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Mac Series f(x)=sinh (x), remainder theorem help

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    (a)Use Definition 10.8.1 to find the Maclaurin series for f(x) = sinh x. Express your answer using Σ notation.

    (b) Find the interval of convergence for the series found in part (a).

    (c) Use the Remainder Theorems 10.7.4 and 10.9.2 to show that the series found in part (a) converges to f(x) = sinh x on the interval of convergence found in part (b).


    3. The attempt at a solution

    So I got a) and b) I'm sure, but I'm having trouble with c).

    For (a) I got [tex]\sum\frac{x^{2k+1}}{(2k+1)!}[/tex] 0-->inf

    for (b) I used the abs ratio test to find the limit = 0 therefore the interval is [tex](-\infty,+\infty)[/tex]

    10.7 is mac/taylor polys
    10.8 is mac/taylor series (power series)
    10.9 is convergence of taylor series.
    The remainder theorm of 10.7 states that "If the function f can be differentiated n+1 times on an interval I containing the number Xo, and if M is an upper bound for [tex]\left|f^{(n+1)}(x)\right|[/tex] on I,
    that is, [tex]\left|f^{(n+1)}(x)\right|\leq M[/tex]
    for all x in I, then [tex]\left|R_{n}(x)\right|\leq\frac{M}{(n+1)!}\left|x-x_{0}\right|^{n+1}[/tex]

    But I don't understand how to use that theorem or 10.9.2 that says The equality: [tex]f(x)=\sum\frac{f^{(k)}(x_{0}}{k!}(x-x_{0})^k[/tex]

    holds at a point x if and only if lim of Rn(x)=0 as n approaches +inf.
     
    Last edited: Nov 29, 2009
  2. jcsd
  3. Nov 29, 2009 #2

    LCKurtz

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    Suppose you are given any x. Consider an interval [-c,c] large enough to contain x. For f(x) = sinh(x) you obviously know all the derivatives are either sinh(x) or cosh(x).

    1. What is a bound M for your |f(n+1)(x)| derivative on that interval?
    2. Can you show |Rn(x)| --> 0 as n --> infinity. Remember, you have a bound on x too.

    If the remainder goes to zero, the series equals the function.
     
    Last edited: Nov 29, 2009
  4. Nov 29, 2009 #3
    so could I choose x=1 with an interval of (0,2)?

    Well, is there an upper bound? as x---> inf, so does sinh(x) and cosh(x)
     
  5. Nov 29, 2009 #4

    LCKurtz

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    You don't get to choose x. You have to show that for each x, Rn(x) --> 0. It is n that is going to infinity, not x. Read my post a little more carefully and see if you can answer the questions I asked.
     
  6. Nov 29, 2009 #5
    I still don't see how the function can have a bound on an infinite interval when the function goes to infinite with the interval.

    And as n--> +inf, the derivative is either cosh(x) or sinh(x)

    After looking back into my notes, I see Rn(x) = f(x)-Pn(x)

    I'm really confused on what this is saying.

    Am I supposed to be finding this remainder? I still don't understand M..
     
  7. Nov 29, 2009 #6

    LCKurtz

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    So if you could show the left side of that goes to zero as n --> infinity that would mean the right side does too:

    [tex]0 = \lim_{n\rightarrow \infty}(P_n(x) - f(x) = \lim_{n\rightarrow \infty}\sum_{k=0}^{n}\frac{x^{2k+1}}{(2k+1)!}-f(x)=\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)!}-f(x)[/tex]
    That means the series converges to the function (equals the function)
    That is why you are trying to show the remainder goes to zero. You still need to address my first post.
     
  8. Nov 30, 2009 #7
    So M would be c?

    Rn(x) = f(x)-Pn(x) I can see how Rn goes to zero and the right side does too as n ---> inf because Pn gets more and more accurate.

    why wouldn't the remainder go to zero? or is this a tool to confirm our own work?

    Sorry, it still hasn't clicked as to how I should do this.

    Should I set M to inf? It didn't give me an x.
     
  9. Nov 30, 2009 #8

    LCKurtz

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    Why would M be c? In your original post you stated what M is in general. So what is M for your specific problem?
    You have it just backwards. Showing that the remainder goes to zero is how we show the series converges to the function. It is possible to have a MacLaurin series not equal the function on any interval.
    I give you the x. You have to show that no matter what x I give you, you can show the remainder goes to zero for that x as n-->infinity. You have an n! in the denominator and a power of n in the numerator. If you can find a bound for how large everything else is maybe you can show Rn(x) goes to zero for that x. Start by figuring out an M that works as an upper bound on the interval.
     
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