HookThe sum that defeated the Bernoullis

In 1650 the Italian mathematician Pietro Mengoli asked what looks like a warm-up exercise: find the exact value of \(1+\tfrac{1}{4}+\tfrac{1}{9}+\tfrac{1}{16}+\dots\), the sum of the reciprocals of the square numbers. The Bernoullis — the most formidable mathematical dynasty in Europe — attacked the problem for decades and admitted defeat. In 1735 Leonhard Euler, aged 28, produced the answer with an argument of outrageous nerve. He wrote \(\sin x\) as the infinite polynomial \(x-\tfrac{x^3}{3!}+\tfrac{x^5}{5!}-\dots\), noticed that this 'polynomial' has roots at \(0,\pm\pi,\pm2\pi,\dots\), and applied the relationship between a polynomial's roots and its coefficients as if the rules for quadratics stretched to degree infinity. Comparing coefficients delivered \(1+\tfrac{1}{4}+\tfrac{1}{9}+\dots=\tfrac{\pi^2}{6}\) — and made Euler famous across Europe.

Both of Euler's weapons live in this section, the largest pure block of AQA 7367. You will relate roots to coefficients up to quartics and build new polynomials whose roots are transformations of old ones (D1–D2), sum series with the standard results and collapse them with the method of differences (D3–D4), and expand functions as Maclaurin series, using them — or l'Hôpital's rule — to evaluate limits (D5–D7). Then the graphing arsenal: polynomial, rational and modulus inequalities (D8–D10), reciprocal and rational graphs with their asymptotes (D11–D13), ranges found by quadratic theory with no calculus at all (D14), the four conics (D15), and the transformations that move every one of them around the plane (D16).

ModelRoots and coefficients — a polynomial's DNA

For \(ax^2+bx+c=0\) with roots \(\alpha,\beta\): \(\alpha+\beta=-\tfrac{b}{a}\) and \(\alpha\beta=\tfrac{c}{a}\). For the cubic \(ax^3+bx^2+cx+d=0\) with roots \(\alpha,\beta,\gamma\): \(\sum\alpha=-\tfrac{b}{a}\), \(\sum\alpha\beta=\tfrac{c}{a}\) and \(\alpha\beta\gamma=-\tfrac{d}{a}\). The quartic continues the pattern — \(\sum\alpha=-\tfrac{b}{a}\), \(\sum\alpha\beta=\tfrac{c}{a}\), \(\sum\alpha\beta\gamma=-\tfrac{d}{a}\), \(\alpha\beta\gamma\delta=\tfrac{e}{a}\) — with the signs alternating as you read down. Divide by \(a\) every single time: forgetting the leading coefficient is the most common error in the whole topic.

Two workhorse identities turn these into answers: \(\alpha^2+\beta^2+\gamma^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta\), and \(\tfrac{1}{\alpha}+\tfrac{1}{\beta}+\tfrac{1}{\gamma}=\tfrac{\sum\alpha\beta}{\alpha\beta\gamma}\). For a polynomial whose roots are a linear transformation of the old ones (D2), skip the symmetric-function bookkeeping entirely and substitute: if the new roots are \(y=2\alpha-1\), write \(x=\tfrac{y+1}{2}\) and feed it into the original equation. Every root moves at once, in one line of algebra.

Worked example

The roots of \(2x^3-6x^2+5x-7=0\) are \(\alpha,\beta,\gamma\). Read off \(\sum\alpha=3\), \(\sum\alpha\beta=\tfrac{5}{2}\), \(\alpha\beta\gamma=\tfrac{7}{2}\). Then \(\alpha^2+\beta^2+\gamma^2=3^2-2\times\tfrac{5}{2}=4\).

Now the cubic with roots \(2\alpha-1\), \(2\beta-1\), \(2\gamma-1\): set \(y=2x-1\), so \(x=\tfrac{y+1}{2}\). Substituting, \(2\cdot\tfrac{(y+1)^3}{8}-6\cdot\tfrac{(y+1)^2}{4}+5\cdot\tfrac{y+1}{2}-7=0\); multiplying through by 4 gives \((y+1)^3-6(y+1)^2+10(y+1)-28=0\), which expands to \[y^3-3y^2+y-23=0.\] Self-check with the sum of roots: the new sum should be \(2\sum\alpha-3=3\), and the new coefficients give \(-\tfrac{-3}{1}=3\). It matches — that ten-second check catches nearly every slip in this topic.

