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trouty323
02-26-2012, 02:11 PM
Derive the following formulas using Taylor series and then establish the error terms for each.

f ' (x) = (1/2*h) [4*f(x+h) - 3*f(x) - f(x+2h)]

f ' (x) = (1/4*h) [f(x+2h) - f(x-2h)]

I honestly have no idea how to go about solving this. I worked with Taylor series a long time ago, but I am stuck on this. Any help would be greatly appreciated! :D

Subhotosh Khan
02-26-2012, 02:14 PM
Derive the following formulas using Taylor series and then establish the error terms for each.

f ' (x) = (1/2*h) [4*f(x+h) - 3*f(x) - f(x+2h)

f ' (x) = (1/4*h) [f(x+2h) - f(x-2h)

I honestly have no idea how to go about solving this. I worked with Taylor series a long time ago, but I am stuck on this. Any help would be greatly appreciated! :D

What does your text-book say?

Did you try to use Google - to find out about error-term in Taylor's series?

Please tell us what you found.

trouty323
02-26-2012, 02:44 PM
Well, the text book does not go into much detail about this. It is very concise and does not show the steps. Also, my teacher's accent is very hard to understand. That is why I came here :-). I did try to Google this, but had a hard time finding examples that are like mine. Even if I did find an example, it just looks like a bunch of letters and numbers to me. I do not remember Taylor series being this hard, which is why I'm very frustrated and confused.

Subhotosh Khan
02-26-2012, 02:58 PM
Please tell us the Taylor's series expansion of:

f(x+h) and f(x+2h)

in terms of f(x), f'(x) and h.

We will take care of the error terms as we progress.

trouty323
02-26-2012, 03:15 PM
Is this what you mean?

f(h) + (f '(x+h)) + ((f ''(x+h)*x2)/(2)) + ((f '''(x+h)*x3)/(6)) + ... + ((f k(x+h)*xk)/(k!))

f(2h) + (f '(x+2h)) + (( f ''(x+2h)*x2)/(2)) + ((f '''(x+2h)*x3)/(6)) + ... + ((f k(x+2h)*xk)/(k!))

Subhotosh Khan
02-26-2012, 03:24 PM
Is this what you mean?

f(h) + (f '(x+h)) + ((f ''(x+h)*x2)/(2)) + ((f '''(x+h)*x3)/(6)) + ... + ((f k(x+h)*xk)/(k!)) = ?

f(2h) + (f '(x+2h)) + (( f ''(x+2h)*x2)/(2)) + ((f '''(x+2h)*x3)/(6)) + ... + ((f k(x+2h)*xk)/(k!))= ?



Yes and NO...

Read my question carefully....

also in your original post you have several open "[" - please correct those

trouty323
02-26-2012, 03:40 PM
I'm not sure what you mean by = ?

Should it be:

f(h) + (f '(x+h)) + ((f ''(x+h)*x2)/(2)) + ((f '''(x+h)*x3)/(6)) + ... + ((f k(x+h)*xk)/(k!)) = f(x+h)

f(2h) + (f '(x+2h)) + (( f ''(x+2h)*x2)/(2)) + ((f '''(x+2h)*x3)/(6)) + ... + ((f k(x+2h)*xk)/(k!)) = f(x+2h)

Subhotosh Khan
02-26-2012, 04:17 PM
The form you want is:

f(x+h) = f(x) + h * f'(x).....

trouty323
02-26-2012, 04:47 PM
I'm sorry, I'm not following. I don't see how this coincides with my original formulas.

Subhotosh Khan
02-26-2012, 05:12 PM
Then probably you need a face-to-face tutor - beyond me.....

trouty323
02-26-2012, 05:27 PM
All I really need is to see it worked out. I can only learn through examples. I'm not too concerned with fully understanding every aspect down to the last detail. This homework is due tomorrow, and after that we are moving on to different things.

trouty323
02-26-2012, 05:58 PM
f(x+h) = f(x) +h*f'(x) + (1/2!)(h2)f''(x) + (1/3!)(h3)f'''(x) + ...

Is this correct?