bstoev Posted November 19, 2013 Posted November 19, 2013 (edited) The problem states that: [math]y_0 \neq 0[/math] [math]|y - y_0| < \frac{\epsilon|y_0|^2}{2}[/math] And I must use those to prove that: [math]y \neq 0[/math] [math]|\frac{1}{y} - \frac{1}{y_0}| < \epsilon[/math] My professor told me to utilize the inverse triangle inequality: [math]|a| - |b| \leq |a - b|[/math] Solving the first part was easy - I changed one of the expressions and used the triangle inequality:I change [math]|y - y_0|[/math] to [math]|y_0 - y|[/math] Then I use the inequality:[math]|y_0 - y| \geq |y_0| - |y| < \frac{|y_0|}{2} => -|y| < \frac{|y_0|}{2} - |y_0| => -|y| < -\frac{|y_0|}{2} => |y| > \frac{|y_0|}{2}[/math]So we know [math]|y| > 0[/math] Now for the second part, I put the expression [math]|\frac{1}{y} - \frac{1}{y_0}|[/math] into the form [math]|\frac{y_0 - y}{yy_0}|[/math] , but I don't know what else to do.Can someone help me? Edited November 19, 2013 by bstoev
mathematic Posted November 19, 2013 Posted November 19, 2013 |(yo-y)/yyo| < ε|yo/y|/2 (using given condition) < ε (using ineqality you obtained)
bstoev Posted November 19, 2013 Author Posted November 19, 2013 |(yo-y)/yyo| < ε|yo/y|/2 (using given condition) < ε (using ineqality you obtained) how did you derive that [math]\frac{y_0}{y}[/math] part in [math]\frac{\epsilon|\frac{y_0}{y}|}{2}[/math] ?Did you somehow prove that [math]|\frac{y_o}{y}|[/math] is greater than [math]|y_0|^2[/math] ?
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