Hi mathlovers ,

Obtain a function whose first difference is $ e^x $. Can any member give me hint to solve this problem?

Printable View

- Mar 4th 2014, 05:29 AMVinodProblem in finite differences
Hi mathlovers ,

Obtain a function whose first difference is $ e^x $. Can any member give me hint to solve this problem? - Mar 4th 2014, 06:21 AMPlatoRe: Problem in finite differences
- Mar 4th 2014, 06:59 AMVinodRe: Problem in finite differences
- Mar 4th 2014, 07:42 AMPlatoRe: Problem in finite differences
- Mar 5th 2014, 01:14 AMBobPRe: Problem in finite differences
Call your $\displaystyle x$ values $\displaystyle x_{0},x_{1},x_{2},x_{3}, \dots \equiv x_{0},x_{0}+h,x_{0}+2h,x_{0}+3h,\dots$

and the corresponding function values

$\displaystyle f(x_{0}),f(x_{1}),f(x_{2}),f(x_{3}),\dots \equiv f_{0},f_{1},f_{2},f_{3},\dots,$

then the entries in the first difference column will be

$\displaystyle f_{1}-f_{0},f_{2}-f_{1},f_{3}-f_{2},\dots.$ .

If you were to divide each of these by $\displaystyle h$ you would get a discrete version of the derivative of $\displaystyle f(x).$

If this is to be/represent $\displaystyle e^{x},$ what does that suggest that $\displaystyle f(x)$ should be ? Think calculus.

Make that (obvious) choice for $\displaystyle f(x)$ and construct the first difference column, (a little algebraic simplification is required).

You should find that that doesn't quite work, but after tweaking $\displaystyle f(x)$ slightly it does. - Mar 5th 2014, 04:33 AMVinodRe: Problem in finite differences
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} $ - Mar 5th 2014, 05:11 AMBobPRe: Problem in finite differences
Why $\displaystyle 1-e$ rather than $\displaystyle e-1$ ?

Doesn't your choice produce $\displaystyle -e^{x}$ in the first difference column ?

Also, you assume a step length of 1, does that have to be the case ? - Mar 5th 2014, 05:29 AMVinodRe: Problem in finite differences