Revival-Math Q&A

agentredlum

Math Team
Jul 2011
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North America, 42nd parallel
The Fiboncci sequence begins with the numbers 1,1 and then each subsequent number is the sum of the two previous numbers.

1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , ...

Prove that every 5th Fibonacci number is divisible by 5

Q&A RULES

1) Anyone can post an answer.

2) The one who posted the problem decides the winner.

3) The winner must post the next question or relinquish control.

:)
 
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Jul 2010
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I will use induction. Our base case $P_1$ is obviously true since:

\(\displaystyle F_{5}=5\)

Thus, our induction hypothesis $P_n$ is:

\(\displaystyle F_{5n}=5m\) where \(\displaystyle m,n\in\mathbb{N}\)

Now, we if use the recursive definition of the sequence and observe that:

\(\displaystyle F_{5(n+1)}-F_{5n}=F_{5n+4}+F_{5n+3}-F_{5n}\)

\(\displaystyle F_{5(n+1)}-F_{5n}=F_{5n+3}+2F_{5n+2}+F_{5n+1}-F_{5n}\)

\(\displaystyle F_{5(n+1)}-F_{5n}=3F_{5n+2}+2F_{5n+1}-F_{5n}\)

\(\displaystyle F_{5(n+1)}-F_{5n}=5F_{5n+1}+2F_{5n}\)

Adding this to the hypothesis, we obtain:

\(\displaystyle F_{5(n+1)}=5F_{5n+1}+3\cdot5m=5\left(F_{5n+1}+3m \right)\)

We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.
 
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Apr 2014
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\(\displaystyle F_{5(n+1)}-F_{5n}=3_{5n+2}+2F_{5n+1}-F_{5n}\)

\(\displaystyle F_{5(n+1)}-F_{5n}=5_{5n+1}+F_{5n}\)
Sorry to nitpick ;) but it’s \(\displaystyle F_{5(n+1)}-F_{5n}=5_{5n+1}+\color{red}2\color{black}F_{5n}\) and we should have
$$F_{5(n+1)}\ =\ 5F_{5n+1}+3F_{5n}\ =\ 5\left(F_{5n+1}+3m \right)$$
For example,
$$F_{10}\ =\ 55\ =\ 40+15\ =\ 5F_6+3F_5$$
$$F_{15}\ =\ 610\ =\ 445+165\ =\ 5F_{11}+3F_{10}$$
 
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Jul 2010
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Sorry to nitpick ;) but it’s \(\displaystyle F_{5(n+1)}-F_{5n}=5_{5n+1}+\color{red}2\color{black}F_{5n}\) and we should have
$$F_{5(n+1)}\ =\ 5F_{5n+1}+3F_{5n}\ =\ 5\left(F_{5n+1}+3m \right)$$
For example,
$$F_{10}\ =\ 55\ =\ 40+15\ =\ 5F_6+3F_5$$
$$F_{15}\ =\ 610\ =\ 445+165\ =\ 5F_{11}+3F_{10}$$
Yes, I have corrected my post. :D
 
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agentredlum

Math Team
Jul 2011
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North America, 42nd parallel
Excellent , MarkFL is the winner!

Olinguito gets a bonus point for proofreading.

MarkFL , you have the next question.

:)
 
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Jul 2010
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Okay...here is a fun exercise:

Problem: The double integral
\[\int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy\]
is an improper integral and could be defined as the limit of double integrals over the rectangle:

$[0,t]\times[0,t]$ as $t\to 1^{-}$.

1.) Expand the integrand as a geometric series to show that \[\int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy = \sum_{n=1}^{\infty}\frac{1}{n^2}\]
2.) Leonhard Euler proved that \[\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}\]
Prove this fact by evaluating the integral found in (1).
 
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agentredlum

Math Team
Jul 2011
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North America, 42nd parallel
This proof is from Brendan W. Sullivan who gives Tom Apostol credit for inventing it.

http://docs.google.com/viewer?url=http://www.google.com/url?q=http://math.cmu.edu/~bwsulliv/basel-problem.pdf&sa=U&ei=DOM-U5uSOq-E2gXM94CQBQ&ved=0CCUQFjAB&sig2=HUZ3u0fnm-Fio2Pc7JbngQ&usg=AFQjCNHAmIG41l9OTt6V3Rznk0-l89V8rw

I will add some details.

