ISI2016-DCG-47

Question

The Taylor series expansion of $f(x)=\ln(1+x^{2})$ about $x=0$ is

• A. $\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{n}}{n}$
• B. $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n}}{n}$
• C. $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n+1}}{n+1}$
• D. $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n+1}}{n+1}$

Answer 1

We know that the Taylor series expansion of any function  $f(x)$  centered around any general point  $x=a$  is given as :

 

$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...$  (Refer: Taylor Series Definition)

 

So, for the given function  $f(x)=ln(1+x^{2})$,  at   $x=0$, the taylor series expansion would look like:

 

$f(x)=f(0)+\frac{f'(0)}{1!}(x-0)+\frac{f''(0)}{2!}(x-0)^{2}+\frac{f'''(0)}{3!}(x-0)^{3}+\frac{f''''(0)}{4!}(x-0)^{4}+...$  -------(1)

 

$f(0)=0$

 

Now,  $f'(x)=\frac{2x}{1+x^{2}}$    $\Rightarrow$   $f'(0)=0$

 

$f''(x)=\frac{2-2x^{2}}{(1+x^{2})^{2}}$   $\Rightarrow$    $f''(0)=2$

 

So, the series would be :  $f(x)=0+\frac{0}{1!}(x)+\frac{2}{2!}(x^{2})+...$

 

Here, Option A gives the first term as   $(-1)^{1}\times \frac{x^{1}}{1}$  , which is clearly wrong as the first term of the taylor series of this function is  $x^{2}$.

 

Option B gives the first term of the series as  $(-1)^{2}\times \frac{x^{2}}{1}$, which is correct and hence the correct answer

 

Option C gives the first term of the series as  $(-1)^{2}\times \frac{x^{3}}{2}$, which clearly is wrong

 

Option D gives the first term of the series as  $(-1)^{2}\times \frac{x^{2}}{2}$, which clearly is wrong

 

Hence Option B is the correct answer. To be more specific one can also check the second term of the series but that would require calculating the fourth derivative(as third derivative at  $x=0$  turns out to be  $0$) of this function which would take more time.