This Further Maths roots of polynomials, series and Maclaurin revision notes is written for pupils aged 16-18 and follows the National Curriculum. In short, it shows you how to relate roots and coefficients, use summation formulae and the method of differences and find Maclaurin series. Before you start, it helps to revise matrices and linear transformations and inverse matrices and systems of equations, because this lesson builds on them.
Further Maths roots of polynomials, series and Maclaurin revision notes: what you will learn
This chapter links the roots of a polynomial to its coefficients, sums standard series by formula and by the method of differences, and builds Maclaurin series for functions.
Starter activity
Think of a quadratic equation such as \(x^2-7x+10=0\). Solve it, then add the two roots and multiply them. Do the same for \(x^2+3x-18=0\). What do you notice about the sum and the product compared with the coefficients? Now try to guess what happens for a cubic like \((x-1)(x-2)(x-4)=0\) once it is expanded. In the second half of this chapter we ask a related question in reverse: if we know the values of a function and its derivatives at a single point, how much of the function can we rebuild as a polynomial?
Roots and coefficients
Rule: Vieta's formulae
For \(ax^2+bx+c=0\) with roots \(\alpha,\beta\): \(\alpha+\beta=-\dfrac ba\) and \(\alpha\beta=\dfrac ca\).
For \(ax^3+bx^2+cx+d=0\): \(\sum\alpha=-\dfrac ba\), \(\sum\alpha\beta=\dfrac ca\), \(\alpha\beta\gamma=-\dfrac da\).
For \(ax^4+bx^3+cx^2+dx+e=0\): \(\sum\alpha=-\dfrac ba\), \(\sum\alpha\beta=\dfrac ca\), \(\sum\alpha\beta\gamma=-\dfrac da\), \(\alpha\beta\gamma\delta=\dfrac ea\). The signs alternate.
Key fact: symmetric functions
\(\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta\) and \(\sum\dfrac1\alpha=\dfrac{\sum\alpha\beta}{\alpha\beta\gamma}\).
Example 1
The roots of \(x^3-4x^2+x+6=0\) are \(\alpha,\beta,\gamma\). Find \(\sum\alpha^2\).
\(\sum\alpha=4\), \(\sum\alpha\beta=1\), \(\alpha\beta\gamma=-6\). So \(\sum\alpha^2=4^2-2(1)=14\).
Method: equation with transformed roots
- Write \(y=g(x)\) for the new roots.
- Solve for \(x\) in terms of \(y\).
- Substitute into the original equation and simplify.
- Rename \(y\) as \(x\).
For roots \(\alpha+1\) put \(x=y-1\); for \(2\alpha\) put \(x=\frac y2\); for \(\frac1\alpha\) put \(x=\frac1y\) and multiply by \(y^n\).
Example 2
The roots of \(x^3-6x^2+11x-6=0\) are \(1,2,3\). Find the equation with roots \(\alpha-1\).
Put \(x=y+1\): \((y+1)^3-6(y+1)^2+11(y+1)-6=y^3-3y^2+2y=0\), with roots \(0,1,2\) as expected.
Note: special cases
Roots in arithmetic progression \(a-d,a,a+d\) give \(a\) from the sum of roots. Roots in geometric progression \(\frac ar,a,ar\) give \(a^3\) from the product. If the coefficients are real, a complex root brings its conjugate.
Watch out
Sign errors in Vieta's formulae are common: for a cubic, \(\alpha\beta\gamma=-d/a\). Also remember to rename \(y\) back to \(x\) in a transformed equation.
Sums of series
Rule: standard sums
\[\sum_{r=1}^nr=\tfrac12n(n+1),\quad\sum_{r=1}^nr^2=\tfrac16n(n+1)(2n+1),\quad\sum_{r=1}^nr^3=\tfrac14n^2(n+1)^2.\]To sum from \(r=a\) to \(b\), subtract: \(\sum_1^b-\sum_1^{a-1}\). Always factorise the answer.
Example 3
Find \(\sum_{r=1}^nr(r+3)\).
\(\tfrac16n(n+1)(2n+1)+\tfrac32n(n+1)=\tfrac16n(n+1)(2n+10)=\tfrac13n(n+1)(n+5)\).
Example 4
Find \(\sum_{r=1}^n(3r^2-1)\) and test it.
\(\tfrac12n(n+1)(2n+1)-n=\tfrac n2\left[(n+1)(2n+1)-2\right]=\tfrac12n(2n^2+3n-1)\). For \(n=2\) the sum is \(2+11=13\) and the formula gives \(\tfrac12\cdot2\cdot13=13\).
Method of differences
Definition: method of differences
Write the general term as \(f(r)-f(r+1)\) (or \(f(r+1)-f(r)\)). The series then telescopes: all but a few terms cancel.
Method: telescoping a sum
- Use partial fractions to write the general term as a difference.
- Write out the first three and last three terms.
