Roots of polynomials, series and Maclaurin – further maths worksheet answers to download as PDF

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1. Answer to Question 1

   Method: for \(ax^2+bx+c=0\), \(\alpha+\beta=-\dfrac ba\) and \(\alpha\beta=\dfrac ca\); for a cubic \(ax^3+bx^2+cx+d=0\), \(\sum\alpha=-\dfrac ba\), \(\sum\alpha\beta=\dfrac ca\), \(\alpha\beta\gamma=-\dfrac da\). Other symmetric expressions are rewritten in terms of these, for example \(\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta\).

   1. Here \(a=2\), \(b=-5\), \(c=1\). For \(ax^2+bx+c=0\): \(\alpha+\beta=-\dfrac ba=\) \(\dfrac52\) and \(\alpha\beta=\dfrac ca=\) \(\dfrac12\).
   2. Express each quantity in terms of \(\alpha+\beta\) and \(\alpha\beta\). \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\dfrac{25}4-1=\) \(\dfrac{21}4\). \(\dfrac1\alpha+\dfrac1\beta=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{5/2}{1/2}=\) 5.
   3. Here \(a=1\), \(b=2\), \(c=-3\), \(d=5\). \(\sum\alpha=-\dfrac21=\) \(-2\), \(\sum\alpha\beta=\) \(-3\), \(\alpha\beta\gamma=-\dfrac51=\) \(-5\).
   4. Use the identity \((\alpha+\beta+\gamma)^2=\sum\alpha^2+2\sum\alpha\beta\). \(\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta=4-2(-3)=\) 10.

2. Answer to Question 2

   1. Let \(w=2x\), so \(x=\tfrac w2\): \(\tfrac{w^3}8-\tfrac{3w^2}4+\tfrac{5w}2-2=0\). Multiply by 8: \(w^3-6w^2+20w-16=0\).
   2. Let \(w=x+1\), so \(x=w-1\): \((w-1)^3-3(w-1)^2+5(w-1)-2=w^3-3w^2+3w-1-3w^2+6w-3+5w-5-2\), so \(w^3-6w^2+14w-11=0\).
   3. Let \(w=\tfrac1x\), so \(x=\tfrac1w\): \(\tfrac1{w^3}-\tfrac3{w^2}+\tfrac5w-2=0\). Multiply by \(w^3\): \(1-3w+5w^2-2w^3=0\), i.e. \(2w^3-5w^2+3w-1=0\).
   4. The original sum of roots is 3, so the new sum should be \(2\times3=6\), and the equation in (a) gives \(-\tfrac{-6}1=6\) ✓

3. Answer to Question 3

   Method: for \(ax^2+bx+c=0\), \(\alpha+\beta=-\dfrac ba\) and \(\alpha\beta=\dfrac ca\); for a quartic the pattern continues with alternating signs; for a cubic \(ax^3+bx^2+cx+d=0\), \(\sum\alpha=-\dfrac ba\), \(\sum\alpha\beta=\dfrac ca\), \(\alpha\beta\gamma=-\dfrac da\). Other symmetric expressions are rewritten in terms of these, for example \(\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta\).

   1. \(\sum\alpha=\) 2, \(\sum\alpha\beta=\) 3, \(\sum\alpha\beta\gamma=-(-1)=\) 1, \(\alpha\beta\gamma\delta=\) 4.
   2. \(\sum\alpha^2=(\sum\alpha)^2-2\sum\alpha\beta=4-6=\) \(-2\)
   3. If all four roots were real, their squares would be non-negative and \(\sum\alpha^2\ge0\). Since \(\sum\alpha^2=-2<0\), at least two roots are not real.
   4. \(\sum\dfrac1\alpha=\dfrac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}=\) \(\dfrac14\)

4. Answer to Question 4

   Method: split the sum into multiples of \(\sum r^3\), \(\sum r^2\), \(\sum r\) and \(\sum1=n\), substitute the standard formulae, and factorise out common factors such as \(n(n+1)\) as early as possible.

