Numerical Methods: past paper questions & worked solutions

The function $f(x)=x^2-5x+e^x$ is continuous on $[1,2]$.

Since the endpoint values have opposite signs, there is a root strictly between 1 and 2.

Using the given iteration with $x_1=1$,

Continue with unrounded values:

Subsequent iterates are $1.7340023\ldots,1.7339945\ldots,1.7339960\ldots$, which settle to $1.7339957\ldots$. Thus (i) $x_2=1.5105$ and (ii) $\alpha=1.7340$, each to 4 decimal places.

The polynomial $g$ is continuous on $[3,4]$.

There is a sign change, so the intermediate value theorem gives a root strictly between 3 and 4.

(i) With $x_1=3$, $x_2=\sqrt[6]{994}=3.1591074477\ldots$. (ii) Retaining unrounded values gives

Further iteration leaves the displayed four decimal places unchanged.

Let $g(x)=3x-5+5\sin(x^2)$. Using radians,

Since $g$ is continuous and changes sign, a root $\alpha$ lies between $0.7$ and $0.8$.

On the stated root interval, $0.49\lt x^2\lt 0.64\lt \pi/2$, so the principal inverse sine and positive square root give

Starting with $x_1=0.7$,

Continue $x_{n+1}=\sqrt{\arcsin(1-0.6x_n)}$ in radians, retaining the full calculator values. The iterates converge to $0.758924378\ldots$, so

Work in radians. Substituting the endpoints into $f(x)=2\sec x+6x-3$ gives

The function is continuous on $[0.1,0.2]$ since $\cos x\ne0$ there. The sign change therefore places a root between the endpoints. Since the given root is unique, this root is $\alpha$.

Set $f(x)=0$ and use $\sec x=1/\cos x$:

Thus $\alpha$, which satisfies $f(\alpha)=0$, satisfies this rearranged equation.

With $x_1=0.15$ and angles in radians,

Round only the displayed result to four decimal places.

Continuing the iteration with unrounded internal values gives

The iterates settle at $0.16223153\ldots$, giving the root to four decimal places.

For $f(x)=x^4-5x^2+4x-7$,

The polynomial is continuous on $[2,3]$ and the endpoint values have opposite signs. Therefore a root lies strictly between 2 and 3.

Rearrange $x^4-5x^2+4x-7=0$ to $x^4=5x^2-4x+7$. Since $f(0)=-7$, zero is not a root, so division by $x$ is valid at a root:

Taking the real cube root gives the requested form.

With $x_1=2$,

Continue using unrounded intermediate values: $x_3=2.146846021\ldots$, $x_4=2.154063147\ldots$, $x_5=2.155869187\ldots$, $x_6=2.156321484\ldots$. Further iteration gives $x_{10}=2.156472046\ldots$ and $x_{13}=2.156472632\ldots$.

The function is continuous on $[1,2]$ and the endpoint values have opposite signs, so a root $\alpha$ lies in $(1,2)$.

Multiplying $f(x)=0$ by $x^2e^x$, which is nonzero on the stated interval,

As $2x+3\gt 0$ here, divide and take the real cube root:

Starting with $x_1=1$ and retaining unrounded values,

Repeated iteration converges to $1.181331\ldots$, giving $\alpha=1.1813$ to four decimal places.

Using the product and chain rules,

At a stationary point, divide by $x/2$, since $x\gt 0$, to obtain $4\cos(x/2)=x\sin(x/2)$. The endpoint $x=\pi$ does not satisfy this equation, so at P we have $0\lt x\lt \pi$ and $\cos(x/2)\gt 0$. Hence $\tan(x/2)=4/x$. The principal arctangent is valid because $0\lt x/2\lt \pi/2$:

Use radian mode and retain full calculator precision between iterations:

Thus the requested second and sixth terms, not the limiting value, are as below.

Apply the product rule to the exponential term:

The derivative is continuous on $[0,1]$, and

The change of sign therefore places a zero of $f'$ between 0 and 1. Since the zero is unique, $\boxed{\alpha\in(0,1)}$, and hence it is in $[0,1]$.

Start at $x_1=0$, retaining unrounded values in each substitution:

Thus $\boxed{x_3=0.3260}$ to 4 decimal places.

Continuing the supplied iteration gives

Further iteration stabilises at $0.367471265\ldots$, so $\boxed{\alpha=0.3675}$ to 4 decimal places.

By the product and chain rules, $f'(x)=\cos(x/3)-(x/3)\sin(x/3)$.

Set $f'(x)=0$: $\cos(x/3)=(x/3)\sin(x/3)$. At a solution, cosine cannot vanish (that would leave a nonzero right side), so $\tan(x/3)=3/x$. On the principal positive branch $0\lt x/3\lt \pi/2$, this gives

This is the branch used by the supplied iteration for the root near 2.581. The original domain $x\gt 0$ also admits later tangent branches: in general $x=3\arctan(3/x)+3n\pi$. The principal-branch rearrangement must not be interpreted as listing every positive stationary point.

