CP-4 Further algebra and functions — printable coverage pack

CP-4.1 · Understand and use the relationship between roots and coefficients of polynomial equations up to quartic equations.

For a monic cubic with roots $\alpha,\beta,\gamma$ , the coefficients give $\alpha+\beta+\gamma=-a$ , $\alpha\beta+\beta\gamma+\gamma\alpha=b$ and $\alpha\beta\gamma=-c$ in $x^3+ax^2+bx+c=0$ .
For a monic quartic, the elementary symmetric sums have alternating signs: sum of roots, pair products, triple products and the product are $-a,b,-c,d$ .
Derived sums are built from these relations; for example, $\sum\alpha^2=(\sum\alpha)^2-2\sum\alpha\beta$ .
A common error is to ignore the leading coefficient or lose the alternating signs when the polynomial is not monic.

Tier 1 · Easy

1. The roots of $x^3-5x^2+2x+8=0$ are $\alpha,\beta,\gamma$ . Without solving the equation, find $\alpha+\beta+\gamma$ , $\alpha\beta+\beta\gamma+\gamma\alpha$ and $\alpha\beta\gamma$ .[3 marks]
Answer
- $5$ , $2$ , $-8$
Method: Compare with $x^3-s_1x^2+s_2x-s_3=0$ . Therefore $s_1=5$ , $s_2=2$ and $-s_3=8$ , so $\alpha\beta\gamma=s_3=-8$ .

Tier 2 · Standard

1. The non-zero roots of $2x^4-3x^3-5x^2+7x-4=0$ are $\alpha,\beta,\gamma,\delta$ . Without solving the equation, find $\alpha+\beta+\gamma+\delta$ and $\frac1\alpha+\frac1\beta+\frac1\gamma+\frac1\delta$ .[4 marks]
Answer
- $\frac32$
- $\frac74$
Method: After dividing by $2$ , the sum of roots is $-(-3/2)=3/2$ . The sum of triple products is $-7/2$ and the product is $-2$ . Hence the sum of reciprocals is $\frac{\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta}{\alpha\beta\gamma\delta}=\frac{-7/2}{-2}=\frac74$ .

Tier 3 · Hard

1. The roots of $x^3-4x^2+px-6=0$ are $\alpha,\beta,\gamma$ , and $\alpha^2+\beta^2+\gamma^2=10$ . Determine $p$ and then find $\alpha^3+\beta^3+\gamma^3$ without solving the cubic.[5 marks]
Answer
- $p=3$
- $\alpha^3+\beta^3+\gamma^3=46$
Method: $\alpha+\beta+\gamma=4$ and $\alpha\beta+\beta\gamma+\gamma\alpha=p$ . Thus $10=4^2-2p$ , giving $p=3$ . Also $\alpha\beta\gamma=6$ . Using $\sum\alpha^3=(\sum\alpha)(\sum\alpha^2)-(\sum\alpha\beta)(\sum\alpha)+3\alpha\beta\gamma$ gives $4(10)-3(4)+3(6)=46$ .

CP-4.2 · Form a polynomial equation whose roots are a linear transformation of the roots of a given polynomial equation (of at least cubic degree).

If a new root is $y=a\alpha+b$ , rearrange to $\alpha=(y-b)/a$ and substitute this expression into the original polynomial.
After substitution, multiply by a suitable power of $a$ to clear denominators, then collect powers of the new variable.
For a shift $y=\alpha+b$ , replace the old variable by $y-b$ ; for a scale $y=a\alpha$ , replace it by $y/a$ .
A common error is to substitute the forward transformation $a y+b$ instead of expressing the old root in terms of the new one.

Tier 1 · Easy

1. The roots of $t^3-3t+1=0$ are $\alpha,\beta,\gamma$ . Form a polynomial equation whose roots are $\alpha+2,\beta+2,\gamma+2$ .[3 marks]
Answer
- $x^3-6x^2+9x-1=0$
Method: Let a new root be $x=t+2$ , so $t=x-2$ . Substitute into the original equation: $(x-2)^3-3(x-2)+1=0$ . Expanding and collecting terms gives $x^3-6x^2+9x-1=0$ .

Tier 2 · Standard

1. The roots of $2t^3-t^2+4t-3=0$ are $\alpha,\beta,\gamma$ . Form a polynomial equation whose roots are $3\alpha-1$ , $3\beta-1$ and $3\gamma-1$ .[4 marks]
Answer
- $2x^3+3x^2+36x-46=0$
Method: Let $x=3t-1$ , so $t=(x+1)/3$ . Substitute into the original polynomial and multiply by $27$ : $2(x+1)^3-3(x+1)^2+36(x+1)-81=0$ . Expanding gives $2x^3+3x^2+36x-46=0$ .

