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Find the first 4 terms of the Taylor series for the following functions:

(a) $\ln x$ centered at $a=1$,
(b) $\frac{1}{x}$ centered at $a=1$,
(c) $\sin x$ centered at $a=\frac{\pi}{4}$.
in Mathematics by Bronze Status (9,486 points) | 22 views

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(a)

$f(x)=\ln x .$ So $f^{(1)}(x)=\frac{1}{x}, f^{(2)}(x)=-\frac{1}{x^{2}}, f^{(3)}(x)=\frac{2}{x^{3}}, f^{(4)}(x)=-\frac{6}{x^{4}}$ and so
$\ln x=\ln 1+(x-1) \times 1+\frac{(x-1)^{2}}{2 !} \times(-1)+\frac{(x-1)^{3}}{3 !} \times(2)+\frac{(x-1)^{4}}{4 !} \times(-6)+\ldots$
$\quad=(x-1)-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3}-\frac{(x-1)^{4}}{4}+\ldots$

(b)

$f(x)=\frac{1}{x} .$ So $f^{(1)}(x)=-\frac{1}{x^{2}}, f^{(2)}(x)=\frac{2}{x^{3}}, f^{(3)}(x)=-\frac{6}{x^{4}}$ and so
$\frac{1}{x}=1+(x-1) \times(-1)+\frac{(x-1)^{2}}{2 !} \times(2)+\frac{(x-1)^{3}}{3 !} \times(-6)+\cdots$
$=1-(x-1)+(x-1)^{2}-(x-1)^{3}-\cdots$

(c)

$f(x)=\sin x .$ So $f^{(1)}(x)=\cos x, f^{(2)}(x)=-\sin x, f^{(3)}(x)=-\cos x$ and so
$\begin{aligned} \sin x &=\frac{\sqrt{2}}{2}+\left(x-\frac{\pi}{4}\right) \times\left(\frac{\sqrt{2}}{2}\right)+\frac{\left(x-\frac{\pi}{4}\right)^{2}}{2 !} \times\left(-\frac{\sqrt{2}}{2}\right)+\frac{\left(x-\frac{\pi}{4}\right)^{3}}{3 !} \times\left(-\frac{\sqrt{2}}{2}\right)+\cdots \\ &=\frac{\sqrt{2}}{2}\left(1+\left(x-\frac{\pi}{4}\right)-\frac{\left(x-\frac{\pi}{4}\right)^{2}}{2}-\frac{\left(x-\frac{\pi}{4}\right)^{3}}{6}+\cdots\right] \end{aligned}$
by Bronze Status (9,486 points)

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