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 November 6th, 2010, 11:42 AM #1 Joined: Sep 2010 Posts: 73 Thanks: 0 determine pairs of functions independent Reviewing for an exam, so I'll probably be posting a lot of questions. I'll try my best to figure it out myself but when I've spent too much time on it, I'll come here. So, first question: Determine whether the pairs of functions are linearly independent or dependent on the real line. f(x) = x^2 g(x) = x^2 * |x| I need to take the Wronskian... but how do I do the derivative of an absolute value?
 November 6th, 2010, 12:51 PM #2 Global Moderator   Joined: May 2007 Posts: 4,701 Thanks: 257 Re: determine pairs of functions independent Let f(x) = |x|. f'(x)=1, x > 0 f'(x)= -1, x <0 f'(0) does not exist.
 November 6th, 2010, 01:54 PM #3 Joined: Sep 2010 Posts: 73 Thanks: 0 Re: determine pairs of functions independent Well then, how do I know whether I use 1 or -1 for f'(x)? Either way, I get the result does not equal zero, so they are linearly dependent. What if I come across an absolute value where f'(x) = 1 would make it dependent and f'(x) = -1 would make it independent? Then what?
 November 7th, 2010, 12:20 AM #4 Joined: Dec 2009 Posts: 150 Thanks: 0 Re: determine pairs of functions independent The following representation of $|x|$ may be of help to you: $|x| \,= \, \sqrt{ x^2 }$ Squaring x gives a positive number, even if x was negative. Then the square root acts on the positive result to give back a positive result. So whether or not x is negative, $\sqrt{ x^2 }$ returns the positive magnitude of $x$. Using the chain rule we differentiate the function: $\frac{d}{dx} |x| \,= \, \frac{d}{dx} \sqrt{x^2} \, = \, \frac{1}{2\sqrt{x^2}} (2x) \, = \, \frac{x}{\sqrt{x^2}} \, = \, \frac{x}{|x|}$ Thus the product rule gives us that: $g'(x) \, = \, (2x)|x| + x^2 (\frac{x}{|x|}) \, = \, 2x|x| + \frac{x^3}{|x|} \, = \, (2x + \frac{x^3}{|x|^2})|x| \, = \, (2x + \frac{x^3}{x^2})|x| \, = \, 3x|x|$ I used the fact that $|x|^2 \,= \, x^2$ to get to the second to last step. So let's compute the Wronskian: $W(x) \,= \, f(x)g'(x) - f#39;(x)g(x) \,=\, (x^2)(3x|x|) - (2x)(x^{2} |x|) \, = \, x^{3} |x|$ We can see that $W(x) \,= \, 0$ if and only if $x \,= \, 0$ AN ASIDE: In the alternative definition of $|x|$ I gave above, it should be noted that I am saying that $|x|$ is the composition of the squaring function $x^2 \, : \, \mathbb_{R} \to [0,\infty)$ with the square root function $\sqrt{x} \, : \, [0 , \infty) \to [0, \infty)$. Reversing the order in which you compose the two functions (square rooting before squaring) would require restricting the domain of $(\sqrt{x})^{2}$ to $[0, \infty)$ , which by virtue of its smaller domain is not equal to the absolute value function. If you were to consider $(\sqrt{x})^{2}$ on the whole real line $\mathbb_{R}$ you would have to have the range of $(\sqrt{x})^{2}$ by the union of the axis of the complex plane (all pure imaginary and purely real numbers), and the result for negative x is not what we want either: Let $x \, < \, 0$ , then $(\sqrt{x})^{2} \,= \, (\sqrt{-|x|})^{2} \, = \, (i \sqrt{|x|})^{2} \,=\, -|x|$ , where $i \,= \, \sqrt{-1}$
November 8th, 2010, 12:28 PM   #5
Math Team

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Re: determine pairs of functions independent

Nice post, forcesofodin!
Quote:
 Originally Posted by forcesofodin [latex]g'(x) \, = \, (2x)|x| + x^2 (\frac{x}{|x|}) \,
From here on, you can also use that $\frac{x}{|x|}=\frac{|x|}{x}$
So that
$(2x)|x| + x^2 (\frac{|x|}{x})\,=\,(2x+x)|x|=3x|x|$

Hoempa

 November 9th, 2010, 05:18 AM #6 Joined: Sep 2010 Posts: 73 Thanks: 0 Re: determine pairs of functions independent Wow, that was a lovely post! Thanks for clearing it up

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