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Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.
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I just wanted to evaluate

$$ \sum_{k=0}^n \cos k\theta $$

and I know that it should give

- Floor function properties: $ = + $ and $ = \sum_{k = 0}^{n - 1} $
- Given a Fourier series $f(x)$: What's the difference between the value the expansion takes for given $x$ and the value it converges to for given $x$?
- Prove by induction: $2^n = C(n,0) + C(n,1) + \cdots + C(n,n)$
- Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$
- Euler-Maclaurin summation for $e^{-x^2}$
- Relation between $\gcd$ and Euler's totient function .

$$ \cos\left(\frac{n\theta}{2}\right)\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin(\theta / 2)} $$

I tried to start by writing the sum as

$$ 1 + \cos\theta + \cos 2\theta + \cdots + \cos n\theta $$

and expand each cosine by its series representation. But this soon looked not very helpful so I need some clue about how this partial sum is calculated more efficiently …

- Does this double series converge?
- proving an invloved combinatorial identity
- Splitting an infinite unordered sum (both directions)
- Proving $\sum_{k=1}^n k\cdot k! = (n+1)!-1$ without using mathematical Induction.
- How to prove the sum of combination is equal to $2^n - 1$
- Proof by induction that $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$
- Is $H(\theta) = \sum \limits_{k=1}^{\infty} \frac{1}{k} \cos (2\pi n_k \theta)$ for a given sequence $n_k$ equal a.e. to a continuous function?
- How to prove a formula for the sum of powers of $2$ by induction?
- Power series summation
- proof by induction: sum of binomial coefficients $\sum_{k=0}^n (^n_k) = 2^n$

Use:

$$

\sin \frac{\theta}{2} \cdot \cos(k \theta) = \underbrace{\frac{1}{2} \sin\left( \left(k+\frac{1}{2}\right)\theta\right)}_{f_{k+1}} – \underbrace{\frac{1}{2} \sin\left( \left(k-\frac{1}{2}\right)\theta\right)}_{f_{k}}

$$

Thus

$$ \begin{eqnarray}

\sin \frac{\theta}{2} \cdot \sum_{k=1}^n \cos(k \theta) &=& \sum_{k=1}^n \left(f_{k+1} – f_k\right) = f_{n+1}-f_1 \\ &=& \frac{1}{2} \underbrace{\sin\left(\left(n+\frac{1}{2}\right)\theta\right)}_{\sin(\alpha+\beta)}-\frac{1}{2} \underbrace{\sin \frac{\theta}{2}}_{\sin(\alpha-\beta)} = \cos(\alpha) \sin(\beta) \\

&=& \cos\left(\frac{n+1}{2}\theta\right) \sin\left(\frac{n}{2} \theta\right)

\end{eqnarray}

$$

where $\alpha = \frac{n+1}{2} \theta$ and $\beta = \frac{n}{2} \theta$.

Note that $\cos(n\theta) = \Re({e^{in\theta}})$. Thus, our sum can be thought of as $\Re(\sum_{n = 0}^{N}{e^{in\theta}})$. Now, $$\sum_{n = 0}^N{e^{in\theta}} = \frac{e^{i(N+1)\theta}-1}{e^{i\theta} – 1}.$$ Now $$\frac{e^{i(N+1)\theta}-1}{e^{i\theta} – 1} = \frac{e^{i(N+1)\theta/2}}{e^{i\theta/2}}\frac{e^{i(N+1)\theta/2} – e^{-i(N+1)\theta/2}}{e^{i\theta/2} – e^{-i\theta/2}}.$$ Take the real part of the right hand side of the equality and simplify and you will get the result you want. The tricky part is knowing to break up $\frac{e^{i(N+1)\theta}-1}{e^{i\theta} – 1}$ as we did in the above equation. Let me know if you get stuck or don’t understand.

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