# Thread: Maclaurin expansion of e^x

1. ## Maclaurin expansion of e^x

Here is the question Use the Maclaurin expansion of e^x to find the value e of correct to four decimal places
here is the series
e^x=1+x+(x^2)/2!+...
here is the answer

Problem
Whilst I obviously have the answer, and I have the sequence to follow, I cannot see how the answer is derived from the series, nor can I reproduce the answer on my calculator (ti89 using solve)
Can someone please point out thru the progression how the answer is achieved. Thanks

2. Hi!

To find 'e' to four decimal places you have to take the first terms of the series for x=1 , i.e.:

$e^x=\displaystyle{\sum_{n=0}^{\infty}}\frac{x^n}{n !}$

for $x=1$ we have

$e^1= \displaystyle{\sum_{n=0}^{\infty}}\frac{1^n}{n!}$

$\displaystyle{\sum_{n=0}^{8}}\frac{1}{n!}=1+1+\fra c{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\fr ac{1}{720}+\frac{1}{5040}=2,718253968\approx2,7183$

3. Originally Posted by dazza67
Here is the question Use the Maclaurin expansion of e^x to find the value e of correct to four decimal places
here is the series
e^x=1+x+(x^2)/2!+...
here is the answer

Problem
Whilst I obviously have the answer, and I have the sequence to follow, I cannot see how the answer is derived from the series, nor can I reproduce the answer on my calculator (ti89 using solve)
Can someone please point out thru the progression how the answer is achieved. Thanks
you can use the Remainder Estimation Theorem:

$\left|R_n(x)\right| \leq \frac{M}{(n+1)!}\left|x-x_0\right|^{n+1}$

4. Originally Posted by Marine
Hi!

To find 'e' to four decimal places you have to take the first terms of the series for x=1 , i.e.:

$e^x=\displaystyle{\sum_{n=0}^{\infty}}\frac{x^n}{n !}$

for $x=1$ we have

$e^1= \displaystyle{\sum_{n=0}^{\infty}}\frac{1^n}{n!}$

$\displaystyle{\sum_{n=0}^{8}}\frac{1}{n!}=1+1+\fra c{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\fr ac{1}{720}+\frac{1}{5040}=2,718253968\approx2,7183$
true! but that is if you already know that the value of $n$ should be 8..

5. Originally Posted by Marine
Hi!

To find 'e' to four decimal places you have to take the first terms of the series for x=1 , i.e.:

$e^x=\displaystyle{\sum_{n=0}^{\infty}}\frac{x^n}{n !}$

for $x=1$ we have

$e^1= \displaystyle{\sum_{n=0}^{\infty}}\frac{1^n}{n!}$

$\displaystyle{\sum_{n=0}^{8}}\frac{1}{n!}=1+1+\fra c{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\fr ac{1}{720}+\frac{1}{5040}=2,718253968\approx2,7183$
I see that now, it really is frusrating being so close, but yet so far away. Thanks

6. Originally Posted by kalagota
you can use the Remainder Estimation Theorem:

$\left|R_n(x)\right| \leq \frac{M}{(n+1)!}\left|x-x_0\right|^{n+1}$
Can you explain this to me please, I haven't come up against this yet. Thanks

7. Originally Posted by dazza67
Here is the question Use the Maclaurin expansion of e^x to find the value e of correct to four decimal places
here is the series
e^x=1+x+(x^2)/2!+...
here is the answer

Problem
Whilst I obviously have the answer, and I have the sequence to follow, I cannot see how the answer is derived from the series, nor can I reproduce the answer on my calculator (ti89 using solve)
Can someone please point out thru the progression how the answer is achieved. Thanks
You are told that:

$
e^x=1+x+(x^2)/2!+...
$

and you want $e$, so we are interested in:

$
e=e^1=1+1+1/2!+...
$

Put:

$S_n=1+1+1/2!+ .. + 1/(n-1)!$

$S_1=1$
$S_2=2$
$S_3=S_2+1/2!=2.5$
$S_4=S_3+1/3!=2.666667$
$S_5=S_4+1/4!=2.708333$
$S_6=S_5+1/5!=2.716667$
$S_7=S_6+1/6!=2.718056$

and so on.

RonL

8. ok.
Remainder Estimation Theorem: If $f$ can be differentiated $n+1$ times on an interval $I$ containing $x_0$ and if $M$ is an upper bound for $\left|f^{(n+1)}(x)\right|$ on $I$, that is $\left|f^{(n+1)}(x)\right| \leq M \, \forall \, x \in I$, then
$\left|R_n(x)\right| \leq \frac{M}{(n+1)!}\left|x-x_0\right|^{n+1}$

as an example, you can read this..
Error Estimation

9. Brilliant thanks Guys, I am going over the tread from kalagota hoping to find and edge in the next lot of exams, Thanks again for you time..