The MATLAB Notebook v1.5.2

We get a one-parameter family of answers depending on the variable y. In fact the three planes determined by the three linear equations are not independent, because the first equation is the sum of the second and third. The locus of points that satisfy the three equations is not a point, the intersection of three independent planes, but rather a line, the intersection of two distinct planes. Once again we check.

B = [3, -9, 8; 2, -3, 7; 1, -6, 1]; B*[x; y; z]

ans =

[ 2]

[ -1]

[ 3]

clear x y; syms x y; factor(x^4 - y^4)

ans =

(x-y)*(x+y)*(x^2+y^2)

(a)

simplify(1/(1 + 1/(1 + 1/x)))

ans =

(x+1)/(2*x+1)

(b)

simplify(cos(x)^2-sin(x)^2)

ans =

2*cos(x)^2-1

Let's try simple instead.

[better, how] = simple(cos(x)^2 - sin(x)^2)

better =

cos(2*x)

how =

combine

4.1067e+143

410674437175765127973978082146264947899391086876012309414440570235106991532497229781400618467066824164751453321793982128440538198297087323698003

But note the following:

sym('3^301')

ans =

3^301

This does not work because sym, by itself, does not cause an evaluation.

(a)

solve('8*x + 3 = 0','x')

ans =

-3/8

(b)

vpa(ans, 15)

ans =

-.375000000000000

(c)

syms p q; solve('x^3 + p*x + q = 0', 'x')

ans =

[ 1/6*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)-2*p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)]

[ -1/12*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+1/2*i*3^(1/2)*(1/6*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+2*p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3))]

[ -1/12*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)-1/2*i*3^(1/2)*(1/6*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+2*p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3))]

(d)

ezplot('exp(x)'); hold on; ezplot('8*x - 4')

fzero('exp(x) - 8*x + 4', 1)

ans =

0.7700

fzero('exp(x) - 8*x + 4', 3)

ans =

2.9929

(a)

hold off; ezplot('x^3 - x', [-4 4])

(b)

ezplot('sin(1/x^2)', [-2 2])

X = -2:0.1:2;

plot(X, sin(1./X.^2))

Warning: Divide by zero.

This picture is incomplete. Let's see what happens if we refine the mesh.

X = -2:0.01:2; plot(X, sin(1./X.^2))

Warning: Divide by zero.

Because of the wild oscillation near x = 0, neither plot nor ezplot gives a totally accurate graph of the function.

(c)

ezplot('tan(x/2)', [-pi pi]); axis([-pi, pi, -10, 10])

(d)

X = -2:0.05:2;

plot(X, exp(-X.^2), X, X.^4 - X.^2)

10.

Let's plot 2^x and x⁴ and look for points of intersection. We plot them first with ezplot just to get a feel for the graph.

ezplot('x^4'); hold on; ezplot('2^x'); hold off

Note the large vertical range. We learn from the plot that there are no points of intersection between 2 and 6 or -6 and -2; but there are apparently two points of intersection between -2 and 2. Let's change to plot now and focus on the interval between -2 and 2. We'll plot the monomial dashed.

X = -2:0.1:2; plot(X, 2.^X); hold on; plot(X, X.^4, '--'); hold off

We see that there are points of intersection near -0.9 and 1.2. Are there any other points of intersection? To the left of 0, 2^x is always less than 1, whereas x⁴ goes to infinity as x goes to -¥. On the other hand, both x⁴ and 2^x go to infinity as x goes to ¥, so the graphs may cross again to the right of 6. Let's check.

X = 6:0.1:20; plot(X, 2.^X); hold on; plot(X, X.^4, '--'); hold off

We see that they do cross again, near x = 16. If you know a little calculus, you can show that the graphs never cross again (by taking logarithms, for example), so we have found all the points of intersection. Now let's use the fzero command to find these points of intersection numerically. This command looks for a solution near a given starting point. To find the three different points of intersection we will have to use three different starting points. The above graphical analysis suggests appropriate starting points.

r1 = fzero('2^x - x^4', -0.9)

r2 = fzero('2^x - x^4', 1.2)

r3 = fzero('2^x - x^4', 16)

r1 =

-0.8613

r2 =

1.2396

r3 =

Let's check that these "solutions" satisfy the equation.

subs('2^x - x^4', 'x', r1)

subs('2^x - x^4', 'x', r2)

subs('2^x - x^4', 'x', r3)

ans =

2.2204e-016

ans =

-8.8818e-016

ans =

So r1 and r2 very nearly satisfy the equation, and r3 satisfies it exactly. It is easily seen that 16 is a solution. It is also interesting to try solve on this equation.

symroots = solve('2^x - x^4 = 0')

symroots =

[ -4*lambertw(-1/4*log(2))/log(2)]

[ 16]

[ -4*lambertw(-1/4*i*log(2))/log(2)]

[ -4*lambertw(1/4*log(2))/log(2)]

[ -4*lambertw(1/4*i*log(2))/log(2)]

In fact we get the three real solutions already found and two complex solutions.

format short; double(symroots)

ans =

1.2396

16.0000

-0.1609 + 0.9591i

-0.8613

-0.1609 - 0.9591i

Only the real solutions correspond to points where the graphs intersect.