ISI2015-DCG-47

Question

The Taylor series expansion of $f(x)= \text{ln}(1+x^2)$ 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. $\sum _{n=0}^{\infty} (-1)^{n+1} \frac{x^{n+1}}{n+1}$

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  $(-1)^{1}\times \frac{x^{1}}{1}$, 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.

 

 

Answer 2

They are asking for taylor series about $x=0$, which is nothiing but maclurain series of $Ln(1+x^2)$, they are two methods to solve problem,

 

1) using integration for power series method to solve this,

in short how it works is find the antiderivate of f(x) [ let F(x) be antiderivative of f(x)] and then integrate the series of F(x).

in this case we f(x) = $ln({x^2+1})$ , now our task is to find an expression so that by integrating it we’ll get f(x),

now we know that $\int \frac{1}{u}du = lnu$, in our case we are having $u=1+x^{2}$, and du = 2x,

now for this $\int \frac{2x}{1+x^2}dx$ if i take $u=1+x^{2}$, then by using substitution rule we get $\int \frac{1}{u}du$,

so finally we got the expression which gives us $ln({x^2+1})$, now our task is to find the series corresponds to $\frac{2x}{x^2+1}$.

we know that,

$\frac{1}{1-x}$ = $1+x+x^2+x^3+x^4……..$ = $\sum_{k=0}^{n}x^k$ ,this one converges iff $|x| < 1$ because its a geogetric series..

substitute $-x^2$  in place of x in the above series then we’ll get $\frac{1}{1+x^2}$ = $1 -x^2 +x^4 -x^6 …...=$ $\sum_{k=0}^{n}(-1)^kx^{2k}$,

then multiply one both sides with 2x,

$\frac{2x}{1+x^2} = 2x-2x^3 +2x^5 -2x^7 +2x^9…. = 2\sum_{k=0}^{n}(-1)^kx^{2k+1}$

now finally we got the series for $\frac{2x}{1+x^2}$ now just integrate the series expression:-

$\int \frac{2x}{1+x^2}dx = 2\int (\sum_{k=0}^{n} (-1)^k x^{2k+1}dx$

$=2\int [ x – x^3 + x^5 – x^7 + x^9 ……..]dx$

$=2[ \frac{x^2}{2} – \frac{x^4}{4} + \frac{x^6}{6} – \frac{x^8}{8} …... ]$

$= 2 \sum_{k=0}^{n} (-1)^k \frac{x^{2k+2}}{2k+2}$

$ = \sum_{k=0}^{n} (-1)^k \frac{x^{2k+2}}{k+1}$

$ = \sum_{k=1}^{n} (-1)^{k+1} \frac{x^{2k}}{k} $

 Method 2 :-

series expansion of $ln(1+x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} ……. = \sum_{k=1}^{n}(-1)^{k+1} \frac{x^k}{k}$

now just substitute $x^2$ in place of x in above summation form,

=$\sum_{k=1}^{n}(-1)^{k+1} \frac{((x^2)^k}{k}$

=$\sum_{k=1}^{n}(-1)^{k+1} \frac{ x^{2k}}{k}$

=$\sum_{k=1}^{n}(-1)^{k+1} \frac{x^{2k}}{k}$

so option B will be the correct one .

note: whenever i was written n in series summation, consider it as infinity i was unable to find the infinity symbol so.