Tier 3 · Hard

1. The roots of $t^4-2t^3+t^2+3t-1=0$ are $\alpha,\beta,\gamma,\delta$ . Form a polynomial equation whose roots are $2-3\alpha$ , $2-3\beta$ , $2-3\gamma$ and $2-3\delta$ .[6 marks]
Answer
- $x^4-2x^3-3x^2-77x+85=0$
Method: Let $x=2-3t$ , so $t=(2-x)/3$ . Substitute into the original polynomial and multiply by $3^4=81$ : $(2-x)^4-6(2-x)^3+9(2-x)^2+81(2-x)-81=0$ . Expanding gives $x^4-2x^3-3x^2-77x+85=0$ .

CP-4.3 · Understand and use formulae for the sums of integers, squares and cubes and use these to sum other series.

Use $\sum_{r=1}^n r=\frac{n(n+1)}2$ , $\sum_{r=1}^n r^2=\frac{n(n+1)(2n+1)}6$ and $\sum_{r=1}^n r^3=\left[\frac{n(n+1)}2\right]^2$ .
Expand a polynomial expression in $r$ , split the summation term by term, and then substitute the standard formulae.
For example, $\sum(3r^2-2r+4)=3\sum r^2-2\sum r+4n$ before simplification.
A common error is to treat $\sum 1$ as $1$ instead of $n$ , or to change the summation limits while expanding.

Tier 1 · Easy

1. Evaluate $\sum_{r=1}^{20}r(r+1)$ .[3 marks]
Answer
- $3080$
Method: $r(r+1)=r^2+r$ , so the sum is $\frac{20(21)(41)}6+\frac{20(21)}2=2870+210=3080$ .

Tier 2 · Standard

1. Show that $\sum_{r=1}^{n}(3r^2-2r+4)=\frac{n(2n^2+n+7)}2$ .[4 marks]
Answer
- $\frac{n(2n^2+n+7)}2$
Method: Use standard sums: $3\sum r^2-2\sum r+4n=\frac{n(n+1)(2n+1)}2-n(n+1)+4n$ . Factoring $n/2$ gives $\frac n2[(n+1)(2n+1)-2(n+1)+8]=\frac{n(2n^2+n+7)}2$ .

Tier 3 · Hard

1. Prove that $\sum_{r=1}^{n}r(r+1)(2r+1)=\frac{n(n+1)^2(n+2)}2$ . Hence evaluate the sum when $n=15$ .[6 marks]
Answer
- $\frac{n(n+1)^2(n+2)}2$
- $32640$
Method: Expand $r(r+1)(2r+1)=2r^3+3r^2+r$ . Therefore the sum is $2[\frac{n(n+1)}2]^2+3\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2$ . Factoring and simplifying gives $\frac{n(n+1)^2(n+2)}2$ . At $n=15$ , this is $\frac{15(16)^2(17)}2=32640$ .

CP-4.4 · Understand and use the method of differences for summation of series including use of partial fractions.

The method of differences rewrites a term as $f(r)-f(r+1)$ , so intermediate terms cancel when the series is written out.
Partial fractions often reveal the required difference, such as $\frac1{r(r+1)}=\frac1r-\frac1{r+1}$ .
Write the first few and last few terms before cancelling; the surviving boundary terms determine the finite sum.
A common error is to cancel the final terms as well as the intermediate ones or to omit a constant factor from the partial fractions.

Tier 1 · Easy

1. Use the method of differences to find $\sum_{r=1}^{n}\frac1{(r+2)(r+3)}$ .[3 marks]
Answer
- $\frac n{3(n+3)}$
Method: $\frac1{(r+2)(r+3)}=\frac1{r+2}-\frac1{r+3}$ . Hence the sum is $(\frac13-\frac14)+(\frac14-\frac15)+\cdots+(\frac1{n+2}-\frac1{n+3})=\frac13-\frac1{n+3}=\frac n{3(n+3)}$ .

Tier 2 · Standard

1. Find a closed form for $\sum_{r=1}^{n}\frac1{r(r+2)}$ . Hence evaluate the sum for $n=10$ .[5 marks]
Answer
- $\frac{n(3n+5)}{4(n+1)(n+2)}$
- $\frac{175}{264}$
Method: $\frac1{r(r+2)}=\frac12(\frac1r-\frac1{r+2})$ . After cancellation, the sum is $\frac12(1+\frac12-\frac1{n+1}-\frac1{n+2})=\frac{n(3n+5)}{4(n+1)(n+2)}$ . Substituting $n=10$ gives $\frac{10(35)}{4(11)(12)}=\frac{175}{264}$ .

Tier 3 · Hard

1. Use partial fractions and the method of differences to prove that $\sum_{r=1}^{n}\frac1{r(r+1)(r+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ . Hence find the corresponding infinite sum.[6 marks]
Answer
- $\frac{n(n+3)}{4(n+1)(n+2)}$
- Infinite sum $=\frac14$
Method: $\frac1{r(r+1)(r+2)}=\frac12[\frac1{r(r+1)}-\frac1{(r+1)(r+2)}]$ . The finite sum is therefore $\frac12[\frac1{1\cdot2}-\frac1{(n+1)(n+2)}]=\frac14-\frac1{2(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ . Letting $n\to\infty$ gives $\frac14$ .

CP-4.5 · Find the Maclaurin series of a function including the general term.

