Find a power series for the function

[arl146]

Homework Statement

Find a power series representation for the function and determine the radius of convergence.

f(x) = arctan(x/3)

Homework Equations

not really any equations ... just intervals

The Attempt at a Solution

here's what i did:

f'(x) = $\frac{1}{1+(x/3)^2}$ = $\frac{1}{1+(x^2/9)}$

arctan(x/3) = $\int\frac{1}{1+(x^2/9)}dx$ = $\int\frac{1}{1-(-x^2/9)}dx$ = ∫ (Ʃ (-1)ⁿ* $\frac{x^(2n)}{9}$)dx = ∫ (1/9 - x²/9 + x⁴/9 - x⁶/9 + ...)dx

= C + x/9 - x³/27 + x⁵/45 - x⁷63 + ...

To find C:

make x = 0 so that C = arctan(0) = 0.
so, arctan(x/3) = x/9 - x³/27 + x⁵/45 - x⁷63 + ...

= Ʃ (-1)ⁿ*(x^(2n+1) / (18n+9))
from n=0 to infinity


is this all right ??

 

[Dick]

In your power series it isn't just (x^2)^n. It's (x^2/9)^n.

 

[arl146]

I'm not sure if I know where/what you mean. Can you be a little more specific?

 

[Dick]

1/(1+(x^2/9)) is equal to the sum of (-x^2/9)^n. Not the sum of (-x^2)^n/9.

 

[arl146]

Oh ok. So you mean the 9 has an exponent of n too?

So you get arctan(x/3) = x - x^3/27 + x^5/405 - x^7/5103 + ...

so it would be summation (-1)^n * x^(2n+1) / ?
I don't know about the bottom part I can't figure it out

 

[Dick]

Oh ok. So you mean the 9 has an exponent of n too?

So you get arctan(x/3) = x - x^3/27 + x^5/405 - x^7/5103 + ...

so it would be summation (-1)^n * x^(2n+1) / ?
I don't know about the bottom part I can't figure it out

Just keep track of how you are getting the numbers on the bottom. There is another error as well. arctan(x/3) isn't equal to $\int \frac{1}{1+(\frac{x}{3})^2} dx$. To do the integral you do a substitution of u=x/3. There's another factor coming from the du.

 

[arl146]

oh =[

ok so when you have arctan(x/3) = ∫ $\frac{1}{1+(x/3)^2}$dx and then you do the u=x/3 and then 3*du=dx so then you have 3*∫ $\frac{1}{1+u^2}$dx
what do i do next?

 

[Dick]

You don't have arctan(x/3)=$\int \frac{1}{1+(\frac{x}{3})^2} dx$. That's what I'm telling you. You need to fix that before you proceed. Use the u substitution to figure out what the integral actually is. Then use that to write arctan(x/3) correctly as an integral.

 

[arl146]

is the correct integral just the same thing just with just a 3 in front of the integral?

 

[Dick]

is the correct integral just the same thing just with just a 3 in front of the integral?

If $\int \frac{1}{1+(\frac{x}{3})^2} dx$=3*arctan(x/3), what's arctan(x/3)?

 

[arl146]

there'd be a 1/3 times the integral

 

[Dick]

there'd be a 1/3 times the integral

Yes, the 1/3 is the factor you are missing in your power series.

 

[arl146]

so its this summation (-1)^n * x^(2n+1) / ?
just with a 3 on the bottom? but there has to more to the bottom, i can't figure that out, there doesn't seem to be a pattern that i can catch for the bottom

 

[Dick]

so its this summation (-1)^n * x^(2n+1) / ?
just with a 3 on the bottom? but there has to more to the bottom, i can't figure that out, there doesn't seem to be a pattern that i can catch for the bottom

There are going to be three things on the bottom, the 3, a power of 3^2 and something that comes from integrating x^(2n). This is really not that hard.

 

[arl146]

So there should be the 3, 2n+1, and I don't get the other part

 

[Dick]

So there should be the 3, 2n+1, and I don't get the other part

I was thinking of the 9^n. What do you get? Compare with the series you gave in post 5 after you add the extra 1/3.

 

[arl146]

So you get arctan(x/3) = x - x^3/27 + x^5/405 - x^7/5103 + ...

Wait, are the numbers on the bottom still right? Or does the 1/3 affect that?

 

[Dick]

Wait, are the numbers on the bottom still right? Or does the 1/3 affect that?

No, they are not right yet. You have to multiply by 1/3.

 

[arl146]

ok so just

arctan(x/3) = x/3 - x^3/81 + x^5/1215 - x^7/15309 + ...

so like we said, there is a 3, 2n+1, and you said something about something that is related to the 9^n is on the bottom too?