For MarkFL's 1) We observe

$0 \le xy \le 1$

So we let $xy = r$ without fear in order to use the formula for a geometric series.

$$\dfrac{1}{1-xy} = \dfrac{1}{1-r} = 1 + r + r^2 + r^3 + ... = \\ 1 + xy + (xy)^2 + (xy)^3 + ... = \sum_{n = 1}^{\infty}(xy)^{n-1}$$

Now you can do this

$$ \int_{0}^1 \int_{0}^1 ( 1 + xy + x^2y^2 + x^3y^3 + ... ) dxdy$$

Work from the inside out , integrate term by term with respect to x keeping y constant.

$$ \int_{0}^1 (x + \dfrac{x^2y}{2} + \dfrac{x^3y^2}{3} + \dfrac{x^4y^3}{4} + ...) |_{0}^1)dy$$

Plug in the upper limit x = 1 (the lower limit x = 0 vanishes making no contribution)

$$ \int_{0}^1 (1 + \dfrac{y}{2} + \dfrac{y^2}{3} + \dfrac{y^3}{4} + ... ) dy $$

Integrate with respect to y

$$ (y + \dfrac{y^2}{2 \cdot 2} + \dfrac{y^3}{3 \cdot 3} + \dfrac{y^4}{4 \cdot 4} + ...) |_{0}^1 $$

Plug in the upper limit y = 1 (the lower limit y = 0 vanishes),

$$ 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + ... + \dfrac{1 }{n^2} + ... = \sum_{n = 1}^{\infty } \dfrac{1 }{n^2} $$

This confirms MarkFL's 1)

For MarkFL's 2) I refer the reader to the link given above because giving details here would turn this into a very long post.

:)
 
Jul 2010
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522
St. Augustine, FL., U.S.A.'s oldest city
This problem was given as a university level problem of the week at MHB, and this is the solution I posted:

1.) Expanding the integrand as a geometric series, we may write:

\(\displaystyle \frac{1}{1-xy}= \sum_{n=0}^{\infty}(xy)^n\)

Hence the integral becomes:

\(\displaystyle \int_0^1\int_0^1 \sum_{n=0}^{\infty}(xy)^n\,dx\,dy=\int_0^1\left[\sum_{n=0}^{\infty}\frac{x^{n+1}y^n}{n+1} \right]_0^1\,dy=\)

\(\displaystyle \int_0^{1}\sum_{n=1}^{\infty}\frac{y^{n-1}}{n}\,dy=\left[\sum_{n=1}^{\infty}\frac{y^{n}}{n^2} \right]_0^1=\sum_{n=1}^{\infty}\frac{1}{n^2}\)

2.) Using the change of variables:

\(\displaystyle (x,y)=(u-v,u+v)\)

we obtain:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\iint_{R}\frac{1}{1-u^2+v^2}\left|\frac{\partial (x,y)}{\partial (u,v)} \right|\,du\,dv\)

Calculating the Jacobian matrix, we find:

\(\displaystyle \frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\\\end{vmatrix}=\begin{vmatrix}1&-1\\1&1\\\end{vmatrix}=1(1)-(-1)(1)=2\)

Thus, we have:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=2\iint_{R}\frac{1}{1-u^2+v^2}\,du\,dv\)

Remapping the boundaries in terms of the new variables, we find $R$ is a square in the $uv$-plane with vertices:

\(\displaystyle (0,0),\,\left(\frac{1}{2},-\frac{1}{2} \right),\,\left(\frac{1}{2},\frac{1}{2} \right),\,(1,0)\)

Reversing the order of integration and using the symmetry of the square, we obtain:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\int_0^{ \frac{1}{2}}\int_0^u \frac{dv\,du}{1-u^2+v^2}+\int_{ \frac{1}{2}}^1\int_0^{1-u} \frac{dv\,du}{1-u^2+v^2} \right)\)

Next, we may compute:

\(\displaystyle \int_0^u\frac{dv}{1-u^2+v^2}=\left[\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{v}{\sqrt{1-u^2}} \right) \right]_0^u=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}} \right)=\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\)

\(\displaystyle \int_0^{1-u}\frac{dv}{1-u^2+v^2}=\left[\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{v}{\sqrt{1-u^2}} \right) \right]_0^{1-u}=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}} \right)=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)\)

If we let:

\(\displaystyle \tan(\theta)=\sqrt{\frac{1-u}{1+u}}\)

Squaring, we obtain:

\(\displaystyle \tan^2(\theta)=\frac{1-u}{1+u}\)

Add through by 1:

\(\displaystyle \tan^2(\theta)+1=\frac{1-u}{1+u}+1\)

Apply a Pythagorean identity on the left and combine terms on the right:

\(\displaystyle \sec^2(\theta)=\frac{2}{1+u}\)

Invert both sides:

\(\displaystyle \cos^2(\theta)=\frac{u+1}{2}\)

Solving for $u$, we find:

\(\displaystyle u=2\cos^2(\theta)-1=\cos(2\theta)\)

Hence, we find:

\(\displaystyle \tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)=\theta=\frac{1}{2}\cos^{-1}(u)\)

Using the identity \(\displaystyle \sin^{-1}(u)+\cos^{-1}(u)=\frac{\pi}{2}\) we finally have:

\(\displaystyle \tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)=\frac{1}{2}\left(\frac{\pi}{2}-\sin^{-1}(u) \right)\)

Utilizing these results, we now may state:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\int_0^{ \frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du+\frac{1}{2}\int_{ \frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\left(\frac{\pi}{2}-\sin^{-1}(u) \right)\,du \right)\)

Now, let's look at the first integral:

\(\displaystyle \int_0^{ \frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du\)

Using the substitution:

\(\displaystyle \alpha=\sin^{-1}(u)\,\therefore\,d\alpha=\frac{1}{\sqrt{1-u^2}}\,du\)

we now have:

\(\displaystyle \int_0^{\frac{\pi}{6}}\alpha\,d\alpha=\frac{1}{2} \left[\alpha^2 \right]_0^{\frac{\pi}{6}}= \frac{1}{2}\left(\frac{\pi}{6} \right)^2=\frac{\pi^2}{72}\)

Next, let's break the second integral into two parts:

i) \(\displaystyle \frac{\pi}{4}\int_{\frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\,du=\frac{\pi}{4}\left[\sin^{-1}(u) \right]_{\frac{1}{2}}^1=\frac{\pi}{4}\left(\frac{\pi}{2}-\frac{\pi}{6} \right)=\frac{\pi^2}{12}\)

ii) \(\displaystyle -\frac{1}{2}\int_{\frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du\)

Using the substitution:

\(\displaystyle \alpha=\sin^{-1}(u)\,\therefore\,d\alpha=\frac{1}{\sqrt{1-u^2}}\,du\)

we now have:

\(\displaystyle -\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \alpha\,d\alpha=-\frac{1}{4}\left[\alpha^2 \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}=-\frac{1}{4}\left(\left(\frac{\pi}{2} \right)^2-\left(\frac{\pi}{6} \right)^2 \right)=-\frac{\pi^2}{18}\)

Thus, putting these results together, there results:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\frac{\pi^2}{72}+\frac{\pi^2}{12}-\frac{\pi^2}{18} \right)=4\left(\frac{\pi^2}{24} \right)\)

And, we may then state:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}\)

Shown as desired.

So...agentredlum, you now have the floor. :cool:
 
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agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
Thanx , it would have taken me 2 hours + to write in rhe details using mathjax with my PS3 controller.

Are you concerned at all that when $$x = y = 1$$ we get $$ \dfrac{1}{1-1}$$ and the Geometric series does not apply?

:)
 
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Jul 2010
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Thanx , it would have taken me 2 hours + to write in rhe details using mathjax with my PS3 controller.

Are you concerned at all that when $$x = y = 1$$ we get $$ \dfrac{1}{1-1}$$ and the Geometric series does not apply?

:)
I see $x$ and $y$ as "approaching" 1 from beneath as given by the definition of the rectangle. :D
 
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