Using radians and $x_1=2.5$, the successive values are

Keep full precision between steps and round the requested terms only.

Define $F(x)=\cos(x/3)-(x/3)\sin(x/3)$, which is continuous near 2.581. The rounding interval endpoints give

By continuity a root lies strictly between these endpoints; every number in this open interval rounds to 2.581 to three decimal places.

Apply the product and chain rules:

Since $e^{-x^2}\gt 0$, a stationary point satisfies

Neither endpoint $x=0,\pi/3$ is stationary. In the interior $\sin3x\gt 0$ and $x\gt 0$, so $\cos3x\gt 0$ at a stationary point and $0\lt 3x\lt \pi/2$. Therefore division and the principal inverse tangent are valid:

Work in radians and retain calculator precision between iterations:

Thus (i) $\boxed{x_2=0.4367}$ and (ii) $\boxed{x_4=0.4307}$.

Let $h(x)=3\cos3x-2x\sin3x$, which is continuous. At the rounding boundaries,

The sign change places a root strictly between $0.4295$ and $0.4305$. On $0\lt x\lt \pi/6$, $h'(x)=-11\sin3x-6x\cos3x\lt 0$, so the stationary root is unique; for $\pi/6\leq x\leq\pi/3$, $h(x)\lt 0$. Hence the $x$ coordinate of $P$ is $\boxed{0.430\text{ to 3 decimal places}}$.

Let $F(x)=6\ln(2x+3)-x^2/2+4$. Then $F(-1.25)=-0.94013308\ldots$ and $F(-1.2)=0.21504626\ldots$. The log argument is positive on this interval, so F is continuous there. The change of sign proves a zero between these endpoints, giving the stated bracket for P.

Apply the given recurrence with $x_1=6$:

Continuing without rounding the internal values,

Further iteration settles at $6.41417418\ldots$. The positive square-root branch corresponds to Q.

Differentiate: $F'(x)=12/(2x+3)-x$. For a stationary point, $12=x(2x+3)$, or $2x^2+3x-12=0$. The quadratic formula gives $x=(-3\pm\sqrt{105})/4$. The negative choice is less than $-3/2$ and outside the original domain. Also $F”(x)=-24/(2x+3)^2-1\lt 0$, confirming the admissible point is a maximum.

Write $f(x)=(x^3-4x)e^{-x/2}$. By the product rule,

At O, $f'(0)=-4$, so the normal gradient is $1/4$ and its equation is $y=x/4$. At the other intersection P, x is nonzero; equating curve and normal and dividing by x gives $(x^2-4)e^{-x/2}=1/4$. Thus $x^2=4+e^{x/2}/4=(16+e^{x/2})/4$. Since P lies in the third quadrant, select the negative square root.

Starting from $x_1=-2$, $x_2=-\tfrac12\sqrt{16+e^{-1}}=-2.0228617996\ldots$. Further unrounded iterates are $x_3=-2.0226034084\ldots$, $x_4=-2.0226063126\ldots$ and $x_5=-2.0226062799\ldots$. The continuing sequence settles at $-2.0226062803\ldots$.

At P, $18=5e^{x-1}+3$, so $e^{x-1}=3$. Taking natural logarithms gives $x-1=\ln3$, hence $x=1+\ln3=\ln e+\ln3=\ln(3e)$.

Define $F(x)=5e^{x-1}+x^2-7$; its zeros are the intersection abscissae, and F is continuous. At the endpoints of the three-decimal rounding interval,

Continuity guarantees a root between them. Moreover $F'(x)=5e^{x-1}+2x\gt 0$ on this positive interval, so that root is unique there. It is the positive intersection $\alpha$, and rounds to 1.134 to three decimal places.

Start at $x_1=-3$ and keep full precision in $x_{n+1}=-\sqrt{7-5e^{x_n-1}}$:

The later iterates agree to six decimal places. The negative root corresponds to $\beta$ in the figure.

At the first landing, $h=0$ and $x=d\gt 0$:

Taking logarithms, $0.02d=\ln3$, hence $d=50\ln3=54.930614\ldots$. The root $x=0$ is the launch point, not the landing point.

By the product and chain rules,

At a stationary point,

Since $x+50\gt 0$, logarithms give $0.02x=\ln(150/(x+50))$. Also $h”(x)=-(0.02+0.0002x)e^{0.02x}\lt 0$ on $[0,d]$. Thus the stationary point is the unique maximum; the endpoints both have zero height.

(i) With $x_1=30$,

(ii) Continuing with full intermediate precision gives $x_3=30.54431108\ldots$, $x_4=31.09139065\ldots$, $x_5=30.75292479\ldots$ and $x_6=30.96205562\ldots$. Continued iteration converges to $30.88212\ldots$, which lies in $(0,d)$. This is the horizontal distance at maximum height, not the maximum height itself.

