# ISI2015-DCG-47

52 views

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

1. $\sum _{n=1}^{\infty} (-1)^n \frac{x^n}{n}$
2. $\sum _{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n}}{n}$
3. $\sum _{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n+1}}{n+1}$
4. $\sum _{n=0}^{\infty} (-1)^{n+1} \frac{x^{n+1}}{n+1}$
in Calculus
recategorized

1 vote

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  $(-1)^{1}\times \frac{x^{1}}{1}$, 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.

selected by

## Related questions

1
44 views
The Taylor series expansion of $f(x)=\ln(1+x^{2})$ about $x=0$ is $\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{n}}{n}$ $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n}}{n}$ $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n+1}}{n+1}$ $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n+1}}{n+1}$
In the Taylor expansion of the function $f(x)=e^{x/2}$ about $x=3$, the coefficient of $(x-3)^5$ is $e^{3/2} \frac{1}{5!}$ $e^{3/2} \frac{1}{2^5 5!}$ $e^{-3/2} \frac{1}{2^5 5!}$ none of the above
The value of $\underset{x \to 0}{\lim} \dfrac{\tan ^2 x – x \tan x }{\sin x}$ is $\frac{\sqrt{3}}{2}$ $\frac{1}{2}$ $0$ None of these
Let $I=\int (\sin x – \cos x)(\sin x + \cos x)^3 dx$ and $K$ be a constant of integration. Then the value of $I$ is $(\sin x + \cos x)^4+K$ $(\sin x + \cos x)^2+K$ $- \frac{1}{4} (\sin x + \cos x)^4+K$ None of these