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Homework Help: Differential calculus questions

  1. May 26, 2005 #1
    Please help.

    For the function 1/(1-x) the Taylor polynomial of degree 3 about x=0 is:
    p(x) = 1 + x + x^2 + x^3

    a. Find an upper bound on l R(x) l if x = 0.5

    b. Write down the first three non-zero terms of the Taylor series for
    g(x) = 1/(4-x^2)

    Is there an easier way to do b other than finding the 2nd, 3rd, etc derivatives, because I'm getting completely confused with this?

    2. Let f(x) = x^5 - 20x + 5

    The real solutions are +/-sqrt2

    Find the maximum and minimum value of f(x) for 0 is less than or equal to x, which is less than or equal to 2.

    I just can't remember how to do this.

    Thank you
  2. jcsd
  3. May 26, 2005 #2
    b) Given is the taylor series for [itex] \frac{1}{1-x} [/itex]. Can you fit g(x) so that it is in a similar form?

    2. Do you recall the mean value theorem?
    It says that in a closed interval, a continuous curve f(x) takes on a max and min value either at its critical points or endpoints. You can find the critical points by solving f'(x) = 0, as these points represent where the curve changes direction (either at a high or low point).
  4. May 26, 2005 #3


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    1. The "error" in using a Taylor polynomial of order n is the same form as the next, n+1, term in the Taylor series with x replaced by some unknown point. Here, the next term is just x^4 so the error is |c^4| where c is between 0 and 0.5. What is the largest possible value of that?

    whozum, if you mean that the fact that a differentiable function takes on its extreme values on an interval either at the enpoints or in the interior where the derivative is 0 is result of the mean value theorem, you are right- but that is not the mean value theorem itself.
  5. May 26, 2005 #4
    Correct, it is one of the consequences of the mean value theorem.
  6. May 29, 2005 #5
    For 2, to find the max and min value, you simply solve for f'(x)=0, right? Then you get +/-sqrt2. Then you determine which is max and which is min. Then you sub both into f(x) to find the max and min values. But for the answer to this question ,the min value is sqrt2 and the max is 0. I don't see how that works out, because -sqrt2 gives a y value of 27.6, which is clearly larger than 5.

    For question 1.a., the answer is R(0.5) = 0.5/(1-c)^5. I don't get how they get the "1-c" part. When I get the 4th derivative, I get 24/(1-x)^5. Then you use f(4)(c)x^4/4! (the remainder term), then I end up with (x^4*(c))/((1-x)^5). So I don't know how that works out.
  7. May 29, 2005 #6
    2: Remember the domain you are working on. 0 < x < 2 is the interval whose max and min you are interested in.
  8. May 29, 2005 #7
    Then how do you know 0 is a maximum? I mean, you couldn't do trial and error. You said that a max or a min occurs etiher at its critical points and/or endpoints right? +/-sqrt2 are critical points, but sqrt2 is the only one in that domain. So either 0 or 2 (which I'm assuming are hte endpoints) must be a max. Is this right?
  9. May 29, 2005 #8
    0,2 and sqrt(2) are the only possibilities for max and min. Plug them into f(x) and see which ones biggest and smallest. The respective value will give you the max/min.
  10. May 29, 2005 #9
    Ok, I get you. Do you know 1.a. at all?
  11. May 29, 2005 #10
    Sorry, I'm not certain on how to do those.
  12. May 29, 2005 #11
    Thats ok...I think I worked it out anyway.

    Thanks for your help. :smile:
  13. May 29, 2005 #12


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    That's not quite the error. You'll need to also consider how the (n+1)st derivative behaves (see any of the usual remainder formulas). In this case you can also work out the remainder exactly, since it's just a geometric series and see it's outside what considering |c^4| would suggest.
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