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November 6th, 2010, 11:42 AM  #1 
Member 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: 5,603 Thanks: 365  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 
Member 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 
Senior Member Joined: Dec 2009 Posts: 150 Thanks: 0  Re: determine pairs of functions independent
The following representation of may be of help to you: 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, returns the positive magnitude of . Using the chain rule we differentiate the function: Thus the product rule gives us that: I used the fact that to get to the second to last step. So let's compute the Wronskian: We can see that if and only if AN ASIDE: In the alternative definition of I gave above, it should be noted that I am saying that is the composition of the squaring function with the square root function . Reversing the order in which you compose the two functions (square rooting before squaring) would require restricting the domain of to , which by virtue of its smaller domain is not equal to the absolute value function. If you were to consider on the whole real line you would have to have the range of 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 , then , where 
November 8th, 2010, 12:28 PM  #5  
Math Team Joined: Apr 2010 Posts: 2,765 Thanks: 352  Re: determine pairs of functions independent
Nice post, forcesofodin! Quote:
So that Hoempa  
November 9th, 2010, 05:18 AM  #6 
Member 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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