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worzel
2004-Oct-14, 05:55 PM
I've been trying to understand a proof of the Taylor Series in a teach-yourself calc book but I can't. All the proof's I've found online use a different type of proof which I can't understand deeply enough to correlate to the proof I've got.

The proof in brief

f(x+h) = A0 + A1.h + A2.h^2 + A3.h^3 + ...

as this is true for all h take h=0 then

f(x) = A0

continually differentiate both sides with respect to h

f'(x+h) = A1 + 2.A2.h + 3.A3.h^2 + 4.A4.h^3 + ...
f''(x+h) = 2.A2 + 6.A3.h + 12.A4.h^2 + ...
f'''(x+h) = 6.A3 + 24.h + ...

then choosing h=0 we find that

f'(x) = A1
f''(x) = 2.A2
f'''(x) = 6.A3
...

or

A0 = f(x)
A1 = f'(x)
A2 = f''(x)/2
A3 = f'''(x)/6
...

so

f(x+h) = f(x) + (f'(x).h)/1! + (f''(x).h^2)/2! + (f'''(x).h^6)/6! + ...

This all seems fine to me expect that when you see the Taylor series used they use the series of partial derivatives of f(x) w.r.t x, not h. Obviously I'm an idiot but if someone could help me I'd be very appreciative.

Disinfo Agent
2004-Oct-14, 06:05 PM
Since f only depends on one variable, the independent variable is a dummy. You can call it x, y, z... or h, and write f(x), f(y), f(z) ... or f(h), as long as you're consistent in your notation. Here, they've decided to call the independent variable h.

They regard x as a fixed quantity and h as the independent variable. (Some authors write x_0+h, to emphasize that x_0 is a particular value.) If this is still confusing to you, write y=x+h, and differentiate both sides with respect to y.

[Edited to, hopefully, be clearer.]

worzel
2004-Oct-14, 06:17 PM
They're regarding x as fixed and h as the independent variable. (Some authors write x_0+h, to emphasize that x_0 is a particular value.) If this is confusing to you, write y=x+h, and differentiate both sides with respect to y.
I understand differentiating it wrt h. But this then gives us

d f(x_0 + h) / d.h = A1 + 2.A2.h^2 + ...

and when h=0

d f(x_0) / d.h = A1

so A1 is the rate of change of a function of a constant w.r.t to a variable not even in the function any more.

Am I just being too literal?

[ Edit: did you anticipated my last question or did I anticipate your last edit? :) ]

Disinfo Agent
2004-Oct-14, 06:19 PM
I've added a couple of words to my post. Does it help?

worzel
2004-Oct-14, 06:22 PM
I've added a couple of words to my post. Does it help?
Yes, thank you. You've put the finger on where I am confused. I'll have a think about it on the tube and make sure I really understand it.

Disinfo Agent
2004-Oct-14, 06:25 PM
[ Edit: did you anticipated my last question or did I anticipate your last edit? :) ]
We'll just never know... :wink:

Evan
2004-Oct-14, 06:36 PM
This page may be of help.

http://www005.upp.so-net.ne.jp/polyprocessor/chap2/sec1_.html

Normandy6644
2004-Oct-14, 07:01 PM
Just wait till you use Laurent Series. Loads of fun! #-o

A Thousand Pardons
2004-Oct-14, 07:24 PM
and when h=0

d f(x_0) / d.h = A1

so A1 is the rate of change of a function of a constant w.r.t to a variable not even in the function any more.

Am I just being too literal?
No, it is one of the disadvantages of Liebniz's notation, although his was generally considered stronger than Newton's. It suggested relationships that were handy in solving problems, but in this case it seems confusing.

"d f( ) / dh" is a differentiation function operator, not a division.

So, if f(x) = x^3, then d f(x) / dx = 3 * x^2, and then d f(2) / dx = 12, and you can apply the same comments to that as you have about d f(x_0) / d.h = A1

That's a large reason why you'll often see that d f(2) / dx = 12 expressed as f'(2) = 12 instead.

Normandy6644
2004-Oct-14, 08:10 PM
and when h=0

d f(x_0) / d.h = A1

so A1 is the rate of change of a function of a constant w.r.t to a variable not even in the function any more.

Am I just being too literal?
No, it is one of the disadvantages of Liebniz's notation, although his was generally considered stronger than Newton's. It suggested relationships that were handy in solving problems, but in this case it seems confusing.

"d f( ) / dh" is a differentiation function operator, not a division.

So, if f(x) = x^3, then d f(x) / dx = 3 * x^2, and then d f(2) / dx = 12, and you can apply the same comments to that as you have about d f(x_0) / d.h = A1

That's a large reason why you'll often see that d f(2) / dx = 12 expressed as f'(2) = 12 instead.

That's how I always did it, expressing the derivate evaluated at a point with the f'(x) notation. You can always do (df/dx)|2 or something to show that it's being evaluated at 2 or any other number.

worzel
2004-Oct-14, 08:35 PM
"d f( ) / dh" is a differentiation function operator, not a division.


That's how I always did it, expressing the derivate evaluated at a point with the f'(x) notation. You can always do (df/dx)|2 or something to show that it's being evaluated at 2 or any other number.

Yeah the book uses f'(x) notation. I only did that to make explicit that the function was being differentiated w.r.t h. Maybe I should have written

fh(x) = A1

I'm gonna write it out replacing x+h with some other variable like disinfo suggested just to make sure I do really see it (fell asleep on the tube so didn't get much thinking done).

A Thousand Pardons
2004-Oct-16, 11:05 PM
I'm gonna write it out replacing x+h with some other variable like disinfo suggested just to make sure I do really see it (fell asleep on the tube so didn't get much thinking done).
You might want to try h = (x - a), I see that alot. That's where a = x_0. So, your "f(x + h)" in the OP would be "f(a + h)" which is the same thing then as f(x) but it makes explicit that you are expanding the function about a particular value a for x. But then d f(a + h)/dh and d f(x)/dx are the same thing.