# ISI2015-MMA-59

98 views

In the Taylor expansion of the function $f(x)=e^{x/2}$ about $x=3$, the coefficient of $(x-3)^5$ is

1. $e^{3/2} \frac{1}{5!}$
2. $e^{3/2} \frac{1}{2^5 5!}$
3. $e^{-3/2} \frac{1}{2^5 5!}$
4. none of the above
in Calculus
recategorized

The general expansion of a function using taylor series at any given point  $x=a$  for a function  $f(x)$  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 the coefficient of the term  $(x-3)^{5}$  would be  $\frac{f'''''(3)}{5!}$

Here,  $f(x)=e^{\frac{x}{2}}$    $\Rightarrow$   $f'(x)=e^{\frac{x}{2}}\times (\frac{1}{2})$

$\Rightarrow$   $f''(x)=e^{\frac{x}{2}}\times (\frac{1}{2})^{2}$

$\Rightarrow$   $f'''(x)=e^{\frac{x}{2}}\times (\frac{1}{2})^{3}$

So, similarly   $f'''''(x)=e^{\frac{x}{2}}\times (\frac{1}{2})^{5}$

$\therefore$   at  $x=3$,  the value of the coefficient would be :   $e^{\frac{3}{2}}\times \frac{1}{2^{5}5!}$.

Option B is the correct answer.

## Related questions

1
69 views
The Taylor series expansion of $f(x)= \text{ln}(1+x^2)$ about $x=0$ is $\sum _{n=1}^{\infty} (-1)^n \frac{x^n}{n}$ $\sum _{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n}}{n}$ $\sum _{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n+1}}{n+1}$ $\sum _{n=0}^{\infty} (-1)^{n+1} \frac{x^{n+1}}{n+1}$
The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is $2$ $2e$ $2 \pi$ $2i$
The limit $\underset{n \to \infty}{\lim} \left( 1- \frac{1}{n^2} \right) ^n$ equals $e^{-1}$ $e^{-1/2}$ $e^{-2}$ $1$
Let $a_n= \bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1- \frac{1}{\sqrt{n+1}} \bigg), \: \: n \geq1$. Then $\underset{n \to \infty}{\lim} a_n$ equals $1$ does not exist equals $\frac{1}{\sqrt{\pi}}$ equals $0$