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Proof Taylor series of (1-x)^(-1/2) converges to function

  1. Jun 7, 2015 #1
    Hello,

    I want to prove that the taylor expansion of [itex]f(x)={\frac{1}{\sqrt{1-x}}}[/itex] converges to ƒ for -1<x<1. If I didn't make a mistake the maclaurin series should look like this:
    [itex]Tf(x;0)=1+\sum_{n=1}^\infty{\frac{(2n)!}{(2^n n!)^2}}x^n[/itex]
    My attempt is to use the lagrange error bound, which is given by [itex]E(x)\leq M\frac{|x|^{n+1}}{(n+1)!}[/itex], where M is the upper bound for [itex]|f^{(n+1)}(x)|[/itex] for the interval [0,x] and show it tends to zero for [itex]\lim_{n\rightarrow \infty}[/itex].
    My problem is the following: For functions like ex or sin(x) M is independent of n and thus the limit is easy to compute, but here M is given by [itex]f^{(n+1)}(x)=\frac{(2n)!}{4^n n!}(1-x)^{-\frac{2n+1}{2}}[/itex] (which I conclude from graphing the function). So I need to prove that the following holds true (for -1<x<1):
    [itex]\lim_{n\rightarrow \infty}(\frac{(2n)!}{4^n n!}(1-x)^{-\frac{2n+1}{2}}\frac{|x|^{n+1}}{(n+1)!})[/itex]=0
    I have no clue how to solve such a limit and would appreciate any help! Of course I don't expect anyone to work it out, but maybe someone has an advice as how to tackle complicated limits like this one... .

    Thanks in advance!
     
    Last edited: Jun 7, 2015
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  3. Jun 7, 2015 #2

    micromass

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    Here's another way that does not involve the limit:

    1) Prove the series ##\sum_{n=1}^{+\infty} \frac{(2n)!}{2^{2n} (n!)^2}x^n## converges to a differentiable function ##g(x)##.
    2) Prove that ##\frac{d}{dx} g(x)\sqrt{1+x}## is ##0##, and hence that ##g(x) = \frac{c}{\sqrt{1+x}}##
    3) Prove that ##c=1##.
     
  4. Jun 7, 2015 #3

    micromass

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    If you do want to evaluate the limit, then you should consider the ratio test.
     
  5. Jun 7, 2015 #4

    FactChecker

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    Is it a requirement to directly prove that the expansion converges? Instead, can you use the fact that it is the composition of functions whose radii of convergence are well known? (or easy to prove)
     
  6. Jun 7, 2015 #5
    Thanks to both of you for your replies!
    @micromass I will try to get my head around your method after I had some sleep. I just tried to work it out using the ratio test and got a wrong result (I have fairly little experience with computing limits), but maybe I will find the error tomorrow :-)
    @FactChecker Not necessarily, however I not only want to prove that the expansion converges, but also that it converges to the function ƒ. So I am curious, is one able to prove the latter using the fact it is a composite function?
     
  7. Jun 7, 2015 #6

    FactChecker

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    I am having second thoughts. It is possible to prove convergence around x=0, but it's not clear to me, without a lot of thinking, what it would take to prove it especially for the full range -1<x<1.
     
  8. Jun 8, 2015 #7
    I decided to work out the limit and keep getting the same result, namely [itex]\displaystyle\lim_{n\rightarrow +\infty} {\frac{a_{n+1}}{a_n}}=\frac{|x|}{1-x}[/itex]. But instead it should be zero or am I wrong here? Clearly the result I got is wrong, since that limit doesn't exist for x>0,5. So is it true that I should get [itex]\displaystyle\lim_{n\rightarrow +\infty} {\frac{|x|}{n(1-x)}}=0[/itex] if [itex]x\neq0[/itex] for the proof to succeed?
    Also I want to mention that the limit for E(x) I wrote down in my original post is wrong since I mixed up [itex]f^{(n+1)}(x)[/itex] and [itex]f^{(n)}(x)[/itex].
     
    Last edited: Jun 8, 2015
  9. Jun 8, 2015 #8

    micromass

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    The ratio tests states for a series ##\sum a_n## that if ##\lim_{n\rightarrow +\infty}\left|\frac{a_{n+1}}{a_n}\right|<1##, then the series converges (and hence the sequence of terms converges to ##0##).

    So you don't want the limit to go to ##0##, you want it to be ##<1##. So indeed, it seems you have a problem for ##x>1/2##. This might be because of a computational error, or because the specific bound was to rough.
     
  10. Jun 8, 2015 #9
    Mhh... I didn't consider the possibility that the error bound could cause problems and it really seems that it does, since I checked my result several times (and proving the taylor expansion converges for 1/2<x is not really satisfying). Thus I am considering your method, but I have a few questions regarding the individual steps: 1) I did the ratio test and the result is that the series converges for |x|<1. Is that already proving the series converges to a differentiable function in that domain? 2) How would I prove that without knowing how [itex]g(x)[/itex] exactly looks like? 3) How would I determine the value of the integration constant?

    Sorry if I'm asking too many questions at a time, I am very grateful for your input :-)
     
  11. Jun 8, 2015 #10

    micromass

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    Not in general, but for power series it does. You should look for a good theorem on power series which says exactly that. Any analysis book should have it.

    Evaluate both sides for a suitable concrete ##x## (for example ##x=0##).
     
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