Agglomeration Of Tricks

Calculus Level 5

1 4 ∫ 0 1 t ( 1 − t ) 1 + t 2 ( 1 − t ) 2 d t = ℜ ( − z tan ⁡ − 1 z ) \large \dfrac14 \int_0^1 \dfrac{t (1-t)}{1+t^2(1-t)^2} \, dt = \Re(-z \tan^{-1} z )

If the equation above holds true, where z z is a zero of a polynomial with integer coefficients, find ∣ z ∣ − 4 |z|^{-4} .

Clarification : ℜ ( x ) \Re(x) denotes the real part of the complex number x x . For example, ℜ ( 3 + 4 i ) = 3 \Re(3+4i) = 3 .


The answer is 17.

Jake Lai by Jake Lai , 21, Singapore

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1 solution

Jake Lai
May 22, 2016

We tackle the integral directly using the series x 1 + x 2 = − ∑ n = 1 ∞ ( − 1 ) n x 2 n − 1 \displaystyle \frac{x}{1+x^2} = -\sum_{n=1}^\infty (-1)^n x^{2n-1} :

∫ 0 1 t ( 1 − t ) 1 + t 2 ( 1 − t ) 2 d t = − ∑ n = 1 ∞ ( − 1 ) n ∫ 0 1 t 2 n − 1 ( 1 − t ) 2 n − 1 d t = − ∑ n = 1 ∞ ( − 1 ) n B ( 2 n − 1 , 2 n − 1 ) = − ∑ n = 1 ∞ ( − 1 ) n n ( 4 n 2 n ) . \begin{aligned} \int_0^1 \frac{t(1-t)}{1+t^2(1-t)^2} \ dt &= -\sum_{n=1}^\infty (-1)^n \int_0^1 t^{2n-1}(1-t)^{2n-1} \ dt \\ &= -\sum_{n=1}^\infty (-1)^n \mathrm{B}(2n-1,2n-1) \\ &= -\sum_{n=1}^\infty \frac{(-1)^n}{n\binom{4n}{2n}}. \end{aligned}

It is known that ∑ n = 1 ∞ x n n ( 2 n n ) = 2 x 4 − x tan ⁡ − 1 x 4 − x \displaystyle \sum_{n=1}^\infty \frac{x^n}{n\binom{2n}{n}} = 2\sqrt{\frac{x}{4-x}} \tan^{-1} \sqrt{\frac{x}{4-x}} (see comments for proof). Hence,

∑ n = 1 ∞ u 2 n n ( 4 n 2 n ) = ∑ n = 1 ∞ u n n ( 2 n n ) + ∑ n = 1 ∞ ( − u ) n n ( 2 n n ) = 2 u 4 − u tan ⁡ − 1 u 4 − u + 2 − u 4 + u tan ⁡ − 1 − u 4 + u . \begin{aligned} \sum_{n=1}^\infty \frac{u^{2n}}{n\binom{4n}{2n}} &= \sum_{n=1}^\infty \frac{u^n}{n\binom{2n}{n}} + \sum_{n=1}^\infty \frac{(-u)^n}{n\binom{2n}{n}} \\ &= 2\sqrt{\frac{u}{4-u}} \tan^{-1} \sqrt{\frac{u}{4-u}} + 2\sqrt{\frac{-u}{4+u}} \tan^{-1} \sqrt{\frac{-u}{4+u}}. \end{aligned}

Substituting u = i u = i above, we thus have,

∑ n = 1 ∞ ( − 1 ) n n ( 4 n 2 n ) = 2 i 4 − i tan ⁡ − 1 i 4 − i + 2 − i 4 + i tan ⁡ − 1 − i 4 + i . \begin{aligned} \sum_{n=1}^\infty \frac{(-1)^n}{n\binom{4n}{2n}} &= 2\sqrt{\frac{i}{4-i}} \tan^{-1} \sqrt{\frac{i}{4-i}} + 2\sqrt{\frac{-i}{4+i}} \tan^{-1} \sqrt{\frac{-i}{4+i}}. \end{aligned}