 

[Dick]

ok so just

arctan(x/3) = x/3 - x^3/81 + x^5/1215 - x^7/15309 + ...

so like we said, there is a 3, 2n+1, and you said something about something that is related to the 9^n is on the bottom too?

I really think you can work this out by yourself if you starting thinking about it.

 

[arl146]

if i could work it out myself i would have been already knew the answer

 

[Dick]

if i could work it out myself i would have been already knew the answer

I don't think I can help anymore without just plain telling you the answer. You've got all the parts of the denominator listed in post 19 and you've also got four correct terms of the sequence. And you've got the numerator correct in post 13. Just put them together.

 

[arl146]

wait so its JUST 3, 2n+1, and 9^n on the bottom?

so total it's: summation (-1)^n * x^(2n+1) / (3*(2n+1)*9^n)

 

[Dick]

wait so its JUST 3, 2n+1, and 9^n on the bottom?

so total it's: summation (-1)^n * x^(2n+1) / (3*(2n+1)*9^n)

Check that against the terms in post 19. Does it work?

 

[arl146]

yea...

 

[arl146]

Hey can I just get a confirmation that is right?

 

[Dick]

yea...

I thought "yea..." meant you already knew that. Any particular reason to doubt that's it's right?

 

[arl146]

No I just like to make sure. Also I have to find the radius of convergence .. Could I use the ratio test for that? I started to do it and I got |x^2|* lim (2n+1)/(9(2n+3)) = |x^2|*lim (2n+1)/(18n+27) = (I used l'hopitals here) |x^2|*lim(2/18) or just 1/9 to simplify. Where do I go next? None of the examples in the book have an x involved. I wanted to do the part of the test where if L>1 it diverges or less than 1 converges. How do I go about that. Or if not use that way, how do I get the radius of Convergence ?

 

[Dick]

No I just like to make sure. Also I have to find the radius of convergence .. Could I use the ratio test for that? I started to do it and I got |x^2|* lim (2n+1)/(9(2n+3)) = |x^2|*lim (2n+1)/(18n+27) = (I used l'hopitals here) |x^2|*lim(2/18) or just 1/9 to simplify. Where do I go next? None of the examples in the book have an x involved. I wanted to do the part of the test where if L>1 it diverges or less than 1 converges. How do I go about that. Or if not use that way, how do I get the radius of Convergence ?

Yes, use the ratio test. And do it more carefully than you just did it. Show your steps if you want me to check it. I'm surprised none of the examples in you book have an x involved in the ratio test. That's sort of what radius of convergence is all about.

 

[arl146]

Yea there's only 2 examples and they don't have any x's involved like mine. The book really doesn't show good examples in my opinion for reasons like that. The one examples is "test the series summation ((-1)^n * n^3)/(3^n) for absolute convergence." and all they have to do is do the ratio test, in which they get the limit equal to 1/3 which is obviously less than 1. The other examples is the same thing!

Anyways, the a sub n+1 part would equal x^(2(n+1)+1) on top so just x^(2n+3) and 3*(2(n+1)+1)*(9^n+1) on bottom. And that is multiplied times the a sub n part which is 3*(2n+1)*(9n) on top and x^(2n+1) on bottom. Sorry I can't throw this stuff all together in a better way, still on my phone. I simplified all that to ((x^2)*(2n+1))/(9*(2n+3)) and that's how I got my first part I gave you

 

[Dick]

Yea there's only 2 examples and they don't have any x's involved like mine. The book really doesn't show good examples in my opinion for reasons like that. The one examples is "test the series summation ((-1)^n * n^3)/(3^n) for absolute convergence." and all they have to do is do the ratio test, in which they get the limit equal to 1/3 which is obviously less than 1. The other examples is the same thing!

Anyways, the a sub n+1 part would equal x^(2(n+1)+1) on top so just x^(2n+3) and 3*(2(n+1)+1)*(9^n+1) on bottom. And that is multiplied times the a sub n part which is 3*(2n+1)*(9n) on top and x^(2n+1) on bottom. Sorry I can't throw this stuff all together in a better way, still on my phone. I simplified all that to ((x^2)*(2n+1))/(9*(2n+3)) and that's how I got my first part I gave you

Ok, so if you get x^2/9 for the limiting ratio, then what values of x will give you a convergent series?

 

[arl146]

Negative 3 and positive 3. Is this where I have to then check the end points which is just +3 and -3 like I said?

 

[Dick]

Negative 3 and positive 3. Is this where I have to then check the end points which is just +3 and -3 like I said?

Yes, you should check the endpoints.

 

[arl146]

Ok and so for that I just have to plug in the endpoints into the series we found and I have to use the alternating series test for the convergence test of the endpoints right?

 

[Dick]

Ok and so for that I just have to plug in the endpoints into the series we found and I have to use the alternating series test for the convergence test of the endpoints right?

Yes, plug them in. Then use any test you have that applies to the series.