Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 11

The first four terms form the cubic Taylor polynomial about $x_0$:With a continuous fourth derivative nearby, Taylor's theorem gives a remainder $O((x-x_0{)}^4)$. Continuous derivatives through third order suffice for the weaker remainder $o((x-x_0{)}^3)$. These are local approximations; equality with an infinite Taylor series requires additional convergence information.

For $f(x)=x^{1/4}$, the first two derivatives are $f^{\prime }(x)=\frac{1}{4}{x}^{-3/4}$ and $f^{\prime \prime }(x)=-\frac{3}{16}{x}^{-7/4}$. At $16$ they giveUse the first three Taylor expansion terms with increment $17-16=1$:The positive third derivative shows that this quadratic approximation underestimates the true value. The Taylor remainder is at most $7/262144\approx 2.67\times 10^{-5}$, since $f^{(3)}(x)=(21/64)x^{-11/4}$ decreases on $[16,17]$ and the cubic remainder is $f^{(3)}(\xi )/6$.

Multiply the Maclaurin series for the exponential and the sine, keeping terms through degree three:Only $x$, $x^2$, and $(1/2-1/6)x^3$ contribute at these orders, so

Put $v=x+x^2$ in the logarithmic series $\ln (1+v)=v-v^2/2+v^3/3+O(v^4)$. Since $v^2=x^2+2x^3+O(x^4)$ and $v^3=x^3+O(x^4)$,Therefore