Concept 1
How near is near enough?
Build a centred polynomial and justify its error bound.
Write the centre and desired order before differentiating. Taylor coefficients use derivative values at the centre, divided by factorials. Every term uses displacement from that centre, so a centre of one requires powers of x minus one. The polynomial matches derivatives locally but is not automatically equal to the function away from the centre. An error bound needs control of the next derivative over the interval joining the centre and evaluation point. Taylor’s theorem provides an intermediate point, not its known location. A longer polynomial alone does not establish convergence of an infinite Taylor series.
Additional study formulas · NOT PRINTED — LEARN / DERIVE
Taylor remainder · ON THE SUPPLIED EXAM SHEET · View p. 2
If $f$ has continuous third derivative on the joining interval and $|f^{(3)}|\le M$, then $|f(x)-P_2(x)|\le M|x-a|^3/6$.
Worked example
Given $f(x)=\ln x$, centre $a=1$; derivative values are $0,1,-1$.
Substitute $P_2(1.1)=0.1-(0.1)^2/2=0.095$.
Bound On $[1,1.1]$, $|f^{(3)}(x)|=2/x^3\le2$, so error is at most $0.00033334$.
Check $\ln(1.1)\approx0.09531018$; observed error about 0.00031018 is within the bound.
Active recall
The quadratic coefficient is…
Divide by the factorial $2!=2$.
Correct: derivative values are evaluated at the centre.
The order two term uses the second derivative.
Transfer to a new situation
At centre 2, which displacement belongs in the polynomial?
Correct: measure distance from the centre.
Using x alone shifts the centre to zero.
Subtract the centre, rather than adding it.