How do you find the power series for #f'(x)# and #int f(t)dt# from [0,x] given the function #f(x)=Sigma x^n/lnn# from #n=[2,oo)#?

Question

Scarlett Allison · Accepted Answer

See below

For: #f(x)=sum_(n = 2)^(oo) x^n/lnn# We differentiate wrt x: #f'(x)= sum_(n = 2)^(oo) (n x^(n-1))/lnn# And: #int_0^x f(t) \ dt = int_0^x sum_(n = 2)^(oo) t^n/lnn \ dt# #= sum_(n = 2)^(oo) (x^(n+1))/((n+1) lnn )#

Samantha Adkison · Answer

undefined To find the power series for $ f'(x) $ and $ \int f(t) \, dt $ from $ [0,x] $ given the function $ f(x) = \sum_{n=2}^{\infty} \frac{x^n}{\ln n} $, we differentiate and integrate the given function term by term.

1. To find $ f'(x) $:
Differentiate each term of the series $ f(x) $ term by term to get $ f'(x) $.

$$ f'(x) = \sum_{n=2}^{\infty} nx^{n-1} $$

2. To find $ \int f(t) \, dt $:
Integrate each term of the series $ f(x) $ term by term with respect to $ t $ from $ 0 $ to $ x $ to get $ \int f(t) \, dt $.

$$ \int f(t) \, dt = \sum_{n=2}^{\infty} \frac{x^{n+1}}{(n+1) \ln n} $$

So, the power series for $ f'(x) $ is $ \sum_{n=2}^{\infty} nx^{n-1} $, and the power series for $ \int f(t) \, dt $ is $ \sum_{n=2}^{\infty} \frac{x^{n+1}}{(n+1) \ln n} $.