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Homework Help: Taylors Remainder Confusion

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Use Taylor's remainder formula to show that the Taylor series for f(x) is about the point indicated converges to f(x) for all x.

    [tex]f(x) = e^{5x}[/tex] about [tex]x=0[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Since,

    [tex]f^{n}(x) = 5^{n}e^{5x}[/tex],

    Taylor's remainder formula for [tex]e^{5x}[/tex] and c = 0 gives

    [tex]e^{5x} = 1 + 5x + \frac{5^{2}}{2!}x^{2} + \frac{5^{3}}{3!}x^{3} + ... + \frac{5^{n}}{n!}x^{n} + R_{n}[/tex]

    Where [tex]R_{n} = \frac{5^{n+1}e^{5z_{n}}}{(n+1)!}x^{n+1}[/tex]

    There are two cases we must consider:

    Case 1: If x < 0 then,

    ****NOTE: The following line is the line I am confused about,****

    [tex]x < z_{n} < 0,[/tex] and [tex]|R_{n}| < 5^{n+1}\frac{|x|^{n+1}}{(n+1)!}[/tex]

    How do they go about making this conclusion? Do they work from the this in the following manner?

    [tex]x < z_{n} < 0[/tex]

    [tex]5x < 5z_{n} < 0[/tex]

    [tex]e^{5x} < e^{5z_{n}} < e^{0}[/tex]

    [tex]e^{5x}\frac{|x|^{n+1}}{(n+1)!} < e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!} < e^{0}\frac{|x|^{n+1}}{(n+1)!}[/tex]

    And then take the limit as n goes to infinity on the outside and use squeeze theorem to prove the rest?

    Where are they getting the second part of this line:

    Also, they are using squeeze theorem around [tex]lim_{n \rightarrow \infty} |R_{n}|[/tex], how come they are only showing one function around [tex]|R_{n}|[/tex]. (i.e. [tex]5^{n+1}\frac{|x|^{n+1}}{(n+1)!}[/tex]) Where is the 0 term?

    Once I understand this I will continue and finish off the problem, this is the only line that confuses me.

    Can someone please clarify?
     
    Last edited: Sep 20, 2010
  2. jcsd
  3. Sep 20, 2010 #2

    Dick

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    It's because
    [tex]
    0 \leq e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!} \leq e^{0}\frac{|x|^{n+1}}{(n+1)!}
    [/tex]
    The maximum value of e^(5*z_n) for z_n in the interval [x,0] is at z_n=0. Where e^(5*z_n)=e^0=1. That's the largest value R_n could possibly have.
     
  4. Sep 20, 2010 #3
    Where did they get the less than or equal to signs? I don't understand where they're coming from.

    Also, how do you know that,

    [tex]e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!}[/tex] is indeed larger or equal to 0?

    Is it because the minimum value for e^(5*z_n) for z_n in the interval [x,0] is at z_n = x. Where e^(5*z_n) = e^(5x) which is always less than zero because x < 0.
     
    Last edited: Sep 20, 2010
  5. Sep 20, 2010 #4

    Dick

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    I put the less than or equal signs in. That part isn't very important. The point is just that (as you said) z_n is between x and 0. So 0<=exp(5*z_n)<=1. So |R_n| is bounded below by zero and above by the expression replacing exp(5*z_n) with 1. And yes, again. The value of the upper limit goes to 0 as n->infinity.
     
  6. Sep 20, 2010 #5

    Dick

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    You want a bound for |R_n|. An absolute value is always >=0. And z_n is in [x,0]. And exp(z_n) is never negative. Are you trying to make this confusing?
     
  7. Sep 20, 2010 #6
    No I'm not trying to make this confusing, I'm trying to understand it. Didn't you get things wrong when you were learning?

    My concern is that I've been doing problems and I always have trouble proving that,

    [tex]lim_{n \rightarrow \infty} R_{n} = 0[/tex]

    In order to do this I need to make use of the squeeze theorem, and in order for me to use squeeze theorem I have to be able to assign the correct boundaries around [tex] R_{n} [/tex] for a given range of x. (i.e. x < 0)

    I'm trying to understand this so I can consistently show that either [tex]lim_{n \rightarrow \infty} R_{n} = 0[/tex] or not.
     
  8. Sep 20, 2010 #7

    Dick

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    Sure, I got things wrong when learning. But this seems to be getting more confusing. If you can show |R_n|->0, then you have automatically shown R_n->0. You don't need a 'correct lower bound' for |R_n|. 0 will work just fine. Worry about the upper bound. Show that goes to 0.
     
  9. Sep 21, 2010 #8
    Okay I'm going to give this another shot.

    Case 1: x < 0

    So, [tex] 0 > z_{n} > x[/tex]

    Since,

    [tex]|R_{n}| = 5^{n+1}e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!}[/tex]

    We can see that in the range [tex]0 > z_{n} > x[/tex] the largest value we can obtain for [tex]e^{5z_{n}}[/tex] will be obtained when we set [tex]z_{n} = 0[/tex].