Differentiate $y=(3+4e^{x^2})^{1/2}$ using the chain rule:

Since the tangent at $P=(\alpha,\sqrt{3+4e^{\alpha^2}})$ passes through the origin, its gradient is $y_P/\alpha$, with $\alpha\gt 0$. Equate this to the derivative:

Consequently $x=\alpha$ solves the displayed equation.

Let $h(x)=4x^2e^{x^2}-4e^{x^2}-3$, a continuous function. Then $h(1)=-3\lt 0$ and $h(2)=12e^4-3=652.1778\ldots\gt 0$, so a root lies between 1 and 2. Moreover $h'(x)=8x^3e^{x^2}\gt 0$ for $x\gt 0$, so this is the unique positive root corresponding to the tangent abscissa $\alpha$.

From $4x^2e^{x^2}=4e^{x^2}+3$, divide by the positive $e^{x^2}$ to get $4x^2=4+3e^{-x^2}$. The source domain has $x\ge0$, so take the positive square root.

With $x_1=1$, the unrounded iterates start

Continue: $x_6=1.1051212304\ldots$, $x_7=1.1050504628\ldots$, $x_8=1.1050661137\ldots$, $x_9=1.1050626523\ldots$; further iteration settles at $1.1050632792\ldots$. Therefore (i) $x_3=1.0997$ and (ii) $\alpha=1.1051$, both to four decimal places.

At one year,

The negative sign indicates a loss. Its magnitude is about $828058$ pounds, which is approximately $830000$ pounds.

The model is continuous for $t\ge0$. Evaluate

Continuity and the sign change give a zero between 6 and 7.

Set $P=0$ and move the logarithm to the other side:

Multiply by 10 and divide by 4 to obtain the required form.

Starting with $t_1=6$, retain full calculator precision at each substitution:

Therefore the requested values to three decimal places are as follows.

The break-even time is approximately $6.315$ years, or $6.315\times12=75.78$ months. Profit starts just after that time; the first complete number of months is therefore 76, not 75. Indeed $P(75/12)\lt 0$ and $P(76/12)\gt 0$.

Let $f(x)=2\cos3x-3x+4$, using radians. Then

The function is continuous on $[0.8,0.9]$, so this change of sign shows that there is a root between the endpoints. Moreover, $f'(x)=-6\sin3x-3\lt 0$ on this interval, so this root is unique there. Thus the crossing shown at $P$ has $\boxed{0.8\lt \alpha\lt 0.9}$.

Using the inverse cosine in radians,

Hence $\boxed{x_2=0.8327}$ to 4 decimal places.

Continue the iteration, retaining the full calculator values at each step:

Therefore $\boxed{x_5=0.8110}$ to 4 decimal places.

Differentiate the curve equation:

At a stationary point,

For $x\gt 0$, the stationary points therefore belong to the two families

In the first family, $\cos3x=-\frac{\sqrt3}{2}$, giving $y”=9\sqrt3\gt 0$, so these are local minima. In the second family, $\cos3x=\frac{\sqrt3}{2}$ and $y”=-9\sqrt3\lt 0$, so these are local maxima and are excluded. Thus all positive minimum coordinates are

The two smallest are obtained with $n=0$ and $n=1$:

At $t=0$, $H=32+40-20=52$.

As $t\to\infty$, both decaying exponential terms tend to zero.

At the maximum,

Also $H”(t)=1.6e^{-0.2t}-16.2e^{-0.9t}$. At the stationary point, $e^{-0.9T}=(4/9)e^{-0.2T}$, so $H”(T)=-5.6e^{-0.2T}\lt 0$, confirming a maximum.

For $H=37$,

Taking logarithms gives

Using $-\ln u=\ln(1/u)$ for $u\gt 0$ then gives

Starting with $t_1=10$,

Continuing without rounding intermediate values gives $t_3=10.3954778346\ldots$, $t_4=10.3954789826\ldots$ and $t_5=10.3954789844\ldots$. Thus (i) $t_2=10.3947$, and (ii) $M=10.3955$ minutes, both to 4 decimal places.

Differentiate the velocity:

At a stationary velocity, $e^{t-10}=9e^{-0.75t}$, so $e^{1.75t}=9e^{10}$ and $t=(10+\ln9)/1.75=6.9698426\ldots$ seconds. The second derivative $-e^{t-10}-6.75e^{-0.75t}$ is negative, so this is a maximum. Substitute this unrounded time into v to get $11.88728125\ldots$ metres per second.

Integrating the velocity gives $12t-e^{t-10}+16e^{-0.75t}$. Evaluate between 0 and T:

Rearrange to $12T=116-16e^{-0.75T}+e^{T-10}-e^{-10}$ and divide by 12.

With $T_1=10$, substitution gives $T_2=9.7492587708\ldots$. Continuing with unrounded values gives

Further iteration settles at $9.7293319307\ldots$ seconds. Thus (i) $T_2=9.7493$ and (ii) the race time is $9.7293$ seconds, to four decimal places.