Determining linear indepedence/dependence of a set of functions

[chwala]

TL;DR Summary

See attached.

Looking at the wronskian applications- came across this;

Okay, i noted that one can also have this approach(just differentiate directly). Sharing just incase one has more insight.

$-18c \sin 2x -4k\cos x \sin x - 4k\sin x\cos x =0$
$-18c\sin 2x-2k\sin2x-2k\sin 2x=0$
$-18c\sin 2x = 4k\sin2x$
$-18c=4k⇒k=\dfrac{-18}{4}=\dfrac{-9}{2}$ and therefore $c=1$

thus $f(x)$ and $g(x)$ linearly dependent.

...

 

[Office_Shredder]

You haven't written down the Wronskian, you just showed the derivatives are linearly dependent. This isn't enough - e.g. f(x)=x and g(x)=x+1 are linearly independent, but they have linearly dependent derivatives. The Wronskian does not vanish though : fg'-f'g=-1 which is not zero.

In addition you can't use the Wronskian to prove linear dependence anyway for general differentiable functions - you can get a 0 for the Wronskian even if the functions happen to be linearly independent. This is pretty rare (e.g. can't happen if the functions are analytic, so would end up being fine in this case anyway) but the proof is much more subtle than the direction of Wronskian nonzero implies linear independence.

 

[chwala]

You haven't written down the Wronskian, you just showed the derivatives are linearly dependent. This isn't enough - e.g. f(x)=x and g(x)=x+1 are linearly independent, but they have linearly dependent derivatives. The Wronskian does not vanish though : fg'-f'g=-1 which is not zero.

In addition you can't use the Wronskian to prove linear dependence anyway for general differentiable functions - you can get a 0 for the Wronskian even if the functions happen to be linearly independent. This is pretty rare (e.g. can't happen if the functions are analytic, so would end up being fine in this case anyway) but the proof is much more subtle than the direction of Wronskian nonzero implies linear independence.

I think there should be a relationship between wronskian of a set of functions and linear dependence/independence.

I will get back on this...

 

[Office_Shredder]

I think there should be a relationship between wronskian of a set of functions and linear dependence/independence.

I will get back on this...

The relationship is if the Wronskian is nonzero, then the functions are linearly independent.

If the Wronskian is zero, you can't tell anything without additional data about the functions.

 

[chwala]

The relationship is if the Wronskian is nonzero, then the functions are linearly independent.

If the Wronskian is zero, you can't tell anything without additional data about the functions.

Correct for the first paragraph;

then for the second paragraph ...the textbook may be wrong going with your reasoning...

but if a set of functions are not independent then the only other possibility is that they're dependent . The text indicates so, that is when wronskian is equal to $0$.

I will check the literature on this. Thanks.

 

[Mark44]

$-18c \sin 2x -4k\cos x \sin x - 4k\sin x\cos x =0$
$-18c\sin 2x-2k\sin2x-2k\sin 2x=0$
$-18c\sin 2x = 4k\sin2x$
$-18c=4k⇒k=\dfrac{-18}{4}=\dfrac{-9}{2}$ and therefore $c=1$

thus $f(x)$ and $g(x)$ linearly dependent.

The fourth line above is incorrect.
$-18c = 4k \rightarrow k = -\dfrac {18 c} 4 = -\dfrac {9c} 2$
One possible value for c is 1, as is shown in the image you posts. However, from the equation you started with you can't conclude that c = 1 (and k = -9/2) is the only pair of solutions.
You have one equation with two unknowns, so it's not possible to determine unique values for c and k.

 

[chwala]

The fourth line above is incorrect.
$-18c = 4k \rightarrow k = -\dfrac {18 c} 4 = -\dfrac {9c} 2$
One possible value for c is 1, as is shown in the image you posts. However, from the equation you started with you can't conclude that c = 1 (and k = -9/2) is the only pair of solutions.
You have one equation with two unknowns, so it's not possible to determine unique values for c and k.

Agreed @Mark44 ; my point was that there was a different way to come up with those pairs using;

$-18c \sin 2x -4k\cos x \sin x - 4k\sin x\cos x =0$

... to realize,

$k=-\dfrac{9c}{2}$ with this, the other possible pairs can be found.

 

[chwala]

Correct for the first paragraph;

then for the second paragraph ...the textbook may be wrong going with your reasoning...

but if a set of functions are not independent then the only other possibility is that they're dependent . The text indicates so, that is when wronskian is equal to $0$.

I will check the literature on this. Thanks.

I just counter checked @Office_Shredder. The reference is on this link:

https://tutorial.math.lamar.edu/Classes/DE/Wronskian.aspx

You are correct ...Wronskian being zero does not imply that the functions are dependent. It is indeed possible for two functions to have wronskian $=0$ and still be linearly independent.
Bringing me to the question as to why the use of the Wronskian (fact 2) in determining dependence of set of functions with it's controversy in mind.

 

[Office_Shredder]

I just counter checked @Office_Shredder. You are correct ...wronskian being zero does not imply that the functions are dependent. It is indeed possible for two functions to have wronskian $=0$ and still be linearly independent.
Bringing me to the question as to why the use of wronskian in establishing this.

It's almost always the case that you want to prove functions are linearly independent, and it's surprisingly annoying to do so from first principles of knowing what they are.

 

[chwala]

Allow me to ask this here as the question is related to the same reference in my post $8$.

Consider example $4$ in the reference below where the Wronskian of the second order equation was determined using Abel's theorem. My question is can we work backwards to determine the two solutions?

or

is it that one solution must be given in order to determine the other solution.

https://tutorial.math.lamar.edu/Classes/DE/Wronskian.aspx

The specific ode is indicated below;

$t^4y''-2t^3y' - t^8y=0$.

I'll check this using the characteristic method...solve the quadratic that is...

 

[Mark44]

Consider example 4 in the reference below where the Wronskian of the second order equation was determined using Abel's theorem. My question is can we work backwards to determine the two solutions?

No. All you have is the relationship between the two functions; namely, their Wronskian. It's a little like knowing the area of a rectangular plot of land, but this knowledge alone doesn't give you any information about the length and width of the plot.

or

is it that one solution must be given in order to determine the other solution.

I'm not sure, but if you know one of the two solutions it might be possible to find the other.

 

[pasmith]

is it that one solution must be given in order to determine the other solution.

Yes.

In the 2nd order case for $$y'' + p(x)y' + q(x)y = 0$$ the Wronskian $$W[y_1,y_2] = y_1y_2' - y_2y_1'$$ can be written as $$y_1^2\left(\frac{y_2}{y_1}\right)'.$$ It can be shown that if $y_1$ and $y_2$ are solutions then $W[y_1,y_2]$ is a solution of $$W' + p(x)W = 0$$ so that $$W = W_0 \exp\left( -\int p(x)\,dx \right)$$ is non-zero unless $W_0 = 0$. From the other expression for $W$ it follows that if $y_1$ and $y_2$ are linearly dependent then $(y_2/y_1)' = 0$ and their Wronskian vanishes.

It follows that if $y_1$ is a solution and $W \neq 0$ satisfies $W' + pW = 0$ then $$y_2 = y_1 \int \frac{W}{y_1^2}\,dx$$ is a linearly independent solution.