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Solving Equations

This worksheet contains various commented examples that demonstrate the Maple powerful equation solver, solve.

$restart$

An Introduction to the solve Command

The first example, which is the canonical example for this algorithm, is the following:

>	$eqns := \{x^{a} y^{b} - U, 1 + \frac{μ x^{a} a y^{b}}{x}, 1 + \frac{μ x^{a} y^{b} b}{y}\}$

$eqns ≔ \{1 + \frac{μ x^{a} a y^{b}}{x}, 1 + \frac{μ x^{a} y^{b} b}{y}, x^{a} y^{b} - U\}$

(1.1)

The difficulty here lies on $x^{a}$ and x, and also on $y^{b}$ and y, for which an unknown a and b cannot be related together in a polynomial form.

>	$solve (eqns, [x, y, μ])$

$[[x = {&ExponentialE;}^{- \frac{\ln (\frac{b}{a}) b - \ln (U)}{b + a}}, y = {&ExponentialE;}^{\frac{\ln (\frac{b}{a}) a + \ln (U)}{b + a}}, μ = - \frac{{&ExponentialE;}^{\frac{\ln (\frac{b}{a}) a + \ln (U)}{b + a}}}{U b}]]$

(1.2)

The same equation, even when $x^{a}$ and $y^{b}$ are replaced by arbitrary functions, can still be solved in terms of the inverses of these arbitrary functions.

>	$eqns := \{A (x) B (y) - U, 1 + \frac{μ A (x) a B (y)}{x}, 1 + \frac{μ A (x) B (y) b}{y}\}$

$eqns ≔ \{1 + \frac{μ A (x) a B (y)}{x}, 1 + \frac{μ A (x) B (y) b}{y}, A (x) B (y) - U\}$

(1.3)

Solving gives

>	$solve (eqns, [x, y, μ])$

$[[x = - RootOf (- A (-_Z U a) B (-_Z U b) + U) U a, y = - RootOf (- A (-_Z U a) B (-_Z U b) + U) U b, μ = RootOf (- A (-_Z U a) B (-_Z U b) + U)]]$

(1.4)

Carefully Solving Equations

These problems require the correct treatment of trigonometrics and exponentials, because the determinant is zero and no solutions exist.

>	$eqns := \{2 \cdot x \cdot \cos (1) + y = 1, \sin (2) \cdot x + \sin (1) \cdot y = 0\}$

$eqns ≔ \{2 x \cos (1) + y = 1, \sin (2) x + \sin (1) y = 0\}$

(2.1)

>	$solve (eqns, [x, y])$

$[]$

(2.2)

>	$eqns := \{\sin (w) \cdot x + (1 - \cos (w)) \cdot y = 5, (1 + \cos (w)) \cdot x + \sin (w) \cdot y = 3\}$

$eqns ≔ \{(1 + \cos (w)) x + \sin (w) y = 3, \sin (w) x + (1 - \cos (w)) y = 5\}$

(2.3)

>	$solve (eqns, [w, x, y])$

$[[w = - \arctan (\frac{15}{8}) + π, x = - \frac{5 y}{3} + \frac{17}{3}, y = y]]$

(2.4)

Equations and Inequalities Involving Exponentials, Logarithms, and Powers

The following examples involve exponentials or logarithms (where the answer can be expressed as rational expressions), and the use of the LambertW function. For example,;

>	$eqns := x^{2} = 2^{x}$

$eqns ≔ x^{2} = 2^{x}$

(3.1)

>	$solve (eqns, [x])$

$[[x = 2], [x = 4], [x = - \frac{2 LambertW (\frac{\ln (2)}{2})}{\ln (2)}]]$

(3.2)

The following problem stems from an exponential data fit.

>	$e1 := 46 = Ts + {&ExponentialE;}^{0} \cdot A &colon;$

>	$e2 := 39 = Ts + {&ExponentialE;}^{10 a} \cdot A &colon;$

>	$e3 := 33 = Ts + {&ExponentialE;}^{20 a} \cdot A &colon;$

>	$eqns := \{e1, e2, e3\}$

$eqns ≔ \{33 = Ts + {&ExponentialE;}^{20 a} A, 39 = Ts + {&ExponentialE;}^{10 a} A, 46 = Ts + A\}$

(3.3)

>	$solve (eqns, [Ts, A, a])$

$[[Ts = −3, A = 49, a = \frac{\ln (\frac{6}{7})}{10}]]$

(3.4)

Here are some examples of exponentials mixed in inequalities.

