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Math Help - Problem in finite differences

  1. #1
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    Problem in finite differences

    Hi mathlovers ,

    Obtain a function whose first difference is $ e^x $. Can any member give me hint to solve this problem?
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    Re: Problem in finite differences

    Quote Originally Posted by Vinod View Post
    Hi mathlovers ,

    Obtain a function whose first difference is $ e^x $. Can any member give me hint to solve this problem?
    What does "first difference" mean?
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    Re: Problem in finite differences

    Quote Originally Posted by Plato View Post
    What does "first difference" mean?
    Hi,
    The first differences are calculated by subtracting each entry from the next entry.X is a argument and Y=f(x) corresponding to that argument is entry.
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    Re: Problem in finite differences

    Quote Originally Posted by Vinod View Post
    Hi,
    The first differences are calculated by subtracting each entry from the next entry.X is a argument and Y=f(x) corresponding to that argument is entry.
    That makes no sense whatsoever. In a continuum there is no next entry.
    If $x\mapsto e^x$, then given an $x$ there is no meaning for "the next entry".

    I think you have misunderstood something here.
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  5. #5
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    Re: Problem in finite differences

    Call your x values x_{0},x_{1},x_{2},x_{3}, \dots \equiv x_{0},x_{0}+h,x_{0}+2h,x_{0}+3h,\dots
    and the corresponding function values
    f(x_{0}),f(x_{1}),f(x_{2}),f(x_{3}),\dots \equiv f_{0},f_{1},f_{2},f_{3},\dots,
    then the entries in the first difference column will be
    f_{1}-f_{0},f_{2}-f_{1},f_{3}-f_{2},\dots. .
    If you were to divide each of these by h you would get a discrete version of the derivative of f(x).
    If this is to be/represent e^{x}, what does that suggest that f(x) should be ? Think calculus.
    Make that (obvious) choice for f(x) and construct the first difference column, (a little algebraic simplification is required).
    You should find that that doesn't quite work, but after tweaking f(x) slightly it does.
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    Re: Problem in finite differences

    Quote Originally Posted by BobP View Post
    Call your x values x_{0},x_{1},x_{2},x_{3}, \dots \equiv x_{0},x_{0}+h,x_{0}+2h,x_{0}+3h,\dots
    and the corresponding function values
    f(x_{0}),f(x_{1}),f(x_{2}),f(x_{3}),\dots \equiv f_{0},f_{1},f_{2},f_{3},\dots,
    then the entries in the first difference column will be
    f_{1}-f_{0},f_{2}-f_{1},f_{3}-f_{2},\dots. .
    If you were to divide each of these by h you would get a discrete version of the derivative of f(x).
    If this is to be/represent e^{x}, what does that suggest that f(x) should be ? Think calculus.
    Make that (obvious) choice for f(x) and construct the first difference column, (a little algebraic simplification is required).
    You should find that that doesn't quite work, but after tweaking f(x) slightly it does.
    Hi,
    I got the answer to the problem by referring to the solved answers to the similar finite differences problems. Here is the answer. We have $ e^x $ as first difference

    $ e^x =e^x\frac {1-e}{1-e} $
    $ e^x =\frac {e^x}{1-e}-\frac {e^{x+1}}{1-e} $
    Hence,
    the function is$\frac {e^{x+1}}{1-e} $
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    Re: Problem in finite differences

    Why 1-e rather than e-1 ?
    Doesn't your choice produce -e^{x} in the first difference column ?
    Also, you assume a step length of 1, does that have to be the case ?
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  8. #8
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    Re: Problem in finite differences

    Quote Originally Posted by BobP View Post
    Why 1-e rather than e-1 ?
    Doesn't your choice produce -e^{x} in the first difference column ?
    Also, you assume a step length of 1, does that have to be the case ?
    1-e produce $ e^x $ in the first difference column.
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