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 :( )