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The question is that we have to find the real value(s) of $x\in[-1,1]$ satisfying the equation: $${11 \choose 1}x^{10}-{11 \choose 3}x^{8}+{11 \choose 5}x^{6}-{11 \choose 7}x^{4}+{11 \choose 9}x^{2}-{11 \choose 11}=0$$

I couldn't spot any standard series to simplify the expression on the LHS and so I tried writing it as $$\frac{-1}{2}\big[(1+x)^{11}+(1-x)^{11}]$$ but this is obviously not true since it does provide us with the required terms but their signs are not correct.

How can I make the appropriate signs appear on the terms too? I suspect that $i$ may be involved but I'm unsure about how to apply this idea. Any help would be appreciated.

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  • $\begingroup$ You have just obtained the right thing. Just write your 'easier form' like this, $\frac{-1}{2}[(ix+1)^{11}+(ix-1)^{11}]$. Now, this will be $0$ when $\frac{(ix+11)}{(ix+11)}=(-1){\zeta_{11}}^k ,k\leq11$ where, $\zeta_{11}$ is $11$ th root of unity. $\endgroup$
    – Alapan Das
    May 2, 2020 at 14:53

2 Answers 2

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Hint:

$$\dfrac{(1+y)^{2n+1}+(1-y)^{2n+1}}2=\sum_{r=0}^{r\le n}\binom{2n+1}{2r}y^{2r}$$

We have $$\sum_{r=0}^n\binom{2n+1}{2r}(-1)^rx^{2r}=0$$

Comparing $y=ix$

So, $$(1+ix)^{2n+1}=-(1+ix)^{2n+1}$$

$$\iff\left(\dfrac{1+ix}{1-ix}\right)^{2n+1}=-1=e^{(2m+1)i\pi}$$

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Perhaps this will help:

$$(x+i)^{11}={11\choose 0}x^{11}+{11\choose 1}x^{10}i -{11\choose 2}x^{9}- {11\choose 3}x^{8}i+...$$

and

$$(x-i)^{11}={11\choose 0}x^{11}-{11\choose 1}x^{10}i -{11\choose 2}x^{9}+ {11\choose 3}x^{8}i+...$$

so your expresion is:$$ i\cdot {(x+i)^{11} -(x-i)^{11}\over 2}$$

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  • $\begingroup$ Shouldn't it be $\frac{i}{2}$ on the outside? $\endgroup$
    – infinite-blank-
    May 2, 2020 at 14:44

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