Concept preview

What you will be able to do

1. 01Build the degree-$n$ Taylor polynomial $T_n(x)=\sum_{k=0}^{n}\dfrac{f^{(k)}(a)}{k!}(x-a)^k$ centred at $x=a$, and the Maclaurin case $a=0$, for functions such as $e^x$, $\sin x$, $\cos x$, $\dfrac{1}{1-x}$, and $\ln(1+x)$.
2. 02Use a Taylor or Maclaurin polynomial to approximate a function value near the centre, computing each coefficient with the correct derivative order, factorial $k!$, and power $(x-a)^k$.
3. 03Explain geometrically why $T_n$ hugs $f$ near the centre — matching value, slope, and higher derivatives there — and why raising the degree $n$ widens the interval on which the approximation stays close.

Start with a demand and see what it forces. We want a polynomial $T$ centred at $x=a$ whose value and derivatives agree with $f$ at $a$: $T(a)=f(a)$, $T'(a)=f'(a)$, $T''(a)=f''(a)$, and so on. Write the polynomial in powers of $(x-a)$, namely $T(x)=c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3+\cdots$. Setting $x=a$ instantly gives $c_0=f(a)$, because every other term carries a factor of $(x-a)$ that vanishes. Differentiate once and set $x=a$: only the $c_1$ term survives, so $c_1=f'(a)$.