MechanismSumming series — the standard results and the telescope

Three standard results do the heavy lifting (D3): \(\sum_{r=1}^{n}r=\tfrac{1}{2}n(n+1)\), \(\sum_{r=1}^{n}r^2=\tfrac{1}{6}n(n+1)(2n+1)\) and \(\sum_{r=1}^{n}r^3=\tfrac{1}{4}n^2(n+1)^2\) — the last two are printed in the AQA formulae booklet, and the cubes result is famously the square of the first. To sum anything polynomial, expand the summand, split the sum across the standard results, then factorise the answer fully: AQA mark schemes present, and reward, the factorised form. For a sum running from \(r=n+1\) to \(2n\), subtract one standard sum from another.

The method of differences (D4) handles sums the standard results cannot touch. If each term can be written as \(u_r=f(r)-f(r+1)\) — partial fractions are the usual factory for this — then summing makes almost everything cancel in a diagonal cascade, leaving only the leading and trailing edges. When the gap is two, as in \(f(r)-f(r+2)\), two terms survive at each end instead of one. And if \(f(r)\to0\) as \(r\to\infty\), letting \(n\to\infty\) hands you the sum to infinity for free.

Worked example

Find \(\sum_{r=1}^{n}\frac{1}{r(r+2)}\) and its limit as \(n\to\infty\). Partial fractions: \(\frac{1}{r(r+2)}=\frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+2}\right)\). Write the rows: \(r=1\) gives \(\tfrac{1}{2}\left(1-\tfrac{1}{3}\right)\); \(r=2\) gives \(\tfrac{1}{2}\left(\tfrac{1}{2}-\tfrac{1}{4}\right)\); \(r=3\) gives \(\tfrac{1}{2}\left(\tfrac{1}{3}-\tfrac{1}{5}\right)\) — each negative term is killed by a positive term two rows below. Only the first two positives and the last two negatives survive: \[\sum_{r=1}^{n}\frac{1}{r(r+2)}=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)=\frac{3}{4}-\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{n+2}\right).\] Check with \(n=1\): the formula gives \(\tfrac{3}{4}-\tfrac{1}{2}\left(\tfrac{1}{2}+\tfrac{1}{3}\right)=\tfrac{1}{3}\), and directly \(\tfrac{1}{1\times3}=\tfrac{1}{3}\). As \(n\to\infty\) the tail terms vanish, so the sum to infinity is \(\tfrac{3}{4}\).

ModelMaclaurin series — every smooth function as an infinite polynomial

The Maclaurin series (D5) rebuilds a function from its derivatives at zero: \[f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{(r)}(0)}{r!}x^r+\dots\] Differentiate repeatedly, evaluate at \(0\), assemble — and be ready to state the general term, which is where the pattern-spotting marks live. The five expansions in the booklet (D6) come with validity attached: \(e^x\), \(\sin x\) and \(\cos x\) converge for all \(x\); \(\ln(1+x)\) only for \(-1<x\le1\); \((1+x)^n\) for \(|x|<1\) when \(n\) is not a positive integer. Composites come from substitution, not re-differentiation: for \(e^{2x}\) replace \(x\) by \(2x\); for \(\ln(1+3x)\) replace and rescale the validity to \(-\tfrac{1}{3}<x\le\tfrac{1}{3}\).

Limits (D7): for a \(\tfrac{0}{0}\) form, expand top and bottom as series and keep the first surviving power — everything above it vanishes as \(x\to0\). The alternative is l'Hôpital's rule: if \(\frac{f(x)}{g(x)}\) has the form \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\), its limit equals the limit of \(\frac{f'(x)}{g'(x)}\) — differentiate numerator and denominator separately, and repeat only while the form stays indeterminate. The rule says nothing about non-indeterminate limits, where applying it produces confident nonsense.

