Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 9

For an $N+1$ times continuously differentiable function, Taylor's theorem giveswhere $\xi$ lies between $x_0$ and $x$. An infinite Taylor series represents the function when its remainder tends to zero; this holds locally for an analytic function, rather than for every smooth function automatically.

Integrate the geometric series for $1/(1+x)$ from zero to $x$. The first three nonzero terms areThe resulting Taylor series converges for $|x|<1$.

For fixed real $a\ne 0$, write the square root as $|a|\sqrt{1+x^2/a^2}$ and use the binomial series with exponent $-1/2$:The absolute values are necessary for negative $a$. This Taylor series is valid for $|x|<|a|$. If $a=0$, the quotient is $x/|x|$ for $x\ne 0$; it has no continuous extension at zero and therefore has no Taylor expansion about zero.

Factor the Gaussian function as $e^{-(x-a{)}^2}=e^{-a^2}{e}^{2ax}{e}^{-x^2}$. Multiplying the two exponential series givesFor $a\ne 0,\pm 1/\sqrt{2}$, its first three nonzero terms areVanishing coefficients require separate treatment in this Taylor expansion of a translated Gaussian. If $a=0$, the first three nonzero terms are $\boxed{1-x^2+x^4/2}$. If $a=\pm 1/\sqrt{2}$, they areThe latter expression retains the cubic term because the quadratic coefficient is zero.

Use the real logarithm near zero and subtract its two Taylor series:Thus the first three nonzero terms are