   1. \(6\times\tfrac16n(n+1)(2n+1)+2\times\tfrac12n(n+1)=n(n+1)(2n+1)+n(n+1)=n(n+1)(2n+2)=\) \(2n(n+1)^2\) ✓ Factorising \(n(n+1)\) first keeps the algebra short.
   2. Substitute \(n=20\) into the formula from part a). \(2\times20\times21^2=\) 17 640
   3. A sum that starts at \(r=11\) is the sum to 20 minus the sum to 10. \(\displaystyle\sum_{r=1}^{20}r^3-\sum_{r=1}^{10}r^3=\tfrac14(400)(441)-\tfrac14(100)(121)=44\,100-3025=\) 41 075
   4. Add the first two terms directly and compare with the formula. \(n=2\): \((6+2)+(24+4)=36\) and \(2\times2\times9=36\) ✓

5. Answer to Question 5

   1. \(\dfrac1{r(r+2)}=\) \(\dfrac12\left(\dfrac1r-\dfrac1{r+2}\right)\)
   2. \(\tfrac12\left[\left(1-\tfrac13\right)+\left(\tfrac12-\tfrac14\right)+\left(\tfrac13-\tfrac15\right)+\cdots+\left(\tfrac1{n-1}-\tfrac1{n+1}\right)+\left(\tfrac1n-\tfrac1{n+2}\right)\right]\). Every negative term cancels with a positive term two places later, leaving \(1\), \(\tfrac12\), \(-\tfrac1{n+1}\) and \(-\tfrac1{n+2}\).
   3. \(S_n=\tfrac12\left(\tfrac32-\tfrac1{n+1}-\tfrac1{n+2}\right)=\tfrac34-\tfrac12\cdot\dfrac{(n+2)+(n+1)}{(n+1)(n+2)}=\) \(\dfrac34-\dfrac{2n+3}{2(n+1)(n+2)}\) ✓
   4. As \(n\to\infty\) the fraction tends to 0: \(\dfrac34\).

6. Answer to Question 6

   1. \(f=e^{2x}\), \(f'=2e^{2x}\), \(f''=4e^{2x}\), \(f'''=8e^{2x}\); at 0 these are \(1,2,4,8\). So \(e^{2x}=1+2x+\tfrac42x^2+\tfrac86x^3+\cdots=\) \(1+2x+2x^2+\tfrac43x^3+\cdots\)
   2. \(f=\ln(1+x)\): \(f(0)=0\), \(f'=(1+x)^{-1}\to1\), \(f''=-(1+x)^{-2}\to-1\), \(f'''=2(1+x)^{-3}\to2\), \(f^{(4)}=-6(1+x)^{-4}\to-6\). So \(\ln(1+x)=x-\tfrac{x^2}2+\tfrac{x^3}3-\tfrac{x^4}4+\cdots\), valid for \(-1\lt x\le1\).
   3. \(\cos u=1-\tfrac{u^2}2+\tfrac{u^4}{24}-\cdots\) with \(u=3x\): \(1-\tfrac92x^2+\tfrac{27}8x^4-\cdots\)

7. Answer to Question 7

   1. Each dimension appears 4 times: \(4(\alpha+\beta+\gamma)=4\times12=\) 48 dm.
   2. Surface area \(=2(\alpha\beta+\beta\gamma+\gamma\alpha)=2\times44=\) 88 dm². Volume \(=\alpha\beta\gamma=\) 48 dm³.
   3. \(d^2=\alpha^2+\beta^2+\gamma^2=12^2-2\times44=56\), so \(d=\sqrt{56}=2\sqrt{14}\approx7.48\) dm.
   4. \(8-48+88-48=0\) ✓. Dividing: \(x^3-12x^2+44x-48=(x-2)(x^2-10x+24)=(x-2)(x-4)(x-6)\): 2, 4 and 6 dm. Edges \(4\times12=48\) ✓, area \(2(8+24+12)=88\) ✓, volume 48 ✓, diagonal \(\sqrt{4+16+36}=\sqrt{56}\) ✓.

8. Answer to Question 8

   Method: split the sum into multiples of \(\sum r^3\), \(\sum r^2\), \(\sum r\) and \(\sum1=n\), substitute the standard formulae, and factorise out common factors such as \(n(n+1)\) as early as possible.