Worked example

Evaluate \(\lim_{x\to0}\dfrac{x-\sin x}{x^3}\). By series: \(\sin x=x-\tfrac{x^3}{6}+\tfrac{x^5}{120}-\dots\), so \(x-\sin x=\tfrac{x^3}{6}-\tfrac{x^5}{120}+\dots\) and \[\frac{x-\sin x}{x^3}=\frac{1}{6}-\frac{x^2}{120}+\dots\to\frac{1}{6}.\] By l'Hôpital, three passes, confirming the \(\tfrac{0}{0}\) form each time: \(\dfrac{1-\cos x}{3x^2}\), then \(\dfrac{\sin x}{6x}\), then \(\dfrac{\cos x}{6}\to\dfrac{1}{6}\). Same answer; the series route is a line shorter and shows why the limit is \(\tfrac{1}{6}\) — the cubic term of \(\sin x\) is the whole story.

MechanismInequalities — respect what you multiply by

Cubic and quartic inequalities (D8) are sketching problems in disguise: move everything to one side, factorise, mark the roots, read the sign off the shape. \(x(x+2)(x-3)>0\) has roots \(-2\), \(0\), \(3\); a cubic with positive leading coefficient sweeps up from the bottom left, so the product is positive on \(-2<x<0\) and \(x>3\). A sign table — one row per factor, one column per region — does the same job without the drawing, and scales to quartics with repeated roots, where the curve touches and bounces instead of crossing.

Rational inequalities (D9) have one commandment: never multiply by \(cx+d\), whose sign you do not know. Two safe routes. Either multiply both sides by \((cx+d)^2\), which is positive wherever the expression exists — turning \(\frac{ax+b}{cx+d}<ex+f\) into a polynomial inequality, which is D8 again — or bring everything to one side over a common denominator and sign-analyse the single fraction, taking critical values from the numerator and the denominator. Either way \(x=-\tfrac{d}{c}\) is excluded from every answer, because the expression does not exist there.

For moduli (D10): \(|f(x)|<a\) unpacks to \(-a<f(x)<a\), and \(|f(x)|>a\) to \(f(x)<-a\) or \(f(x)>a\) — note the or. When both sides are moduli, square both sides (both are non-negative, so the direction survives), or sketch both graphs and solve the boundary equalities to locate the crossings.

Worked example

Solve \(\dfrac{x+2}{x-1}<4\). One side: \(\dfrac{x+2}{x-1}-4=\dfrac{x+2-4(x-1)}{x-1}=\dfrac{3(2-x)}{x-1}<0\). Critical values: \(x=1\) (denominator) and \(x=2\) (numerator). For \(x<1\): \(2-x>0\) and \(x-1<0\), quotient negative — satisfied. For \(1<x<2\): both positive — fails. For \(x>2\): negative over positive — satisfied. Answer: \(x<1\) or \(x>2\). Spot-checks agree: at \(x=0\) the left side is \(-2\), safely below 4; at \(x=1.5\) it is \(7\), correctly excluded.

Now a two-modulus inequality: \(|2x-1|<|x+3|\). Both sides are non-negative, so square: \(4x^2-4x+1<x^2+6x+9\), i.e. \(3x^2-10x-8<0\), which factorises as \((3x+2)(x-4)<0\), giving \(-\tfrac{2}{3}<x<4\). At \(x=0\): \(1<3\), inside and true. At \(x=5\): \(9<8\) is false — outside the interval, exactly as predicted.

MechanismRational graphs — asymptotes first, then the discriminant trick

Given a graph of \(y=f(x)\), the reciprocal graph \(y=\frac{1}{f(x)}\) (D11) follows mechanical rules: where \(f\) is zero, \(\frac{1}{f}\) has a vertical asymptote, and vice versa — \(f\)'s vertical asymptotes become places where \(\frac{1}{f}\) approaches zero. The sign never changes, because \(f\) and \(\frac{1}{f}\) share it. A local maximum of \(f\) at height \(k\neq0\) becomes a local minimum of \(\frac{1}{f}\) at height \(\frac{1}{k}\), and the two graphs intersect exactly where \(f(x)=\pm1\). Sketch \(f\), apply the rules, done.