>	$eqns := {&ExponentialE;}^{x} + x \leq 1$

$eqns ≔ {&ExponentialE;}^{x} + x \leq 1$

(3.5)

>	$solve (eqns, [x])$

$[[x \leq 0]]$

(3.6)

>	$eqns ≔ \|x\| \cdot x + {&ExponentialE;}^{x} > 0$

$eqns ≔ 0 < |x| x + {&ExponentialE;}^{x}$

(3.7)

>	$solve (eqns, [x])$

$[[- 2 LambertW (\frac{1}{2}) < x]]$

(3.8)

>	$eqns := \frac{1}{2} \leq {&ExponentialE;}^{x} \cdot x^{2}$

$eqns ≔ \frac{1}{2} \leq {&ExponentialE;}^{x} x^{2}$

(3.9)

>	$solve (eqns, [x]) &semi;$

$[[x \leq 2 LambertW (- \frac{\sqrt{2}}{4}), 2 LambertW (−1, - \frac{\sqrt{2}}{4}) \leq x], [2 LambertW (\frac{\sqrt{2}}{4}) \leq x]]$

(3.10)

>	$evalf ()$

$[[x \leq −1.487962064, −2.617866616 \leq x], [0.5398352768 \leq x]]$

(3.11)

You can also solve equations involving powers.

>	$eqns := a^{2 x} - a^{\frac{x}{2}} + a^{\frac{x}{3}} = 2$

$eqns ≔ a^{2 x} - a^{\frac{x}{2}} + a^{\frac{x}{3}} = 2$

(3.12)

>	$solve (eqns, [x])$

$[[x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 1))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 2))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 3))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 4))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 5))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 6))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 7))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 8))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 9))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 10))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 11))}{\ln (a)}], [x = \frac{6 \ln (RootOf ({_Z}^{12} - {_Z}^{3} + {_Z}^{2} - 2, index = 12))}{\ln (a)}]]$

(3.13)

>	$eqns := {(x + 1)}^{x + a} = {(x + 1)}^{3}$

$eqns ≔ {(x + 1)}^{x + a} = {(x + 1)}^{3}$

(3.14)

>	$solve (eqns, [x])$

$[[x = 0], [x = 3 - a], [x = −1]]$

(3.15)

>	$eqns := {(x^{5} - 1)}^{x}$

$eqns ≔ {(x^{5} - 1)}^{x}$

(3.16)

>	$solve (eqns, [x])$

$[[x = 1], [x = RootOf ({_Z}^{4} + {_Z}^{3} + {_Z}^{2} +_Z + 1, index = 1)], [x = RootOf ({_Z}^{4} + {_Z}^{3} + {_Z}^{2} +_Z + 1, index = 4)]]$

(3.17)

abs, signum, and csgn

Some of these examples may cause ranges to be returned.

>	$eqns := {\|{(z + \|z + 2\|)}^{2} - 1\|}^{2} = 9$

$eqns ≔ {|{(z + |z + 2|)}^{2} - 1|}^{2} = 9$

(4.1)

>	$solve (eqns, [z])$

$[[z \leq −2], [z = 0]]$

(4.2)

>	$eqns := \frac{{\|z\|}^{2}}{z - 1} < \frac{{&ExponentialE;}^{2}}{&ExponentialE; - 1}$

$eqns ≔ \frac{{|z|}^{2}}{z - 1} < \frac{{&ExponentialE;}^{2}}{&ExponentialE; - 1}$

(4.3)

