# linear independence

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CAN somebody,please write the definition for the linear independence of the following functions??

$e^x,e^{2x}$

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CAN somebody,please write the definition for the linear independence of the following functions??

$e^x,e^{2x}$

I might possibly, be barking up the wrong tree, but for a given set of vectors...$q_1\bold{v}_1+q_2\bold{v}_2+q_3\bold{v}_3 = \bold{0}$ The set is linearly independent if (trivially) $q=0$. So, surely you'd need to define the interval of $x$, for the set of those two functions ?

EDIT: I misread, you're asking for a definition. I'm a bit confused by this (possibly beyond my scope), plus, why are you asking ? Is this a brain teaser of some sort, or are you expecting somebody to do your work ?

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The problem with this definition is the following:

If we define the functions to be linearly independent,iff ,for all a,b,x $ae^x+be^{2x}=0\Longrightarrow a=b=0$,then in this definition if we put: a=1,b=-1,x=0 we have $ae^x + be^{2x}=0$,but $a\neq 0$and $b\neq 0$.

Which shows that the definition is not applicable.

Any suggestions??

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Your own definition (which you should have been given in the first post, btw) sais "for all x", not "for any x". Try a=1, b=-1 and x=1.

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I am sorry ,but i fail to see the difference between "for all x" and "for any x".

Besides taking the negation of the above formula we get:

there exists a,b ,x such that $ae^x +be^{2x}=0$ and $a\neq 0$ and $b\neq 0$,a formula true for a=1,b=-1 and x=0.

Hence the formula for linear independence is not true ,thus the functions are linearly dependent

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What ? Like you said in your previous post that $a = b = 0$ and in the same sentence that $a \neq 0$ and $b \neq 0$. So ?

It can still be linearly independent if you defined x, despite the coefficents of a and b (providing they don't equal 0.) What's the difference between putting a coefficient on x and defining x, if you want it to be linearly independent ?

However, I don't see the difference between 'all x', and 'any x' either. The problem is you didn't define x !

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Sorry, I think I didn't read your post carefully enough. The definition you gave starts to give me headaches but I think putting the "for all x" behind the equation might fix it.

Anyways, I was thinking about the following definition (like said, it seems I didn't read your post carefully enough - sorry again) that seems to make a bit more sense to me:

- Two functions f(x), g(x) are linearly dependent if there is a pair (a,b) with $a \neq 0$ or $b \neq 0$ such that af(x)+bg(x)=0, where 0 is the zero function, i.e. a function that is zero for all x. This implies that (af+bg)(x) must be zero for all x.

equivalently:

- The negation would be that two functions f(x),g(x) are linearly independent if for any pair (a,b) with $a \neq 0$ or $b \neq 0$ there exists at least one point x such that $af(x) + bg(x) \neq 0$.

Sidenote: By "for any x" I meant "there exists one x such that ...". You hopefully see that this is different from "for all x it is true that ... "

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- The negation would be that two functions f(x),g(x) are linearly independent if for any pair (a,b) with $a \neq 0$ or $b \neq 0$ there exists at least one point x such that $af(x) + bg(x) \neq 0$.

"

Your definition of linear independence would make the functions:

f(x) = x and g(x) =2x linearly independent,because for a=1 ,b=1 there exists,x=2 such that ,$ax +2bx\neq 0$

But are these two functions linearly independent??

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There is no point x such that for the pair a=2, b=-1 af(x)+bg(x) = 2x - 2x = 0 was $\neq 0$. I meant to write "for all pairs (a,b)" in case that really wasn't obvious (like from the sentence before that one).

EDIT: But perhaps it's best for you to just look up a definition from some webpage (Mathworld or Wikipedia), realize that "0" in af(x)+bg(x)=0 means the function O(x)=0 (and not a real number) and actually try to answer your question with that. The big difference is that O(x) is zero for every x, not just for one special choice of x.

Edited by timo

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In wikipedia the proof that i could get is the following one ;

"Let V be the vector space of all functions of a real variable t. Then the functions $e^t$ and $e^{2t}$ in V are linearly independent.

Proof

Suppose a and b are two real numbers such that

$ae^t + be^{2t}$ = 0

for all values of t. We need to show that a = 0 and b = 0. In order to do this, we divide through by $e^t$ (which is never zero) and subtract to obtain

$be^t$ = −a.

In other words, the function $be^t$ must be independent of t, which only occurs when b = 0. It follows that a is also zero."

