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Application of Derivatives

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]p(x)\ =\ a_0\ +\ a_1x\ +\ a_2x^2\ +\ .......\ + a_nx^n[/itex].

    Now if [itex]|p(x)|\ <=\ |e^{x-1}\ -\ 1|[/itex] for all [itex]x\ >=\ 0[/itex] then

    Prove [itex]|a_1\ +\ 2a_2\ +\ .......\ + na_n|\ <=\ 1[/itex].

    2. Relevant Graph( [itex]|e^{x-1}\ -\ 1|[/itex] )
    w8qi9s.jpg

    3. The attempt at a solution
    From the graph we can conclude that
    p(x) should pass through (1,0)
    => [itex]a_1\ +\ a_2\ +\ .......\ + a_n\ =\ 0[/itex]

    Further, I'm not able to apply any other condition given to simplify the expression. Their is of course something to do with the derivative as I found this question in a book of differentiation.
    Any help would be highly appreciated.

    Thanks
     
    Last edited: Jul 10, 2014
  2. jcsd
  3. Jul 10, 2014 #2

    ehild

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    What is the derivative of p(x)?

    Have you copied the problem correctly?



    ehild
     
    Last edited: Jul 10, 2014
  4. Jul 10, 2014 #3
    Sorry, there isn't x in the proof
    I have corrected the problem.

    [itex]p'(x)\ =\ a_1\ +\ 2a_2x\ +\ 3a_3x^2\ +\ .....\ +\ na_nx^{n-1}[/itex]
     
    Last edited: Jul 10, 2014
  5. Jul 10, 2014 #4

    benorin

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    Perhaps it's [itex]p'(1) =\ a_1\ +\ 2a_2\ +\ 3a_3\ +\ .....\ +\ na_n[/itex]
     
  6. Jul 10, 2014 #5

    micromass

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    That is not the correct derivative of ##p(x)##. The derivative should have some variables ##x##.
     
  7. Jul 10, 2014 #6

    benorin

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    wrong. note that p passing through (1,0) means that [itex]p(1)=a_0+a_1+\cdots +a_n=0\Rightarrow a_0=-(a_1+\cdots +a_n)[/itex]
     
    Last edited: Jul 10, 2014
  8. Jul 10, 2014 #7
    right I messed with it. Additionally I don't know why I'm not able to edit the first post.

    I corrected it. Thanks.
     
  9. Jul 11, 2014 #8
    If I could just prove that the inequality of the funciton
    i.e. [itex]|p(x)|\ <=\ |e^{x-1}\ −\ 1|[/itex]
    also holds true for the derivative
    i.e. [itex](|p(x)|)'\ <=\ (|e^{x-1}\ −\ 1|)'[/itex]
    then I could just substitute x = 1 and prove the question.

    But I just couldn't formulate how to deduce from the given conditions to that part.
    Any Ideas ?
     
  10. Jul 13, 2014 #9
  11. Jul 13, 2014 #10

    vela

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    As you noted, p(1) = 0, so you can say that
    $$\lvert p(x) - p(1) \rvert \le \lvert e^{x-1}-1 \rvert.$$ Now consider the definition of a derivative as a limit of a difference quotient and write down the expression for p'(1). Do you see the connection now?
     
  12. Jul 14, 2014 #11
    Right.

    [tex]p'(1)\ =\ lim_{h\rightarrow0}\frac{p(1+h)-p(1)}{h}\\
    \implies a_1+2a_2+3a_3+......+na_n\ =\ lim_{h\rightarrow0}\frac{p(1+h)}{h},\ as\ p(1)\ =\ 0
    [/tex]
    I was able to solve it.
    Thank you for your time
     
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