Cos3xcoseccx + sin3xsecx = 2cot2x

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Ответ:

tylerwoodson

04.04.2022

Mathematics

Step-by-step explanation:

Given a series $\sum_{k=1}^\infty f(z)$ , the Weierstrass M-test tell us that if we find a sequence of positive numbers M_n such that $|f(z)|\leq M_n$ in a certain domain D, and the series $\sum_{n=1}^\infty M_k$ converges, then the series $\sum_{k=1}^\infty f_k(z)$ converges uniformly in the domain D.

So, our objective is to find the so called sequence M_k . The main idea is to bound the sequence of functions $\frac{z^k}{z^k + 1}$ .

Now, notice that the values of z are always positive, so z^k is always positive, so $z^k+1\geq 1$ for all values of z in $\overline{D}$ . Then,

$\Big| \frac{z^k}{z^k + 1}\Big| \leq z^k,$

because if we make the values of the denominator smaller, the whole fraction becomes larger.

Moreover, as z is in the interval [0,r], we have that $z\leq r$ and as consequence $|z^k|\leq r^k$ . With this in addition to the previous bound we obtain

$\Big| \frac{z^k}{z^k + 1}\Big| \leq |z^k|\leq r^k.$

With this, our sequence is M_k = r^k and the corresponding series is $\sum_{k=1}^\infty r^k$ , which is a geometric series with ratio less than 1, hence it is convergent.

Then, as consequence of Weierstrass M-test we have the uniform convergence of the series in the given domain.

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