# Verifying Maclaurin Series for f(x)=ln(1+x)

• Richter915
In summary, the conversation covers finding Maclaurin and power series for various functions. The first function given is f(x) = ln(1+x), which can be represented as a power series \sum_{n=0}^{+\infty} (-1)^{n}\frac{x^{n+1}}{n+1}. The conversation also covers finding power series for functions such as f(x) = \frac{1}{1+9x^2}, which can be represented as 9\sum \{(-1)^n}{x^{2n}, and f(x) = \frac{1+x^2}{1-x^2}, which can be represented as \sum \{(x)^2}
Richter915
I'm supposed to find the following function as a Maclaurin Series. Please check if I'm correct.

f(x) = ln(1+x)

$$\sum \frac{\(x^n)((-1)^{n+1})}{n}$$

and that sum goes from n=1 to $$\infty$$

I also have to find the following functions as power series so please check it for me!

$$f(x) = \frac{1}{1+9x^2} = 9\sum \{(-1)^n}{x^{2n}\\ \\f(x) = \frac{1+x^2}{1-x^2} = \sum \{(x)^2}+{x^{2n+2}}$$

those should be two separate functions but I don't know how to separate them into two lines. both of these series go from n=0 to $$\infty$$

Last edited:
You're saying that

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

which converges for $|x|<1$.It's okay.

Daniel.

Yes it's ok.

thank you both. Would you mind looking at the two power series questions I just posted...I'll probably be adding more questions throughout the day so if you could help me, it would be greatly appreciated. I'm very new to sequence and series so these questions are probably very "basic" but I do need help, thank you again.

Find a Power series representation for the function:

$$\frac{x}{4x+1}=4\sum_{n=0}^{+\infty} (-1)^{n}{x^{n+1}$$

does that look about right to you guys?

U can write the second as

$$\frac{1+x^{2}}{1-x^{2}}=1+2\frac{1}{\frac{1}{x^{2}}-1}$$

and it will be easier to write the series.

Daniel.

ah I just realized that I was missing a plus sign.
$$f(x) = \frac{1+x^2}{1-x^2} = \sum \{(x)^2}+{x^{2n+2}}$$

I'm not too sure of ur method but is my answer correct?

On

$$\frac{x}{4x+1}= x \frac{1}{1 - (-4x)}$$

Cyclovenom said:
On

$$\frac{x}{4x+1}= x \frac{1}{1 - (-4x)}$$
Right that's what I did and I ended up with the series which I displayed.

$$f(x) = x\sum \{(-4x)^n} = 4x\sum {-1^n}{x^n}$$

Richter915 said:
Right that's what I did and I ended up with the series which I displayed.

$$f(x) = x\sum \{(-4x)^n} = 4x\sum {-1^n}{x^n}$$

Should have been

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

Starting with a geometric sum, find the sum of the series:

$$\sum_{n=1}^{\infty} {n}{x^{n-1}}$$

so with this, I realized that the series is just the derivative of the geometric sum. Since I know that the sum of the geometric series is 1/(1-x), I just took the the derivative of that and got 2/(1-x)^2...does that sound about right? I want to make sure that part is right before I move on. Thank you.

Cyclovenom said:
Should have been

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

oh ok, so I can't just move the four out to the front like I did? I'm just glad that your correction makes sense to me, THANK YOU!

On

$$\frac{1+x^2}{1-x^2} = 1 + \frac{2x^2}{1 -x^2}$$

That way is easier because

$$1 + 2x^2 \sum_{n=0}^{\infty} (x^2)^{n}$$

Cyclovenom said:
On

$$\frac{1+x^2}{1-x^2} = 1 + \frac{2x^2}{1 -x^2}$$

That way is easier because

$$1 + 2x^2 \sum_{n=0}^{\infty} (x^2)^{n}$$

I see but the way I did it was to rewrite the function as:

$$(1+x^2) \frac {1}{1-x^2}$$

and then I worked it from there.