   1. Layer \(r\) has \(r^2\) boxes; add each new layer to the previous total. Bottom layers 16, 25; totals 30, 55. So 4 layers need 30 boxes and 5 layers need 55 boxes.
   2. \(\displaystyle\sum_{r=1}^{n}r^2=\) \(\tfrac16n(n+1)(2n+1)\) Check: \(n=3\) gives \(\tfrac16\times3\times4\times7=14\).
   3. Try values of \(n\) near the answer and compare the totals with 1000. 13 layers need \(\tfrac16\times13\times14\times27=819\) boxes; 14 layers need \(1015>1000\). So 13 layers (819 boxes).
   4. Find the number of boxes, then multiply by the mass of each box. \(\tfrac16\times10\times11\times21=385\) boxes; \(385\times0.6=\) 231 kg.

9. Answer to Question 9

   1. \(\sin\theta=\theta-\dfrac{\theta^3}{6}+\dfrac{\theta^5}{120}-\cdots\)
   2. \(\sin0.2=0.198669\). Error \(=\dfrac{0.2-0.198669}{0.198669}\times100=\) 0.67%.
   3. \(0.2-\dfrac{0.008}6=0.2-0.001333=\) 0.198667, which agrees with \(\sin0.2\) to 5 decimal places (error about \(3\times10^{-6}\), the size of the next term \(\tfrac{0.2^5}{120}\)).
   4. \(\theta-\sin\theta=\dfrac{\theta^3}6-\dfrac{\theta^5}{120}+\cdots\), so the quotient is \(\dfrac16-\dfrac{\theta^2}{120}+\cdots\to\) \(\dfrac16\).

10. Answer to Question 10

    1. \(1+0.1+0.005+0.000167=\) 1.105167, error about \(4\times10^{-6}\).
    2. \(0.2-0.02+0.002667-0.0004=\) 0.182267, error about \(5.5\times10^{-5}\).
    3. The terms of the exponential series are divided by factorials, so they become small much faster; in the logarithm series the terms only shrink like \(\tfrac{x^n}n\), so more terms are needed.
    4. \(\ln3\) needs \(x=2\), outside the interval of validity \(-1\lt x\le1\): the series diverges. (One can use \(\ln3=-\ln\tfrac13\) with \(x=-\tfrac23\) instead.)

11. Answer to Question 11

    1. \(\cosh u=1+\dfrac{u^2}2+\dfrac{u^4}{24}+\cdots\)
    2. At \(x=\pm10\), \(y=20\cosh\tfrac12\); at \(x=0\), \(y=20\). Sag \(=20\cosh\tfrac12-20=20\left(\cosh\tfrac12-1\right)\) ✓
    3. \(20\left(\tfrac{0.25}2+\tfrac{0.0625}{24}\right)=20(0.125+0.0026042)=\) 2.5521 m
    4. Exact: \(20(1.1276260-1)=2.5525\) m, so the estimate is within 0.4 mm. The parabola keeps only the \(u^2\) term, giving 2.5 m: the higher terms are very small for a shallow cable.

12. Answer to Question 12

    1. Sum 39 years; product 1989.
    2. \(39^2-2\times491=1521-982=\) 539
    3. Let \(w=x+5\), so \(x=w-5\). Substituting and expanding: \(w^3-54w^2+956w-5544=0\). (Check: the new sum \(54=39+15\) ✓.)
    4. Divide by \((x-9)\): \(x^2-30x+221=(x-13)(x-17)\): 13 and 17. \(81+169+289=539\) ✓

13. Answer to Question 13

    1. Tile \(r\) has area \((2r-1)^2=4r^2-4r+1\); adding for \(r=1\) to \(n\) gives the sum ✓
    2. \(4\times\tfrac16n(n+1)(2n+1)-4\times\tfrac12n(n+1)+n=\tfrac13n\left(2(n+1)(2n+1)-6(n+1)+3\right)=\tfrac13n(4n^2-1)=\) \(\tfrac13n(2n-1)(2n+1)\) ✓
    3. \(\tfrac13\times20\times39\times41=\) 10 660 cm²
    4. \(1\) m² \(=10\,000\) cm². 19 tiles: \(\tfrac13\times19\times37\times39=9139\) cm²; 20 tiles exceed \(10\,000\). 19 tiles.

