## anonymous 5 years ago what is the Taylor series for e^(9x/5)

1. anonymous

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: $P(x)=\sum_{n=0}^{\infty}(9x/5)^n/n!$

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