It is trivial to show i 4 − i = − i 4 + i ‾ \displaystyle \sqrt{\frac{i}{4-i}} = \overline{\sqrt{\frac{-i}{4+i}}} . It is also easy to show that w tan ⁡ − 1 w ‾ = w ‾ tan ⁡ − 1 w ‾ \overline{w \tan^{-1} w} = \overline{w} \tan^{-1} \overline{w} (exercise).

Therefore, since z + z ‾ = 2 R e ( z ) z+\overline{z} = 2\mathrm{Re}(z) , we finally have

1 4 ∫ 0 1 t ( 1 − t ) 1 + t 2 ( 1 − t ) 2 d t = − 1 4 ∑ n = 1 ∞ ( − 1 ) n n ( 4 n 2 n ) = − R e ( i 4 − i tan ⁡ − 1 i 4 − i ) , \begin{aligned} \frac{1}{4} \int_0^1 \frac{t(1-t)}{1+t^2(1-t)^2} \ dt &= -\frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^n}{n\binom{4n}{2n}} \\ &= -\mathrm{Re}(\sqrt{\frac{i}{4-i}} \tan^{-1} \sqrt{\frac{i}{4-i}}), \end{aligned}

so ∣ z ∣ − 4 = ∣ i 4 − i ∣ − 4 = 17 \displaystyle |z|^{-4} = \left| \sqrt{\frac{i}{4-i}} \right|^{-4} = \boxed{17} . (Note that z z can be any root of 17 z 4 + 2 z 2 + 1 = 0 17z^4+2z^2+1 = 0 , eg z = − i 4 + i \displaystyle z = \sqrt{\frac{-i}{4+i}} .)

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Proof of identity:

Using the beta function and geometric series,

∑ n = 1 ∞ x n n ( 2 n n ) = ∑ n = 1 ∞ x n 2 ∫ 0 1 t n − 1 ( 1 − t ) n − 1 d t = x 2 ∫ 0 1 1 1 − x t ( 1 − t ) d t = 1 2 ∫ 0 1 1 ( t − 1 2 ) 2 + 4 − x 4 x d t = 1 2 [ 4 x 4 − x tan ⁡ − 1 ( ( t − 1 2 ) 4 x 4 − x ) ] 0 1 = 2 x 4 − x tan ⁡ − 1 x 4 − x . □ \begin{aligned} \sum_{n=1}^\infty \frac{x^n}{n\binom{2n}{n}} &= \sum_{n=1}^\infty \frac{x^n}{2} \int_0^1 t^{n-1}(1-t)^{n-1} \ dt \\ &= \frac{x}{2} \int_0^1 \frac{1}{1-xt(1-t)} \ dt \\ &= \frac{1}{2} \int_0^1 \frac{1}{(t-\frac{1}{2})^2 + \frac{4-x}{4x}} \ dt \\ &= \frac{1}{2} \left[ \sqrt{\frac{4x}{4-x}} \tan^{-1} \left( (t-\tfrac{1}{2})\sqrt{\frac{4x}{4-x}} \right) \right]_0^1 \\ &= 2 \sqrt{\frac{x}{4-x}} \tan^{-1} \sqrt{\frac{x}{4-x}}. \qquad \square \end{aligned}

Jake Lai - 5 years ago

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another way would be to differentiate the arcsin^2 series arcsin ⁡ 2 ( x ) = 1 2 ∑ n = 1 ∞ ( 2 x ) n n 2 ( 2 n n ) \arcsin^2(x) = \dfrac{1}{2}\sum_{n=1}^\infty \dfrac{(2x)^n}{n^2 \binom{2n}{n}} other then that same method(+1).

