Taylor's Theorem - Chapter 10, page 13

[Benjamin Bickers]

Hi All,

I have a question on the Taylor's Theorem that is given in the notes in Chapter 10, page 13 that I was hoping someone might be able to answer.

The theorem is given to the second order in the notes as follows (sorry for the ugly formulas!):

df(xt) = f'(xt)dxt + 1/2f''(xt)(dxt)^2 + …

The reader is then referred to the tables page 3 if unfamiliar with the above theorem. However, the theorem given in the tables looks very different:

f(x+h) = f(x) + hf'(x) + 1/2h^2f''(x) + …

I can't, through renaming variables and moving parts around, get from the first theorem to the second, does anyone know how these two formulas can be reconciled?

Thanks very much!

 

[studier]

Hi All,

I have a question on the Taylor's Theorem that is given in the notes in Chapter 10, page 13 that I was hoping someone might be able to answer.

The theorem is given to the second order in the notes as follows (sorry for the ugly formulas!):

df(xt) = f'(xt)dxt + 1/2f''(xt)(dxt)^2 + …

The reader is then referred to the tables page 3 if unfamiliar with the above theorem. However, the theorem given in the tables looks very different:

f(x+h) = f(x) + hf'(x) + 1/2h^2f''(x) + …

I can't, through renaming variables and moving parts around, get from the first theorem to the second, does anyone know how these two formulas can be reconciled?

Thanks very much!

let df(x) = f(x+h) - f(x)

then h= dxt in the top formula

 

[Benjamin Bickers]

Hi studier,

Thanks for getting back so quickly!

That was along my line of thinking, however, I don't think we can say that df(xt) = f(xt + dxt) - f(xt).

For example, say f(xt) = xt^2, then the LHS = dxt^2 and RHS = (x+dxt)^2 - xt^2 = d^2.xt^2 + 2dxt^2
Dividing both sides by d gives: LHS = xt^2 and RHS = dxt^2 + 2xt^2
letting d tend to zero gives: LHS = xt^2 and RHS = 2xt^2
which aren't equal - am I missing something obvious here?

Sorry again for the ugly formulas!

Thanks again!

 

[mugono]

Hi,

The below is my reconciliation 'attempt' .

1. f(x+h) = f(x) + hf'(x) + 1/2h^2f''(x) + …
subtracting f(x)
2. f(x+h) - f(x) = hf'(x) + 1/2h^2f''(x) + ...
dividing by h
3. [f(x+h) - f(x)] / h = f'(x) + 1/2hf''(x) + ...
taking the limit as h tends to 0+
4. df / dh = f’(x) + 1/2dhf''(x) + ...
multiplying by dh
5. df = f’(x)dh + 1/2(dh)^2f''(x) + ... [end of rec].

Hope that helps.

 

[Benjamin Bickers]

Ah I see - that makes sense - thanks very much Mugono!

Now I just have to get my head around the rest of this stuff!