- Cancel and list the terms that survive at each end.
- For the sum to infinity, let \(n\to\infty\).
Example 5
Find \(\sum_{r=1}^n\dfrac1{r(r+1)}\) and the sum to infinity.
\(\dfrac1{r(r+1)}=\dfrac1r-\dfrac1{r+1}\), so \(\sum_{r=1}^n=1-\dfrac1{n+1}\), and the sum to infinity is 1. For \(\dfrac1{(r+1)(r+3)}=\tfrac12\left(\dfrac1{r+1}-\dfrac1{r+3}\right)\) two terms survive at each end.
Maclaurin series
Theorem: Maclaurin series
If \(f\) and its derivatives exist at \(0\):
\[f(x)=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^2+\dfrac{f'''(0)}{3!}x^3+\cdots\]Key fact: standard series
- \(e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots\) (all \(x\))
- \(\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\), \(\cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots\) (all \(x\))
- \(\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\cdots\) (\(-1\lt x\le1\))
- \((1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\cdots\) (\(|x|<1\) unless \(n\) is a positive integer)
Method: series of a composite or a solution of an equation
- Substitute \(x\to kx\) or \(x\to x^2\) into a standard series; scale the range of validity (for example \(\ln(1+2x)\) is valid for \(-\tfrac12\lt x\le\tfrac12\)).
- For a product, multiply the series and collect equal powers.
- If \(y\) satisfies a differential equation, differentiate the equation repeatedly and evaluate at \(x=0\).
Example 6
Find the first two non-zero terms of the Maclaurin series of \(\tan x\).
\(y'=1+y^2\), \(y''=2yy'\), \(y'''=2y'^2+2yy''\). At 0: \(y=0\), \(y'=1\), \(y''=0\), \(y'''=2\). So \(\tan x=x+\dfrac{2}{3!}x^3+\cdots=x+\dfrac{x^3}3+\cdots\).
Example 7
Use a series to find \(\lim_{x\to0}\dfrac{1-\cos x}{x^2}\) and approximate \(\sqrt{1.1}\).
\(1-\cos x=\tfrac{x^2}2-\cdots\), so the limit is \(\tfrac12\). And \(\sqrt{1.1}=(1+0.1)^{1/2}\approx1+0.05-0.00125+0.0000625=1.0488125\).
Watch out
- Missing the surviving terms in a method of differences: list enough terms.
- Omitting the factorials in a Maclaurin series.
- Quoting a series without its range of validity.
More methods and worked examples
Method: finding symmetric functions of the roots
- Write down \(\sum\alpha\), \(\sum\alpha\beta\), \(\sum\alpha\beta\gamma\) and the product of the roots from the coefficients, taking care with the alternating signs.
- Express the required quantity in terms of these sums. Useful identities are \(\sum\alpha^2=(\sum\alpha)^2-2\sum\alpha\beta\) and \(\sum\dfrac1\alpha=\dfrac{\sum\alpha\beta}{\alpha\beta\gamma}\) for a cubic.
- For higher powers, use the fact that each root satisfies the equation, for example \(\alpha^3=-b\alpha^2-c\alpha-d\) for a monic cubic, and add over all the roots.
- Substitute the numerical values and simplify. Check, if possible, with roots that you can find directly.
Worked example 8: sum of reciprocals of squares
The roots of \(x^3-4x^2+2x+3=0\) are \(\alpha\), \(\beta\), \(\gamma\). Find \(\dfrac1{\alpha^2}+\dfrac1{\beta^2}+\dfrac1{\gamma^2}\).
From the coefficients, \(\sum\alpha=4\), \(\sum\alpha\beta=2\) and \(\alpha\beta\gamma=-3\). The reciprocals \(\tfrac1\alpha\), \(\tfrac1\beta\), \(\tfrac1\gamma\) have sum \(\dfrac{\sum\alpha\beta}{\alpha\beta\gamma}=-\dfrac23\) and pairwise products with sum \(\dfrac{\sum\alpha}{\alpha\beta\gamma}=-\dfrac43\). So the required sum of squares is \(\left(-\tfrac23\right)^2-2\left(-\tfrac43\right)=\tfrac49+\tfrac83=\tfrac{28}9\). Treating the reciprocals as new roots in this way avoids messy fractions.
Worked example 9: a sum using the standard results
Find \(\sum_{r=1}^{n}(2r-1)^3\) in a factorised form.
Expand: \((2r-1)^3=8r^3-12r^2+6r-1\). Using the standard sums, the total is \(8\cdot\tfrac14n^2(n+1)^2-12\cdot\tfrac16n(n+1)(2n+1)+6\cdot\tfrac12n(n+1)-n\). This simplifies to \(2n^2(n+1)^2-2n(n+1)(2n+1)+3n(n+1)-n=n^2(2n^2-1)\). Check with \(n=2\): \(1+27=28\) and \(4\times7=28\). Always test a formula of this kind with a small value of \(n\), because a slip in the algebra is easy to make and very easy to detect.
