Taylor Series Question?

Question

If the Taylor Series expansion for f(x) = ln(1+x/1-x) is:

2x + 2/3x^3 + 2/5 x^5 + 2/7 x^7...

how do I find f'''''(0)? (fifth derivative)

The answer to this problem is 48 and I don't know how they got there...

Answer 1

The definition of Taylor expansion around a is:

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

So, in this case we know a = 0, and f(0) = 0. It should also be obvious we need the coefficient of the x⁵ term:

f'''''(0)/5! = 2/5
f'''''(0) = 2*4! = 48

Answer 2

your function is
ln((1+x)/(1-x)) and not ln(1 +x/(1-x))

Taylor Series about x =0 (using 8 terms) is:

f(x) = 2x+(2x^3)/3+(2x^5)/5+(2x^7)/7
f(x) =f(0) + xf'(x) +....+ (x^5)*f ""'(0)/5! +....

to find f ""'(0), we solve
(x^5)*f ""'(0)/5! = (2x^5)/5
f ""'(0) = 2*5!/5 = 2*2*3*4 = 48

Answer 3

I don´know how many´´´´´have you written but in general
the coefficient of x^n is fn(0)/n!
so for the third derivative
2/3=f3(0)/3! so f3(0)= 4
f4(0)=0 and f5(0)= 2/5*5! =48

Answer 4

Taylor series can be differentiated term-by-term anywhere they are convergent.

So you can just take the derivative of 2x, and of 2/3 x^3, and so on...

Answer 5

Compare the coeffiecients of x^5 terms,
2/5 x^5 = f'''''(0)x^5/5!
f'''''(0) = (2/5)(5!) = 48