>	$solve (eqns, [z])$

$[[z < 1], [z < \frac{{&ExponentialE;}^{2} + \sqrt{{({&ExponentialE;}^{2})}^{2} - 4 {&ExponentialE;}^{2} &ExponentialE; + 4 {&ExponentialE;}^{2}}}{2 (&ExponentialE; - 1)}, - \frac{- {&ExponentialE;}^{2} + \sqrt{{({&ExponentialE;}^{2})}^{2} - 4 {&ExponentialE;}^{2} &ExponentialE; + 4 {&ExponentialE;}^{2}}}{2 (&ExponentialE; - 1)} < z]]$

(4.4)

>	$eqns := \{x^{2} < 1, y^{2} \leq 1, x + y < \frac{1}{2}\}$

$eqns ≔ \{y^{2} \leq 1, x^{2} < 1, x + y < \frac{1}{2}\}$

(4.5)

>	$solve (eqns, [x, y])$

$[[x < - \frac{1}{2}, −1 < x, y \leq 1, −1 \leq y], [x = - \frac{1}{2}, y < 1, −1 \leq y], [x < 1, - \frac{1}{2} < x, y = −1], [x < 1, - \frac{1}{2} < x, y < - x + \frac{1}{2}, −1 < y]]$

(4.6)

>	$eqns := x + y + \frac{4}{x + y} < 10$

$eqns ≔ x + y + \frac{4}{x + y} < 10$

(4.7)

>	$solve (eqns, [x])$

$[[x < - y], [x < - y + 5 + \sqrt{21}, - y + 5 - \sqrt{21} < x]]$

(4.8)

>	$eqns := signum (x) x < \|x - 4\| \|4 - x\| x$

$eqns ≔ signum (x) x < {|x - 4|}^{2} x$

(4.9)

>	$solve (eqns, [x])$

$[[5 < x], [x < 3, 0 < x]]$

(4.10)

>	$eqns := csgn (x) = \|x\|$

$eqns ≔ csgn (x) = |x|$

(4.11)

>	$solve (eqns, [x])$

$[[x = 0], [x = 1]]$

(4.12)

The following example is verified for six different intervals in x.

>	$eqns := \frac{1}{\|x^{2} - 1\|} < \|x^{3} - 2 x^{2} - 5 x + 1\|$

$eqns ≔ \frac{1}{|x^{2} - 1|} < |x^{3} - 2 x^{2} - 5 x + 1|$

(4.13)

>	$solve (eqns, [x])$

$[[x < RootOf ({_Z}^{5} - 2 {_Z}^{4} - 6 {_Z}^{3} + 3 {_Z}^{2} + 5_Z - 2, index = 4), RootOf ({_Z}^{5} - 2 {_Z}^{4} - 6 {_Z}^{3} + 3 {_Z}^{2} + 5_Z - 2, index = 5) < x], [RootOf ({_Z}^{5} - 2 {_Z}^{4} - 6 {_Z}^{3} + 3 {_Z}^{2} + 5_Z - 2, index = 3) < x], [x < RootOf ({_Z}^{4} - 2 {_Z}^{3} - 6 {_Z}^{2} + 3_Z + 5, index = 4)], [x < RootOf ({_Z}^{4} - 2 {_Z}^{3} - 6 {_Z}^{2} + 3_Z + 5, index = 2), RootOf ({_Z}^{4} - 2 {_Z}^{3} - 6 {_Z}^{2} + 3_Z + 5, index = 1) < x], [x < 0, RootOf ({_Z}^{4} - 2 {_Z}^{3} - 6 {_Z}^{2} + 3_Z + 5, index = 3) < x], [x < RootOf ({_Z}^{5} - 2 {_Z}^{4} - 6 {_Z}^{3} + 3 {_Z}^{2} + 5_Z - 2, index = 2), RootOf ({_Z}^{5} - 2 {_Z}^{4} - 6 {_Z}^{3} + 3 {_Z}^{2} + 5_Z - 2, index = 1) < x]]$

(4.14)

>	$evalf ()$

$[[x < −1.187054561, −1.468115597 < x], [3.394451740 < x], [x < −1.640158839], [x < 3.382444102, 1.086913325 < x], [x < 0., −0.8291985883 < x], [x < 0.8738032298, 0.3869151880 < x]]$

(4.15)

