Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 15

For $a\ne 0$, factor out $|a|$, not $a$, and apply the binomial series with exponent $1/2$:For each fixed nonzero $a$, these are the first three nonzero terms of the Taylor series about zero; the full binomial series converges when $|x|<|a|$.

For $a=0$, the function is $|x|$. Its right derivative at zero is $1$ and its left derivative is $-1$, so no Taylor series about zero exists when $a=0$. The one-sided formulas are exactly $x$ for $x\ge 0$ and $-x$ for $x\le 0$; they do not form a single two-sided Taylor expansion, nor can they supply three nonzero terms. The parameter case explicitly included in the question must therefore be treated as this exception.

Write $t_j=t^{(j)}(0)$ and $s=1+t_0>0$. The chain rule givesThe condition $t_3{t}_1<0$ ensures $t_1\ne 0$. Since $t_0\ne 0$, also $s\ne 1$ and $\ln s\ne 0$. Hence the first two nonzero terms are the constant and linear terms:The inequality $s>0$, together with continuity of $t$, makes the logarithm defined on a neighbourhood of zero. The remainder is written in Big O notation.

The second derivative chain rule givesWith $t_0=0$, the constant term vanishes, $f^{\prime }(0)=t_1\ne 0$, and $f^{\prime \prime }(0)=t_2-t_1^2\ne 0$ by the specified case. Thus the first two nonzero terms in the Taylor expansion are

Here $t_0=0$ and $t_2=t_1^2$, so the constant and quadratic terms vanish, while the linear coefficient remains $t_1\ne 0$. A further derivative givesConsequently $f^{(3)}(0)=t_3-3t_1{t}_2+2t_1^3=t_3-t_1^3$. Since $t_3$ and $t_1$ have opposite signs, $t_3$ and $t_1^3$ also have opposite signs, and their difference is nonzero. The first two nonzero terms are thereforeThis is where the full sign condition $t_3{t}_1<0$ rules out a further cancellation in the Taylor expansion.