Derivation of 3pt forward formula

[jkthejetplane]

Homework Statement

3 Point forward derivation using Taylor expansion

Relevant Equations

Hi,
My homework is to derive the formula for 3 point forward difference using Taylor expansion.
I have just gotten back into school after some years off and I feel like I’m overthinking a piece to get me started.
Should I start with taking the Taylor expansion of f(x+2h) and f(x+h), subtract the later from the former, then solve for f’(x)?
I have been trying to look up how get started but I feel like I’m setting it up wrong. Anyone point me in the correct direction with an explanation?
Thanks

f(x+2h) = f(x) + 2hf’(x) + .5(2h)^2f”(x)
f(x+h) = f(x) + hf’(x) + .5h^2f”(x)

f(x+2h) - f(x+h)?

f(x+2h) - 2f(x+h)?

 

[Steve4Physics]

Try this: https://www.physicsforums.com/attachments/sdfdfs-jpg.111906/

The formula at the top is the 3 point forward distance formula.
So you would start by finding f(x+h) and f(x+2h) and then the derivation is similar to the 3 point backward distance formul which is shown.

 

[jkthejetplane]

Try this: https://www.physicsforums.com/attachments/sdfdfs-jpg.111906/

The formula at the top is the 3 point forward distance formula.
So you would start by finding f(x+h) and f(x+2h) and then the derivation is similar to the 3 point backward distance formul which is shown.

why is the formula in the red 4f(x-h)?
I was given the formula and the way I tried to derive it I felt like I was missing terms. Maybe from the 4 but I’m not sure why it’s there

 

[Mark44]

why is the formula in the red 4f(x-h)?

Whoever solved it noticed that 4f(x - h) - f(x - 2h) led to an equation involving f(x) and f'(x) and some other terms that could be ignored. It's not at all an obvious thing to do.

 

[Chestermiller]

What you do is write the 3 term expansions for f(x+h) and f(x+2h) and then eliminate the f'' term between the two equations. This will automatically deliver the 3 point 2nd order accurate forward formula for f'(x)

 

[Steve4Physics]

why is the formula in the red 4f(x-h)?
I was given the formula and the way I tried to derive it I felt like I was missing terms. Maybe from the 4 but I’m not sure why it’s there

It’s an old image I found. The red ringing must have been associated with an acompanying explanation: it is linked to the elimination of the *second* derivative terms (see below).

Pardon the lack of Latex for equations.

In the derivation in the image, the Taylor expansions required are:

f(x-h) = f(x) – hf’(x) + f’’(x)h²/4 ①

f(x-2h) = f(x) – 2hf’(x) + f’’(x)(2h)²/4
= f(x) – 2hf’(x) + f’’(x)h² ②

Note we include 2nd derivatives. Having 3 (rather than 2) points improves accuracy because we can take account both 1st and 2nd derivatives. That’s what makes a 3 point method more accurate than a 2 point method

We need to eliminate f’’(x)h² terms from ① and ②. So we multiply ① by 4 to give:
4f(x-h) = 4f(x) – 4hf’(x) + f’’(x)h² ③

Then ③ – ② gives:
4f(x-h) – f(x-2h) = 4f(x) – 4hf’(x) + f’’(x)h² - (f(x) – 2hf’(x) + f’’(x)h²)
= 4f(x) – 4hf’(x) - f(x) + 2hf’(x)
This eliminates the f’’(x)h² terms. You should now understand the red ringing.

After a bit of simple algebra we are left with
2hf’(x) = 3f(x) - 4f(x-h) + f(x-2h)
Q.E.D.

That the backwards difference. The forwards difference is very similar.

 

[Chestermiller]

It’s an old image I found. The red ringing must have been associated with an acompanying explanation: it is linked to the elimination of the *second* derivative terms (see below).

Pardon the lack of Latex for equations.

In the derivation in the image, the Taylor expansions required are:

f(x-h) = f(x) – hf’(x) + f’’(x)h²/4 ①

f(x-2h) = f(x) – 2hf’(x) + f’’(x)(2h)²/4
= f(x) – 2hf’(x) + f’’(x)h² ②

Note we include 2nd derivatives. Having 3 (rather than 2) points improves accuracy because we can take account both 1st and 2nd derivatives. That’s what makes a 3 point method more accurate than a 2 point method

We need to eliminate f’’(x)h² terms from ① and ②. So we multiply ① by 4 to give:
4f(x-h) = 4f(x) – 4hf’(x) + f’’(x)h² ③

Then ③ – ② gives:
4f(x-h) – f(x-2h) = 4f(x) – 4hf’(x) + f’’(x)h² - (f(x) – 2hf’(x) + f’’(x)h²)
= 4f(x) – 4hf’(x) - f(x) + 2hf’(x)
This eliminates the f’’(x)h² terms. You should now understand the red ringing.

After a bit of simple algebra we are left with
2hf’(x) = 3f(x) - 4f(x-h) + f(x-2h)
Q.E.D.

That the backwards difference. The forwards difference is very similar.

Isn't this the unabridged version of what I said in my post?

 

[jkthejetplane]

THANK YOU ALL. I guess i just didn't notice that the goal was to eliminate the f" term as i assumed it was the same as two point and it was the error term. I see now the error term is the f"' term which now makes sense looking back at the final equation that i was trying to derive.
Thank you again to all who contributed
-jk

 

[Steve4Physics]

Isn't this the unabridged version of what I said in my post?

Hi. Apologies. I didn’t see your post. I saw the OP’s query about the image I attached (“why is the formula in the red 4f(x-h)? ”) and just jumped in. Your post was below so I missed it.

The OP seemed to be struggling, so I expanded on the working in the image and gave the rationale. Probably too much direct help in hindsight.

I’ll take care to read all replies in future. FWIW I’ve given #5 a ‘Like’.