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Taylor expansion of function

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Expand ##f(x) = \sqrt{2x+1}## into a Taylor series around point ##c=1##. Find the interval of convergence.

    2. Relevant equations


    3. The attempt at a solution
    I do know that ##f(x) = \sum\frac{1}{n!}f^{(n)}(c)(x-c)^n## assuming the function is representable as a Taylor series. How do I figure out the series for this particular problem?
    I have calculated some of the first derivatives of ##f(1)##:
    [tex]f(x) = 3^\frac{1}{2} + 3^{-\frac{1}{2}}(x-1) - \frac{1}{2!}3^{-\frac{3}{2}}(x-1)^2+\frac{1}{3!}3\cdot 3^{-\frac{5}{2}}+ ... = \sum\limits_{n=0}^\infty \frac{1}{n!} ??? (x-1)^n[/tex]
    What do I do to come up with the general term for ##???##
     
  2. jcsd
  3. May 16, 2015 #2

    Ray Vickson

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    I think the easiest way is to write ##x = 1 + t/2## and expand around ##t = 0##.
     
  4. May 16, 2015 #3
    What is the idea behind this substitution? What is your reason for picking so-and-so substitution? What are you trying to simplify?
     
  5. May 16, 2015 #4

    Ray Vickson

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    The expansion of ##(1+u)^p## about ##u = 0## is well known, even if ##p## is fractional and/or negative. If you are not familiar with it, you should learn it; basically, it is the binomial expansion extended to arbitrary powers, with an appropriate definition of ##C(p,k)## ("p choose k") for integer ##k \geq 0## and arbitrary real ##p \neq 0## (Newton, circa 1665).

    Of course, you could use ##x = 1 + t## instead of ##x = 1 + t/2##. Try these and see what you get.
     
  6. May 17, 2015 #5
    Well for ##(1+x)^p = 1 + px +\frac{p(p-1)}{2!}x^2 + \frac{p(p-1)(p-2)}{3!}x^3 + ...## Then for this problem it would become:

    [tex](1+2x)^{\frac{1}{2}} = 1 + \frac{1}{2}(2x) + \frac{1}{2!}\cdot\frac{1}{2}\cdot-\frac{1}{2}(2x)^2 + \frac{1}{3!}\cdot\frac{1}{2}\cdot -\frac{1}{2}\cdot-\frac{3}{2}(2x)^3 + \frac{1}{4!}\cdot\frac{1}{2}\cdot -\frac{1}{2}\cdot -\frac{3}{2}\cdot -\frac{5}{2}(2x)^4 + ... = 1 + x -\frac{1}{2}x^2 + \frac{1}{2}x^3 - \frac{5}{8}x^4[/tex]
    If I substitute ##x = t+\frac{1}{2}## then
    [tex]\sqrt{2}(1+t)^{\frac{1}{2}} = \sqrt{2}\left (1 + \frac{1}{2}t -\frac{1}{8} + \frac{1}{16}t^3 -\frac{5}{128}t^4 + ...\right ) , |t|<1[/tex]
    up until the 5th summand everything looks ok and then I get a factor of ##\frac{5}{128}##. How should I proceed?

    I also tried ##x = t+1## the problem is the same, at the 5th summand the logic breaks down.

    Can I say that if ##|t|< 1## then substituting back: ##|x-\frac{1}{2}|< 1##?
     
    Last edited: May 17, 2015
  7. May 17, 2015 #6
    (1+t)^p , expand that then replace p by 1/2 and t = 2x
    (1+t)^p = Σ(p)n*(t^n)/n! Where (p)n is the falling factorial so just change t with 2x
    (P)n = p(p-1)(p-2)....
     
  8. May 17, 2015 #7
    Can you just write the series as [tex](1+2x)^{\frac{1}{2}} = \sum_{n=0}^\infty \frac{P(n)}{n!}(2x)^n[/tex]
     
  9. May 17, 2015 #8
  10. May 17, 2015 #9
    [itex](1+2x)^{\frac{1}{2}} = \sum_{n=0}^\infty \frac{(P)_{\frac{1}{2}}}{n!}(2x)^n[/itex]? Does the index represent the power?

    Anyhow, if ##|t|<1##, this is around point ##c=0##. If I substitute back ##|x-\frac{1}{2}|<1## would this be around point ##c=1##?
    If so then [itex]\begin{cases}x>-\frac{1}{2}, &-1< x-\frac{1}{2}<0\\x<\frac{3}{2}, &0\leq x-\frac{1}{2}<1 \end{cases}[/itex]
     
    Last edited: May 17, 2015
  11. May 17, 2015 #10
    In your case p = 1/2 so (1/2)n = ((-1)^(n+1))*(2n-1)!!/2^n for n>1 when n = 0 (1/2)n = 1, i think you can expand it now,1+ Σ[((-1)^n+1)*(2n-1)!!*x^n]/n!
     