Linear Programming and Simplex Problems

>	$eqns := \{0 \leq y, 0 \leq x - y, x + y \leq 0\}$

$eqns ≔ \{0 \leq y, 0 \leq x - y, x + y \leq 0\}$

(5.1)

>	$solve (eqns, [x, y])$

$[[x = 0, y = 0]]$

(5.2)

>	$eqns := \{2 \leq x + y, x - 2 y \leq 1, 0 \leq x - y, y - \frac{x}{2} \leq - \frac{1}{2}\}$

$eqns ≔ \{0 \leq x - y, 2 \leq x + y, x - 2 y \leq 1, y - \frac{x}{2} \leq - \frac{1}{2}\}$

(5.3)

>	$solve (eqns, [x, y])$

$[[\frac{5}{3} \leq x, y = \frac{x}{2} - \frac{1}{2}]]$

(5.4)

Other Modes of Solving

>	$eqns := identity (\frac{1}{x^{2} - 1} = \frac{A}{C x + 1} + \frac{B}{D x - 1}, x)$

$eqns ≔ identity (\frac{1}{x^{2} - 1} = \frac{A}{C x + 1} + \frac{B}{D x - 1}, x)$

(6.1)

>	$solve (eqns, [A, B, C, D])$

$[[A = - \frac{1}{2}, B = \frac{1}{2}, C = 1, D = 1], [A = - \frac{1}{2}, B = \frac{1}{2}, C = −1, D = −1]]$

(6.2)

>	$eqns := identity (\sin (x) = a \cdot {&ExponentialE;}^{b x} + c \cdot {&ExponentialE;}^{d x}, x)$

$eqns ≔ identity (\sin (x) = a {&ExponentialE;}^{b x} + c {&ExponentialE;}^{d x}, x)$

(6.3)

>	$solve (eqns, [a, b, c, d])$

$[[a = \frac{I}{2}, b = −I, c = - \frac{I}{2}, d = I], [a = - \frac{I}{2}, b = I, c = \frac{I}{2}, d = −I]]$

(6.4)

>	$eqns := {&ExponentialE;}^{x} < 3 x or 5 x < {&ExponentialE;}^{x} and {&ExponentialE;}^{x} < 6 x$

$eqns ≔ {&ExponentialE;}^{x} < 3 x or 5 x < {&ExponentialE;}^{x} < 6 x$

(6.5)

>	$solve (eqns, [x])$

$[[x < - LambertW (−1, - \frac{1}{3}), - LambertW (- \frac{1}{3}) < x], [x < - LambertW (- \frac{1}{5}), - LambertW (- \frac{1}{6}) < x], [x < - LambertW (−1, - \frac{1}{6}), - LambertW (−1, - \frac{1}{5}) < x]]$

(6.6)

>	$evalf ()$

$[[x < 1.512134552, 0.6190612867 < x], [x < 0.2591711018, 0.2044814493 < x], [x < 2.833147892, 2.542641358 < x]]$

(6.7)

>	$_EnvExplicit := true &colon;$

>	$eqns := \{x^{6} + 8, x^{2} + 2 \neq 0\}$

$eqns ≔ \{x^{6} + 8, x^{2} + 2 \neq 0\}$

(6.8)

>	$solve (eqns, [x])$

>	$_EnvExplicit :='_EnvExplicit' &colon;$

$[[x = \sqrt{1 - I \sqrt{3}}], [x = - \sqrt{1 - I \sqrt{3}}], [x = \sqrt{1 + I \sqrt{3}}], [x = - \sqrt{1 + I \sqrt{3}}]]$

(6.9)

Special Functions

>	$eqns ≔ Ψ (3 x - 99) - Ψ (3 x - 100) + \frac{3}{x^{2}}$

$eqns ≔ Ψ (3 x - 99) - Ψ (3 x - 100) + \frac{3}{x^{2}}$

(7.1)

>	$solve (eqns, [x])$

$[[x = - \frac{9}{2} + \frac{\sqrt{1281}}{2}], [x = - \frac{9}{2} - \frac{\sqrt{1281}}{2}]]$

(7.2)

