Is {sinhx,cosh,e^x,1} linearly independent

Original post by hannnahx

i think that sinhx and coshx are linearly independent and also that sinhx, coshx and e^x are too.

In bold: Are you just guessing? I suspect so.

How do you show a set is linearly independent? What's your definition?

Reply 2

TeeEm

19

Original post by hannnahx

i think that sinhx and coshx are linearly independent and also that sinhx, coshx and e^x are too. But I'm not sure how to prove that all four sinhx,coshx, e^x and 1 are linearly independent?

if it works like vectors ( if my memory serves me well ) then

Acoshx + Bsinhx + Ce^x +D = 0

they are linearly independent if the only way you can satisfy this relationship is iff A=B=C=D=0

is this the case here?
find a counterexample and you proved linear dependence

wait for the purists

Reply 3

Original post by hannnahx

i think that sinhx and coshx are linearly independent and also that sinhx, coshx and e^x are too. But I'm not sure how to prove that all four sinhx,coshx, e^x and 1 are linearly independent?

You probably want to use the Wronskian. This is a determinant whose form you can google. Its rows are formed by the functions, and the first, second, etc derivatives of the functions.

If the Wronskian is non-zero for some

x \in D

where

D

is the domain of the functions, then the functions are linearly independent. (The converse is not true; if the Wronskian is identically zero over the domain, you can't conclude that the functions are linearly dependent BUT SEE BELOW)

(edited 8 years ago)

Reply 4

Original post by ghostwalker

In bold: Are you just guessing? I suspect so.

Given that

e^{-x}

is not there, it looks right, I think.

Reply 5

ghostwalker

Original post by atsruser

Given that

e^{-x}

is not there, it looks right, I think.

I don't see the problem - perhaps there's a definition of linearly independence that I'm not aware of, particularly as you talk about using the Wronskian.

Edit: Just did a bit of digging, and first and third rows of the Wronskian would be identical, hence it's zero for all x. Or have I misunderstood?

(edited 8 years ago)

Reply 6

Original post by ghostwalker

I don't see the problem - perhaps there's a definition of linearly independence that I'm not aware of, particularly as you talk about using the Wronskian.

No, your hint was right. I managed to forget cosh was there too. Sorry. In fact, you can still use the Wronskian but it requires an extra condition to be true to show l.i. and it's probably easier just to do it from the definition (of which there is only one, of course.)

Reply 7

Original post by ghostwalker

I don't see the problem - perhaps there's a definition of linearly independence that I'm not aware of, particularly as you talk about using the Wronskian.

Edit: Just did a bit of digging, and first and third rows of the Wronskian would be identical, hence it's zero for all x. Or have I misunderstood?

I don't know. If you are thinking that you can claim l.d. from that, then you can't. However, I've just done a bit of digging too, and discovered that if the functions are analytic and the Wronskian is identically zero, then that condition guarantees l.d. I was aware of another condition. So I think here, the Wronskian works fine to show l.d.

Reply 8

ghostwalker

Original post by atsruser

I don't know. If you are thinking that you can claim l.d. from that, then you can't. However, I've just done a bit of digging too, and discovered that if the functions are analytic and the Wronskian is identically zero, then that condition guarantees l.d. I was aware of another condition. So I think here, the Wronskian works fine to show l.d.

Thanks. I'm not familiar with the finer points of the Wronskian and as such wouldn't use it. I looked as you mentioned it. However, all's well.

Reply 9

Zhy

Original post by TeeEm

they are linearly independent if the only way you can satisfy this relationship is iff A=B=C=D=0

Careful -- the implication does not go both ways. Otherwise, every system would be linearly independent.

The Wronskian also seems excessive here: sinh and cosh are linear combinations of e^x and e^(-x). What happens when you add them?

Reply 10

TeeEm

19

Original post by Zhy

Careful -- the implication does not go both ways. Otherwise, every system would be linearly independent.

The Wronskian also seems excessive here: sinh and cosh are linear combinations of e^x and e^(-x). What happens when you add them?

thank you
I will try to be careful in the future
(definitely will not use the Wronskian)

Reply 11

DFranklin

18

Original post by atsruser

You probably want to use the Wronskian. This is a determinant whose form you can google. Its rows are formed by the functions, and the first, second, etc derivatives of the functions.I feel I have to say that for *this* question, that's a sledgehammer - a few seconds of thought should suffice.

Reply 12