years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $$x$$ and the lower limit is not a constant.

So, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.

Suppose the lower limit is $$L(x)$$ and the upper limit is $$U(x)$$ of the integral. Define

$$g(x) = \int_{L(x)}^{U(x)}f(t)\text{ d}t\text{.}$$

Suppose $$F$$ is an antiderivative of $$f$$. By part II of the Fundamental Theorem of Calculus, you know that $$g(x) = \int_{L(x)}^{U(x)}f(t)\text{ d}t = F(U(x)) - F(L(x))\text{.}$$ Then, the derivative of $$g$$ is given by, assuming differentiability of $$U$$ and $$L$$,

$$\dfrac{\text{d}}{\text{d}x}[g(x)] = F^{\prime}(U(x))U^{\prime}(x)-F^{\prime}(L(x))L^{\prime}(x)$$ after making use of the chain rule for derivatives. But, $$F$$ is an antiderivative of $$f$$, so $$F^{\prime} = f$$, hence $$\dfrac{\text{d}}{\text{d}x}[g(x)] = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x)\text{.}$$ In other words, the main result is $$\boxed{ \dfrac{\text{d}}{\text{d}x}\int_{L(x)}^{U(x)}f(t)\text{ d}t = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x)\text{.}}$$

Applying to this problem, we have $$f(\theta) = \theta \tan(\theta)$$, $$U(x) = \dfrac{\pi}{4}$$, and $$L(x) = \sqrt{x}$$. The derivatives are $$U^{\prime}(x) = 0$$ and $$L^{\prime}(x) = \dfrac{1}{2\sqrt{x}}$$. Hence, the derivative of $$g$$ is $$g^{\prime}(x) = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x) = f\left(\dfrac{\pi}{4}\right)(0) - f\left(\sqrt{x}\right) \cdot \dfrac{1}{2\sqrt{x}}$$ which simplifies to $$g^{\prime}(x) = - f\left(\sqrt{x}\right) \cdot \dfrac{1}{2\sqrt{x}} = -\sqrt{x}\tan(\sqrt{x}) \cdot \dfrac{1}{2\sqrt{x}} = -\dfrac{1}{2}\tan(\sqrt{x})\text{.}$$

• Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again! – James Ronald Nov 24 '18 at 23:11


# Infinite cyclic group isomorphism.

I have one probably basic question, but still it bothers me how to show it.

Namely, if two groups are isomorphic (i.e. there is a bijective group homomorphism between them) and if one of them is infinite cyclic (precisely, in my case it is discussed about $(\mathbb Z,+)$), does the other group necessary have to be infinite cyclic?

Precisely, I'm in algebraic topology and I know that the fundamental group of cycle $\mathbb S^1$ is isomorphic to integers, but does this imply that the fundamental group of cycle is infinite cyclic as well?

Any sketch of the proof would be welcome.

• All cyclic groups are isomorphic to either $\mathbb{Z}$ or $\mathbb{Z}/n\mathbb{Z}$ for a natural number $n$. – Student Mar 6 '17 at 21:20

Indeed, this is true and is known as the fundamental theorem of finitely generated abelian groups..

In particular, we know that every finitely generated abelian group is of the form:

$$\mathbb Z^n \oplus \mathbb Z_p\oplus\cdots \oplus \mathbb Z_q$$

so if your group is infinite and cyclic, then it is abelian, of rank $1$ and torsion-free, meaning that it is indeed the same thing as $\mathbb Z$.

Edit: on the other hand, this now feels like a lot of overkill, despite being the way that I think about it. Here is a proof sketch:

Let $G$ be infinite cyclic $\langle a^n \mid n \in \mathbb Z \rangle$. Suppose further that $\phi: \mathbb Z \to G$ is the map $n \mapsto a^n$.

You should check that this is indeed a homomorphism. Surjectivity is clear. Injectivity follows from the fact that if $a^n=a^m$ for $n >m$, then $a^n (a^m)^{-1}=e \implies a^{n-m}=e$ while the order of $a$ was supposed to be infinite.

Hence, this is an isomorphism.

Edit 2: Let me try to address your question in the comments. Let $\pi_1(S^1)$ be the group consisting of loop classes, equipped with the usual multiplication. We already have that there exists an isomorphism $$\rho: \pi_1(S^1) \to (\mathbb Z,+)$$

and we have further that

$$\phi:\mathbb Z \to G$$ is an isomorphism when $G$ is infinite cyclic. By the composition of isomorphisms,

$$\phi \circ \rho:\pi_1(S^1) \to G$$ is an isomorphism as well. Hence, we can conclude that it is infinite cyclic.

This seems like a more general problem, isomorphism is a transitive property, if $A \cong B \cong C$, then $A \cong C$ as well.

Edit 3: We claim that $(\mathbb Z,+)$ is an infinite cyclic group. To see this, consider $1 \in \mathbb Z$. Clearly, every element $n \in \mathbb N$ can be written as $\underbrace{1+1+\dots+1}_{n \, \mathrm{times}}$ and each inverse can be given by $-1$, which is the additive inverse of $1 \in \mathbb Z$. In other words, $1$ generates the group and has infinite order. Perhaps the notation is confusing, in additive notation:

A group $(G,+)$ is said to be infinite cyclic if $$G=\langle n \cdot a \mid n \in \mathbb Z\rangle.$$

Hence if you already know that $\pi_1(S^1) \cong \mathbb Z$, then it must be infinite cyclic, since $\mathbb Z$ already was.

• Thanks. I understand what you are trying to say, but isn't this (in some sense) the reversed statement, i.e. for a given infinite cyclic group say $(G,*)$, one can construct isomorphism between $G$ and $\mathbb Z$. But I think I have the opposite problem; I already know that my (fundamental) group $\pi_1(\mathbb S^1,*)$ is isomorphic to $(\mathbb Z,+)$ where for $(\mathbb Z,+)$ "infinite cyclic" behaviour is known and what I am trying to show (if possible) is that the first group must be infinite cyclic as well. – edward_scissorhands Mar 7 '17 at 8:39
• It's unclear what you are confused about to me. What I have just demonstrated is that it does not matter whether you say that the fundamental group is infinite cyclic of is the additive group on the integers, since the two of them are the same group, up to symbolic manipulation. – Andres Mejia Mar 7 '17 at 8:44
• the fundamental group is infinite cyclic. I included another argument. However, I think that this argument is backwards in the way of intuition. The thing to understand is that $\pi_1(S^1)$ is a free group given by a single generator, so it is in fact cyclic, so it is abelian, and hence is $\mathbb Z$. In particular, any free group modulo its commutator subgroup is again $\mathbb Z^n$ for some $n$. To see an example where it is a free group (infinite) on two generators, but not $\mathbb Z$, consider $\pi_1(S^1 \vee S^1)$. – Andres Mejia Mar 7 '17 at 8:54
• If it is isomorphic to $(\mathbb Z,+)$ then it is infinite cyclic, since $\mathbb Z$ is itself infinite cyclic. For example, it is generated by a single element $1$ that has infinite order. – Andres Mejia Mar 7 '17 at 8:57
• @Eurydice see my edits – Andres Mejia Mar 7 '17 at 9:01