what is the Taylor series for e^(9x/5)

Question

anonymous · Answer

What value of c is being asked for?  Since I'm not given one, I'll just use c = 0.  Anyway, start by taking a bunch of derivatives and evaluating them at x = 0.

$$f(x)=e ^{9x/5}$$
$$f'(x)=9/5e ^{9x/5}$$
$$f''(x)=(9/5)^2e ^{9x/5}$$
$$f'''(x)=(9/5)^3e ^{9x/5}$$

When you evaluate these functions at x = 0, you will get f(0) = 1, f'(0) = 9/5, f''(0) = (9/5)^2, and f'''(0) = (9/5)^3.  Use these as coefficients of your infinite sum of polynomial terms.

$$P(x)=(9/5)^0x^0/0! + (9/5)^1x^1/1! + (9/5)^2x^2/2! + ...$$

Rewrite this as an infinite series...

$$P(x)=\sum_{n=0}^{\infty}(9/5)^nx^n/n!$$

Then, you can clean up a little bit to get your answer:
$$P(x)=\sum_{n=0}^{\infty}(9x/5)^n/n!$$