How to find the series rapresentation in $t=0$ of the following expression
$$f(t)=\frac{1}{(2t+1)^{2}}$$
Can someone show me also the steps. Thank you so much!
Series rapresentation $f(t)=\frac{1}{(2t+1)^{2}}$
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$\begingroup$
sequences-and-series
power-series
taylor-expansion
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2 Answers
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$\displaystyle \frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n\enspace$ for $\enspace-1 One derivation for $x$ gives $\enspace\displaystyle \frac{1}{(1-x)^2}=\sum\limits_{n=1}^\infty nx^{n-1}$ . With $x:=-2t$ and therefore $\enspace -\frac{1}{2}< t< \frac{1}{2}$ follows: $$\displaystyle \frac{1}{(1+2t)^2}=\sum\limits_{n=1}^\infty n(-2t)^{n-1}$$
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Writing f(t) as power instead of as quotient it is easier to calculate the successive derivatives: $$f(t) = (2t+1)^{-2}$$ $$f'(t) = (-2)(2t+1)^{-3}2$$ $$f''(t) = (-3)(-2)(2t+1)^{-3}2^2$$ $$f^{3}(t) = (-4)(-3)(-2)(2t+1)^{-3}2^3$$ $$\cdots$$ $$f^{n}(t) = \cdots$$ Can you continue?