Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 9

Taylor's theorem expands about $a$ in powers of $h=x-a$. With the usual differentiability assumptions, its finite form with Lagrange remainder iswhere $\xi$ lies between $a$ and $x$. If the remainder tends to zero as $n\to \infty$, this yields the Taylor series $f(x)={\sum }_{k\ge 0}{f}^{(k)}(a)(x-a{)}^k/k!$. Smoothness alone does not guarantee that infinite-series equality.

Put $h=x-1$. The Taylor series of the natural logarithm gives the first four nonzero terms:The corresponding infinite series converges for $|h|<1$.

For the arctangent, put $h=x-1$. Its derivative is $(d/dh)\arctan (1+h)=1/(2+2h+h^2)=\frac{1}{2}-\frac{h}{2}+\frac{h^2}{4}+O(h^4)$. Integrating and using $\arctan 1=\pi /4$ givesThe constant counts as the first nonzero term. The quartic coefficient vanishes; the next nonzero term is $-h^5/40$.

The removable value at zero is one. Expand the denominator as $e^x-1=x(1+x/2+x^2/6+\cdots )$ and determine the reciprocal coefficients by multiplication. The first four nonzero terms areThis is the exponential generating function of the Bernoulli numbers, with convention $B_1=-1/2$. The coefficients of $x^3$ and $x^5$ are zero.