Taylor Series

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Homework Statement



Give the Taylor Series for exp(x^3) around x = 2.

Homework Equations



f(x) = Sum[f(nth derivative)(x-2)^n]/n!

The Attempt at a Solution



I know the solution for e^x but can't seem to find a formula for the nth derivative of exp(x^3) around x = 2.

Thanks for any help.
 

Answers and Replies

  • #2
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You know the formula for the taylor series expansion of e^y right?

Just substitute y=x^3 into that expansion and voila you have the formula for e^(x^3)
 
  • #3
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I do now that the trick you mentioned works around x = 0; however, I am highly suspicious that this works for x around other than 0. Please show me explicitely why if you still think as you said.
 
  • #4
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It works just fine, but you get a Taylor series expansion in (x^3-2)^n, not (x-2)^n. It's still a Taylor series though.

e^(x^3)=e^2 sum(n=0..infinity) (1/n!)(x^3-2)^n

Edit- the correct expression about x=2 is
e^(x^3)=e^8 sum(n=0..infinity) (1/n!)(x^3-8)^n
 
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  • #5
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It works just fine, but you get a Taylor series expansion in (x^3-2)^n, not (x-2)^n. It's still a Taylor series though.

e^(x^3)=e^2 sum(n=0..infinity) (1/n!)(x^3-2)^n
I think your series would actually give the expansion for e^[(x-2)3] and not for e^[x3-2].
 
  • #6
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I think your series would actually give the expansion for e^[(x-2)3] and not for e^[x3-2].
Neither, it gives the T series expansion for e^(x^3)

Edit- sorry I should have replaced 2 with 8 in the above equation. Then it works.
 
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  • #7
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christian how did you get the e^2?
 
  • #8
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Neither, it gives the T series expansion for e^(x^3)

Edit- sorry I should have replaced 2 with 8 in the above equation. Then it works.
No it still does not (x-2)^3 is not the same as x3-8.
 
  • #9
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christian how did you get the e^2?
d/dn exp(x) = exp(x)

so when x=2, exp(x)=e^2

However, when you substitute x=(x')^3, you have to change to e^8
 
  • #10
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No it still does not (x-2)^3 is not the same as x3-8.
I never said it was!
 
  • #11
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oh, I see. But then are saying that exp(2^3) is the nth derivative of f around x = 2? Seems strange since for instance, first f derivative = 3x^2exp(x^3). When you substitute x = 2, the above equation is unequal to exp(2^3) as you imply. Not to mention the other derivatives: all unequal to exp(2^3).
 
  • #12
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Here- I'll go through the steps

e^x = e^a sum (x-a)^n / n!

so

e^(x^3)=e^a sum(x^3-a)^n / n!

= e^(b^3) sum (x^3-b^3)^n / n!

when x=b the RHS is equal to e^(b^3), thus the T series is evaluated about x=b. So for b=2 we have

e^(x^3)= e^8 sum (x^3-8)^n / n!
 
  • #13
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"The derivative of e^(x^3) is not e^(x^3) it is (3x^2)(e^(x^3)"

Where did I say otherwise?
 
  • #14
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I see what you are doing but doesn't the last equation you give has the (x^3-8)^n, when it should be somthing like (x^3 - 2)^n, for T be around x = 2?
 
  • #15
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I see what you are doing but doesn't the last equation you give has the (x^3-8)^n, when it should be somthing like (x^3 - 2)^n, for T be around x = 2?
No, for the reasons I gave above. Look at the value of the expansion when x=2.

The reason for all the confusion is because I'm doing a T series expansion in x^3, and not in x. To do a T series expansion in x might be much more difficult, but the Q doesn't require an expansion in powers of x, though it may be implied.
 
  • #16
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yeah, you might be right, but it is definitely the first time I see it as
(x^3 - 2^3)^n for T be about x = 2. Seems unusual.
 
  • #17
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yeah, you might be right, but it is definitely the first time I see it as
(x^3 - 2^3)^n for T be about x = 2. Seems unusual.
First time I've seen it too! It works though, looking at some graphing software.
 
  • #18
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Thanks a lot for all the help. I am convinced.
 
  • #19
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Aha, I had a brainwave!

x^3=(x-2+2)^3
x'=x-2, and x' is small near x=2
so x^3=(x'+2)^3
=8(x'/2+1)^3
=(approx)=8(3x'/2+1)
=12x'+8
=12(x-2)+8

so given that x^3 = approx= 12(x-2)+8 about x=2

and

e^(x^3)=e^8 sum(n=0..infinity) (1/n!)(x^3-8)^n

By substitution we have

e^(x^3)=e^8 sum(n=0..infinity) (1/n!)(12(x-2))^n

Which, again looking at my graphing program, works. So, this is a nicer answer, because the expansion is in (x-2). However, it would be nice to have a more concise proof.
 
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  • #20
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OK, here's the concise version....

e^(x^3) about x=2
let x'=x-2
then e^(x^3)=e^[(x'+2)^3]
=e^[8(x'/2+1)^3]
=e^[12x'+8] (approximately)
=e^8 e^12x'
=e^8 sum{n=0..infinity} (1/n!)[12(x-2)]^n

Edit: Aaarrgghh! it's not quite exact, because you also need to expand (x'/2+1)^3 to higher powers in x'. So you really end up with a sum over n and m, where m is a Taylor series expansion of (x'/2+1)^3 about x'=0 in powers of x'^m.

It's messy, but doable.

I wonder if there's a way to only sum over one index.
 
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  • #21
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OK, doing the full expansion of (x'/2+1)^3 I get this for the final? answer

e^(x^3) about x=2

= e^8 sum{l,m,n}12^n 6^m (x-2)^(n+2m+3l) / l!m!n!

Edit: well I guess you can write it in powers of (x-2)^k by rewriting the above as

= e^8 sum{k} [sum{l,m,n} delta(3l+2m+n,k) 12^n 6^m (x-2)^(n+2m+3l) / l!m!n!]
 
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