Aareyan Manzoor - 5 years ago

Sketch of alternative proof:

We use the fact that if f ( x ) = ∑ n a n x n \displaystyle f(x) = \sum_n a_nx^n , then ∑ n a m n x m n = 1 m ∑ k = 1 m f ( ρ k x ) \displaystyle \sum_n a_{mn}x^{mn} = \frac{1}{m} \sum_{k=1}^m f(\rho^kx) , where ρ = e 2 i π / m \rho = e^{2i\pi/m} .

Let f ( x ) = 2 x 4 − x tan ⁡ − 1 x 4 − x = ∑ n = 1 ∞ x n n ( 2 n n ) \displaystyle f(x) = 2 \sqrt{\frac{x}{4-x}} \tan^{-1} \sqrt{\frac{x}{4-x}} = \sum_{n=1}^\infty \frac{x^n}{n\binom{2n}{n}} . Through the above fact, we know that

∑ n = 1 ∞ 1 n ( 8 n 4 n ) = f ( i ) + f ( − 1 ) + f ( − i ) + f ( 1 ) \sum_{n=1}^\infty \frac{1}{n\binom{8n}{4n}} = f(i)+f(-1)+f(-i)+f(1)

and

∑ n = 1 ∞ 1 n ( 4 n 2 n ) = f ( 1 ) + f ( − 1 ) . \sum_{n=1}^\infty \frac{1}{n\binom{4n}{2n}} = f(1)+f(-1).

Also, since ∫ 0 1 2 t 4 n − 1 ( 1 − t ) 4 n − 1 d t = 1 n ( 8 n 4 n ) \displaystyle \int_0^1 2t^{4n-1}(1-t)^{4n-1} \ dt = \frac{1}{n\binom{8n}{4n}} and ∫ 0 1 t 2 n − 1 ( 1 − t ) 2 n − 1 d t = 1 n ( 4 n 2 n ) \displaystyle \int_0^1 t^{2n-1}(1-t)^{2n-1} \ dt = \frac{1}{n\binom{4n}{2n}} ,

∑ n = 1 ∞ 1 n ( 8 n 4 n ) = ∫ 0 1 2 t 3 ( 1 − t ) 3 1 − t 4 ( 1 − t ) 4 d t \sum_{n=1}^\infty \frac{1}{n\binom{8n}{4n}} = \int_0^1 \frac{2t^3(1-t)^3}{1-t^4(1-t)^4} \ dt

and

∑ n = 1 ∞ 1 n ( 4 n 2 n ) = ∫ 0 1 t ( 1 − t ) 1 − t 2 ( 1 − t ) 2 d t . \sum_{n=1}^\infty \frac{1}{n\binom{4n}{2n}} = \int_0^1 \frac{t(1-t)}{1-t^2(1-t)^2} \ dt.

Hence,

∫ 0 1 t ( 1 − t ) 1 + t 2 ( 1 − t ) 2 d t = ∫ 0 1 t ( 1 − t ) 1 − t 2 ( 1 − t ) 2 d t − ∫ 0 1 2 t 3 ( 1 − t ) 3 1 − t 4 ( 1 − t ) 4 d t = − f ( i ) − f ( − i ) . \begin{aligned} \int_0^1 \frac{t(1-t)}{1+t^2(1-t)^2} \ dt &= \int_0^1 \frac{t(1-t)}{1-t^2(1-t)^2} \ dt - \int_0^1 \frac{2t^3(1-t)^3}{1-t^4(1-t)^4} \ dt \\ &= -f(i)-f(-i). \end{aligned}

Jake Lai - 5 years ago

Question was an overhead bouncer. :P ,very elegant solution.btw.

