Problem of the Week #80 - October 7th, 2013

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: For $n\geq 0$, show that
\[\int_0^1 (1-x^2)^n\,dx = \frac{2^{2n}(n!)^2}{(2n+1)!}.\]

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Hint: [sp]Start by showing that if $I_n$ denotes the integral, then
\[I_{k+1}=\frac{2k+2}{2k+3}I_k.\][/sp]

 

[Chris L T521]

This week's question was correctly answered by anemone and MarkFL. You can find both of their solutions below.

anemone's solution: [sp]We're asked to prove that \(\displaystyle \int_0^1(1-x^2)^n dx=\frac{2^{2n}(n!)^2}{(2n+1)!}\).

First, let's examine the LHS expression, the definite integral, \(\displaystyle \int_0^1(1-x^2)^n dx\),

Using the following trigonometric substitution,

\(\displaystyle x=\sin \theta\) \(\displaystyle \rightarrow dx=\cos \theta d\theta\)

The integral is now

\(\displaystyle \int_0^{\frac{\pi}{2}} \cos^{2n} \theta \cos \theta d\theta=\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta\)

Using the integration by parts with the following substitution:

\(\displaystyle u=\cos^{2n} \theta\;\;\rightarrow\;\;\frac{du}{d\theta}=2n (\cos^{2n-1} \theta)(-\sin \theta)\)

\(\displaystyle \frac{dv}{d\theta}=\cos \theta\;\;\rightarrow\;\;v=\sin \theta\)

The integral is then

\(\displaystyle \int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta=\left[(\cos^{2n} \theta)(\sin \theta) \right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 2n (\cos^{2n-1} \theta)(-\sin \theta)(\sin \theta)d \theta\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(0)+2n\int_0^{\frac{\pi}{2}} (\sin^2 \theta)(\cos^{2n-1} \theta)d \theta\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=2n\int_0^{\frac{\pi}{2}} (1-\cos^2 \theta)(\cos^{2n-1} \theta)d \theta\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=2n\int_0^{\frac{\pi}{2}} (\cos^{2n-1} \theta)d \theta-2n\int_0^{\frac{\pi}{2}} (\cos^{2n+1} \theta)d \theta\)

\(\displaystyle (2n+1)\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta =2n\int_0^{\frac{\pi}{2}} (\cos^{2n-1} \theta)d \theta\)

\(\displaystyle \int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta =\frac{2n}{2n+1}\int_0^{\frac{\pi}{2}} (\cos^{2n-1} \theta)d \theta\)

If we let \(\displaystyle I_n=\int_0^1(1-x^2)^n dx=\int_0^{\frac{\pi}{2}} \cos^{2n+1} \theta d\theta\), we now have

\(\displaystyle I_n=\frac{2n}{2n+1}I_{n-1}\)

\(\displaystyle I_n=\frac{2n}{2n+1}I_{n-1}\)

\(\displaystyle \;\;\;\;=\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}I_{n-2}\)

\(\displaystyle \;\;\;\;=\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}\cdot\frac{2(n-2)}{2(n-2)+1}I_{n-3}\)

\(\displaystyle \;\;\;\;=\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}\cdot\frac{2(n-2)}{2(n-2)+1}\cdots\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}\)

\(\displaystyle \;\;\;\;=\left(\frac{2n}{2n+1}\cdot\frac{2(n-1)}{2(n-1)+1}\cdot\frac{2(n-2)}{2(n-2)+1}\cdots\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3} \right)\left(\frac{2n}{2n}\cdot\frac{2(n-1)}{2(n-1)}\cdot\frac{2(n-2)}{2(n-2)}\cdots\frac{6}{6}\cdot\frac{4}{4}\cdot\frac{2}{2} \right)\)

\(\displaystyle \;\;\;\;=\frac{((2n)(2(n-1))(2(n-2))\cdots(6)(4)(2))^2}{(2n+1)(2n))\cdots(6)(5)(4)(3)(2)(1)}\)

\(\displaystyle \;\;\;\;=\frac{(2^n(n)(n-1)(n-2)\cdots(3)(2)(1))^2}{(2n+1)!}\)

\(\displaystyle \;\;\;\;=\frac{(2^n(n!))^2}{(2n+1)!}\)

\(\displaystyle \;\;\;\;=\frac{2^{2n}(n!)^2}{(2n+1)!}\)

and we're done![/sp]

MarkFL's solution: [sp]Let:

\(\displaystyle I_n=\int_0^1\left(1-x^2 \right)^n\,dx\)

Apply integration by parts, where:

\(\displaystyle u=\left(1-x^2 \right)^n\,\therefore\,du=n\left(1-x^2 \right)^{n-1}(-2x)\,dx\)

\(\displaystyle dv=dx\,\therefore\,v=x\)

And we may state:

\(\displaystyle I_n=\left[x\left(1-x^2 \right)^n \right]_0^1+2n\int_0^1 x^2\left(1-x^2 \right)^{n-1}\,dx\)

\(\displaystyle I_n=0-2n\int_0^1 \left(1-x^2-1 \right)\left(1-x^2 \right)^{n-1}\,dx\)

\(\displaystyle I_n=-2n\left(\int_0^1 \left(1-x^2 \right)^{n}\,dx-\int_0^1 \left(1-x^2 \right)^{n-1}\,dx \right)\)

Now, using the definition of the definite integral we may write:

\(\displaystyle I_n=2n\left(I_{n-1}-I_n \right)\)

Solving for $I_n$, we find:

\(\displaystyle (2n+1)I_n=2nI_{n-1}\)

\(\displaystyle I_n=\frac{2n}{2n+1}I_{n-1}\)

Now, we should observe that:

\(\displaystyle I_0=\int_0^1\left(1-x^2 \right)^0\,dx=1\)

Iterating all the way down to $n=0$, we have:

\(\displaystyle I_n=\frac{(2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2}{(2n+1)(2n-1)(2n-3)\cdots7\cdot5\cdot3}\cdot1\)

Multiplying by \(\displaystyle 1=\frac{(2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2\cdot1}{(2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2\cdot1}\) we have:

\(\displaystyle I_n=\frac{\left((2n)(2(n-1)(2(n-2))\cdots6\cdot4\cdot2\cdot1 \right)^2}{(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)\cdots7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}\)

\(\displaystyle I_n=\frac{\left(2^n\cdot n! \right)^2}{(2n+1)!}\)

\(\displaystyle I_n=\frac{2^{2n}(n!)^2}{(2n+1)!}\)

Shown as desired.[/sp]