# Thread: Problem in finite differences

1. ## 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?

2. ## Re: Problem in finite differences

Originally Posted by Vinod
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?

3. ## Re: Problem in finite differences

Originally Posted by Plato
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.

4. ## Re: Problem in finite differences

Originally Posted by Vinod
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.

5. ## Re: 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.

6. ## Re: Problem in finite differences

Originally Posted by BobP
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.
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$\frac {e^{x+1}}{1-e}$

7. ## Re: 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 ?

8. ## Re: Problem in finite differences

Originally Posted by BobP
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 ?
1-e produce $e^x$ in the first difference column.