Question

Write the Maclaurin Series (including the general term) for f(x)=ln(e+x).

Answer 1

Hmn do you know what is Maclaurin series?

Answer 2

Yes, I do

Answer 3

Ok! If I explain you how to get the MacLaurin series for f(x) = ln(1+x) , will you be able to get the Maclaurin series for f(x) = ln(e+x) ?

Answer 4

Yes, since e is just a constant

Answer 5

Ok so our new question is : Write the Maclaurin series for f(x) = ln (1+x) .

Answer 6

First of all, the center of our Taylor series will be a = 0 .

So , let us construct a table that will give us derivatives and enable us to calculate \(f^{(n)}(a)\)

Answer 7

For ln(1+x)

Answer 8

Please wait, it will take time for me to create the table.

Answer 9

Okay :D

Answer 10

OS does not support table latex here :(

Answer 11

Aww, have any other way?

Answer 12

Yes , please wait for 1minute more.

Answer 13

Use print screen?

Answer 14

Sorry

Answer 15

It's fine

Answer 16

\(\begin{array}{|c|c|c|}
\hline

\textbf{n} & f^{(n)}(x)& f^{(n)}(a)(0)\\ \hline
n & f^{(n)}(x) & f^{(n)}(a)(0) \\ \hline
0 & ln(1+x) & ln(1)= 0 \\ \hline
1 & (1+x)^{-1} & 1 \\ \hline
2 & -(1+x)^{-2} & -1 \\ \hline
3 & +2(1+x)^{-3} & +2 \\ \hline
4 & -2.3(1+x)^{-4} & -2.3 \\ \hline
. & . & . \\ \hline
. & . & . \\ \hline
. & . & . \\ \hline
n \ne 0 & (-n)^{n+1}(n-1)! \cfrac{1}{(1+x)^n} & (-1)^{n+1}(n-1)! \\ \hline
\end{array}\)

Answer 17

@ashleynguyenx3 sorry for being so late. but finally I made it :)

Answer 18

YAY! :D

Answer 19

Excellent job @mathslover

Answer 20

So it'll basically be the same right?

Answer 21

So now the general form is : \(f^{(n)}(2) = (-1)^{n+1}(n-1)! \) if \(n \ne 0\) , and and \(f^{(0)}(0)=0\) . Since the form hanged, we will have to pull the n = 0 term out of our sum .

Answer 22

Thank you, hero.

Answer 23

\(\large{c_n = \cfrac{f^{(n)}(0)}{n!} = \cfrac{(-1)^{n+1}(n-1)!}{n!} = \cfrac{(-1)^{n+1}}{n} , \space n \ne 0 ; \space c_0 = 0}\)

Answer 24

So the taylor series is given by :

\(\large{\ln (1+x) = 0 + \sum_{n=1}^{{\infty}} \cfrac{f^{(n)}(0)}{n!}(x-0)^n = \sum_{n=1}^{\infty}\cfrac{(-1)^{n+1}}{n} x^n , |x|< R }\)

Answer 25

Can you do it now, @ashleynguyenx3 ?

Answer 26

You just have to calculate radius of the convergence, R. You can do it by using the ratio test :

\(\large{a_n = \cfrac{(-1)^{n+1}}{n} x^n}\)

Answer 27

Yeah, I know that's the second part to my problem.

Answer 28

So you got it?

Answer 29

Um yeah I think

Answer 30

I'm gonna try to solve it right now

Answer 31

Good! Let us know what you get, and also if you have any problem , we will try to solve it.

Answer 32

Okays thank you! :D

Answer 33

You're welcome ashley. \(\textbf{Welcome!} \space \ddot \smile \)

Answer 34

Very complete solution mathslover :) do u know how to do this with geometric series? have u learnt integral yet?

Answer 35

im asking ashley too :)

Answer 36

Thank you mukushla , ok you're asking that to me and ashley ?

Answer 37

I thought you use geometric series to just find the convergence or divergence..

Answer 38

Well, I have started learning integral already . though, not in soo deep, but thanks to some sites and wikipedia that provided me with some extra knowledge. I saw such a type of question in openstudy too, just saw it and when I scrolled up and down the wikipedia page of MacLaurin series, i remembered everything.

Answer 39

very well, then i refuse to bring that solution here :) because it needs integral calculus :)

Answer 40

@mathslover I don't know if I did this right..

Answer 41

There's a better solution with integrals as @mukushla said. But I'd just have done what mathslover has :-D

Answer 42

How would you use integrals?

Answer 43

Friends, please post your opinions too, it will be better if the user learns something easier and better soln for this question. @mukushla @ParthKohli

Answer 44

I don't know how to do it exactly, but I read on MSE about solving an integral through Maclaurin Series. I'm pretty sure Mukushla is talking just the vice-versa.

But as I said, I would choose what Mathslover did.

Answer 45

I guess we can kinda integrate \(\ln(e + x)\), find the Maclaurin Series for that, and then differentiate it.

Answer 46

What's the anti derivative of lnx?

Answer 47

\[x(\ln(x) - 1)\]I guess

Answer 48

\[+C\]

Answer 49

So it'd be (e+x)(ln(e+x)-1)+c

Answer 50

so its like this friends, geometric series of \(\frac{1}{1+x}\) for \(|x|<1\)\[\frac{1}{1+x}=1-x+x^2-x^3+...\]take integral of both sides\[\int_{0}^{x} \frac{1}{1+x} \ \text{d}x=\int_{0}^{x} (1-x+x^2-x^3+...) \ \text{d}x\]\[\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...+\frac{(-1)^{n+1}}{n}x^n+...\]

Answer 51

This looks something that I have studied somewhere

Answer 52

Great work Mukushla too

Answer 53

;) Tnx :)