Question

Find the expansion of f(x) = ^e^x ⁄1+e^x, given ∫f(x)dx = ln⁡(2), for x = 0

(a) \(\frac{1}{2}-\frac{x}{4}-\frac{x^3}{48}-…\)

(b) \(\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+….\)

(c) \(\frac{1}{2}+\frac{x}{4}+\frac{x^3}{48}+….\)

(d) \(\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+….\)

I had been asked this question in homework.

Question is from Taylor Mclaurin Series topic in portion Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (b) \(\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+….\)

Best explanation: Given,f(x)=\(\frac{e^x}{1+e^x}\)

Differentiating it we get

\(f^1 (x)=ln⁡(1+e^x)+C\), now putting x=0 we get,c=0

Hence,

\(f^1 (x)=ln⁡(1+e^x)\)

Now,\(f^1 (x)=ln⁡(1+e^x)=ln⁡(2)+\frac{x}{2}+\frac{x^2}{8}-\frac{x^4}{192}+..\)(By,Mclaurin’s expansion)

Hence, Differentiating it we get,

f(x)=\(\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+….\).