# Gradient of an Unknown Function!!!

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• Jul 22nd 2006, 10:01 AM
mubashirmansoor
Hello, The latest research of mine concerning the derivative of a function without using diffrentiation and even without having the knowledge of what the equation of the funtion is, is attached to the post, please send me your comments on this latest document of mine.
With regards.
Mubashir
• Jul 22nd 2006, 10:12 AM
CaptainBlack
Quote:

Originally Posted by mubashirmansoor
Hello, The latest research of mine concerning the derivative of a function without using diffrentiation and even without having the knowledge of what the equation of the funtion is, is attached to the post, please send me your comments on this latest document of mine.
With regards.
Mubashir

Over complicated compared to classical numerical methods.

RonL
• Jul 24th 2006, 05:30 AM
mubashirmansoor
Complicated But....
Well Captain, It is complicated compared with diffrentiation But the good thing about it is that we dont need to know the function with which the terms are related, The main Uses of this formula is to find the accleration of the heavenly bodies etc...
Now what did you really mean by numerical methods? Do we needto know the polynomial for finding the gradient????

• Jul 24th 2006, 06:44 AM
CaptainBlack
Quote:

Originally Posted by mubashirmansoor
Well Captain, It is complicated compared with diffrentiation But the good thing about it is that we dont need to know the function with which the terms are related, The main Uses of this formula is to find the accleration of the heavenly bodies etc...
Now what did you really mean by numerical methods? Do we needto know the polynomial for finding the gradient????

No complicated compared to

$\displaystyle f'(x)\approx \frac{f(x+\Delta)-f(x-\Delta)}{2\Delta}$

RonL
• Jul 25th 2006, 03:28 AM
mubashirmansoor
Editted
Well yes but this formula is applicable to all curves while there are diffrent types of the diffrentiated formula you are talking about for diffrent graphs...
For clarification and more accuracy I've done some changes written in the attached file.
The ultimate formula is;

dy/dx =10^10*tan (½(tan-1((f(x+1)–f(x))/10^10)+tan-1((f(x)–f(x-1))/10^10)))

Regards
Mubashir
• Jul 25th 2006, 05:32 AM
CaptainBlack
Quote:

Originally Posted by mubashirmansoor
Well yes but this formula is applicable to all curves while there are diffrent types of the diffrentiated formula you are talking about for diffrent graphs...
For clarification and more accuracy I've done some changes written in the attached file.
The ultimate formula is;

dy/dx =10^10*tan (½(tan-1((f(x+1)–f(x))/10^10)+tan-1((f(x)–f(x-1))/10^10)))

Regards
Mubashir

For your example, what is the numerical value that your method gives for
the estimate of the derivativeof $\displaystyle x^2$ at $\displaystyle x=2$?

RonL
• Jul 25th 2006, 06:12 AM
mubashirmansoor
Exact....
It's answers are abseloutly the same as diffrentiation technique,
I've tried it in all types of graphs & works for all.
Do you still call it Useless???
• Jul 25th 2006, 07:28 AM
CaptainBlack
Quote:

Originally Posted by mubashirmansoor
It's answers are abseloutly the same as diffrentiation technique,
I've tried it in all types of graphs & works for all.
Do you still call it Useless???

No it is exact for 2-nd degree curves only. Also I beleive it is identical in
practice with what I gave above.

Try it for $\displaystyle x^3$ at $\displaystyle x=2$.

to indicate to you that the calculation is not simple. For instance how
many transendental function evaluations are involved? (transendental
functions include the trig functions and their inverses, as well as logs
and exponential functions).

RonL
• Jul 28th 2006, 05:41 AM
mubashirmansoor
What's going on???
Sorry for the delay in my response,

You have written that the answers are not accurate in the formula provided, There is something happening which I don’t understand, (I’m a student of O’levels so I’d be glad if you help me figure out the problem) if we use the formula with small gaps between the terms, it works very accurately & there is just .00000001 difference in the answers compared to differentiation. I mean the formula given below;

dy/dx = tan ( 0.5 ( tan-1 (( f(x+0.1) – f(x) ) / 0.01) + tan-1 (( f(x) – f(x-0.01) / 0.01)))

This equation is even applicable for x^20 or higher degree equations… To get rid of inserting these extremely nearby points, I thought to decrease the size of the curve (not the dimensions) by 10^10, in this way 1 unit would turn to 10 billionth of a unit and even the points with huge gaps between each other would become extremely near. So when the calculation was taking place the curve was small enough to get good approximations, and after the evaluation the curve was brought to its original size hence multiplied by 10^10.
I don’t feel that the technique has a problem but its answers are absolutely the same as differentiation for degree 2 equations but as the degrees increase the answers become larger and larger by about 1 to 3 units for the degrees 3 and 13/15 respectively.

We know that differentiation is itself not accurate and that we say “delta x” approaches zero which is impossible, and the real answers should be a little bit more than that of differentiation. What’s the problem with the last part of my technique???

I’ll be really thankful if you help me figure out what’s going on…
• Jul 28th 2006, 11:59 AM
CaptainBlack
Quote:

Originally Posted by mubashirmansoor

We know that differentiation is itself not accurate and that we say “delta x” approaches zero which is impossible, and the real answers should be a little bit more than that of differentiation. What’s the problem with the last part of my technique???

Differentiation is exact, it gives the exact value of the slope at a point on a curve.

You will have to learn some calculus before you will understand what
is going on in detail.

A good (cheap-ish) book to learn calculus from is "Calculus: An Intuitive and
Physical Approach" by Morris Kline (see here )

Then a good book to learn some numerical analysis from is "Numerical
Methods that Work" by Forman Acton (see here )

RonL
• Jul 29th 2006, 06:10 AM
mubashirmansoor
If diffrentiation is exact then why do we say dy/dx is aproximately equal to delta y/delta x??? :confused:
• Jul 29th 2006, 06:45 AM
topsquark
Quote:

Originally Posted by mubashirmansoor
If diffrentiation is exact then why do we say dy/dx is aproximately equal to delta y/delta x??? :confused:

Because $\displaystyle \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$. They are not the same.

-Dan
• Jul 29th 2006, 07:10 AM
CaptainBlack
Quote:

Originally Posted by mubashirmansoor
If diffrentiation is exact then why do we say dy/dx is aproximately equal to delta y/delta x??? :confused:

$\displaystyle \Delta y / \Delta x$ is used as a stage in the development of the derivative.

The derivative is the slope of a curve at a point, we approximate this by
calculating the slope of a secant of the curve near the point of interest.
Then by letting the points at which the secant meets the curve near the
point both approach the point of interest in the limit obtain the exact slope
of the curve at the point if this should happen to exist. The slope of the
secant is $\displaystyle \Delta y / \Delta x$.

If you consult a reference on calculus as I have suggested earlier you will
learn what is going on better than you will form a discussion here.

RonL
• Jul 29th 2006, 02:12 PM
mubashirmansoor
The link provided is to be used for having a look at the detailed discusion done on this topic;

http://www.physicsforums.com/showthr...65#post1044865
• Jul 30th 2006, 07:39 AM
topsquark
Quote:

Originally Posted by mubashirmansoor
The link provided is to be used for having a look at the detailed discusion done on this topic;

http://www.physicsforums.com/showthr...65#post1044865

I'm not sure what your point is. They seem to be having the same issues with your programme as we are...

-Dan
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