>	$eqns := \max (x, 3 x - 12) = \min (10 x + 8, 22 - x)$

$eqns ≔ \max (x, - 12 + 3 x) = \min (22 - x, 10 x + 8)$

(7.3)

>	$solve (eqns, [x])$

$[[x = \frac{17}{2}], [x = - \frac{8}{9}]]$

(7.4)

>	$eqns := LambertW (3 x) = \ln (x)$

$eqns ≔ LambertW (3 x) = \ln (x)$

(7.5)

>	$solve (eqns, [x])$

$[[x = {&ExponentialE;}^{3}]]$

(7.6)

>	$eqns := RootOf (a x^{2} = {&ExponentialE;}^{x}, x) = a$

$eqns ≔ RootOf (a {_Z}^{2} - {&ExponentialE;}^{_Z}) = a$

(7.7)

>	$solve (eqns, [a])$

$[[a = - 3 LambertW (- \frac{1}{3})], [a = - 3 LambertW (−1, - \frac{1}{3})], [a = - 3 LambertW (\frac{1}{6} - \frac{I \sqrt{3}}{6})], [a = - 3 LambertW (\frac{1}{6} + \frac{I \sqrt{3}}{6})]]$

(7.8)

Trigonometric Functions

This particular problem describes the positioning of a simple two-part robot arm. The answer is a little concentrated and has a degree-2 algebraic function.

>	$eq1 := x - (l1 \cos (t1) + l2 \cos (t1 + t2)) &colon;$

>	$eq2 := y - (l1 \sin (t1) + l2 \sin (t1 + t2)) &colon;$

>	$eqns := \{eq1, eq2\}$

>	$solve (eqns, [t1, t2])$

$eqns ≔ \{x - l1 \cos (t1) - l2 \cos (t1 + t2), y - l1 \sin (t1) - l2 \sin (t1 + t2)\}$

$[[t1 = \arctan (- \frac{2 x RootOf ({l1}^{4} - 2 {l1}^{2} {l2}^{2} + 2 {l1}^{2} x^{2} - 2 {l1}^{2} y^{2} + {l2}^{4} - 2 {l2}^{2} x^{2} - 2 {l2}^{2} y^{2} + x^{4} + 2 x^{2} y^{2} + y^{4} + (- 4 {l1}^{2} x + 4 {l2}^{2} x - 4 x^{3} - 4 x y^{2})_Z + (4 x^{2} + 4 y^{2}) {_Z}^{2}) - {l1}^{2} + {l2}^{2} - x^{2} - y^{2}}{2 l1 y}, \frac{RootOf ({l1}^{4} - 2 {l1}^{2} {l2}^{2} + 2 {l1}^{2} x^{2} - 2 {l1}^{2} y^{2} + {l2}^{4} - 2 {l2}^{2} x^{2} - 2 {l2}^{2} y^{2} + x^{4} + 2 x^{2} y^{2} + y^{4} + (- 4 {l1}^{2} x + 4 {l2}^{2} x - 4 x^{3} - 4 x y^{2})_Z + (4 x^{2} + 4 y^{2}) {_Z}^{2})}{l1}), t2 = \arctan (\frac{RootOf ({l1}^{4} - 2 {l1}^{2} {l2}^{2} + 2 {l1}^{2} x^{2} - 2 {l1}^{2} y^{2} + {l2}^{4} - 2 {l2}^{2} x^{2} - 2 {l2}^{2} y^{2} + x^{4} + 2 x^{2} y^{2} + y^{4} + (- 4 {l1}^{2} x + 4 {l2}^{2} x - 4 x^{3} - 4 x y^{2})_Z + (4 x^{2} + 4 y^{2}) {_Z}^{2}) (2 {l1}^{2} x - 2 {l2}^{2} x + 2 x^{3} + 2 x y^{2}) - {l1}^{4} + 2 {l1}^{2} {l2}^{2} - 2 {l1}^{2} x^{2} + 2 {l1}^{2} y^{2} - {l2}^{4} + 2 {l2}^{2} x^{2} + 2 {l2}^{2} y^{2} - x^{4} - 2 x^{2} y^{2} - y^{4}}{4 l1 y l2 RootOf ({l1}^{4} - 2 {l1}^{2} {l2}^{2} + 2 {l1}^{2} x^{2} - 2 {l1}^{2} y^{2} + {l2}^{4} - 2 {l2}^{2} x^{2} - 2 {l2}^{2} y^{2} + x^{4} + 2 x^{2} y^{2} + y^{4} + (- 4 {l1}^{2} x + 4 {l2}^{2} x - 4 x^{3} - 4 x y^{2})_Z + (4 x^{2} + 4 y^{2}) {_Z}^{2})}, \frac{- {l1}^{2} - {l2}^{2} + x^{2} + y^{2}}{2 l1 l2})]]$

