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

[teleport]

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.

 

[christianjb]

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)

 

[teleport]

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.

 

[christianjb]

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

 

[d_leet]

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].

 

[christianjb]

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.

 

[teleport]

christian how did you get the e^2?

 

[d_leet]

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.

 

[christianjb]

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

 

[christianjb]

No it still does not (x-2)^3 is not the same as x³-8.

I never said it was!

 

[teleport]

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).

 

[christianjb]

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!

 

[christianjb]

"The derivative of e^(x^3) is not e^(x^3) it is (3x^2)(e^(x^3)"

Where did I say otherwise?

 

[teleport]

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?

 

[christianjb]

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.

 

[teleport]

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.

 

[christianjb]

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.

 

[teleport]

Thanks a lot for all the help. I am convinced.

 

[christianjb]

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.

 

[christianjb]

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.

 

[christianjb]

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!]