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B Taylor series

  1. Jun 5, 2017 #1
    Hello! Can someone explain to me, in real analysis, what is the difference in expanding a function as a Taylor series around 2 different point. So we have ##f(x)=\sum c_k (z-z_1)^k = \sum d_k (z-z_2)^k## and as ##k \to \infty## the series equals f in both cases, but why would one choose a point in the favor of the other?
     
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  3. Jun 5, 2017 #2

    PeroK

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    If you are interested in the function at ##z_1## you choose ##z_1## and if you are interested in the function at ##z_2## you choose ##z_2##.
     
  4. Jun 5, 2017 #3

    FactChecker

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    There are a few reasons to expand around different points. One is as @PeroK says, that you are interested in a region around a different point. Another is if you have information about all the derivatives around a different point. A third is that the rate of convergence of the Taylor series may be better at a different point. A fourth is that the expansion around another point may have a circle of convergence that is not included in the region of convergence of the first point. (The fourth reason may be considered related to the third reason.)
     
  5. Jun 5, 2017 #4

    mathwonk

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    to illustrate one of the previous answers, suppose you are interested in a taylor expansion of the real logarithm function ln(x). it is defined for all positive x, but if you expand around the center a>0, it will only converge on the interval of radius a about that center, i.e. on (0,2a). so if you are interested in the values at some larger number than 2a you need to re expand at a larger center.
     
    Last edited by a moderator: Jun 5, 2017
  6. Jun 5, 2017 #5
    it can happen that this formula does not make sense. For example
    $$f(z)=\frac{1}{z^2+1}+\frac{1}{(z-1)^2+1}$$ and ##z_1=-10,\quad z_2=10##
     
    Last edited: Jun 5, 2017
  7. Jun 5, 2017 #6

    FactChecker

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    Good point. A simpler example is 1/z at z1=-1 and z2 = +1 since there is no value of z where both series converge.
     
  8. Jun 5, 2017 #7
    I wanted the function f to be defined on ##\mathbb{R}## :)
     
  9. Jun 5, 2017 #8

    FactChecker

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    Oh. I missed that point. Good example.
     
  10. Jun 6, 2017 #9
    Thank you for your reply! However I am still a bit confused. If I have let's say ##f(x)=e^x## and I want to calculate ##e^7##, I am interested in the function at ##z=7##, not at ##z_1## or ##z_2##. My question is, what information do ##z_1## and ##z_2## give me?
     
  11. Jun 6, 2017 #10

    FactChecker

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    The Taylor series expansion at each point z1 and z2 has a radius of convergence which may or may not include the point z=7. Even if both converge at z=7, they may have different rates of convergence there. In the case of the exponential function, all Taylor series converge at z=7, so it is not the best example. Still, I think it would be obvious that an expansion at z1 = 6 will converge faster at z=7 than an expansion at z2=0.
    Consider the example that @zwierz gave. The expansion at z1=-10 will not converge at z=7 but the expansion at z2=+10 will.
     
  12. Jun 6, 2017 #11

    mathwonk

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    e^x of course will converge everywhere, no matter what center of expansion is chosen. But the best center to choose for computation is one for which you actually know the coefficients, namely x=0. Expanding e^x about the point x=6 would require first making an infinite number of coefficient calculations, each comparable to the one computation you want.
     
  13. Jun 6, 2017 #12

    FactChecker

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    Right. But I was trying to make a general point that applied to more than the exponential function. Also, you can expand the series once and, even if you need to calculate a lot of coefficients, it could give you a tool to use for general values of z.
     
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