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Math Help - Help Fast on Series/Sigma problem!!!

  1. #1
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    Help Fast on Series/Sigma problem!!!

    the question is:

    Show that if

    (a(sub 0)/1) + (a(sub 1)/2) +... (a(sub n)/n+1) = 0

     a_0 /1 + a_1/2 +a_n/(n+1) = 0

    then

    a(sub 0) + a(sub 1)*x +... a(sub n)*(x^n) = 0

      a_0 + x a_1 +.... x^n a_n = 0

    for some x in the interval [0,1]


    I have not leaned anything about infinite series, so the solution doesnt pertain to that. I am guessing it has something to do with Sigma notation, and i can transfer the given equations to Sigma notation, but i have no idea where to go from there.

    This is my guess at what it is saying in Sigma Notation, but I am not very good with it so I may be wrong

    show that if

     \sum_{k=1}^{n}\,\,\frac{a_k}{k+1} =0

    then

     \sum_{k=1}^{n} a_k x^k =0

    for some x in [0,1]

    I need help ASAP! thank you
    Last edited by thunderhockey48; April 30th 2009 at 08:39 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by thunderhockey48 View Post
    the question is:

    Show that if

    (a(sub 0)/1) + (a(sub 1)/2) +... (a(sub n)/n+1) = 0

     a_0 /1 + a_1/2 +a_n/(n+1) = 0

    then

    a(sub 0) + a(sub 1)*x +... a(sub n)*(x^n) = 0

     a_0 + x a_1 +.... x^n a_n = 0

    for some x in the interval [0,1]


    I have not leaned anything about infinite series, so the solution doesnt pertain to that. I am guessing it has something to do with Sigma notation, and i can transfer the given equations to Sigma notation, but i have no idea where to go from there.

    This is my guess at what it is saying in Sigma Notation, but I am not very good with it so I may be wrong

    show that if

     \sum_{k=1}^{n}\,\,\frac{a_k}{k+1} =0

    then

     \sum_{k=1}^{n} a_k x^k =0

    for some x in [0,1]

    I need help ASAP! thank you
    Let  f(x)=\sum_{k=1}^{n} a_k x^k

    Then  \int_0^1 f(x)dx =\sum_{k=1}^{n} a_k \int_0^1x^kdx=\sum_{k=1}^{n}\,\,\frac{a_k}{k+1} =0

    Hence, if a function has zero 'area' it must be zero for at least one value in that interval.
    Last edited by matheagle; April 30th 2009 at 10:04 PM.
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  3. #3
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    how did you get  \int_0^1 f(x)dx =\sum_{k=1}^{n} a_k \int_0^1x^kdx ? then to get  \sum_{k=1}^{n} a_k \int_0^1x^kdx=\sum_{k=1}^{n}\,\,\frac{a_k}{k+1} did you just integrate  \int_0^1x^kdx to get  x^{k+1} / k+1 , then you just multiplied  \sum_{k=1}^{n} a_k , and since  \sum_{k=1}^{n}\,\,\frac{a_k}{k+1} =0 then all of it =0? I can kinda follow how you got it, but not completely... thanks for the immediate reply!
    Last edited by thunderhockey48; April 30th 2009 at 09:47 PM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Look at what I called f(x) and pass the integral through the FINITE sum
    Next integrate AND plug in the value 0 and 1 for x.
    This is a definite integral.

    Ignore the summation...

    Let  f(x)= a_1x+a_2x^2+\cdots +a_nx^n

    Then  \int_0^1 f(x)dx = a_1 \int_0^1xdx +a_2 \int_0^1x^2 dx +\cdots +a_n \int_0^1x^ndx

    Integrate each one and plug in the 0 and the 1.

    You will get what I stated earlier tonight, that...

    Then  \int_0^1 f(x)dx =\sum_{k=1}^{n}\,\,\frac{a_k}{k+1} =0

    Hence, if a continuous function has zero 'area' it must be zero for at least one value in that interval.
    (This can be proved via contradiction. If a function doesn't have a zero, then it's strictly postive or strictly negative.
    In those two case the 'area' will be postive or negative, BUT NOT zero.)
    Last edited by matheagle; April 30th 2009 at 10:08 PM.
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  5. #5
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    Oh, thank you so much! I understand it now. I didn't realize that you could pass the summation through the integral, but your proof below makes it clear. I was forgetting that it was a definite integral. You are a lifesaver!
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by thunderhockey48 View Post
    Oh, thank you so much! I understand it now. I didn't realize that you could pass the summation through the integral, but your proof below makes it clear. I was forgetting that it was a definite integral. You are a lifesaver!
    You can switch the order since this is a finite summation.
    IF it was an infinite summation then you need to worry, because you're switching the order of two limits.
    One limit is your infinite summation, the other is the integral.
    But here you have just one limit, your integral.
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  7. #7
    Super Member redsoxfan325's Avatar
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    Did you get this problem out of Rudin's "Principles of Mathematical Analysis" (Chapter 5, Question 4)?

    Also, it's not a big deal, but I think your summations should be going from 0 to n and not 1 to n.
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  8. #8
    MHF Contributor matheagle's Avatar
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    I ignored the first stuff and used his/her TeX.
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