Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 9

The Taylor expansion about $a$ has coefficients given by the successive derivatives at $a$:When $f$ is $N+1$ times continuously differentiable on the interval between $a$ and $x$, Taylor's theorem gives $R_N(x)=f^{(N+1)}(\xi )(x-a{)}^{N+1}/(N+1)!$ for an intermediate point $\xi$. If the remainders tend to zero, this yields the infinite Taylor series $f(x)={\sum }_{k=0}^{\infty }{f}^{(k)}(a)(x-a{)}^k/k!$. The three functions below are real analytic near zero after taking the continuous extension in part (i).

First expand the exponential function and divide by $x$:Seek its reciprocal as $1+Ax+B{x}^2+C{x}^3+O(x^4)$. Equating coefficients in the product with this expression givesHence $A=1/2$, $B=1/12$, $C=0$. The first three nonzero terms of the Taylor expansion areAlthough the quotient is undefined at zero as written, its limit is one, and this Taylor expansion belongs to its continuous, real analytic extension.

Let $g(x)=\tan (x+\pi /4)$. The derivative identity for tangent gives $g^{\prime }=1+g^2$, so $g(0)=1$, $g^{\prime }(0)=2$, and $g^{\prime \prime }(0)=2g(0)g^{\prime }(0)=4$. Its first three nonzero Taylor expansion terms are thereforeThe factor $1/2!$ on the second derivative is what turns $g^{\prime \prime }(0)=4$ into the coefficient $2$.

Use $d(\ln \sec x)/dx=\tan x$ and $\ln \sec 0=0$. To obtain enough terms of tangent, write $t(x)=x+a{x}^3+b{x}^5+O(x^7)$; it is an odd function. The identity $t^{\prime }=1+t^2$ givesso $a=1/3$, $b=2/15$. Integrate this Taylor expansion term by term from zero to $x$:There is no constant term, and only even powers occur because $\ln \sec x$ is an even function near zero. Thus reaching the first three nonzero terms requires expansion through degree six.