How can I numerically solve for f(q) using only its derivative, f'(x)?

[eman7613]

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 f(x), the equation is not known. Its derivative, f'(x), is known.

I want to find the value of at q, f(q), 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 f'(x)=x^n, where n is a constant, the equation
f(q)=\int_{0}^{q}x^n
is accurate. And indeed, for any derivative f'(x) where f(0)=0, this is true. And I know this since
\int_{a}^{b}f'(x)=f(b)-f(a)
However this does not work on all f'(x), and I can not test if f(0)=0 since I do not have its equation, and am solving numerically.

After some thinking and work, I cam up with
f(q)=(\int_{0}^{q}x^n)+f'(0)-1
This works on many other equations. Such as
f'(x)=e^{-x^{2}}
&
f'(x)=cos(x).

However it does not work over all, and believe it will fail when f(0)=f'(0)=0. As I've tested a few such as
f'(x)=x^3
which will fail under my new formula (f'(0)=f(0)=0).
f'(x)=atan(x)
will also fail, (f'(0)=f(0)=0)

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

 

[HallsofIvy]

?? I have no idea what you are doing here! You can't find f from its derivative because an infinite number of different functions have the same derivative. If you are also assuming that f(0)= 0, please say that.

But even then, you seem to be saying that \int_0^q x^n dx+ f'(0)- 1 will approximate n. For what n? How do you determine what n to use?

If f'(x)= cos(x), and f(0)= 0, then f(x) is -sin(x). I don't see how you can argue that \int_0^q x^n dx will approximate -sin(x) over any reasonable interval for any n.

 

[eman7613]

As I said, n is just some constant in that particular equation. And I am saying that equation works, when f(0)=0, and indeed you can try it.
Of course you can find f from its derivatives... its called an integral, and I am not searching for the integral, but looking to approximate its value at q that is why
\int_{a}^{b}f(x)=f(b)-f(a)
is useful, becasue when f(0)=0
i can do this
\int_{0}^{b}f(x)=f(b)-0
which gives me the value at f(b) and in turn I can give it q, and find f(q).
ie: try it for the x^n that i mentioned.

f(q)=\int_{0}^{q}x^n=\frac{q^{n+1}}{n+1}-\frac{0^{n+1}}{n+1}=\frac{q^{n+1}}{n+1}
Giving me the value of f(q)
However, I do not know when f(0)=0, which is why I am asking about a general formula, since I am trying to do this numerically, (similar to Simpson's rule).

I don't think you finished reading my post, I never said \int_0^q x^n dx+ f(0)- 1, I said it works for other equations, under the situations I specifed, please read my post before replying! (I even showed x^3, same as x^n, which does not work - again, please read!)

And indeed, I have discoverd a way to solve the problem using the equations I've given, but if someone knows of a formal or already well proven algorithm (or ideas for one) I would still appreciate it! (My algorithm is a heuristic, i have not proven it :( )