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Help with Taylor Series problem

  1. Jul 5, 2011 #1
    Now although this is a problem for my EE course, it is more of a calculus question so I figured I would receive the best answers by posting it in this section. I have just started on the problem but could use some input on my thoughts. So here we go (there are two parts):
    (problem screenshot is attatched)

    1. The problem statement, all variables and given/known data

    I'm given the function f(x) = e^ax.

    a) Derive an expression for the first three nonzero terms of the Taylor Series for f(x) about x=0 (Maclaurin Series).

    b) Complete the following table assuming that a=0.1. The function f(x) is the exact function, the function fts2(x) uses the first two nonzero terms of the Taylor Series, and the function fts3(x) uses the first three nonzero terms of the Taylor Series.


    2. Relevant equations



    3. The attempt at a solution

    a) Now I know (correct me if wrong) f(x) = f(0) + f ' (0) + ((f ''(0)/2!) x^2) + .... But am not sure if that's the correct answer to the question? Or do I need to take this further?

    b) There is then a table with x=0, 0.01, 0.1, 0.2, 0.5, 1.0, 2.0, 5.0, & 10. Do I plug these values into the first three nonzero terms found in a? For example, first f(0), then f(0) + f ' (0), then f(0) + f ' (0) + ((f ''(0)/2!) x^2)?

    Any help/input would be greatly appreciated!

    Thank you
     

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    Last edited: Jul 5, 2011
  2. jcsd
  3. Jul 5, 2011 #2

    berkeman

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    Thread closed temporarily for review...
     
  4. Jul 6, 2011 #3

    berkeman

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    Thread re-opened. Thanks for your patience.
     
  5. Jul 6, 2011 #4
    Thank you! I'll continue to work on the problem and post any progress, if any. Was more looking for if my beginning steps were correct so I could base my work off of that rather than learning later on that the whole problem is incorrect. Any input would be greatly appreciated!
     
  6. Jul 6, 2011 #5

    SammyS

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    f(x) = f(0) + f ' (0) x + ((f ''(0)/2!) x^2) + ...

    You left out the x in the second term, the one with the first derivative.
     
  7. Jul 6, 2011 #6

    vela

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    In part (a), you're supposed to explicitly evaluate those derivatives to get the first few terms of the series.
     
  8. Jul 6, 2011 #7
    Thank you! My mistake, must of missed that.

    So correct me if I'm wrong, what you're saying is that f(0) = e^ax, f ' (0) = the first derivative, and f '' (0) = the second derivative, and that is what I need to do in order to solve for part a?
     
  9. Jul 6, 2011 #8

    vela

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    Yup. Find the derivatives, evaluate them at x=0, and plug those numbers into the formula.
     
  10. Jul 6, 2011 #9

    SammyS

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    f(0) = ea(0) = 1 , etc.

    f ' (0) = ?
     
  11. Jul 7, 2011 #10
    f ' (0) = a
    f '' (0) = a2??
     
  12. Jul 7, 2011 #11

    vela

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    It would help if you'd show your work if you're unsure about a calculation.
     
  13. Jul 7, 2011 #12
    f(0) = ea(0) = e0 = 1
    f ' (0) = a*ea(0) = a*1 = a
    f '' (0) = a2*ea(0) = a2*1 = a2

    Does this help? This is the derivatives and breaking down to first, second, and third nonzero terms
     
  14. Jul 7, 2011 #13

    vela

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    Looks good.
     
  15. Jul 7, 2011 #14
    So then I would simply evaluate f(x) = 1 + a + a2 for the first [f(x) = 1], second [f ' (x) = 1 + a], and third [f '' (x) = 1 + a + a2] non zero terms with the values x=0, 0.01, 0.1, 0.2, 0.5, 1.0, 2.0, 5.0, & 10?
     
  16. Jul 7, 2011 #15

    vela

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    No. I have no idea where you came up with those expressions. You seem to be pushing symbols around and ignoring basic algebra. This is largely a plug-and-chug problem.

    You have a function f(x)=eax. If you want to evaluate f(anything), you plug anything into it to get ea(anything). If you want f'(anything), you differentiate eax and then plug anything into the result. That's what you did so far to find f(0)=1, f'(0)=a, and f''(0)=a2.

    So why did you do this? It's because f(0), f'(0), and f''(0) appear in the expression for the Maclaurin series for f(x)
    [tex]f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots[/tex]
    So now plug your previous results into that expression to get
    [tex]1+ax+\frac{(ax)^2}{2!}+\cdots[/tex]
    If this infinite series converges (which it will for all x), it will converge to the value of f(x), so you can write
    [tex]f(x) = e^{ax} = 1+ax+\frac{(ax)^2}{2!}+\cdots[/tex]
    In part (b), you're told
    \begin{align*}
    f_\mathrm{TS2}(x) &= 1+ax \\
    f_\mathrm{TS3}(x) &= 1+ax+\frac{(ax)^2}{2!}
    \end{align*}
    You want to calculate f(x), fTS2(x), and fTS3(x) for various values of x with a=0.1.
     
  17. Jul 7, 2011 #16
    Thank you for the explanation, which helps me to realize I correctly did the other two taylor series problems I was assigned! I definitely understand what you're saying and just think I wasn't explaining my self very well at all with where I was going... I am going to stop working on homework for the day and pick up early in the morning before work. When finished I will scan and post my final solution for any input you may have. Thanks again!
     
  18. Jul 8, 2011 #17
    Just completed the problem and I feel it's correct. Can you see any errors or something that looks out of order?
     

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  19. Jul 8, 2011 #18

    vela

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    You wrote f(0)=a and f(0)=a2, which is wrong. Also, the problem asked for x=5 and x=10 in the table.
     
  20. Jul 8, 2011 #19
    Do you mean I wrote them wrong in my hand work on the left?? I only did that because the example he gave us had that for his problem, if that's what you meant. And also, thanks for catching the five and ten!! I based this excel template off my other two completed problems so I missed that :redface: thank you!!
     
  21. Jul 8, 2011 #20

    vela

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    What I mean is that f(0) isn't equal to a or a2. "f(0)" means you plug 0 into the function f(x), which was defined in this problem to be eax. It doesn't mean "plug 0 into the last function I happened to be thinking of."

    I (and the grader) can infer from what you wrote down that what you were probably thinking was correct, but what you wrote down doesn't accurately reflect what you were thinking. It's similar to having spelling and grammatical errors in an essay. Readers may be able to figure out what you meant, but they shouldn't have to put in the extra effort.
     
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