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Help finding Constants for Taylor Series

  • Thread starter jtleafs33
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Homework Statement


The Taylor expansion of ln(1+x) has terms which decay as 1/n.
Show, that by choosing an appropriate constant 'c', the Taylor series of
(1+cx)ln(1+x)
can be made to decay as 1/n2
(assuming expansion about x=0)

Homework Equations



f(x)=[itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex] f(n)(0) [itex]\frac{x^{n}}{n!}[/itex]

The Attempt at a Solution



I used Maple to differentiate this function and find values at x=0 for several derivative:

f(0)(0) = 0
f(1)(0) = 1
f(2)(0) = 2c-1
f(3)(0) = -3c+2
f(4)(0) = 8c-6
f(5)(0) = -30c+24

f(x)=[itex]\frac{(0)x^{0}}{0!}[/itex]+[itex]\frac{(1)x^{1}}{1!}[/itex]+[itex]\frac{(2c-1)x^{2}}{2!}[/itex]+[itex]\frac{(-3c+2)x^{3}}{3!}[/itex]+[itex]\frac{(8c-6)x^{4}}{4!}[/itex]+[itex]\frac{(-30c+24)x^{5}}{5!}[/itex] ....

This is where I'm stuck... In order to get the terms decaying as 1/n2, I get different values of c for each term...

c0=1
c1=1
c2=[itex]\frac{3}{4}[/itex]
c3=[itex]\frac{8}{9}[/itex]
c4=[itex]\frac{15}{16}[/itex]
c5=[itex]\frac{24}{25}[/itex]

And I need one constant c that will do it all. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
Simon Bridge
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Did you construct the Taylor series for ln(1+x)?
In what way do the coefficients decay as 1/n?
 
  • #3
jbunniii
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Try simplifying the coefficients in your Taylor series to see if you can find a trend. Also, you can avoid the laborious task of calculating derivatives if you approach this another way. Hint: try substituting the Taylor series for ln(1+x) into the expression (1+cx)ln(1+x).

P.S. There's no point trying to set each coefficient equal to 1/n^2. You won't get the coefficients to EQUAL 1/n^2, but you should be able to get them to decay at the same rate as 1/n^2.
 
  • #4
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I did the differentiation method to more easily see the trend myself.
I know ln(1+x)=[itex]\sum[/itex][itex]^{n=\infty}_{n=1}[/itex](-1)[itex]^{n+1}[/itex][itex]\frac{x^{n}}{n}[/itex]

Also, the taylor expansion of a polynomial is just that polynomial
So, (1+cx)ln(1+x)=[itex]\sum[/itex][itex]^{n=\infty}_{n=1}[/itex](-1)[itex]^{n+1}[/itex][itex]\frac{(1+cx)x^{n}}{n}[/itex]

Rearranging this, I can get a general expression for each coefficient:
an=(-1)n+1([itex]\frac{c}{n-1}[/itex]-[itex]\frac{1}{n}[/itex])

But I'm stuck and don't know how to go about choosing this c. I'd imagine I'm going to need an equation which somehow relates an to an+1 and then solve for c, but I don't know what to do.

I need to figure this out and really understand it, because I also have to do the same thing to make the function
(1+ax+bx2)ln(1+x) decay as 1/n3
 
  • #5
jbunniii
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Rearranging this, I can get a general expression for each coefficient:
an=(-1)n+1([itex]\frac{c}{n-1}[/itex]-[itex]\frac{1}{n}[/itex])
OK, this looks promising. Let's rearrange it a bit:
[tex]a_n = (-1)^{n+1}\left( \frac{nc - (n-1)}{n(n-1)} \right) =
(-1)^{n+1}\left(\frac{n(c-1) + 1}{n^2 - n}\right)[/tex]
This should give you a pretty good idea what [itex]c[/itex] should be.
 
Last edited:
  • #6
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So, now I'm trying to solve this expression, substituting your equation into

[itex]\frac{1/n^2}{1/(n+1)^2}[/itex]=an/an+1

But this will still give different C's for different n's. I don't understand how I can find one exact c that will work for all n's. I still don't understand what to do with the equations I have.
 
  • #7
jbunniii
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So, now I'm trying to solve this expression, substituting your equation into

[itex]\frac{1/n^2}{1/(n+1)^2}[/itex]=an/an+1
You won't be able to achieve this for every [itex]n[/itex], but you don't need to. "Decays as [itex]1/n^2[/itex]" means that this is true asymptotically (in the limit).

Suppose I take [itex]c = 1[/itex]. Then
[tex]|a_{n}| = \frac{1}{n^2 - n}[/tex]
Clearly if [itex]n[/itex] is large, then the [itex]n^2[/itex] in the denominator is the dominant term, so asymptotically, [itex]|a_n|[/itex] decays like [itex]\frac{1}{n^2}[/itex].

Equivalently,
[tex]|a_{n}| = \frac{1}{n(n-1)}[/tex]
For large [itex]n[/itex], the distinction between [itex]n[/itex] and [itex]n-1[/itex] is negligible, so [itex]\frac{1}{n(n-1)}[/itex] is almost the same as [itex]\frac{1}{n^2}[/itex].

Now see if you can make this precise. Hint: try to quantify the relative error between [itex]\frac{1}{n(n-1)}[/itex] and [itex]\frac{1}{n^2}[/itex].
 
Last edited:
  • #8
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Okay, that's exactly what I needed.

When I started the post and found a few values of c for various n, I immediately saw that c approached 1 as n approached infinity. Basically, I've been trying to make things work exactly, but I didn't realize I really just need to make things approach that behavior in the limit.

Thanks!
 
  • #9
jbunniii
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Okay, that's exactly what I needed.

When I started the post and found a few values of c for various n, I immediately saw that c approached 1 as n approached infinity. Basically, I've been trying to make things work exactly, but I didn't realize I really just need to make things approach that behavior in the limit.

Thanks!
Right, terminology like "such and such decays as so and so" pretty much universally means "in the limit". I edited my above comment with a hint regarding how to quantify this.
 
  • #10
Simon Bridge
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Wasn't there a clue in the expansion for ln(1+x) to that effect? Does the expansion decay exactly as 1/n or just sort-of (in the limit) like that?
 

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