14. Answer to Question 14

    Method: the Maclaurin series is \(f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^2+\cdots\). Often it is quicker to substitute into a known series (for \(e^x\), \(\sin x\), \(\cos x\), \(\ln(1+x)\), \((1+x)^n\)) and to keep the range of validity of that series.

    1. Use Pythagoras and take out the factor 4 from under the root. \(d=\sqrt{4+0.16}=\sqrt{4(1+0.04)}=2\sqrt{1.04}\) ✓
    2. Differentiate \((1+x)^{1/2}\) twice, or use the binomial series with \(n=\tfrac12\). \(f=(1+x)^{1/2}\), \(f'(0)=\tfrac12\), \(f''(0)=-\tfrac14\): \(\sqrt{1+x}=1+\tfrac12x-\tfrac18x^2+\cdots\)
    3. Substitute \(x=0.04\): \(\tfrac12x=0.02\) and \(\tfrac18x^2=0.0002\). \(2\left(1+0.02-\tfrac18(0.0016)\right)=2(1.0198)=\) 2.0396 m
    4. Exact: \(\sqrt{4.16}=2.039608\) m, so the error is about \(0.000008\) m \(=\) 0.008 mm, negligible for a builder.

15. Answer to Question 15

    1. \(\sum\alpha=\) 0, \(\sum\alpha\beta=\) \(-7\), \(\alpha\beta\gamma=\) \(-6\)
    2. \(0^2-2(-7)=\) 14
    3. Each root satisfies the equation, so \(\alpha^3-7\alpha+6=0\). Adding for all three roots: \(\sum\alpha^3=7\sum\alpha-18=\) \(-18\).
    4. \(x^3-7x=-6\) gives \(x(x^2-7)=-6\). Square: \(x^2(x^2-7)^2=36\), so \(w(w-7)^2=36\): \(w^3-14w^2+49w-36=0\).
    5. \(x=1\) is a root; \(x^3-7x+6=(x-1)(x^2+x-6)=(x-1)(x-2)(x+3)\): roots \(1, 2, -3\). Checks: \(\sum\alpha^2=1+4+9=14\) ✓, \(\sum\alpha^3=1+8-27=-18\) ✓, and \(w\)-roots \(1,4,9\) satisfy \((w-1)(w-4)(w-9)=w^3-14w^2+49w-36\) ✓

16. Answer to Question 16

    1. The conjugate \(1-2i\) is a root. \((x-1-2i)(x-1+2i)=(x-1)^2+4=\) \(x^2-2x+5\).
    2. Dividing, \(f(x)=(x^2-2x+5)(x^2-2x-3)=(x^2-2x+5)(x-3)(x+1)\). Roots \(1\pm2i\), \(3\), \(-1\).
    3. Sum \(=(1+2i)+(1-2i)+3-1=4=-\tfrac{-4}1\) ✓. Product \(=(1+4)(3)(-1)=-15\) ✓.
    4. \(\sum\alpha^2=4^2-2\times6=4\). From the roots: \((1+2i)^2+(1-2i)^2+9+1=(-3+4i)+(-3-4i)+10=4\) ✓
    5. \((x-1)^4=x^4-4x^3+6x^2-4x+1\), so \(f(x)=(x-1)^4-16\) ✓. Then \((x-1)^4=16\) gives \(x-1=2,-2,2i,-2i\): \(x=3,-1,1+2i,1-2i\).

17. Answer to Question 17

    1. First row: \(\tfrac67\), \(\tfrac89\); suggest \(\dfrac{2n}{2n+1}\). Second row: \(5+18=23\), \(23+96=119\); suggest \((n+1)!-1\).
    2. \(\dfrac1{2r-1}-\dfrac1{2r+1}=\dfrac{(2r+1)-(2r-1)}{(2r-1)(2r+1)}=\dfrac2{(2r-1)(2r+1)}\) ✓. Sum: \(\left(1-\tfrac13\right)+\left(\tfrac13-\tfrac15\right)+\cdots+\left(\tfrac1{2n-1}-\tfrac1{2n+1}\right)=1-\dfrac1{2n+1}=\) \(\dfrac{2n}{2n+1}\) ✓
    3. \((r+1)!-r!=r!\,(r+1-1)=r\cdot r!\) ✓. Sum: \((2!-1!)+(3!-2!)+\cdots+((n+1)!-n!)=\) \((n+1)!-1\) ✓
    4. \(1-\dfrac1{2n+1}>0.999\) needs \(2n+1>1000\), so \(n=500\).
    5. The first sum converges to 1; the second grows without limit (like a factorial), so it diverges.