  12. May 17, 2015 #11
    Read wha you've written !
    Can absolute value be less than zero ?
    Edit, if you have evaluated (p)n you'd find that -1<=x<=1 because the 2^n goes by the falling factorial, so -1/2 <=x<=1
     
    Last edited: May 17, 2015
  13. May 17, 2015 #12
    I am attempting to read what you have written :D (highly recommend TeXing mathematical text, I put more emphasis in deciphering your post compared to actually focusing on its meaning in the context of the current problem)
    EDIT: nonsensical
     
    Last edited: May 17, 2015
  14. May 17, 2015 #13

    Ray Vickson

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    Unless I misunderstood your original question, I thought you were asked to find the expansion of ##f(x) = \sqrt{2x+1}## about the point ##x = 1##, which means you want a series involving ##(x-1)^n##, not ##x^n##. That is the reason I suggested writing ##x = 1 + t##, so ##f(x) = \sqrt{3+2t}##. You want the expansion of that in ##t^n## terms.
     
  15. May 17, 2015 #14
    Right you are, I am getting side-tracked and confused :S.
    If ##x = t+1## then [tex]\sqrt{3}(1+\frac{2}{3}t)^{\frac{1}{2}} = 1 + \left [\frac{1}{2}\left (\frac{2}{3}t\right )\right ] + \left [\frac{1}{2!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (\frac{2}{3}t\right )^2\right ] + \left [\frac{1}{3!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (-\frac{3}{2}\right )\left (\frac{2}{3}t\right )^3\right ] + \left [\frac{1}{4!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (-\frac{3}{2}\right )\left (-\frac{5}{2}\right )\left (\frac{2}{3}t\right )^4\right ][/tex]

    However, to come up with the general term. I understand I can express this in the form ##1 + \sum_{n=1}^\infty(-1)^{n+1} \frac{1}{n!} \left (\frac{2}{3}\right )^nt^n##. Problem is what comes in between there? The falling factorial seems viable, but it isn't mentioned in my textbook, I suspect I am expected to solve this problem differently.
     
    Last edited: May 17, 2015
  16. May 17, 2015 #15

    Ray Vickson

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    Are you not allowed to go to the library and look things up, learning about and using tools not in your textbook? Back when I was attending classes that would have been praiseworthy behavior, unless specifically forbidden for some reason----and that never happened. The only requirement was to cite sources. Of course, nowadays going on-line replaces a walk to the library.

    Alternatively, you could just go ahead and compute ##\left. d^n f(x)/dx^n \right|_{x=1}## to get the Taylor expansion coefficients directly.
     
  17. May 17, 2015 #16

    vela

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    The second form you wrote isn't correct. Perhaps that's why it's not making sense to you.

    Forget about the falling factorial. Just look for patterns and simplify. For example, how many powers of 2 do you get in each term in the denominator? How many in the numerator? Do they follow a predictable pattern so you can cancel them out? What about the sign of each term?
     
  18. May 17, 2015 #17
    Yes, the falling factorial is forgotten. This is what I have come up with.

    [tex]1 + \frac{1}{1!}\frac{1}{3^1}t^1 -\frac{1}{2!}\frac{1}{3^2}t^2 + \frac{1}{3!}\frac{3}{3^3}t^3 - \frac{1}{4!}\frac{3\cdot 5}{3^4}t^4 + \frac{1}{5!}\frac{3\cdot 5\cdot 7}{3^5}t^5[/tex]
    I know what the pattern is, but I don't know how to express the ##3\cdot 5\cdot 7\cdot 9 ...## as a general term.
     
    Last edited: May 17, 2015
  19. May 17, 2015 #18

    vela

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    Notation-wise, that's sometimes written using the double-factorial, e.g., ##7!! = 1\cdot 3\cdot 5\cdot 7##. If you define (-1)!! = 1, you can write your series as
    $$1 + \sum_{k=1}^\infty (-1)^{k+1}\frac{(2k-3)!!}{k!3^k} t^k.$$ That's just notation though.

    Another way, which is perhaps more useful to know, is to use the fact that, for example,
    $$ 1\cdot 3\cdot 5\cdot 7 = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{2 \cdot 4 \cdot 6} = \frac{7!}{2^3 3!}$$
     
  20. May 17, 2015 #19
    Yes, I was experimenting around with that, too. I would like to avoid unnecessarely adding new concepts to the solution. Thank you for pointing me in the right direction.
     
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