# Calculus Taylor series help!!?

Obtain the Taylor series for e^2x about x=1

Relevance
• Anonymous

Recall that a Taylor series for f(x) about the point x = a is given by:

f(a) + f'(a)(x - a)/1! + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + f''''(a)(x - 1)^4/4! + ...

We have that:

f(x) = e^(2x)

f'(x) = 2e^(2x)

f''(x) = 4e^(2x)

...

f^(n)(x) = 2^n*e^(2x)

So we have that the Taylor Series for a = 1 is:

f(1) + f'(1)(x - 1)/1! + f''(a)(x - a)^2/2! + f'''(1)(x - a)^3/3! + f''''(1)(x - 1)^4/4! + ...

= e^2 + 2e^2(x - 1) + 4e^2(x - 1)^2/2 + 8e^2(x - 1)^3/6 + 16e^2(x - 1)^4/24 + ...

= ∑ 2^n*e^2(x - 1)^n/n! (from n=0 to infinity) <== ANSWER

I hope this helps!

• Login to reply the answers
• we are managing a place as a function of time x(t). enable's improve this function some element a utilising a Taylor sequence: x(t) = x(a) + x'(a) (t - a) + x''(a) (t - a)^2 / 2 + x'''(a) (t - a)^3 / 3! + ... enable the element a be a = a million the element in time that we are given each and all of the information then x(a) = 3m x'(a) = -1m/s x''(a) = 3m/s^2 x'''(a) = -2m/s^3 x(t=2) = 3 + -a million(2 - a million) + 3(2 - a million)^2 / 2 + -2(2 - a million)^3 / 6 = 3 - a million + 3/2 - a million/3 = 19/6 m

• Login to reply the answers