Richter915 said:
Starting with a geometric sum, find the sum of the series:

$$\sum_{n=1}^{\infty} {n}{x^{n-1}}$$

so with this, I realized that the series is just the derivative of the geometric sum. Since I know that the sum of the geometric series is 1/(1-x), I just took the the derivative of that and got 2/(1-x)^2...does that sound about right? I want to make sure that part is right before I move on. Thank you.

Where you got that two from? in your derivative.

$$\frac{d}{dx} \sum_{n=0}^{\infty} x^{n} = \frac{d}{dx} \frac{1}{1-x}$$

$$\frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2}$$

Last edited:
Cyclovenom said:
Where you got that two from? in your derivative.

$$\frac{d}{dx} \sum_{n=0}^{\infty} x^{n} = \frac{d}{dx} \frac{1}{1-x}$$

$$\frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2}$$
oh I made a mistake...sorry...but does what I said look right?

Yes, what you said is true.

ok but now here's the problem...it says to use what I found to find the sum of:

$$\sum_{n=1}^{\infty} {n}{x^{n}}$$

$$\sum_{n=1}^{\infty} \frac{n}{2^{n}}$$

I honestly have no idea how to apply what I found to solve those series. If you could lead me in the right direction, it'd be greatly appreciated.

Richter915 said:
ok but now here's the problem...it says to use what I found to find the sum of:

$$\sum_{n=1}^{\infty} {n}{x^{n}}$$

$$\sum_{n=1}^{\infty} \frac{n}{2^{n}}$$

I honestly have no idea how to apply what I found to solve those series. If you could lead me in the right direction, it'd be greatly appreciated.

For the first one it's basicly the same except we just use a little trick!, we rewrite it!

$$\sum_{n=1}^{\infty} {n}{x^{n}} = x \sum_{n=1}^{\infty} {n}{x^{n-1}}$$

For the second one think of it this way

$$\sum_{n=1}^{\infty} \frac{n}{2^{n}} = \sum_{n=1}^{\infty} n (\frac{1}{2})^{n}}$$

Cyclovenom said:
For the first one it's basicly the same except we just use a little trick!, we rewrite it!

$$\sum_{n=1}^{\infty} {n}{x^{n}} = x \sum_{n=1}^{\infty} {n}{x^{n-1}}$$

For the second one think of it this way

$$\sum_{n=1}^{\infty} \frac{n}{2^{n}} = \sum_{n=1}^{\infty} n (\frac{1}{2})^{n}}$$
ohhh I see. I knew it was going to be some rewriting trick but I wasn't sure what to do. so for the first one I end up with x/(1-x)^2 and the second one, I just substitute x with 1/2 and I end up with 2. Alright, thanks a lot. Expect more questions soon!

ok so here's a new problem (though it relates to what I previously asked):

Find the sum of the following:

$$\sum_{n=2}^{\infty} {n}{(n-1)}{(x^{n})$$

I rewrote that like this:

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

I think this is the right direction but I have no clue where to go after this...please help, again!

Richter915 said:
ok so here's a new problem (though it relates to what I previously asked):

Find the sum of the following:

$$\sum_{n=2}^{\infty} {n}{(n-1)}{(x^{n})$$

I rewrote that like this:

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

I think this is the right direction but I have no clue where to go after this...please help, again!

You are on the right track!, except you don't know what do because you made a mistake

$${x^2}\sum_{n=2}^{\infty} {n}{(n-1)}{(x^{n-2})$$

Last edited:
Can someone also verify this infinite series for me?:

Evaluate the indefinite integral as an infinite series:

$$\int {e^{x^3}} = \int \sum \frac {x^{3n}}{n!} = C + \sum_{n=0}^{\infty} \frac {x^{3n+1}}{(n! (3n+1))}$$

I posted my answer in there, does that look about right? Thanks.

