Further calculus (Further Pure 1)

Explanation

• The Taylor expansion of $f$ about $x=a$ is $f(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+\dfrac{f'''(a)}{3!}(x-a)^3+\cdots$, and the equivalent shifted form is $f(a+h)=f(a)+hf'(a)+\dfrac{h^2}{2!}f''(a)+\cdots$.
• Taking $a=0$ recovers the Maclaurin series already met in A-level Mathematics.
• In practice you build a short table of $f$, $f'$, $f''$, $f'''$, evaluate every entry at $x=a$, and then assemble the series; keep the powers as $(x-a)^k$ throughout rather than expanding brackets, because the question almost always asks for the answer in ascending powers of $(x-a)$.
• Where a function is a product or composition of standard series it is usually far quicker to multiply or substitute known Maclaurin series than to differentiate repeatedly, and the two routes are a useful check on each other.

Explanation

• To evaluate an indeterminate limit by series, replace each function by its Maclaurin expansion, keep enough terms to reach the lowest power that survives, and then cancel.
• The five expansions that cover almost every question are $\sin x=x-\tfrac{x^3}{6}+\cdots$, $\cos x=1-\tfrac{x^2}{2}+\tfrac{x^4}{24}-\cdots$, $\tan x=x+\tfrac{x^3}{3}+\cdots$, $\mathrm{e}^{x}=1+x+\tfrac{x^2}{2}+\cdots$ and $\ln(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}-\cdots$; $\arctan x=x-\tfrac{x^3}{3}+\cdots$ is also worth knowing.
• Decide in advance how many terms you need: if the denominator is $x^3$ you must expand every numerator to at least $x^3$, and stopping a term early is the standard way to produce a wrong limit.
• Differences of two singular expressions such as $\tfrac1{x^2}-\tfrac1{\sin^2x}$ should first be combined over a common denominator, then expanded.

Explanation

• Leibnitz's theorem gives the $n$th derivative of a product: $\dfrac{\mathrm{d}^n}{\mathrm{d}x^n}(uv)=\sum_{r=0}^{n}\binom{n}{r}u^{(n-r)}v^{(r)}$, where $u^{(k)}$ means the $k$th derivative.
• It is the differentiation analogue of the binomial expansion, and the binomial coefficients are the same ones.
• The theorem is most useful when one factor is a polynomial, because that factor's derivatives stop after finitely many terms and the sum truncates: a quadratic factor leaves only three terms whatever the value of $n$.
• Its second use is to differentiate a differential relation $n$ times, producing a recurrence between successive derivatives at a point; this is the standard route to a Maclaurin series for functions such as $\arcsin x$ and $\mathrm{e}^{\arctan x}$.
• Take care to attach the derivative counts to the correct factor: $u^{(n-r)}$ pairs with $v^{(r)}$, not the other way round.

Explanation

• If $f(a)=g(a)=0$, or if both $|f|$ and $|g|$ tend to infinity, and $g'$ is non-zero near $a$, then $\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}$ whenever the second limit exists.
• The rule may be applied repeatedly while the quotient remains indeterminate, but you must check the $\tfrac00$ or $\tfrac{\infty}{\infty}$ condition again before every application; applying it to a quotient that is not indeterminate is a guaranteed way to get the wrong answer.
• Indeterminate forms of the types $0\times\infty$, $\infty-\infty$, $0^0$, $1^{\infty}$ and $\infty^0$ must first be rewritten: products become quotients by moving one factor to the denominator, differences are combined over a common denominator, and powers are handled by taking logarithms, finding the limit of $\ln y$ and exponentiating at the end.

Explanation

• The Weierstrass, or tangent half-angle, substitution turns any rational function of $\sin x$ and $\cos x$ into a rational function of $t$.
• Put $t=\tan\tfrac{x}{2}$, so that $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$ and, differentiating $t$ with respect to $x$, $\dfrac{\mathrm{d}t}{\mathrm{d}x}=\tfrac12\sec^2\tfrac{x}{2}=\tfrac12\left(1+t^2\right)$, giving $\mathrm{d}x=\dfrac{2\,\mathrm{d}t}{1+t^2}$.
• That factor of $\dfrac{2}{1+t^2}$ almost always cancels against the $1+t^2$ produced by the trigonometric terms, leaving a straightforward rational integral.
• For a definite integral, change the limits with the substitution rather than converting back: $x=0$ gives $t=0$ and $x=\tfrac{\pi}{2}$ gives $t=1$.
• The substitution is invalid across $x=\pi$, where $t$ becomes infinite, so an integral spanning that point must be split.