Maclaurin Series

1. May 1, 2005

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

Last edited: May 1, 2005
2. May 1, 2005

dextercioby

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.

3. May 1, 2005

Yes it's ok.

4. May 1, 2005

Richter915

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.

5. May 1, 2005

Richter915

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?

6. May 1, 2005

dextercioby

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.

7. May 1, 2005

Richter915

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?

8. May 1, 2005

Pyrrhus

On

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

9. May 1, 2005

Richter915

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

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

10. May 1, 2005

Pyrrhus

Should have been

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

11. May 1, 2005

Richter915

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.

12. May 1, 2005

Richter915

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!

13. May 1, 2005

Pyrrhus

On

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

That way is easier because

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

14. May 1, 2005

Richter915

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.

15. May 1, 2005

Pyrrhus

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: May 1, 2005
16. May 1, 2005

Richter915

oh I made a mistake...sorry...but does what I said look right?

17. May 1, 2005

Pyrrhus

Yes, what you said is true.

18. May 1, 2005

Richter915

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.

19. May 1, 2005

Pyrrhus

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

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

20. May 1, 2005

Richter915

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!