Difference between revisions of "Assume Command"

Latest revision as of 16:17, 31 January 2020

Assume( <Condition>, <Expression> )

Evaluates the expression according to the condition

Examples:

Assume(a > 0, Integral(exp(-a x), 0, infinity)) yields 1 / a.
Assume(x>0 && n>0, Solve(log(n^2*(x/n)^lg(x))=log(x^2), x)) yields {x = 100, x = n}
Assume(x<2,Simplify(sqrt(x-2sqrt(x-1)))) yields -sqrt(x - 1) + 1
Assume(x>2,Simplify(sqrt(x-2sqrt(x-1)))) yields sqrt(x - 1) - 1
Assume(k>0, Extremum(k*3*x^2/4-2*x/2)) yields \left\{ \left(\frac{2}{3 k}, -\frac{1}{3 k} \right) \right\}
Assume(k>0, InflectionPoint(0.25 k x^3 - 0.5x^2 + k)) yields \left\{ \left(\frac{2}{3 k}, \frac{27 k^{3} - 4}{27 k^{2}} \right) \right\}

Note: See also Solve Command.

@@ Line 3: / Line 3: @@
 ;Assume( <Condition>, <Expression> )
 :Evaluates the expression according to the condition
-:{{example|1=<code><nowiki>Assume(a > 0,Integral(exp(-a x), 0, infinity))</nowiki></code> yields ''1 / a''.}}
+:{{examples|<div>
-*See also [[Solve Command]].
+:*<code><nowiki>Assume(a > 0, Integral(exp(-a x), 0, infinity))</nowiki></code> yields <code>1 / a</code>.
+:*<code><nowiki>Assume(x>0 && n>0, Solve(log(n^2*(x/n)^lg(x))=log(x^2), x))</nowiki></code> yields <code>{x = 100, x = n}</code>
+:*<code><nowiki>Assume(x<2,Simplify(sqrt(x-2sqrt(x-1))))</nowiki></code> yields <code>-sqrt(x - 1) + 1</code>
+:*<code><nowiki>Assume(x>2,Simplify(sqrt(x-2sqrt(x-1))))</nowiki></code> yields <code>sqrt(x - 1) - 1</code>
+:*<code><nowiki>Assume(k>0, Extremum(k*3*x^2/4-2*x/2))</nowiki></code> yields <math> \left\{ \left(\frac{2}{3  k}, -\frac{1}{3 k} \right) \right\} </math>
+:*<code><nowiki>Assume(k>0, InflectionPoint(0.25 k x^3 - 0.5x^2 + k))</nowiki></code> yields <math> \left\{ \left(\frac{2}{3  k}, \frac{27  k^{3} - 4}{27  k^{2}} \right) \right\} </math>
+</div>}}
+{{note|1=See also [[Solve Command]].}}