Question

are the functions f1=x and f=sinx form a linearly independent set of vectors

Answer 1

on any interval?

Answer 2

we can pick an interval and show that if
\[ax+b\sin(x)=0\] on that interval, then
\[a=0\] and
\[b=0\]

Answer 3

In \[\mathbb{R}^2\] I don't believe they do. For example, if you let x=0 you have the zero vector. Which I believe makes them dependent.

Answer 4

of course
\[ax+b\sin(x)=0\] means for all values of x

Answer 5

they are linearly independent for sure

Answer 6

you certainly cannot write one as a linear combination of the other. that would be a miracle.

Answer 7

But like I said, if you let x=0 then you have a solution to ax+bsin(x)=0 that is NON trivial a and b. :/

Answer 8

to say that
\[ax+b\sin(x)=0\] on say the interval \[(-\pi,\pi)\] means
\[ax+b\sin(x)=0\] for all x in that interval.

Answer 9

yes but must be for all x, not for some x

Answer 10

Yeah, so since x=0 makes the a and b non-trivial doesn't that one value make them dependent automatically?

Answer 11

surely you can find a value of x for which
\[ax+b\sin(x)=0\] but this is an identity of functions, not a value

Answer 12

not the way i understand it, no. we are considering x and sin(x) as functions.

Answer 13

Okay. Then obviously they are linearly independent :P

Answer 14

But i'm assuming sin(x) would have to be on an interval since it is periodic?

Answer 15

if they were dependent one would be a linear combination of the other yes? in other words you could write sine as a linear combination of the identity function. fat chance.

Answer 16

on the interval C'(-infinity, infinity) where C' shows first deravative

Answer 17

this means in the space of functions with first derivatives i believe. of course both x and sin(x) are infinitely differentiable.

Answer 18

i think all you have to do is pick a number for x and show that if
\[ax+b\sin(x)=0\] for that number then both a and b are zero. you can pick
\[\frac{\pi}{3}\] or anything really

Answer 19

how are they independent

Answer 20

or you can use "wronkian" try here
http://tutorial.math.lamar.edu/Classes/DE/Wronskian.aspx

Answer 21

Let x=0 and that falls apart satellite. Which is what I was trying to say.

Answer 22

Because that gives you a solution where a and b=/=0

Answer 23

ok perhaps i am wrong. but i don't think so. for example it is well known that sine and cosine are linearly independent. but it is certainly the case that
\[\cos(\frac{\pi}{4})-\sin(\frac{\pi}{4})=0\] this does not make them dependent. one is not a linear combination of the other

Answer 24

the question is not "is there SOME value of x for which you can solve
\[ax+b\sin(x)=0\]

Answer 25

ah ok. we can in fact use wronskian. set up the 2 by 2 matrix
x sin(x)

1 cos(x)

Answer 26

take the determinant and get
\[x\cos(x)-\sin(x)\]

Answer 27

and i believe theorem says if we can find SOME value for which this determinant is non-zero, then they are independent.

Answer 28

http://en.wikipedia.org/wiki/Wronskian

Answer 29

And doing the wronskian you have:
\[\det\left[\begin{matrix}x & \sin(x) \\ 1 & \cos(x)\end{matrix}\right]=x \cos(x)-\sin(x)\]
Exactly, so they are independent IFF w(x)=/=0 so you have:
xcos(x)-sin(x)=0
xcos(x)=sin(x)
x=tan(x)
Which is only true at x=0
Which is IN THE INTERVAL. Therefore they are dependent from this point of view.

Answer 30

but xcosx(x)-sinx has value of zero for x=0

Answer 31

Which is what I was trying to say the whole time.

Answer 32

but this is linear indepenent in howard and anton linear algebra book

Answer 33

wow are we seeing the same thing but saying different things. i read it to say that if there is SOME value for which the determinant is not zero then they are independent

Answer 34

not that if there is some value for which it is zero they are dependent.

Answer 35

of course they are independent! one is not a linear combination of the other that is for sure. just trying for a "proof"

Answer 36

Hmmm...Maybe I misunderstood. I'm trying to recall the wronskian from diffeq lol.

Answer 37

the fact that they are 0 when x = 0 does not make them dependent, any more than the fact that cosine of pi/4 is the same as sine of pi/4 makes them dependent.

Answer 38

Okay satellite, I concede haha.

Answer 39

The Wronskian can be used to determine whether a set of differentiable functions is linearly independent on an interval. This is useful in many situations. For example, if we wish to verify that two solutions of a second-order differential equation are independent, we may use the Wronskian. Note that if the Wronskian is zero everywhere in the interval, the functions may or may not be linearly independent. A common misconception is that W = 0 everywhere implies linear dependence; the third example below shows that this is not true. However, if all of f1, ..., fn are analytic, then W = 0 everywhere implies linear dependence.

Answer 40

i have confusion about x=0

Answer 41

the wronskian vanishes at x = 0 but it does not vanish everywhere. that is the point.

Answer 42

a set is linearly Independent if the set of solutions only contains the trivial solution x = 0, where x is the zero vector. If ANY other solution exist then they are linearly dependent

Answer 43

They are independent.

Answer 44

ok i got idea

Answer 45

hello math girl. yes of course they are independent. but just because it is obvious doesn't mean you can't prove it

Answer 46

satellite math girl is new comer

Answer 47

Well, here's something that might help: if zero is a vector, k cannot be 0. If k is 0, the vector cannot be zero.

Answer 48

i think this is done via wronskian or sheer obviousness. but i guess obviousness doesn't count, so wronskian does. we come up with some x for which
\[x\cos(x)-\sin(x)\neq 0\] and we are finished

Answer 49

i pick
\[\frac{\pi}{3}\] but you can pick something else