Determining whether two functions are linear independent via wronskian


by ysebastien
Tags: determining, functions, independent, linear, wronskian
ysebastien
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#1
Mar4-12, 12:59 PM
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1. The problem statement, all variables and given/known data

Determine which of the following pairs of functions are linearly independent.

(a) [itex]f(t)=3t,g(t)=|t|[/itex]

(b) [itex]f(x)=x^{2},g(x)=4|x|^{2}[/itex]

2. Relevant equations

the Wronskian is defined as,

W=Det{{f(u),g(u)},{f'(u),g'(u)}}

if {f(u),g(u)} are linearly dependent, W=0

if W=/=0, {f(u),g(u)} are linearly independent

3. The attempt at a solution

The assumed interval for the independent variables t,x are [itex] x,t \in (-\infty,\infty) [/itex]

for (a),

I determined [itex] W(t)=3t-3|t| [/itex], which for x>0 is [itex] W(t)=3t-3t=0 [/itex].
for x<0 we have [itex] W(t)=-3t-3t=-6t [/itex]. Since for some value of [itex] t \in (-\infty,\infty)[/itex] I found a [itex]W(t) \neq 0[/itex] I can conclude that the functions f(t) and g(t) are linearly independent.

Now for (b),

Similarly to (a), I find a [itex] W(x)=8x^{2}|x|-8|x|^{2}x[/itex]

for [itex] x>0 : W(x)=8x^{3}-8x^{3}=0[/itex]
for [itex] x<0 : W(x)=8(-x)^{2}|-x|-8|-x|^{2}(-x)=8x^{3}+8x^{3}=16x^{3}\neq 0[/itex]

Similarly I conclude that f(x) and g(x) are linearly independent since I found values of x which make the wronskian not equal to 0.

However, while my conclusion is correct for (a), (b) is supposedly linearly dependent.

Is my method correct? if so what mistake did I make in concluding that the functions of (b) were linearly independent?

Thanks
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Dick
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Mar4-12, 01:36 PM
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|x|^2 is equal to x^2.
ysebastien
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Mar4-12, 01:43 PM
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Yes,

I use that fact, but I still find that for
[itex] x<0 : W(x)=8(-x)^{2}|-x|-8(-x)^{3}=8x^{3}+8x^{3}=16x^{3}[/itex]

Am I just completely missing something here?

Dick
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Mar4-12, 03:53 PM
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Determining whether two functions are linear independent via wronskian


Your calculation of [itex]W(x)=8x^{2}|x|-8|x|^{2}x[/itex] is wrong. Just use that |x|^2=x^2 from the beginning, so g(x)=4x^2.
LCKurtz
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Mar4-12, 03:57 PM
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The specific place where your calculations are incorrect is$$
\frac d {dx}|x|^2 = 2|x|$$is false.
ysebastien
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Mar4-12, 03:57 PM
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Ah, I see. Thank you


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