search
Log In
0 votes
60 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 by
60 views

1 Answer

0 votes

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

0 votes
1 answer
1
53 views
The Taylor series expansion of $f(x)= \text{ln}(1+x^2)$ about $x=0$ is $\sum _{n=1}^{\infty} (-1)^n \frac{x^n}{n}$ $\sum _{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n}}{n}$ $\sum _{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n+1}}{n+1}$ $\sum _{n=0}^{\infty} (-1)^{n+1} \frac{x^{n+1}}{n+1}$
asked Sep 18, 2019 in Calculus gatecse 53 views
0 votes
1 answer
2
112 views
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 $2$ $2e$ $2 \pi$ $2i$
asked Sep 23, 2019 in Calculus Arjun 112 views
1 vote
1 answer
3
89 views
The limit $\underset{n \to \infty}{\lim} \left( 1- \frac{1}{n^2} \right) ^n$ equals $e^{-1}$ $e^{-1/2}$ $e^{-2}$ $1$
asked Sep 23, 2019 in Calculus Arjun 89 views
0 votes
1 answer
4
96 views
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 $0$
asked Sep 23, 2019 in Calculus Arjun 96 views
...