Cyclovenom said:
You are on the right track!, except you don't know what do because you made a mistake

$${x^2}\sum_{n=1}^{\infty} {n}{(n+1)}{(x^{n-2})$$
wait I don't see why it's x^(n-2)...when you change the series from n=2 to n = 1...x^n becomes x^(n+1)...so then when you extract x^2...it becomes x^(n-1)

also, if I continue the problem with my numbers...I get to a spot like this:

$$x^2 \sum_{n=1}^{\infty} ({n^2x^{n-1}) + ({nx^{n-1})$$

and then I split that up into two series and one of the series equals x^2/(1-x)2 while the other (the one with the n^2) is still stuck as a series...hope that helps!

Last edited:
Richter915 said:
wait I don't see why it's x^(n-2)...when you change the series from n=2 to n = 1...x^n becomes x^(n+1)...so then when you extract x^2...it becomes x^(n-1)

Yes i made a mistake, i meant to keep n=2, don't change it to n=1, because you will see something interesting when it's n=2.

$${x^2}\sum_{n=2}^{\infty} {n}{(n-1)}{(x^{n-2})$$

ohhh ok, Let me try...thank you so much!

Richter915 said:
Can someone also verify this infinite series for me?:

Evaluate the indefinite integral as an infinite series:

$$\int {e^{x^3}} = \int \sum \frac {x^{3n}}{n!} = C + \sum_{n=0}^{\infty} \frac {x^{3n+1}}{(n! (3n+1))}$$

I posted my answer in there, does that look about right? Thanks.

Yes it is correct.

AH!

$${x^2}\sum_{n=2}^{\infty} {n}{(n-1)}{(x^{n-2})$$

is just x^2(f''(1/1-x)) hah, thank you so much!

cyclovenom...you were a great help...I am now done with my math homework and confident on the unsure questions...I am sure that you'll be helping me out again in the near future as I am quickly approaching finals time. Thanks again, take care.

Richter915 said:
cyclovenom...you were a great help...I am now done with my math homework and confident on the unsure questions...I am sure that you'll be helping me out again in the near future as I am quickly approaching finals time. Thanks again, take care.

Thanks! , it's always good to be of help.

## 1. What is a Maclaurin series?

A Maclaurin series is a type of Taylor series, which is a representation of a function as an infinite sum of terms. It is centered at the point x=0 and is used to approximate a function near that point.

## 2. How do you verify a Maclaurin series for a given function?

To verify a Maclaurin series for a given function, you need to compare the terms of the series to the terms of the function. This can be done by taking derivatives of the function and evaluating them at x=0, and then comparing them to the coefficients of the Maclaurin series.

## 3. Why is it important to verify a Maclaurin series?

Verifying a Maclaurin series is important because it ensures that the series is an accurate representation of the function. Without verification, there may be errors in the series that could lead to incorrect approximations of the function.

## 4. What is the Maclaurin series for ln(1+x)?

The Maclaurin series for ln(1+x) is given by:
ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ... = Σ (-1)^(n+1) * x^n / n
This series can also be written in sigma notation as:
Σ (-1)^(n+1) * (x^n / n) where n goes from 1 to infinity.

## 5. How many terms of the Maclaurin series for ln(1+x) are needed for a good approximation?

The number of terms needed for a good approximation of ln(1+x) depends on the desired level of accuracy. Generally, the more terms included in the series, the more accurate the approximation will be. However, for most practical purposes, using the first few terms (around 3-5) is sufficient for a good approximation.

• Introductory Physics Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
493
• Introductory Physics Homework Help
Replies
2
Views
369
• Introductory Physics Homework Help
Replies
6
Views
857
• Introductory Physics Homework Help
Replies
2
Views
288
• Calculus and Beyond Homework Help
Replies
3
Views
616
• Math POTW for University Students
Replies
16
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
576
• Introductory Physics Homework Help
Replies
3
Views
1K