(8.1)

This problem was posed by H. Melenk.

>	$eqns := \{\cos (x1) \cdot x3 - \sin (x1) \cdot x2 + 5 \cdot \sin (x1), - 2 \cdot \cos (x1) + 2 \cdot {x3}^{2} \cdot x2 + 2 \cdot {x2}^{3} + x2 - 5, - 2 \cdot \sin (x1) + 2 \cdot {x3}^{3} + 2 \cdot x3 \cdot {x2}^{2} + x3\}$

$eqns ≔ \{\cos (x1) x3 - \sin (x1) x2 + 5 \sin (x1), - 2 \sin (x1) + 2 {x3}^{3} + 2 x3 {x2}^{2} + x3, - 2 \cos (x1) + 2 {x3}^{2} x2 + 2 {x2}^{3} + x2 - 5\}$

(8.2)

>	$solve (eqns, [x1, x2, x3])$

$[[x1 = π, x2 = 1, x3 = 0], [x1 = π, x2 = RootOf (2 {_Z}^{2} + 2_Z + 3), x3 = 0], [x1 = 0, x2 = RootOf (2 {_Z}^{3} +_Z - 7), x3 = 0], [x1 = \arctan (\frac{21 RootOf ({_Z}^{2} + 1)}{20}, - \frac{29}{20}), x2 = \frac{21}{10}, x3 = \frac{21 RootOf ({_Z}^{2} + 1)}{10}]]$

(8.3)

Inverse Trigonometric Functions

The following examples involve inverse trigonometrics, and some were originally proposed by H. Melenk.

>	$eqns ≔ 2 \cdot \arcsin (x) - \arccos (3 x)$

$eqns ≔ 2 \arcsin (x) - \arccos (3 x)$

(9.1)

>	$solve (eqns, [x])$

$[[x = \frac{\sin (2 \arctan (\frac{3 \sqrt{2}}{2 (\frac{3 \sqrt{2}}{2} + 2 RootOf (12 \sqrt{2} {_Z}^{3} + 8 {_Z}^{4} - 12_Z \sqrt{2} + 10 {_Z}^{2} - 9, index = 1))}))}{3}]]$

(9.2)

>	$eqns ≔ 2 \cdot \arctan (x) - \arctan (\frac{2 x}{1 - x^{2}})$

$eqns ≔ 2 \arctan (x) - \arctan (\frac{2 x}{- x^{2} + 1})$

(9.3)

>	$solve (eqns, [x])$

$[[x = x]]$

(9.4)

>	$eqns := \arccos (x) - \arctan (x)$

$eqns ≔ \arccos (x) - \arctan (x)$

(9.5)

>	$solve (eqns, [x])$

$[[x = \frac{\sqrt{- 2 + 2 \sqrt{5}}}{2}]]$

(9.6)

Equations Requiring Branch Selection

The following examples require branch selection and generally involve ln, radicals, LambertW, inverse trig, and so on.

