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Taylor's Series

  1. Sep 19, 2007 #1
    Taylor's theorem states that, for any reasonable function f(x), the value of f at a point (x+[tex]\delta[/tex]) can be expressed as an infinite series involving f and its derivatives at the point of x:

    f(x+[tex]\delta[/tex]) = f(x) + f '(x)[tex]\delta[/tex] + [tex]\frac{1}{2!}[/tex]f ''(x)[tex]\delta[/tex][tex]^{2}[/tex] + ....

    where the primes denote successive derivatives of f(x). (Depending on the function this series may converge for any increment [tex]\delta[/tex] or only for values of [tex]\delta[/tex] less then some nonzero 'radius of convergence'.) This theorem is enormously useful, especially for small values of [tex]\delta[/tex], when the first one or two terms of the series are often an excellent approximation. Find the taylor's series for ln(1+[tex]\delta[/tex]). Do the same for cos [tex]\delta[/tex].


    We just started Taylor's theorems now, and it seems like there are many of them. Why exactly are they good and what purpose do they serve?

    Is finding the Taylor's series for ln (1+[tex]\delta[/tex]) a matter of substituting it into the above equation or is there more to it?

    Thanks again!
     
    Last edited: Sep 19, 2007
  2. jcsd
  3. Sep 19, 2007 #2

    learningphysics

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    In lots of problems, you may have to find f(x+delta) - f(x)... where delta is very small... your calculator might only give you a rounded figure of 0 because the numbers are too close together...

    However if you take the taylor series of f(x+delta)... then subtract f(x)... you get: f '(x)delta + f ''(x)delta^2. This will give a good approximation of f(x+delta) - f(x).

    That's just one example of why taylor series are useful. I'm sure there are many more.

    Yeah, just substitute into the equation, using x = 1. That should work.
     
  4. Sep 19, 2007 #3
    Effectively solving for delta?
     
  5. Sep 19, 2007 #4

    learningphysics

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    I don't understand what you mean.
     
  6. Sep 19, 2007 #5
    When everything is subbed in, am I solving for delta?
     
  7. Sep 19, 2007 #6

    learningphysics

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    No... you'll just be getting a polynomial formula for ln(1+delta)... in terms of delta...
     
  8. Sep 19, 2007 #7
    If I'm subbing x=1, my ln (1+delta) goes where in the formula?
     
  9. Sep 19, 2007 #8

    learningphysics

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    ln (x + delta) is the left hand side... right hand side is f(x) + f'(x)delta... where f(x) = ln(x)

    ln(x + delta)

    ln(x+delta) = ln(x) + (1/x)delta + ....

    So ln(1+delta) = ln(1) + (1/1)delta + ....
     
  10. Sep 19, 2007 #9
    So ln(1+delta) = ln(1) + (1/1)delta + ....

    is the bolded delta in the denominator?
     
  11. Sep 19, 2007 #10

    learningphysics

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    No. I'm using the formula from your original post exactly... delta doesn't go in the denominator...

    I got the 1/1.... because the deriative of lnx = 1/x... so f'(1) = 1/1
     
  12. Sep 19, 2007 #11
    Ok, just making sure I was reading it right.

    Is that as far as you can go? Since the 2nd derivative would be 0?
     
  13. Sep 19, 2007 #12

    learningphysics

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    How do you get 0?

    f(x) = ln(x)

    f'(x) = 1/x

    f''(x) = -1/x^2

    So f''(1) = -1 not zero...
     
  14. Sep 19, 2007 #13
    Oh, sub afterwards.
    I was deriving 1/1.
     
  15. Sep 19, 2007 #14
    I assume I don't need to do all that many though?
     
  16. Sep 19, 2007 #15
    Is this far enough to do the problem?

    ln (x+[tex]\delta[/tex]) = ln(x) + [tex]\frac{1}{x}[/tex][tex]\delta[/tex] + ([tex]\frac{1}{2!}[/tex])[tex]\frac{1}{x^{2}}[/tex]([tex]\delta^{2}[/tex])
     
  17. Sep 19, 2007 #16

    learningphysics

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    Actually I think the question wants you to keep going... and find a general formula for the entire series... do you see a pattern to the terms of ln(1+delta) ?
     
  18. Sep 19, 2007 #17
    denominator and delta are increasing by powers of 1. along with that 1/2!, and its denominator increasing..
     
  19. Sep 19, 2007 #18

    learningphysics

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    careful... it needs to be (-1/x^2) for the third term...
     
  20. Sep 19, 2007 #19
    ... i cant say i see a pattern
     
  21. Sep 19, 2007 #20

    learningphysics

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    yes, and also, you should have alternating signs...

    f''(x) = -1/x^2

    f'''(x) = 2/x^3

    I recommend finding the first 4 or 5 terms of the series... then you should see the pattern...
     
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