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Thread: Manipulation of the constant C that ocurres after differentiating

  1. #1
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    Manipulation of the constant C that ocurres after differentiating

    Hi!

    I have a question about something I find a bit peculiar. The textbook of my math-course is a bit inconistent in the use of the constant (here $\displaystyle C$) that comes from differentiating an expression.

    I have an expression that has been differentiated, and it looks like this:

    $\displaystyle y = \sqrt{e^{\frac{1}{2}x^2+C}}$

    Sometimes this kind of expression is used to calculate further, but other times it is manipulated into this:

    $\displaystyle y = \sqrt{Ce^{\frac{1}{2}x^2}}$

    My question is this:
    Can I always perfrom this manipulation, or is it just in certain circumstances? I know the constant $\displaystyle C$ is a constant that can be changed if it is only affected by other constants (like $\displaystyle e$ in this situation), but the whole thing is very confusing...

    Any clarifying help will be greatly appreciated!

    Thomas
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  2. #2
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    Quote Originally Posted by luckytommy View Post
    Hi!

    I have a question about something I find a bit peculiar. The textbook of my math-course is a bit inconistent in the use of the constant (here $\displaystyle C$) that comes from differentiating an expression.

    I have an expression that has been differentiated, and it looks like this:

    $\displaystyle y = \sqrt{e^{\frac{1}{2}x^2+C}}$

    Sometimes this kind of expression is used to calculate further, but other times it is manipulated into this:

    $\displaystyle y = \sqrt{Ce^{\frac{1}{2}x^2}}$

    My question is this:
    Can I always perfrom this manipulation, or is it just in certain circumstances? I know the constant $\displaystyle C$ is a constant that can be changed if it is only affected by other constants (like $\displaystyle e$ in this situation), but the whole thing is very confusing...

    Any clarifying help will be greatly appreciated!

    Thomas
    $\displaystyle e^{f(x) + C} = e^{f(x)} e^C = A e^{f(x)}$.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    $\displaystyle e^{f(x) + C} = e^{f(x)} e^C = A e^{f(x)}$.
    Yes, I just skipped those steps in my post. The thing is that the book we use sometimes gives;
    $\displaystyle e^{\frac{1}{2}x^2+C}$
    as an eligible solution, while other times;
    $\displaystyle Ce^{\frac{1}{2}x^2}$
    is preferred. I still don't know why the authors would do this. Maybe it is caused by the simple fact that I am writing authors instead of author ??

    Anyway: I have understood that this "maneuver" is always allowed (at least from problems I have seen ), and that I prefer the latter form.
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  4. #4
    MHF Contributor chisigma's Avatar
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    As explained in...

    http://www.mathhelpforum.com/math-he...tegration.html

    ... the problem is when in the expression...

    $\displaystyle c\cdot e^{\frac{x^{2}}{2}}$ (1)

    ... is $\displaystyle c=0$ [or also $\displaystyle c\le 0$...]. Clearly in this case doesn't hold the identity...

    $\displaystyle c\cdot e^{\frac{x^{2}}{2}}= e^{\frac{x^{2}}{2}+\ln c}$ (2)

    ... because $\displaystyle \ln c$ doesn't exist ...



    Merry Christmas from Italy

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
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    Okay, I understand. So in these situations the "maneuver" would give the one form a smaller specter of solutions than the other...
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