Author Topic: Q5 TUT 5101  (Read 495 times)

Victor Ivrii

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Q5 TUT 5101
« on: November 02, 2018, 03:33:05 PM »
Find the first four terms in power-series expansion about the given point for the given function; find the largest disc in which the series is valid:
 $$\tan (z)\qquad\text{about}\; z_0 = 0.$$

Tianfangtong Zhang

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Re: Q5 TUT 5101
« Reply #1 on: November 02, 2018, 04:15:18 PM »
$$\tan(z) = \frac{\sin(z)}{\cos (z)} = \frac{z - \frac{1}{3!}z^3 + \frac{z^5}{5!} - \frac{z^7}{7!}...}{1 - \frac{z^2}{2!}+\frac{z^4}{4!}...} = a_0 + a_1z + a_2z^2 +a_3z^3 + ...$$

Thus
$(a_0 + a_1z + a_2z^2 +a_3z^3 + ...)(1 - \frac{z^2}{2!}+\frac{z^4}{4!}...) = z - \frac{1}{3!}z^3 + \frac{z^5}{5!} - \frac{z^7}{7!}...$

Thus we get

$a_0 = 0$

$a_1 = 1$

$a_2 = 0$

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Therefore $\tan(z) = z + \frac{1}{3}z^3 + \frac{2}{15}z^5 ....$
« Last Edit: November 04, 2018, 09:47:30 PM by Victor Ivrii »

Ende Jin

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Re: Q5 TUT 5101
« Reply #2 on: November 03, 2018, 01:46:29 PM »
The largest disc is $\{z : |z| < \frac{\pi}{2}\}$