Finding power series representation of functions

Question

I was wondering how to find the power series representation of

$$sinh(x) = \frac{{e^x}-{e^{-x}}}{2}$$ in sigma notation.

Thank you.

Answer 1

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^k}{k!}+....ad-inf ~~(1)$$ And $$e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+...+(-1)^k\frac{x^k}{k!}+....ad-inf ~~~(2)$$ Suntracting (2) from (1) you get $$\sinh x =\frac{e^x-e^{-x}}{2}=x+\frac{x^3}{3!}+\frac{x^5}{5!}+...\frac{x^{2k-1}}{(2k-1)!}+....ad-inf ~~~(3)$$

Answer 2

Since $$e^x=\sum_{k=0}^\infty {x^k\over k!}$$ when we substitute $-x$ for $x$ we get $$e^{-x}=\sum_{k=0}^\infty {(-x)^k\over k!}=\sum_{k=0}^\infty {(-1)^kx^k\over k!}$$ (You were careless with this in your comment.) Now, $$e^x-e^{-x}=\sum_{k=0}^\infty {1-(-1)^k\over k!}x^k$$ by term-by-term addition.

Can you proceed from here?

Answer 3

There is a special way to do it with complex number.

As $\sinh z=-i\sin(iz)$, $$\sinh z=-i\sin(iz)=-i\left[(iz)-\dfrac{(iz)^3}{3!}+\dfrac{(iz)^5}{5!}+\cdots\right]=z+\dfrac{z^3}{3!}+\dfrac{z^5}{5!}+\cdots$$