determine pairs of functions independent

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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?
Last edited by mbradar2 on Sat Nov 06, 2010 3:04 pm, edited 3 times in total.
Queen of Hearts

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Joined: Thu Sep 23, 2010 7:30 am

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.
mathman
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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?
Queen of Hearts

Posts: 73
Joined: Thu Sep 23, 2010 7:30 am

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:
$W(x) \, = \, f(x)g'(x) - f'(x)g(x) \,=\, (x^2)(3x|x|) - (2x)(x^{2} |x|) \, = \, x^{3} |x|$
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 $i \, = \, \sqrt{-1}$

forcesofodin
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Re: determine pairs of functions independent

Nice post, forcesofodin!
forcesofodin wrote:$g'(x) \, = \, (2x)|x| + x^2 (\frac{x}{|x|}) \,

From here on, you can also use that [latex]\frac{x}{|x|}=\frac{|x|}{x}$

So that
$(2x)|x| + x^2 (\frac{|x|}{x})\,=\,(2x+x)|x|=3x|x|$

Hoempa
Hoempa
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Re: determine pairs of functions independent

Wow, that was a lovely post! Thanks for clearing it up