I need to prove that $\cos(x)^2+\sin(x)^2=1$ Here’s how I started (using the Cauchy product): \begin{align} \cos(x)^2+\sin(x)^2 &=\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^l\frac{x^{2l}}{(2l)!}(-1)^{k-l}\frac{x^{2(k-l)}}{(2(k-l))!}+(-1)^l\frac{x^{2(l+1)}}{(2(l+1))!}(-1)^{k-l}\frac{x^{2(k-l)+1}}{(2(k-l)+1)!}\\ &=\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^k\frac{x^{2l}}{(2l)!}\frac{x^{2k}}{x^{2l}(2(k-l))!}\\ &\quad +(-1)^k\frac{x^{2l+1}}{(2l+1)!}\frac{x^{2k+1}}{x^{2l}(2(k-l)+1)!}\\ &=\sum_{k=0}^{\infty}\sum_{l=0}^k(-1)^k\frac{x^{2k}}{(2l)!(2k-2l))!}\\ &\quad +(-1)^k\frac{x^{2(k+1)}}{(2l+1)!(2k-2l+1)!} \end{align} Here, I was told I had to use the equality $\sum_{k=0}^n(-1)^k\binom{n}{k}=0$ But I can’t since the $x$ term is in the way. How do I proceed? EDIT: I have to prove […]

Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is an integrable function with Fourier coefficients given by $\hat{f}$. Then, since $|f|^2 = f \cdot \bar{f}$, we have: $$\displaystyle \int_{0}^{2\pi} |f(k)|^2 \mathrm{d}k = \int_{0}^{2\pi}\left( \sum_{a=0}^{\infty} \hat{f}(a)e^{iak} \right)\left( \sum_{b=0}^{\infty} \hat{f}(b)e^{-ibk} \right)\mathrm{d}k.$$ Using the Cauchy product, we obtain: $$\displaystyle \int_{0}^{2\pi} \sum_{c=0}^{\infty}\sum_{l=0}^{c}\hat{f}(l)\hat{f}(c-l)e^{i(l – c + l)k}\mathrm{d}k.$$ Now, the integral is zero […]

I’m having some trouble proving that the Taylor series about the origin of the function $[Log(1-z)]^2$ to be $$\sum_{n=1}^\infty \frac{2H_n}{n+1}z^{n+1}$$ where $$H_n = \sum_{j=1}^n \frac{1}{j}$$ So far, I’ve been trying to use the definition of Log(1-z) = log|1-z| + iArg(1-z). I’ve also noted that the Taylor expansion for log|1-z| = $\sum_{k=0}^\infty \frac{-z^{k+1}}{k+1}$ from integrating the […]

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