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## Main Question or Discussion Point

I have tried searching and have not come up with an answer to this question - but have come close in my own work (i think). Note: I want to solve this numerically, or by some formula.

I am trying to solve this problem, save I have function [tex]f(x)[/tex], the equation is not known. Its derivative, [tex]f'(x)[/tex], is known.

I want to find the value of at q, [tex]f(q)[/tex], using its derivative. I have been searching for an algorithm or formula for this and all of my results have been off topic (getting results about general integration rules, which is not what I want).

For situations similar to [tex]f'(x)=x^n[/tex], where n is a constant, the equation

[tex]f(q)=\int_{0}^{q}x^n[/tex]

is accurate. And indeed, for any derivative [tex]f'(x)[/tex] where [tex]f(0)=0[/tex], this is true. And I know this since

[tex]\int_{a}^{b}f'(x)=f(b)-f(a)[/tex]

However this does not work on all [tex]f'(x)[/tex], and I can not test if [tex]f(0)=0[/tex] since I do not have its equation, and am solving numerically.

After some thinking and work, I cam up with

[tex]f(q)=(\int_{0}^{q}x^n)+f'(0)-1[/tex]

This works on many other equations. Such as

[tex]f'(x)=e^{-x^{2}}[/tex]

&

[tex]f'(x)=cos(x)[/tex].

However it does not work over all, and believe it will fail when [tex]f(0)=f'(0)=0[/tex]. As I've tested a few such as

[tex]f'(x)=x^3[/tex]

which will fail under my new formula ([tex]f'(0)=f(0)=0[/tex]).

[tex]f'(x)=atan(x)[/tex]

will also fail, ([tex]f'(0)=f(0)=0[/tex])

However the problem remains that I can not check if [tex]f(0)=0[/tex], especially since I'm solving for f(q) in the first place! (since [tex]0\in\Re[/tex]). As far as i can tell, my new formula will work so long as [tex]f'(0)=f(0)=0[/tex] is not true, which means if I could some how test for [tex]f(0)=0[/tex] I would have the problem solved (well for the most part, I dont have a proof, but would be a well tested heuristic).

I think I've fully explained what I'm trying to do - if it is unclear please ask and I will try to be more specific, and if anyone knows the solutions or has an idea please let me know! (or if this is in the wrong thread)

I am trying to solve this problem, save I have function [tex]f(x)[/tex], the equation is not known. Its derivative, [tex]f'(x)[/tex], is known.

I want to find the value of at q, [tex]f(q)[/tex], using its derivative. I have been searching for an algorithm or formula for this and all of my results have been off topic (getting results about general integration rules, which is not what I want).

For situations similar to [tex]f'(x)=x^n[/tex], where n is a constant, the equation

[tex]f(q)=\int_{0}^{q}x^n[/tex]

is accurate. And indeed, for any derivative [tex]f'(x)[/tex] where [tex]f(0)=0[/tex], this is true. And I know this since

[tex]\int_{a}^{b}f'(x)=f(b)-f(a)[/tex]

However this does not work on all [tex]f'(x)[/tex], and I can not test if [tex]f(0)=0[/tex] since I do not have its equation, and am solving numerically.

After some thinking and work, I cam up with

[tex]f(q)=(\int_{0}^{q}x^n)+f'(0)-1[/tex]

This works on many other equations. Such as

[tex]f'(x)=e^{-x^{2}}[/tex]

&

[tex]f'(x)=cos(x)[/tex].

However it does not work over all, and believe it will fail when [tex]f(0)=f'(0)=0[/tex]. As I've tested a few such as

[tex]f'(x)=x^3[/tex]

which will fail under my new formula ([tex]f'(0)=f(0)=0[/tex]).

[tex]f'(x)=atan(x)[/tex]

will also fail, ([tex]f'(0)=f(0)=0[/tex])

However the problem remains that I can not check if [tex]f(0)=0[/tex], especially since I'm solving for f(q) in the first place! (since [tex]0\in\Re[/tex]). As far as i can tell, my new formula will work so long as [tex]f'(0)=f(0)=0[/tex] is not true, which means if I could some how test for [tex]f(0)=0[/tex] I would have the problem solved (well for the most part, I dont have a proof, but would be a well tested heuristic).

I think I've fully explained what I'm trying to do - if it is unclear please ask and I will try to be more specific, and if anyone knows the solutions or has an idea please let me know! (or if this is in the wrong thread)