For \(y=\frac{ax+b}{cx+d}\) (D12): one vertical asymptote at \(x=-\tfrac{d}{c}\), a horizontal one at \(y=\tfrac{a}{c}\) — the ratio of leading coefficients — one intercept on each axis, and no turning points; the curve is a translated, stretched \(\frac{1}{x}\). With quadratics over quadratics (D13) the horizontal asymptote is still the leading ratio \(y=\tfrac{a}{d}\), the vertical asymptotes sit at real roots of the denominator (two, one or none), and the curve may cross its horizontal asymptote at finite \(x\) — solve \(f(x)=\tfrac{a}{d}\) to find where. When the numerator's degree exceeds the denominator's by one, polynomial division produces an oblique asymptote \(y=mx+c\) plus a remainder that vanishes at infinity.

The range trick (D14) replaces calculus entirely. Set \(y\) equal to the function and cross-multiply into a quadratic in \(x\) whose coefficients contain \(y\). Points on the graph need real \(x\), so the discriminant must satisfy \(b^2-4ac\ge0\) — an inequality in \(y\) whose solution is the range, and whose boundary values are the stationary values of the function. Watch the degenerate case where the \(x^2\) coefficient vanishes: treat that \(y\) separately, because the equation drops to linear.

Worked example

Find the range of \(y=\dfrac{x^2+x+1}{x^2+1}\). Cross-multiply: \(y(x^2+1)=x^2+x+1\), so \((y-1)x^2-x+(y-1)=0\). For \(y\neq1\) this is a quadratic in \(x\), and real \(x\) requires \[(-1)^2-4(y-1)^2\ge0\quad\Rightarrow\quad(y-1)^2\le\tfrac{1}{4}\quad\Rightarrow\quad\tfrac{1}{2}\le y\le\tfrac{3}{2}.\] The degenerate case \(y=1\) gives \(-x=0\), so \(x=0\) works and \(y=1\) is attained. Range: \(\tfrac{1}{2}\le y\le\tfrac{3}{2}\), and the stationary values are the boundaries \(\tfrac{1}{2}\) and \(\tfrac{3}{2}\) — no differentiation performed. Verify: \(x=1\) gives \(y=\tfrac{3}{2}\) and \(x=-1\) gives \(y=\tfrac{1}{2}\), so both extremes are genuinely reached.

ModelConics and transformations — four curves, six moves

Four curves to know at sight (D15). The parabola \(y^2=4ax\): vertex at the origin, opens rightward, symmetric about the \(x\)-axis. The ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\): intercepts \((\pm a,0)\) and \((0,\pm b)\) — a unit circle stretched by \(a\) horizontally and \(b\) vertically. The hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\): vertices \((\pm a,0)\), two branches opening left and right, guided by the asymptotes \(y=\pm\tfrac{b}{a}x\). And the rectangular hyperbola \(xy=c^2\): asymptotes the coordinate axes themselves, branches in the first and third quadrants.

Transformations (D16) are substitutions into the equation. Replace \(x\) by \(x-a\) to translate \(a\) to the right; by \(\tfrac{x}{k}\) to stretch by factor \(k\) parallel to the \(x\)-axis. Reflections: in the \(x\)-axis, negate \(y\); in the \(y\)-axis, negate \(x\); in \(y=x\), swap \(x\) and \(y\); in \(y=-x\), map \((x,y)\to(-y,-x)\). A rotation of \(90^\circ\) anticlockwise about the origin sends \((x,y)\to(-y,x)\); an enlargement centre \(O\) with factor \(k\) replaces \(x\) and \(y\) by \(\tfrac{x}{k}\) and \(\tfrac{y}{k}\). For composites, order matters: translate-then-reflect and reflect-then-translate land in different places, so apply the substitutions strictly in sequence.

Worked example

Start with \(\frac{x^2}{9}-\frac{y^2}{4}=1\): vertices \((\pm3,0)\), asymptotes \(y=\pm\tfrac{2}{3}x\). Translate by \(2\) in the \(x\)-direction and \(1\) in the \(y\)-direction — replace \(x\) by \(x-2\) and \(y\) by \(y-1\): \[\frac{(x-2)^2}{9}-\frac{(y-1)^2}{4}=1.\] Everything travels with the curve: vertices to \((5,1)\) and \((-1,1)\), asymptotes to \(y-1=\pm\tfrac{2}{3}(x-2)\). Transforming the asymptotes along with the equation is the exam's favourite check that you moved the whole geometry, not just the algebra.