>	$eqns := x + {(x + 1)}^{\frac{1}{3}} = 2$

$eqns ≔ x + {(x + 1)}^{\frac{1}{3}} = 2$

(10.1)

>	$solve (eqns, [x])$

$[[x = - \frac{{(324 + 12 \sqrt{741})}^{\frac{1}{3}}}{6} + \frac{2}{{(324 + 12 \sqrt{741})}^{\frac{1}{3}}} + 2]]$

(10.2)

>	$eqns := \sqrt{x} - \sqrt{x - 1} = 3$

$eqns ≔ \sqrt{x} - \sqrt{x - 1} = 3$

(10.3)

>	$solve (eqns, [x])$

$[]$

(10.4)

>	$eqns := x + \sqrt{x} + x^{\frac{1}{3}} = 3$

$eqns ≔ x + \sqrt{x} + x^{\frac{1}{3}} = 3$

(10.5)

>	$solve (eqns, [x])$

$[[x = 1]]$

(10.6)

$eqns := 2^{\frac{1}{2}} \cdot 3^{\frac{1}{2}} \cdot 6^{\frac{1}{2}} - 2^{\frac{1}{2}} \cdot 6^{\frac{1}{2}} \cdot x - 2^{\frac{1}{2}} \cdot x \cdot 3^{\frac{1}{2}} + 3 \cdot 2^{\frac{1}{2}} \cdot x^{2} + x \cdot 3^{\frac{1}{2}} \cdot 6^{\frac{1}{2}} - 6^{\frac{1}{2}} \cdot x^{2} - 2 \cdot x^{2} \cdot 3^{\frac{1}{2}} + 6 \cdot x^{3}$

$eqns ≔ \sqrt{2} \sqrt{3} \sqrt{6} - \sqrt{2} \sqrt{6} x - \sqrt{2} x \sqrt{3} + 3 \sqrt{2} x^{2} + x \sqrt{3} \sqrt{6} - \sqrt{6} x^{2} - 2 x^{2} \sqrt{3} + 6 x^{3}$

(10.7)

>	$solve (eqns, [x])$

$[[x = −1], [x = \frac{\sqrt{6}}{12} - \frac{\sqrt{2}}{4} + \frac{\sqrt{3}}{6} + \frac{1}{2} + \frac{I \sqrt{72 + 12 \sqrt{2} \sqrt{3} + 6 \sqrt{2} \sqrt{6} - 4 \sqrt{6} \sqrt{3} + 36 \sqrt{2} - 24 \sqrt{3} - 12 \sqrt{6}}}{12}], [x = \frac{\sqrt{6}}{12} - \frac{\sqrt{2}}{4} + \frac{\sqrt{3}}{6} + \frac{1}{2} - \frac{I \sqrt{72 + 12 \sqrt{2} \sqrt{3} + 6 \sqrt{2} \sqrt{6} - 4 \sqrt{6} \sqrt{3} + 36 \sqrt{2} - 24 \sqrt{3} - 12 \sqrt{6}}}{12}]]$

(10.8)

>	$eqns := \ln (x) = \frac{75 π I}{11} + 17$

$eqns ≔ \ln (x) = \frac{75 I π}{11} + 17$

(10.9)

>	$[solve (eqns, [x])]$

$[[]]$

(10.10)

>	$eqns := x^{2} \cdot (\ln (8) - \ln (x)) = 8$

$eqns ≔ x^{2} (3 \ln (2) - \ln (x)) = 8$

(10.11)

>	$solve (eqns, [x])$

$[[x = 8 {&ExponentialE;}^{\frac{LambertW (- \frac{1}{4})}{2}}], [x = 8 {&ExponentialE;}^{\frac{LambertW (−1, - \frac{1}{4})}{2}}]]$

(10.12)

>	$evalf ()$

$[[x = 6.690844261], [x = 2.725891034]]$

(10.13)

Equations Requiring Expansion of Linear Operators

>	$S := \sum_{i = 1}^{n} {(a + b {&ExponentialE;}^{x_{i}} - y_{i})}^{2}$

$S ≔ n a^{2} + \sum_{i = 1}^{n} (- 2 b {&ExponentialE;}^{x_{i}} y_{i} + 2 a b {&ExponentialE;}^{x_{i}} + b^{2} {({&ExponentialE;}^{x_{i}})}^{2} - 2 a y_{i} + y_{i}^{2})$

(11.1)

>	$eqns := \{\frac{\partial}{\partial a} S, \frac{\partial}{\partial b} S\} &colon;$

