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Math Help - e^x

  1. #1
    Senior Member Paze's Avatar
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    e^x

    I'm watching some MIT OCW and the professor says in relation with the function y=e^x :

    We start the function at y=1

    This has to be equal to its derivative so dy/dx=1

    But I can't stop here...I must add an x to the function so that its slope will be one...So y=1+x...But now they need to be equal so I must add that x to dy/dx so dy/dx=1+x...

    What I don't understand is...Why do we add that x? I understand why we start the differential equation at 1 but I do not understand why he insists that this x has to be added.

    Thank you.
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    Forum Admin topsquark's Avatar
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    Re: e^x

    Quote Originally Posted by Paze View Post
    I'm watching some MIT OCW and the professor says in relation with the function y=e^x :

    We start the function at y=1

    This has to be equal to its derivative so dy/dx=1

    But I can't stop here...I must add an x to the function so that its slope will be one...So y=1+x...But now they need to be equal so I must add that x to dy/dx so dy/dx=1+x...

    What I don't understand is...Why do we add that x? I understand why we start the differential equation at 1 but I do not understand why he insists that this x has to be added.

    Thank you.
    I'm not quite sure why he's doing that. Is this supposed to be a derivation of a power series for e^x?

    -Dan
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    Senior Member Paze's Avatar
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  4. #4
    Senior Member Paze's Avatar
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    Re: e^x

    So far I've gathered that he is attempting to show how e^x relates to compound interest. That's what he's showing with the calculations...I don't understand it quite fully though.

    Oh, hold up. It's because it's simply the derivative..He just has a weird way of wording it. I think I get it!
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    Re: e^x

    It looks like it is a derivation of the Maclaurin series for e^x.

    If, (second line) \frac{dy}{dx}}=1, then y can't simply equal 1, (first line), it must be x plus some constant, which by some reasoning is taken to be 1.

    But that means that \frac{dy}{dx}}=1+x also which in turn means that y must contain an x^{2}/2 term, etc.. Ugh !
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    Senior Member Paze's Avatar
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    Re: e^x

    Quote Originally Posted by BobP View Post
    It looks like it is a derivation of the Maclaurin series for e^x.

    If, (second line) \frac{dy}{dx}}=1, then y can't simply equal 1, (first line), it must be x plus some constant, which by some reasoning is taken to be 1.

    But that means that \frac{dy}{dx}}=1+x also which in turn means that y must contain an x^{2}/2 term, etc.. Ugh !
    Yup, I got it! Thanks!
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  7. #7
    Forum Admin topsquark's Avatar
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    Re: e^x

    Yes, he's generating a Maclaurin series for e^x. Let me walk you through a few steps.
    1) Definition: \frac{dy}{dx} = y.

    So we can start this equation by posting y(0) = 1, which is the starting point for the compound interest. (I'm assuming any.)

    2) Using 1) we know that \frac{dy}{dx} = y = 1

    3) But note that if we have y = 1 then \frac{dy}{dx} = 0, which is not true. We need it to be 1.

    4) So now we are going to construct a new function y(x) such that \frac{dy}{dx} = 1. If we integrate this equation we find that y = 1 + x (the arbitrary constant is 1).

    5) But again we must have that \frac{dy}{dx} = y = 1 + x

    Recap: We now have that y = 1 + x and \frac{dy}{dx} = 1 + x

    But again we need another term to add to y that will make the above dy/dx correct. Repeat steps 3) to 5). Rinse and repeat.

    He's creating the power series one term at a time. Does this help?

    -Dan
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