VocabularyKey terms the mark scheme pays for

Roots–coefficients relations

For \(ax^3+bx^2+cx+d=0\): \(\sum\alpha=-b/a\), \(\sum\alpha\beta=c/a\), \(\alpha\beta\gamma=-d/a\). Signs alternate; the quartic continues the pattern. Always divide by the leading coefficient.

Method of differences

Writing each term as \(f(r)-f(r+1)\) (often via partial fractions) so the sum telescopes, leaving only the first and last few terms.

Maclaurin series

\(f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\dots\) — the infinite polynomial that matches all of \(f\)'s derivatives at zero.

Interval of validity

The \(x\)-values for which an expansion converges: \(\ln(1+x)\) needs \(-1<x\le1\), \((1+x)^n\) needs \(|x|<1\); \(e^x\), \(\sin x\), \(\cos x\) hold for all \(x\).

l'Hôpital's rule

For limits of type \(0/0\) or \(\infty/\infty\) only: the limit of \(f/g\) equals the limit of \(f'/g'\), differentiating top and bottom separately. Repeat while indeterminate.

Oblique asymptote

The slanted line \(y=mx+c\) a rational function approaches when the numerator's degree exceeds the denominator's by one; found by polynomial division.

Discriminant range method

Set \(y=f(x)\), rearrange to a quadratic in \(x\), demand \(b^2-4ac\ge0\) for real \(x\): the resulting inequality in \(y\) is the range, and its boundaries are the stationary values.

Rectangular hyperbola

\(xy=c^2\), whose asymptotes are the coordinate axes — perpendicular to each other, hence 'rectangular'.

TrapsMisconceptions that cost marks

“The sum of the roots of \(2x^3-6x^2+5x-7=0\) is 6.”

Actually: It is \(-b/a=6/2=3\). The relations always carry the leading coefficient underneath, and dropping it is the single most common error — check any answer by dividing the whole equation by \(a\) first.

“Clear the fraction in \(\frac{x+2}{x-1}<4\) by multiplying both sides by \(x-1\).”

Actually: The sign of \(x-1\) is unknown, so the inequality direction is unknown after multiplying. Multiply by \((x-1)^2\), which is positive, or bring everything to one side and sign-analyse. Doing it the naive way silently loses the whole region \(x<1\).

“A Maclaurin expansion is valid wherever you feel like using it.”

Actually: Each expansion has an interval of validity: \(\ln(1+x)\) diverges for \(x>1\), so you may use \(x=1\) to get \(\ln 2\) but never \(x=2\) to get \(\ln 3\). Quoting the validity is a mark in its own right.

“l'Hôpital's rule means differentiating the quotient.”

Actually: You differentiate numerator and denominator separately — no quotient rule — and only when the form is \(0/0\) or \(\infty/\infty\). Applied to a non-indeterminate limit it gives wrong answers with total confidence.

ExamWhat examiners want

Know what the formulae booklet already gives you: \(\sum r^2\), \(\sum r^3\) and the five standard Maclaurin expansions with their validity intervals are printed. The marks in this section are AO1 marks for deploying them and AO2 marks for the connecting argument, not for memory. When you sum a series, expand the summand, split it across the standard results, then factorise — mark schemes award the final accuracy mark to the fully factorised form. And 'hence' is an instruction, not a suggestion: it means use the part you just proved.

In a method-of-differences question the evidence is the cancellation: write the first two rows and the last two rows with dots between, or you are asking the examiner to take the telescoping on faith — showing the surviving terms is a method mark. For the discriminant range method, the sentence 'for real \(x\), \(b^2-4ac\ge0\)' is itself the method mark; write it before substituting anything. Before invoking l'Hôpital, state that the limit has the form \(\tfrac{0}{0}\) — using the rule on a non-indeterminate form is a straight error, and examiners check.

Inequalities and graphs are where accuracy marks leak. Give inequality answers as exact intervals, decide each boundary deliberately (closed for \(\le\), always open where a denominator dies), and test one value per region — thirty seconds that protects four marks. On any sketch, label every asymptote with its equation and every intercept with coordinates; an unlabelled asymptote is an unmarked one. Roots-and-coefficients answers can always be self-checked: transform the roots, recompute their sum from the new coefficients, and confirm the two numbers agree before moving on.