# Taylor's Theorem - Chapter 10, page 13

Discussion in 'CM2' started by Benjamin Bickers, Jan 30, 2020.

1. 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!

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

then h= dxt in the top formula

Last edited: Jan 30, 2020
3. 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!

4. 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.

Last edited: Feb 3, 2020
5. Ah I see - that makes sense - thanks very much Mugono!

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