Worked example 10: a Maclaurin series by substitution
Find the first three non-zero terms of the series for \(e^{x}\cos x\).
Multiply the series \(1+x+\tfrac{x^2}2+\tfrac{x^3}6+\cdots\) and \(1-\tfrac{x^2}2+\cdots\), keeping only terms up to \(x^3\). The constant is 1, the \(x\) term is \(x\), the \(x^2\) terms are \(\tfrac{x^2}2-\tfrac{x^2}2=0\), and the \(x^3\) terms are \(\tfrac{x^3}6-\tfrac{x^3}2=-\tfrac{x^3}3\). So \(e^x\cos x=1+x-\tfrac13x^3+\cdots\). Multiplying known series is much quicker than differentiating the product three times, and it gives the same result.
Watch out: signs and validity
The signs in the formulae for the roots alternate: for \(ax^3+bx^2+cx+d=0\) the sum of the roots is \(-\tfrac ba\), the sum of pairs is \(+\tfrac ca\) and the product is \(-\tfrac da\). For a quartic the product of the four roots is \(+\tfrac ea\). When you use a Maclaurin series, always state where it is valid: the series for \(e^x\), \(\sin x\) and \(\cos x\) hold for all \(x\), but the series for \(\ln(1+x)\) and \((1+x)^n\) need \(-1\lt x\le1\) or \(|x|<1\). A series used outside its interval of validity gives nonsense, however many terms you take.
How good is a Maclaurin approximation?
A Maclaurin polynomial matches the value of the function and of its first few derivatives at \(x=0\), so it is extremely accurate close to zero and less accurate further away. For an alternating series whose terms decrease in size, the error is smaller than the first term left out, which gives a simple way to decide how many terms you need. This is why small-angle approximations such as \(\sin x\approx x\) and \(\cos x\approx1-\tfrac12x^2\) are used in physics and engineering: for angles of a few degrees the next terms are smaller than the errors of measurement.
Worked example 11: a limit using series
Find \(\lim_{x\to0}\dfrac{\ln(1+x)-x}{x^2}\).
Replace \(\ln(1+x)\) by its series: \(\ln(1+x)-x=-\tfrac{x^2}2+\tfrac{x^3}3-\cdots\). Dividing by \(x^2\) gives \(-\tfrac12+\tfrac x3-\cdots\), and as \(x\) tends to zero every term except the first disappears. So the limit is \(-\tfrac12\). This method is often quicker and safer than other approaches, because the series shows exactly which power of \(x\) dominates near zero. A quick numerical check with \(x=0.01\) gives \(\dfrac{0.00995033-0.01}{0.0001}=-0.4967\), which is close to \(-0.5\) as expected.
Exam technique for roots and series
In questions on roots, examiners usually say "without solving the equation", so you must use the relations between roots and coefficients even if you can see the roots. Show the identity you use, such as the expression for the sum of squares, before substituting numbers. In questions on the method of differences, write out enough terms at the start and at the end of the sum to show clearly which terms cancel, because this display earns a method mark. In questions on Maclaurin series, show the derivatives evaluated at zero when the question asks you to find the series from first principles, and state the general term or the interval of validity when it is requested. Finally, simplify coefficients fully: a factorial left in the denominator, such as \(\tfrac{8}{3!}\), should be written as \(\tfrac43\).
Connecting the three topics
The three parts of this chapter are closer than they first appear. The relations between roots and coefficients tell us about a polynomial from its coefficients, while a Maclaurin series builds a polynomial whose coefficients come from the derivatives of a function at a single point. Sums of series appear in both directions: the standard sums of powers are themselves polynomials in \(n\), and the method of differences is a discrete version of the fundamental theorem of calculus, where a sum of differences collapses to the difference of the end values in the same way as an integral of a derivative does. Seeing these links will help you to choose the right method in a long problem.
Did you know?
The relations between the roots and the coefficients of a polynomial are named after the French lawyer and mathematician François Viète, who found them in about 1600. The series that carry the name of the Scottish mathematician Colin Maclaurin were published in 1742, as a special case of the series of Brook Taylor. Long before that, mathematicians in Kerala in southern India, led by Madhava around 1400, had already found the series for sine, cosine and arctangent, and used them to calculate \(\pi\) to eleven decimal places.
Otto’s tip
Test a general formula with \(n=1\) and \(n=2\). If it fails on a small case you have an algebra slip.
How to use this Further Maths roots of polynomials, series and Maclaurin revision notes
First, read the key points so that you know where the Further Maths roots of polynomials, series and Maclaurin revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the roots of polynomials, series and Maclaurin worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the roots of polynomials, series and Maclaurin test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Further Maths roots of polynomials, series and Maclaurin revision notes gives you everything you need for the chapter.
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