18. Answer to Question 18

    1. \(\left(1+x+\tfrac{x^2}2+\tfrac{x^3}6\right)\left(x-\tfrac{x^3}6\right)=x+x^2+\tfrac{x^3}2-\tfrac{x^3}6+\cdots=\) \(x+x^2+\tfrac13x^3+\cdots\)
    2. The numerator is \(\tfrac13x^3+\cdots\), so the limit is \(\dfrac13\).
    3. \(\ln(1+u)=u-\tfrac{u^2}2+\cdots\). \(u=-\tfrac{x^2}2+\tfrac{x^4}{24}\) and \(u^2=\tfrac{x^4}4+\cdots\). So \(\ln\cos x=-\tfrac{x^2}2+\tfrac{x^4}{24}-\tfrac{x^4}8=\) \(-\tfrac{x^2}2-\tfrac{x^4}{12}+\cdots\) ✓
    4. \(-0.005-0.0000083=\) \(-0.0050083\); calculator: \(\ln(\cos0.1)=-0.0050083\) ✓
    5. \(\dfrac d{dx}\ln\cos x=-\tan x\), so \(-\tan x=-x-\tfrac{x^3}3\), giving \(\tan x=x+\tfrac13x^3+\cdots\).

19. Answer to Question 19

    1. Put \(u=-x^2\) in \(1+u+\tfrac{u^2}2+\tfrac{u^3}6\): \(1-x^2+\tfrac{x^4}2-\tfrac{x^6}6+\cdots\) ✓
    2. \(a-\dfrac{a^3}3+\dfrac{a^5}{10}-\dfrac{a^7}{42}+\cdots\)
    3. \(-0.041667\), \(+0.003125\), \(-0.000186\). Total 0.46127.
    4. Error \(=0.461281-0.461272\approx0.000009\). The next term is \(+\dfrac{a^9}{216}=0.0000090\), which matches the error, as expected for an alternating series with decreasing terms.
    5. Integrals like this have no formula in terms of standard functions, but the series converges quickly, so values can be calculated to any accuracy and printed in tables (or programmed into calculators).

20. Answer to Question 20

    1. \(\dfrac1{1+t^2}=1-t^2+t^4-t^6+\cdots\) for \(|t|<1\). Since \(\displaystyle\int_0^x\frac{dt}{1+t^2}=\arctan x\), integrating term by term gives \(x-\tfrac{x^3}3+\tfrac{x^5}5-\tfrac{x^7}7+\cdots\) ✓
    2. \(\arctan1=\tfrac\pi4\), so \(\pi=4\left(1-\tfrac13+\tfrac15-\tfrac17+\cdots\right)\). Four terms: \(4\times0.72381=\) 2.8952, a poor estimate.
    3. \(\arctan\tfrac1{\sqrt3}=\tfrac\pi6\), and \(\left(\tfrac1{\sqrt3}\right)^{2k+1}=\tfrac1{\sqrt3}\cdot\tfrac1{3^k}\). So \(\tfrac\pi6=\tfrac1{\sqrt3}\left(1-\tfrac1{3\times3}+\tfrac1{5\times9}-\tfrac1{7\times27}+\cdots\right)\); multiply by 6 and use \(\tfrac6{\sqrt3}=2\sqrt3\) ✓
    4. \(1-0.111111+0.022222-0.005291=0.905820\), and \(2\sqrt3=3.464102\), so the estimate is \(3.464102\times0.905820=\) 3.13785, within 0.004 of \(\pi=3.14159\).
    5. In (b) the error after four terms is about 0.25, because the terms only shrink like \(\tfrac1n\). In (c) each term is divided by an extra factor of 3, so the terms shrink geometrically and the error after four terms is only about 0.004 (less than the next term, \(\tfrac{2\sqrt3}{9\times81}\approx0.005\)). A few more terms give many correct decimal places.