First derivative 3 point forward difference formula

[fonseh]

Homework Statement

Can someone explain why for the first derivative 3 point forward difference formula is 4f(x-h) - f(x-2h) ??

Homework Equations

The Attempt at a Solution

Why it's not f(x-h) - f(x-2h) ?
Is there anything wrong with the notes ?

 

[Ray Vickson]

Homework Statement

Can someone explain why for the first derivative 3 point forward difference formula is 4f(x-h) - f(x-2h) ??

Homework Equations

The Attempt at a Solution

Why it's not f(x-h) - f(x-2h) ?
Is there anything wrong with the notes ?

The formula is not $4 f(x-h)-f(x-2h)$, and the notes make no such claim. Please try again with a correct question.

Anyway, as far as I can see the notes explain things in a perfectly clear way, using the Taylor expansion.

 

[fonseh]

The formula is not $4 f(x-h)-f(x-2h)$, and the notes make no such claim. Please try again with a correct question.

Anyway, as far as I can see the notes explain things in a perfectly clear way, using the Taylor expansion.

please refer to the circled part ... it's really $4 f(x-h)-f(x-2h)$

 

[fonseh]

The formula is not $4 f(x-h)-f(x-2h)$, and the notes make no such claim. Please try again with a correct question.

Anyway, as far as I can see the notes explain things in a perfectly clear way, using the Taylor expansion.

here's the full notes

 

[Ray Vickson]

please refer to the circled part ... it's really $4 f(x-h)-f(x-2h)$

OK, but your first message said nothing about the "circled part"---that makes a difference. So, I have a question for you: have you taken the Taylor expansions for $f(x-h)$ and $f(x-2h)$ and performed the calculation of $4f(x-h) - f(x-2h)?$ Did you get a different answer from that in the notes?

 

[Chestermiller]

Homework Statement

Can someone explain why for the first derivative 3 point forward difference formula is 4f(x-h) - f(x-2h) ??

Homework Equations

The Attempt at a Solution

Why it's not f(x-h) - f(x-2h) ?
Is there anything wrong with the notes ?

The reason that your equation is not correct is that this is a first order accurate approximation, and the result they are trying to obtain is supposed to be accurate to terms of 2nd order in h.

 

[fonseh]

OK, but your first message said nothing about the "circled part"---that makes a difference. So, I have a question for you: have you taken the Taylor expansions for $f(x-h)$ and $f(x-2h)$ and performed the calculation of $4f(x-h) - f(x-2h)?$ Did you get a different answer from that in the notes?

No , i get the same as the notes , my question is why should we use 4f(x-h) ? why can't we use f(x-h) ?

 

[fonseh]

The reason that your equation is not correct is that this is a first order accurate approximation, and the result they are trying to obtain is supposed to be accurate to terms of 2nd order in h.

Can you explain further ? I still didnt get it

 

[Chestermiller]

Can you explain further ? I still didnt get it

In my judgment, it is well explained in the visual you provided.

 

[fonseh]

In my judgment, it is well explained in the visual you provided.

Why we should use 4f(x-h) ? It's not explained in the notes

 

[Ray Vickson]

No , i get the same as the notes , my question is why should we use 4f(x-h) ? why can't we use f(x-h) ?

You can use what works, and 4f(x-h) works. Try it for yourself, without the "4", and see what happens.

In fact, suppose you want a finite-difference formula of the form
$$ f'(x) \approx \frac{1}{h} [ a f(x) + b f(x-h) + c f(x-2h)].$$
Expand out the numerator as a series in small $h$ and you will get something of the form $(a+b+c) f(x) + B h f'(x) + C h^2 f''(x) + O(h^3),$
where $B$ and $C$ are some expressions in $a,b,c$. You want a numerator expression of the form $0 f(x) + 1 h f'(x) + O(h^2)$ (so that the ratio is $f'(x)$ to first order in $h$). In other words, you want $a+b+c=0$ and $B = 1$. After evaluating $B$ in terms of $a,b,c$ you will have two equations in the three unknowns $a,b,c$, and can solve for $b,c$ in terms of $a$. If you make different choices for $a$ you will get different "finite-difference" formulas for $f'(x)$. See if you can figure out what value of $a$ gives you $1 f(x-h)$ in your finite-difference formula.

 

[Chestermiller]

Why we should use 4f(x-h) ? It's not explained in the notes

What they do is eliminate the term involving f'' from the two equations, and solve for f'. This results in a finite difference approximation that is accurate to terms if h².

 

[fonseh]

What they do is eliminate the term involving f'' from the two equations, and solve for f'. This results in a finite difference approximation that is accurate to terms if h².

why there is a need to eliminate f'' from the equation ?

 

[fonseh]

(a+b+c)f(x)+Bhf′(x)+Ch2f′′(x)+O(h3),(a+b+c)f(x)+Bhf′(x)+Ch2f″(x)+O(h3),(a+b+c) f(x) + B h f'(x) + C h^2 f''(x) + O(h^3),

I tried f(x-h) - f(x-2h) , i get -h'f(x) + (3 /2!) (f"(x) h^2) + (3/3!)(f"'(x)h^3) + ...

I didnt get (a+b+c)f(x)+Bhf′(x)+Ch2f′′(x)+O(h3) though

 

[Chestermiller]

why there is a need to eliminate f'' from the equation ?

Because otherwise you would find that the final finite difference equation would only be first order in h. Try any other linear combination of the two equations and see what you get for f'

 

[Ray Vickson]

I tried f(x-h) - f(x-2h) , i get -h'f(x) + (3 /2!) (f"(x) h^2) + (3/3!)(f"'(x)h^3) + ...

I didnt get (a+b+c)f(x)+Bhf′(x)+Ch2f′′(x)+O(h3) though

That is because you ignored what I wrote. I wrote $a f(x) + b f(x-h) + c f(x-2h)$. I suggest you do the same, then tell us what you obtain.

 

[fonseh]

That is because you ignored what I wrote. I wrote $a f(x) + b f(x-h) + c f(x-2h)$. I suggest you do the same, then tell us what you obtain.

Do you mean f(x) + f(x-h) - f(x-2h) ?
If so , i get f(x) -h'f(x) + (3 /2!) (f"(x) h^2) + (3/3!)(f"'(x)h^3) + ...

 

[Ray Vickson]

Do you mean f(x) + f(x-h) - f(x-2h) ?
If so , i get f(x) -h'f(x) + (3 /2!) (f"(x) h^2) + (3/3!)(f"'(x)h^3) + ...

Why would you ask that? Since when does $a f(x) + b f(x-h) + c f(x-2h)$ suddenly become $f(x)+f(x-h)-f(x-2h)?$ (Ok, it would happen in the one case where we choose $a=1, b=1, c=-1$, but I said no such thing.) I left $a,b,c$ as unevaluated constants, because I thought you wanted to see why certain choices are made and others rejected. The first step is to actually write down what you would get with general $a,b,c.$

I am now quitting this thread.