The Maclaurin series is $f(x)=\sum_{r=0}^{\infty}\frac{f^{(r)}(0)}{r!}x^r$ wherever the expansion is valid.
Differentiate repeatedly, identify $f^{(r)}(0)$ , and include factorials and powers of $x$ explicitly in the general term.
Products or composites can sometimes be handled more efficiently by combining known series or by using a complex exponential and taking a real part.
A common error is to give only the first few terms when the question asks for a general term, or to omit $r!$ from the denominator.

Tier 1 · Easy

1. Find the Maclaurin series of $f(x)=\frac1{1-x}$ and give its general term.[3 marks]
Answer
- $1+x+x^2+x^3+\cdots=\sum_{r=0}^{\infty}x^r$
Method: $f^{(r)}(x)=r!(1-x)^{-(r+1)}$ , so $f^{(r)}(0)=r!$ . Hence the coefficient of $x^r$ is $f^{(r)}(0)/r!=1$ , giving $\sum_{r=0}^{\infty}x^r$ .

Tier 2 · Standard

1. Find the Maclaurin series of $e^{2x}$ through degree $4$ , and state the general term.[4 marks]
Answer
- $1+2x+2x^2+\frac43x^3+\frac23x^4+\cdots$
- General term $\frac{2^r x^r}{r!}$
Method: For $f(x)=e^{2x}$ , $f^{(r)}(x)=2^r e^{2x}$ , so $f^{(r)}(0)=2^r$ . Thus $e^{2x}=\sum_{r=0}^{\infty}\frac{2^r x^r}{r!}=1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+\frac{16x^4}{4!}+\cdots$ , which simplifies to the stated expansion.

Tier 3 · Hard

1. Find the Maclaurin series of $e^x\cos x$ through degree $5$ , and give a general term for the series.[6 marks]
Answer
- $1+x-\frac{x^3}{3}-\frac{x^4}{6}-\frac{x^5}{30}+\cdots$
- General term $\frac{2^{r/2}\cos(r\pi/4)}{r!}x^r$
Method: $e^x\cos x$ is the real part of $e^{(1+i)x}$ . Since $1+i=\sqrt2e^{i\pi/4}$ , the real part of the $r$ th term $\frac{(1+i)^r x^r}{r!}$ is $\frac{2^{r/2}\cos(r\pi/4)}{r!}x^r$ . Substituting $r=0,1,2,3,4,5$ gives $1+x+0x^2-\frac{x^3}{3}-\frac{x^4}{6}-\frac{x^5}{30}+\cdots$ .

CP-4.6 · Recognise and use the Maclaurin series for e^x, ln(1+x), sin x, cos x and (1+x)^n, and be aware of the range of values of x for which they are valid (proof not required).

$e^x=\sum_{r=0}^{\infty}\frac{x^r}{r!}$ , while $\sin x=x-\frac{x^3}{3!}+\cdots$ and $\cos x=1-\frac{x^2}{2!}+\cdots$ ; all three are valid for every real $x$ .
$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$ for $-1<x\leq1$ .
The binomial series $(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots$ is valid for $|x|<1$ when it does not terminate; a non-negative integer power gives an identity for all $x$ .
A common error is to substitute $ax$ into a standard series without also changing its range of validity.

Tier 1 · Easy

1. Write down the Maclaurin series for $\cos x$ through degree $6$ , and state its range of validity.[3 marks]
Answer
- $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$
- Valid for all real $x$
Method: Use the standard cosine series, which contains even powers with alternating signs: $\cos x=1-x^2/2!+x^4/4!-x^6/6!+\cdots$ . Its interval of convergence is all real $x$ .

Tier 2 · Standard

1. Write the first four terms of the expansion of $(1-2x)^{-1/2}$ , and state the range of values of $x$ for which it is valid.[4 marks]
Answer
- $1+x+\frac32x^2+\frac52x^3+\cdots$
- $|x|<\frac12$
Method: In $(1+u)^n$ , take $n=-1/2$ and $u=-2x$ . The first four terms are $1-\frac12u+\frac38u^2-\frac5{16}u^3$ , giving $1+x+\frac32x^2+\frac52x^3$ . The binomial condition $|u|<1$ becomes $|-2x|<1$ , so $|x|<1/2$ .

Tier 3 · Hard

1. Use standard Maclaurin series to show that $\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$ . State the range of validity and use terms up to $x^5$ to approximate $\ln(3/2)$ .[6 marks]
Answer
- $2(x+x^3/3+x^5/5+\cdots)$
- $-1<x<1$
- $\ln(3/2)\approx0.40546$
Method: $\ln(1+x)=x-x^2/2+x^3/3-x^4/4+x^5/5-\cdots$ , while $\ln(1-x)=-x-x^2/2-x^3/3-x^4/4-x^5/5-\cdots$ . Subtracting cancels the even powers and gives the stated series. Both component series are valid together for $-1<x<1$ . To make $(1+x)/(1-x)=3/2$ , use $x=1/5$ . Then $2(1/5+(1/5)^3/3+(1/5)^5/5)=0.405461\ldots$ , so $\ln(3/2)\approx0.40546$ .