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Math Help - Hard Integration Q

  1. #1
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    Hard Integration Q

    Consider the definite integral D=\int_{-\pi}^{\pi}\cos\lambda x\cos nx\,dx where n is a positive integer and lamda is a positive real number.

    Given also that D=\left\{\begin{matrix}<br />
\pi & $for$\,\lambda =n\\<br />
0 & $for$\,\lambda \, $a positive integer$\,,\lambda \neq n\\ <br />
\frac{(-1)^n2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2} & $for$\,\lambda \,$not an integer$<br />
\end{matrix}\right.

    Show that when 0< \lambda < n,\,|D|< \pi and when \lambda > n+\frac{1}{2},\,|D|< 3

    Help of any kind would be greatly appreciated.
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  2. #2
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    Would the first part of the question have something to do with limits? We can see that 0<|\sin(\lambda \pi )|<1 and so the maximum value for the expression is obviously the limit as lamda approaches n since \frac{2\lambda}{\lambda ^2-n^2} approaches infinite. However, im not sure how to evaluate the limit of the entire expression \frac{(-1)^n2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2}.
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  3. #3
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    I'll attempt the second half.
    Let \lambda > n+0.5

    then

    \lambda^2-n^2 > (n+.5)^2-n^2=n+.25

    therefore,

    |\frac{2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2}|<<br />
|\frac{2(n+.5) \sin(\lambda \pi )}{n+.25}|\leq<br />
|\frac{2(n+.5) }{n+.25}|

    Now
    <br />
\frac{(n+.5) }{n+.25} is a strictly decreasing function and n is an integer greater than or equal to 1, so it is maximum for n=1 or:

    |\frac{2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2}|<<br />
\frac{2(1+.5) }{1+.25}<2.4<3

    You could answer part 1 by assuming \lambda < n-0.5 and following the same procedure, then show that D is increasing on the range n-.5<\lambda\leq n and therefore is less than D at n = lambda.
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  4. #4
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    i think you made a mistake on the third line because: \lambda ^2-n^2>n+0.25\Rightarrow\frac{2}{\lambda ^2-n^2}<\frac{2}{n+0.25} and \lambda >n+0.5
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  5. #5
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    Not solved - inequality problem

    Oops sorry. If you find a solution can you please post it, this is really bugging me!
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