>	$solve (eqns, [a, b])$

$[[a = - \frac{(\sum_{i = 1}^{n} {&ExponentialE;}^{x_{i}} y_{i}) (\sum_{i = 1}^{n} {&ExponentialE;}^{x_{i}}) - (\sum_{i = 1}^{n} {({&ExponentialE;}^{x_{i}})}^{2}) (\sum_{i = 1}^{n} y_{i})}{n (\sum_{i = 1}^{n} {({&ExponentialE;}^{x_{i}})}^{2}) - {(\sum_{i = 1}^{n} {&ExponentialE;}^{x_{i}})}^{2}}, b = \frac{n (\sum_{i = 1}^{n} {&ExponentialE;}^{x_{i}} y_{i}) - (\sum_{i = 1}^{n} {&ExponentialE;}^{x_{i}}) (\sum_{i = 1}^{n} y_{i})}{n (\sum_{i = 1}^{n} {({&ExponentialE;}^{x_{i}})}^{2}) - {(\sum_{i = 1}^{n} {&ExponentialE;}^{x_{i}})}^{2}}]]$

(11.2)

The rational solution is found. (Other solutions cannot be expressed in closed form.)

>	$eqns := \{1 - \frac{2 (\arccos (\frac{x}{2}) - \frac{x \sqrt{4 - x^{2}}}{4})}{π} = 0\}$

$eqns ≔ \{1 - \frac{2 \arccos (\frac{x}{2}) - \frac{x \sqrt{- x^{2} + 4}}{2}}{π} = 0\}$

(11.3)

>	$solve (eqns, [x])$

$[[x = - 2 \sin (RootOf (_Z + \sin (_Z) \sqrt{{\cos (_Z)}^{2}}))]]$

(11.4)

>	$eqns := 2 - \sin (1 - x) = 2 x$

$eqns ≔ 2 + \sin (x - 1) = 2 x$

(11.5)

>	$solve (eqns, [x])$

$[[x = 1]]$

(11.6)

Mixtures of Exponentials, Trigonometrics, and Radicals

The following examples involve solving equations with mixtures of exponential, trigonometrics, radicals, and so on.

>	$eqns := {&ExponentialE;}^{RootOf (1 + {_Z}^{2}) y} - x - RootOf (1 + {_Z}^{2}) {(- (x - 1) (x + 1))}^{\frac{1}{2}}$

$eqns ≔ {&ExponentialE;}^{RootOf ({_Z}^{2} + 1) y} - x - RootOf ({_Z}^{2} + 1) \sqrt{- (x - 1) (x + 1)}$

(12.1)

>	$solve (eqns, [x])$

$[[x = \frac{{({&ExponentialE;}^{−I y})}^{2} + 1}{2 {&ExponentialE;}^{−I y}}], [x = \frac{{({&ExponentialE;}^{I y})}^{2} + 1}{2 {&ExponentialE;}^{I y}}]]$

(12.2)

>	$eqns := \ln (x + \sqrt{x - 1} \sqrt{x + 1}) = y$

$eqns ≔ \ln (x + \sqrt{x - 1} \sqrt{x + 1}) = y$

(12.3)

>	$solve (eqns, [x])$

$[[x = \frac{{({&ExponentialE;}^{y})}^{2} + 1}{2 {&ExponentialE;}^{y}}]]$

(12.4)

>	$eqns := \ln ({&ExponentialE;}^{x} + 1) = 2 x$

$eqns ≔ \ln ({&ExponentialE;}^{x} + 1) = 2 x$

(12.5)

>	$solve (eqns, [x])$

$[[x = \ln (\frac{\sqrt{5}}{2} + \frac{1}{2})]]$

(12.6)

>	$eqns := \ln ({&ExponentialE;}^{x} + 1) = \ln ({&ExponentialE;}^{x} - 5) + \ln ({&ExponentialE;}^{x} - 2)$

$eqns ≔ \ln ({&ExponentialE;}^{x} + 1) = \ln ({&ExponentialE;}^{x} - 5) + \ln ({&ExponentialE;}^{x} - 2)$

(12.7)

>	$solve (eqns, [x])$

$[[x = \ln (4 + \sqrt{7})]]$

(12.8)

For more information, see the solve help page. See also the help pages on LambertW and on ln.

Return to Index for Example Worksheets

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