Find a power series for the function

arl146
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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 ??
 
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In your power series it isn't just (x^2)^n. It's (x^2/9)^n.
 
I'm not sure if I know where/what you mean. Can you be a little more specific?
 
1/(1+(x^2/9)) is equal to the sum of (-x^2/9)^n. Not the sum of (-x^2)^n/9.
 
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
 
arl146 said:
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.
 
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?
 
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.
 
Last edited:
is the correct integral just the same thing just with just a 3 in front of the integral?
 
  • #10
arl146 said:
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)?
 
  • #11
there'd be a 1/3 times the integral
 
  • #12
arl146 said:
there'd be a 1/3 times the integral

Yes, the 1/3 is the factor you are missing in your power series.
 
  • #13
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
 
  • #14
arl146 said:
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.
 
  • #15
So there should be the 3, 2n+1, and I don't get the other part
 
  • #16
arl146 said:
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.
 
Last edited:
  • #17
arl146 said:
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?
 
  • #18
arl146 said:
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.
 
  • #19
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?
 
  • #20
arl146 said:
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.
 
  • #21
if i could work it out myself i would have been already knew the answer
 
  • #22
arl146 said:
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.
 
  • #23
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)
 
  • #24
arl146 said:
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?
 
  • #25
yea...
 
  • #26
Hey can I just get a confirmation that is right?
 
  • #27
arl146 said:
yea...

I thought "yea..." meant you already knew that. Any particular reason to doubt that's it's right?
 
  • #28
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 ?
 
  • #29
arl146 said:
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.
 
  • #30
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
 
  • #31
arl146 said:
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?
 
  • #32
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?
 
  • #33
arl146 said:
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.
 
  • #34
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?
 
  • #35
arl146 said:
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.
 
  • #36
ok so for x=3,we have summation (-1)^n * ((3^(2n+1))/(3*(2n+1)*(9^n)))
and for x=-3 we have summation (-1)^n * (((-3)^(2n+1))/(3*(2n+1)*(9^n)))

dont i have to use alternating series for this series though?

for the alternating series test the series has to satisfy 2 things:

(i.) the series with n+1 in replacement of all n's has to be <= the series when n just equals n
(ii.) limit of the series has to equal 0.

could you help me with completing this limit? i usually tend to go towards L'Hopitals but i know in this case that wouldn't be a good route to take so how do i do the limit? i haven't done limits in quite some time. i know that i can take the limit of the top and bottom 'individually' so starting with the top, the x^(2n+1) part, am i able to just plug in n=infinity for the n? and if that's the case .. for the bottom, do i break that up into 2parts, the (2n+1) and the 9^n ? since i can just 'pull' the 1/3 out in front. and once i separate the bottom can i plug in the n=infinity into that too?

as for part (i.) ... i also don't exactly know how to tell if one is less than or greater than the other ... someone else was helping me with a problem like this and i just didnt understand
 
  • #37
arl146 said:
ok so for x=3,we have summation (-1)^n * ((3^(2n+1))/(3*(2n+1)*(9^n)))
and for x=-3 we have summation (-1)^n * (((-3)^(2n+1))/(3*(2n+1)*(9^n)))

dont i have to use alternating series for this series though?

for the alternating series test the series has to satisfy 2 things:

(i.) the series with n+1 in replacement of all n's has to be <= the series when n just equals n
(ii.) limit of the series has to equal 0.

could you help me with completing this limit? i usually tend to go towards L'Hopitals but i know in this case that wouldn't be a good route to take so how do i do the limit? i haven't done limits in quite some time. i know that i can take the limit of the top and bottom 'individually' so starting with the top, the x^(2n+1) part, am i able to just plug in n=infinity for the n? and if that's the case .. for the bottom, do i break that up into 2parts, the (2n+1) and the 9^n ? since i can just 'pull' the 1/3 out in front. and once i separate the bottom can i plug in the n=infinity into that too?

as for part (i.) ... i also don't exactly know how to tell if one is less than or greater than the other ... someone else was helping me with a problem like this and i just didnt understand

Why haven't you at least tried to put the series in a simpler form? Like by using 3^(2n)=9^n??
 
  • #38
uhhh i don't know .. what does that help?
 
  • #39
arl146 said:
uhhh i don't know .. what does that help?

Simplifying a complicated expression usually helps you answer questions about it.
 
  • #40
well i don't know, seeing it at 3^(2n) doesn't really make anything click in my head anymore than 9^n does
 
  • #41
arl146 said:
well i don't know, seeing it at 3^(2n) doesn't really make anything click in my head anymore than 9^n does

There's a 3^(2n) in the numerator and there's a 9^n in the denominator. If you cancel them then you don't have to look at either one. I'm not sure what this debate about "why should I simplify" is accomplishing.
 
  • #42
ohhhh yea i forgot that i plugged in the 3 for the n on top that's why i was confused. but its a 3^(2n+1) so how do i cancel them.. I am just left with 1 on top and bottom then, right? so then I am left with it being 1/(3*(2n+1)) well really i have (1/3)*lim 1/(2n+1) right?
 
  • #43
arl146 said:
ohhhh yea i forgot that i plugged in the 3 for the n on top that's why i was confused. but its a 3^(2n+1) so how do i cancel them.. I am just left with 1 on top and bottom then, right? so then I am left with it being 1/(3*(2n+1)) well really i have (1/3)*lim 1/(2n+1) right?

Use algebra to figure out how 3^(2n+1) is related to 3^(2n). Are you doing this on your phone again or something? Is that what's going on with this short attention span thing? Get a piece of paper and work it out.
 
  • #44
yea i know, i just subtract the exponents. 2n+1-2n youre just left with 1. which is what i said. i mean unless that's wrong but that's what I've been doing and it worked for previous stuff
 
  • #45
arl146 said:
yea i know, i just subtract the exponents. 2n+1-2n youre just left with 1. which is what i said. i mean unless that's wrong but that's what I've been doing and it worked for previous stuff

1 is in the exponent of something. It's not just 1. This is getting pretty painful.
 
  • #46
i know i know sorrrrrry. i was thinking that the 3's cancel but they dont. so I am left with 3/(3*(2n+1)) ?
 
  • #47
arl146 said:
i know i know sorrrrrry. i was thinking that the 3's cancel but they dont. so I am left with 3/(3*(2n+1)) ?

They DO cancel. Why didn't you cancel them before? So you would get 1/(2n+1) if you weren't forgetting something else.
 
  • #48
yea i saw that but i was talking about something that is irrelevant at this point.

i see the 1/(2n+1) .. and for my endpoints one is + and the other -
im forgetting something else with simplifying?!
 
  • #49
arl146 said:
yea i saw that but i was talking about something that is irrelevant at this point.

i see the 1/(2n+1) .. and for my endpoints one is + and the other -
im forgetting something else with simplifying?!

Not with the simplification, you just left a factor behind several posts ago. Like usual.
 
  • #50
other than the (-1)^n i don't see anything
 

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