# Gradient Of an Unknown Function

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mubashirmansoor said:
The expresion shows a fact when "x" is either minima or maxima;

0 = 10^10*tan(.5(tan-1((f(x+.01)-f(x))/10^8)+tan-1((f(x)-f(x-.01))/10^8)))

-( f(x+0.01) - f(x) ) = ( f(x) - f(x-0.01) )

which means: f(x+0.01)=f(x-0.01)
which is a general truth, (& I think a prove to the validity of the technique)

and since no one has pointed this out yet, this would only be a proof that the technique works when you have an extremum at $x$ (ie. when you already know the derivative is 0). And even then, as we've shown you, your technique still won't give the exact derivative of 0 in many cases, because $f$ isn't always symmetric about turning points.

You are abseloutly right, Well it's just a simple method for some approximations...
You were all very helpfull, I've learnt quite a lot by posting this thread :)

Now I just want to know how you find the formula by considering the point that we don't need the function for getting an approximate for the derivative...

mubashirmansoor said:
You are abseloutly right, Well it's just a simple method for some approximations...
You were all very helpfull, I've learnt quite a lot by posting this thread :)

Now I just want to know how you find the formula by considering the point that we don't need the function for getting an approximate for the derivative...

No problem I am not sure exactly what your question is (can you restate it?).

If you mean "what are some approximation methods for derivatives," well, the easiest one is just the secant approximation that Halls has been mentioning,

$$\frac{df}{dx} \approx \frac{\Delta{f(x)}}{\Delta{x}} = \frac{f(x+\Delta x) - f(x-\Delta x)}{2\Delta x}.$$

(and the smaller $\Delta x$ is, the better)

The angle-averaging method that you looked at will also give a decent approximation but is a lot more impractical to use because you have to calculate tangents and arctangents. Sometimes its results will be better, and sometimes worse.

Keep in mind that you can't approximate derivatives without knowing something about the function you're trying to do it with - you have to be able to compute its values at different points for either of these to work!

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What I meant was; how you find the method (which you have already answered) and when it comes to haveing an idea about the equation of the function before calculating the derivatives, I don't agree with you,
Let me give you an example;
A projectile motion, let's say a ball (doesn't needs to be conected by any polynomial function), we can draw the motion on the cartesian plane and without having an idea about the function of it's motion calculate it's gradient(at least an approxiate gradient)...
Not bad..... :)
Thanks once again, You were very helpfull...

HallsofIvy
Science Advisor
Homework Helper
mubashirmansoor said:
What I meant was; how you find the method (which you have already answered) and when it comes to haveing an idea about the equation of the function before calculating the derivatives, I don't agree with you,
Let me give you an example;
A projectile motion, let's say a ball (doesn't needs to be conected by any polynomial function), we can draw the motion on the cartesian plane and without having an idea about the function of it's motion calculate it's gradient(at least an approxiate gradient)...
Not bad..... :)
Thanks once again, You were very helpfull...
How can you possibly "draw the motion on the cartesian plane" with out know the height of the ball at each t- in other words "knowing the function". (Do you understand the difference between a function and a formula?)

Well, We do know the height, but we dont have any type of polynomial ruling the motion... That was what I meant, using the word function was a mistake :)
I'm sorry

mubashirmansoor said:
Well, We do know the height

Yep, that's what I mean. You have to be able to compute (or measure off of a graph, or something) $f(x+\Delta x )$ and $f(x-\Delta x)$ in order to use those approximations.

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