    We can see that,

    [tex] 0 \leq |R_{n}| \leq 5^{n+1} \frac{|x|^{n+1}}{(n+1)!} [/tex]

    [tex] 0 \leq 5^{n+1}e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!} \leq 5^{n+1} \frac{|x|^{n+1}}{(n+1)!} [/tex]

    Dividing all terms by [tex]5^{n+1}[/tex] ***NOTE: This step I'm not too sure about. I need to get the [tex]5^{n+1}[/tex] terms out so when I take the limit of,

    [tex]5^{n+1} \frac{|x|^{n+1}}{(n+1)!}[/tex] as [tex]n \rightarrow \infty[/tex]

    I will get 0. With the [tex]5^{n+1}[/tex] term there I'll get [tex]\infty \cdot 0[/tex]

    EDIT: I am able to simply state that,

    [tex]lim_{n \rightarrow \infty}5^{n+1} \frac{|x|^{n+1}}{(n+1)!} = 0[/tex] ???

    How is this justified? (This is what the given solution states)

    Plowing on...

    [tex] 0 \leq e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!} \leq \frac{|x|^{n+1}}{(n+1)!} [/tex]


    Applying squeeze theorem,

    EDIT: Now that I think about it, since I removed the, [tex]5^{n+1}[/tex] term it isn't really, [tex]|R_{n}|[/tex] anymore because that term's not there, right? So how do we deal with this?

    [tex]lim_{n \rightarrow \infty} 0 \leq lim_{n \rightarrow \infty} |R_{n}| \leq lim_{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+1)!} }[/tex]

    We can see that,

    [tex]lim_{n \rightarrow \infty} 0 = lim_{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+1)!} } = 0[/tex]

    Therefore we can conclude that,

    [tex]lim_{n \rightarrow \infty} |R_{n}| = 0[/tex]

    So,

    [tex]lim_{n \rightarrow \infty} R_{n} = 0[/tex] for all x < 0.

    How does this look? Again, there's that one annoying [tex]5^{n+1}[/tex] I'm not sure how to get rid of in there, but other than that everything seems to make sense. After I figure that part out I'll move on to case 2: x > 0.
     
    Last edited: Sep 21, 2010
  10. Sep 21, 2010 #9

    Dick

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    DON'T divide by 5^(n+1). You've got
    [tex]
    0 \leq |R_{n}| \leq 5^{n+1} \frac{|x|^{n+1}}{(n+1)!}
    [/tex]
    NOW do the squeeze by showing lim n->infinity |5*x|^(n+1)/(n+1)! is zero. If you can show the limit of |x|^(n+1)/(n+1)! zero you can surely show the limit of |5*x|^(n+1)/(n+1)! is zero.
     
  11. Sep 21, 2010 #10
    So my task it to show that,

    [tex]lim_{n \rightarrow \infty}\frac{|5x|^{n+1}}{(n+1)!} = 0[/tex]

    Well we can do the following,

    Let [tex]y = 5x[/tex].

    Now surely it's trivial to see that,

    [tex]lim_{n \rightarrow \infty}\frac{|y|^{n+1}}{(n+1)!} = 0[/tex]

    In fact, now that I think about it, it was trivial to see that simply,

    [tex]lim_{n \rightarrow \infty}\frac{|5x|^{n+1}}{(n+1)!} = 0[/tex]

    Is true, no?

    I want to make sure what I'm doing is justified and I'm not just assuming the limit as n goes to infinity is 0.

    Is this correct, do I need to show anything more? (As far as this limit is concerned)
     
  12. Sep 21, 2010 #11

    Dick

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    Unless you've already proved it in the course, it probably wouldn't hurt to say why you think lim n->infinity |y|^(n+1)/(n+1)!=0 for all y. You seem pretty confident. And it's certainly true. Why do you think so?
     
  13. Sep 21, 2010 #12
    We did prove it in class. Our professor said we can simply use it as a tool at our disposale without justification.

    So I should be okay stating,

    [tex]lim_{n \rightarrow \infty}\frac{|5x|^{n+1}}{(n+1)!} = 0[/tex]

    for all x < 0.

    If that's okay I'm going to move on to case 2: x > 0 in my next post.
     
  14. Sep 21, 2010 #13

    Dick

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    Why wouldn't it be ok? If it's true for all y, then it's true for y=5x. Good idea to move on.
     
  15. Sep 21, 2010 #14
    Case 2: x > 0

    So,

    [tex]0 < z_{n} < x[/tex]

    Since,

    [tex]|R_{n}| = 5^{n+1}e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!}[/tex]

    We can see that in the range [tex]0 < z_{n} < x[/tex] the largest value we can obtain for [tex]e^{5z_{n}}[/tex] is when [tex]z_{n} = x[/tex].

    So we can see that,

    [tex] 0 \leq |R_{n}| \leq 5^{n+1} e^{5x} \frac{|x|^{n+1}}{(n+1)!} [/tex]

    Applying squeeze theorem,

    [tex]lim_{n \rightarrow \infty} 0 \leq lim_{n \rightarrow \infty} |R_{n}| \leq e^{5x} lim_{n \rightarrow \infty} \frac{|5x|^{n+1}}{(n+1)!} }[/tex]

    Since,

    [tex]lim_{n \rightarrow \infty} 0 = e^{5x} lim_{n \rightarrow \infty} \frac{|5x|^{n+1}}{(n+1)!} } = 0[/tex]

    Therefore,

    [tex]lim_{n \rightarrow \infty} |R_{n}| = 0[/tex]

    and it follows that,

    [tex]lim_{n \rightarrow \infty} R_{n} = 0[/tex] [tex]\forall x[/tex]

    Since we've examined both cases.

    How's this look?
     
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