Difference between revisions of "Roots Command"

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{{example| 1=<div><code><nowiki>Roots[f, -2, 1]</nowiki></code> with the function <code>f(x) = 3x³ + 3x² - x</code> yields <code>A = (-1.264, 0), B = (0, 0), C = (0.264, 0)</code></div>}}
 
{{example| 1=<div><code><nowiki>Roots[f, -2, 1]</nowiki></code> with the function <code>f(x) = 3x³ + 3x² - x</code> yields <code>A = (-1.264, 0), B = (0, 0), C = (0.264, 0)</code></div>}}
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<!-- Roots[3x³ + 3x² - x, -2, 1]  in CAS keeps giving -- ((-2527525234655819) / 2000000000000000, (-4349072213472027 (E) - 4500000000000000) / 500000000000000) even though the rounding is set to 2. Please, try it. The (E) - 4500000000000000 must be replaced by (10)^(- 4500000000000000) at least.-
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with Numeric Value, it gives: "(-1.263762617327909, -8.698144426944054 (E) - 9)" with the cientific notation including the "E".
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->

Revision as of 19:04, 15 December 2012



Roots[ <Function>, <Start x-Value>, <End x-Value> ]
Calculates the roots for function in the given interval. The function must be continuous on that interval. Because this algorithm is numeric, it may not find all the roots in some cases.
Example:
Roots[f, -2, 1] with the function f(x) = 3x³ + 3x² - x yields A = (-1.264, 0), B = (0, 0), C = (0.264, 0)


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