Which very much implies that the definition of inderendence of functions is ;

for all a,b,t : $ae^t + be^{2t}=0\Longrightarrow a=b=0$

But again in this definition if we put a=1,b=-1 ,t=0 ,then the definition gives

a false unswer,because $ae^t +be^{2t}=0$,but$a\neq 0$ and $b\neq 0$.

And since in logic : true implying false is ,false ,the definition brakes down for these values

|

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I am sorry ,but i fail to see the difference between "for all x" and "for any x".

This is the root of your problem. It is what leads you to nonsense like this:

But again in this definition if we put a=1,b=-1 ,t=0 ,then the definition gives

a false unswer,because $ae^t +be^{2t}=0$,but$a\neq 0$ and $b\neq 0$.

You need to understand the difference between "for all" and "for any". Example: $e^x=1$ is true for one particular value of x. It most definitely is not true for all x. In comparison, $e^x>0$ is true for all real x, $e^x>1$ is true for all positive values of x.

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SO if somebody tells you that any bus goes to L.A YOU gonna be waiting for bus No 9, for example.

I am pretty sure you gonna be the last one to arrive in L.A

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triclino, you can either continue to make silly, false analogies or you can try to understand. To use your bus analogy correctly, there are uncountably many buses heading out of town. Only one of them, bus number zero, goes to LA. That is not anywhere close to "all".

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there are uncountably many buses heading out of town. .

Such a town does not exist, at least on earth.

There is no difference between "for all" and "for any","ANY mathematician- logician" can tell you that

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There is no difference between "for all" and "for any","ANY mathematician- logician" can tell you that

The problem here is that you are using "for any" to mean the existential qualifier $\exists$ rather than the universal qualifier $\forall$. There is a huge difference between

$\exists a : \forall x\ af(x) + g(x) = 0$

and

$\forall x \exists a: af(x) + g(x) = 0$

The former means that either $g(x)$ is the zero function or that $f(x)$ and $g(x)$ are linearly independent. The latter is trivially true and has no meaning.

Edited by D H

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Ga ga ra ba dum . Ga ga ra ba dum

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Ga ga ra ba dum . Ga ga ra ba dum

triclino, people actually spend time giving you legible answers to the questions you raise. Either you stop acting like a child or stop posting.

Do go over our rules.

~moo

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First of all i never quoted anybody and write underneath ga ga ra ba dum,thus showing disrespect for his/her post.

Ga ga ra ba dum was intended to ease the tension and cause laughter.

Sense of humor in any human activity,i think is absolutely necessary

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First of all i never quoted anybody and write underneath ga ga ra ba dum,thus showing disrespect for his/her post.

Quotes are a way to put things in context; i quoted YOU and then asked you to start following the rules.

If it wasn't clear, the "request" was made as mycapacity as a moderator in this forum, and is only phrased as a request for the sake of politeness. Following our rules is non-negotiable, and the rules include not being disrespectful.

Ga ga ra ba dum was intended to ease the tension and cause laughter.

A commendable, but failed attempt. This is also not the only post where you are acting disrespectfully to others. This was a note to get the discussion going on track rather than to the gutter.

Posting up "ga ga ra ba dum" after a reply suggests the other poster speaks garbage. You might've not meant it to seem this way, but it does. If your intention was to joke around about your lack of understanding, you are in dire need of rephrasing your jokes.

Sense of humor in any human activity,i think is absolutely necessary

Perhaps, but theres a difference between joking and disrespecting. You seem to be going slightly over that line. I'm assuming it's unintentional, which is why i did not give you any "points" towards a suspension, and I do give you the benefit of the doubt, but please try to step back, read what you write, and before you make a joke see if its appropriate to make one.

If people put the time to answer you and 98% of your replies are jokes, then you're wasting their time.

Welcome to Science Forums. Now please participate in debates where all parties are privvy to your jokes, not just you.

~moo

Now please get this discussion back on track and away from petty arguments about the (very clear) set of rules we have in this forum.

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$\exists$ (exists) means that these is at least one $x$ such that $P(x)$ is true in your wff*.

$\forall$ (for all) means that $P(x)$ is true in your wff for all $x$.

I think every mathematician will say there is a difference between exists and for all.

*Well-Formulated Formula. (I must stress my knowledge of logic is tiny)

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I don't know if this is a proof :

(Contraposition) : $e^x,e^{2x}$ are linearly dependent if there are constant a,b nonzero, such that $ae^x+be^{2x}=0,\forall x$.

x=0, implies $a+b=0$

x=1 implies $a+be=0$ a contradiction since a,b are constant.

So those two functions are lin. independent...

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