Parth Lohomi - 5 years ago

It is much simpler to do integration by parts. You simply write: t ( 1 − t ) 1 + t 2 ( 1 − t ) 2 = 1 2 ( 1 i + t ( 1 − t ) + 1 − i + t ( 1 − t ) ) = 1 2 ( 1 1 4 + i − ( t − 1 2 ) 2 + 1 1 4 − i − ( t − 1 2 ) 2 ) \frac{t(1-t)}{1+{t}^{2} {(1-t)}^{2}}=\frac{1}{2}(\frac{1}{i+t (1-t)}+\frac{1}{-i+t (1-t)})=\frac{1}{2}(\frac{1}{\frac{1}{4}+i-{(t-\frac{1}{2})}^{2}}+\frac{1}{\frac{1}{4}-i-{(t-\frac{1}{2})}^{2}})

You then simply integrate by substituting t − 1 / 2 = u ∗ i 1 / 4 − i t-1/2=u*i \sqrt{1/4-i} , i 1 / 4 − i d u = d t i\sqrt{1/4-i} du=dt and t − 1 / 2 = w ∗ i 1 / 4 + i t-1/2=w*i\sqrt{1/4+i} , i 1 / 4 + i d w = d t i\sqrt{1/4+i} dw=dt . This gives us: 1 2 ∫ ( 1 1 4 + i − ( t − 1 2 ) 2 + 1 1 4 − i − ( t − 1 2 ) 2 ) d t = i 1 4 − i 1 2 ∫ 1 1 − u 2 d u + i 1 4 + i 1 2 ∫ 1 1 − w 2 d w = i 1 − 4 i arctan ⁡ u + i 1 + 4 i arctan ⁡ w = − i 1 − 4 i arctan ⁡ i ( 2 t − 1 ) 1 − 4 i − i 1 + 4 i arctan ⁡ i ( 2 t − 1 ) 1 + 4 i \frac{1}{2}\int(\frac{1}{\frac{1}{4}+i-{(t-\frac{1}{2})}^{2}}+\frac{1}{\frac{1}{4}-i-{(t-\frac{1}{2})}^{2}}) dt=\frac{i}{\sqrt{\frac{1}{4}-i}}\frac{1}{2}\int\frac{1}{1-{u}^{2}} du+\frac{i}{\sqrt{\frac{1}{4}+i}}\frac{1}{2}\int\frac{1}{1-{w}^{2}} dw=\frac{i}{\sqrt{1-4i}}\arctan{u}+\frac{i}{\sqrt{1+4i}}\arctan{w}=-\frac{i}{\sqrt{1-4i}}\arctan{\frac{i(2t-1)}{\sqrt{1-4i}}}-\frac{i}{\sqrt{1+4i}}\arctan{\frac{i(2t-1)}{\sqrt{1+4i}}}

Finally, evaluating the integral we get: [ − i 1 − 4 i arctan ⁡ i ( 2 t − 1 ) 1 − 4 i − i 1 + 4 i arctan ⁡ i ( 2 t − 1 ) 1 + 4 i ] 0 1 = − 2 ( i 1 − 4 i arctan ⁡ i i 1 − 4 i + i 1 + 4 i arctan ⁡ i 1 + 4 i ) = − 4 ℜ ( i 1 + 4 i arctan ⁡ i 1 + 4 i ) [-\frac{i}{\sqrt{1-4i}}\arctan{\frac{i(2t-1)}{ \sqrt{1-4i}}}-\frac{i}{\sqrt{1+4i}}\arctan{\frac{i(2t-1)}{\sqrt{1+4i}}}]_{0}^{1}=-2(\frac{i}{\sqrt{1-4i}}\arctan{\frac{i}{i \sqrt{1-4i}}}+\frac{i}{\sqrt{1+4i}}\arctan{\frac{i}{\sqrt{1+4i}}})=-4\Re {(\frac{i}{\sqrt{1+4i}}\arctan{\frac{i}{\sqrt{1+4i}}})}

Therefore z = i 1 + 4 i z=\frac{i}{\sqrt{1+4i}} .

Jeremi Litarowicz - 4 years, 10 months ago

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