I'm stuck trying to calculate $2C(x)^2 = C(2x)+1$

with $C(x) =\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}$ (Cosine Power series)

So I've seen proofs for this identity before, but never using the Cauchy product formula.

So this is what I've tried:

Plugging in the power series we get:

$2\left(\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}\right)^2$

$=$ $2$$\left(\sum_{n=0}^\infty\sum_{k=0}^n (-1)^{n-k}\frac{x^{2(n-k)}}{(2(n-k))!}(-1)^k\frac{x^{2k}}{(2k)!}\right)$

$=$ $2\left(\sum_{n=0}^\infty\sum_{k=0}^n(-1)^n\binom{2n}{2k}\frac{x^{2n}}{(2n)!}\right)\\$ I hope this is correct at least.

For the other side of the equation I'm actually not quite sure how to write down.

It seems to me that this is going to be quite a long equation in general and I'm not able to calculate this further. Is it possible that I'm missing something or how can I proceed further to prove this? (Using the Cauchy product formula)


By definition, $$C(x) =\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}$$ The series is absolutely convergent, so you can consider the Cauchy product with itself, which gives you :

$$C(x)^2 = \left(\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!} \right)\left(\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!} \right)=\sum_{n=0}^{\infty}\sum_{k=0}^n (-1)^k \frac{x^{2k}}{(2k)!}(-1)^{n-k} \frac{x^{2(n-k)}}{(2(n-k))!}$$

which simplifies to give : $$C(x)^2 =\sum_{n=0}^{\infty}\sum_{k=0}^n (-1)^n \frac{x^{2n}}{(2n)!}{2n \choose 2k}= \sum_{n=0}^{\infty} (-1)^n \left(\sum_{k=0}^n{2n \choose 2k}\right)\frac{x^{2n}}{(2n)!}$$

By developping $(1+i)^{2n}$, you can see that for all $n \geq 1$, one has $$\sum_{k=0}^n{2n \choose 2k}=2^{2n-1}$$

Plugging this in the expression, you get $$C(x)^2 = 1 +\sum_{n=1}^{\infty} (-1)^n \left(\sum_{k=0}^n{2n \choose 2k}\right)\frac{x^{2n}}{(2n)!} = 1 + \sum_{n=1}^{\infty} (-1)^n 2^{2n-1}\frac{x^{2n}}{(2n)!}$$

So $$2C(x)^2 = 2 + \sum_{n=1}^{\infty} (-1)^n 2^{2n}\frac{x^{2n}}{(2n)!} = 2 +\sum_{n=1}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!}$$

Finally, reincorporing the term for $n=0$, you get $$2C(x)^2 = 1 + \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} = 1+ C(2x)$$

  • $\begingroup$ Thanks I completely forgot. I actually just proved this formula in the previous exercise and didn't see that it was "hidden" in this equation. Now since I can use this result, can I do a similar thing for the right side of the equation? $\endgroup$
    – 23408924
    Nov 4 '20 at 10:54
  • $\begingroup$ @23408924 You don't need to. What you get by applying the formula to the left side is exactly the right side, I guess. $\endgroup$ Nov 4 '20 at 11:21
  • $\begingroup$ Alright I'll try. I must admit though, I just plugged it in and then tried to show that they were indeed equal, but I might've made a mistake because it does not seem to be equal $\endgroup$
    – 23408924
    Nov 4 '20 at 11:31
  • $\begingroup$ @23408924 You just have to put the $2^{2n}$ together with the $x^{2n}$ to make appear directly the series $C(2x)$, no ? And be careful with the term $n=0$, since the formula I gave you is only valid for $n \geq 1$. Tell me if you don't succeed, I will add details to my answer. $\endgroup$ Nov 4 '20 at 11:35
  • 1
    $\begingroup$ Thank you very much for your answer and patience! I'll compare it to what I have done and see where I made the mistake(s). $\endgroup$
    – 23408924
    Nov 4 '20 at 13:23

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