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Math Help - Three problems: Integration, inequality and matrice

  1. #1
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    Three problems: Integration, inequality and matrice

    Hi,

    Here I have three problems I can't get solved no matter what I try. Truthfully I don't have a clue on how to begin. Hopefully someone here is able to help me! All other homework in the course was handed in a long time ago..

    Any help is appreciated!

    1. Examine the integral of e^(-x)^2 from 0 to infinity.

    2. Prove the inequality 1 + ln(x+sqrt(1+x^2)) > sqrt(1+x^2), 0<x<1

    3.

    P = [1-a a
    b 1-b]

    S= [-a a
    b -b ]

    a) show that S^n=k^(n-1)*S, n>2, k=-(a+b)
    b) show that P^n=I+((1+k)^n-1)/k*S, where I is the identity matrix.

    I'm sorry, don't know how to use that fancy latex-code for the symbols!
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  2. #2
    Moo
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    Hello,

    What is the real formula ?

    1. Examine the integral of e^(-x)^2 from 0 to infinity.
    Is it e^{(-x)^2} (in which case it's quite strange) or is it (e^{-x})^2 ?


    If it's the second one, you just have to transform it in e^{-2x} and find an antiderivate (it's easy because the derivate of -2x is -2)
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    What is the real formula ?



    Is it e^{(-x)^2} (in which case it's quite strange) or is it (e^{-x})^2 ?


    If it's the second one, you just have to transform it in e^{-2x} and find an antiderivate (it's easy because the derivate of -2x is -2)
    Actually it is e^(-x^2), I got it a bit wrong the first time, sorry.
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  4. #4
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    In my memory, this integral can't be defined with usual functions...

    Maybe what they mean by examine is to check if it converges, that is to say continuous on ]0; + \infty[ and that it has a determined limit at its extremities -> study the limit of the function in the integral in 0 and + \infty




    Yep, take a look at this : Erf -- from Wolfram MathWorld
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  5. #5
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    Quote Originally Posted by ManKKe View Post
    Actually it is e^(-x^2), I got it a bit wrong the first time, sorry.
    The definite integral can be found under certain circumstances. Lucky for you one of those circumstances is when integrating from 0 to infinity. The usual approach is to square the integral and convert to polar coordinates. Read this (note that by symmetry your integral will be half the integral from -infty to +infty).
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  6. #6
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    Quote Originally Posted by ManKKe View Post
    [snip]

    2. Prove the inequality 1 + ln(x+sqrt(1+x^2)) > sqrt(1+x^2), 0<x<1

    [snip]
    Do you know the Mean Value Theorem? If so, then I think the following is OK:

    Let f(x) = 1 + \ln \left( x + \sqrt{1 + x^2} \right), ~ 0 < x < 1.


    Then:

    f'(x) = \frac{1}{(x + \sqrt{1 + x^2})} \, \left( 1 + \frac{x}{\sqrt{1 + x^2}}\right) = \frac{1}{(x + \sqrt{1 + x^2})} \, \left( \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}}\right)


    = \frac{1}{\sqrt{1 + x^2}}.


    Now let x lying in 0 < x < 1 be given. Then from the Mean Value Theorem:


    \frac{f(x) - f(0)}{x - 0} = f'(a) \, for some 0 < a < x.


    But f'(a) = \frac{1}{\sqrt{1 + a^2}} > \frac{1}{\sqrt{1 + x^2}} for 0 < a < x.


    Therefore:

    \frac{f(x) - f(0)}{x - 0} = f'(a)


    \Rightarrow \frac{1 + \ln (x + \sqrt{1 + x^2} - 1}{x - 0} > \frac{1}{\sqrt{1 + x^2}}


    \Rightarrow \ln (x + \sqrt{1 + x^2}) > \frac{x}{\sqrt{1 + x^2}}


    \Rightarrow 1 + \ln (x + \sqrt{1 + x^2}) > 1 + \frac{x}{\sqrt{1 + x^2}}.

    It's not hard to show that 1 + \frac{x}{\sqrt{1 + x^2}} > \sqrt{1 + x^2} ~ for 0 < x < 1 and so the result follows.
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  7. #7
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    Thank you very much, you have been very helpful! It was very hard to figure out the Gaussian integral since I didn't know about it. To use the MVT on the inequality was also something I wouldn't have come up with myself.
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  8. #8
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    The matrix still remains, does anyone have the time to solve it?

    Thank you.
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