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#26 2013-12-11 10:21:09

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 86,449

Re: Wronskian use identities !

That makes v1(x) = v2(x) which makes the determinant 0 that means they are linearly dependent.

Seems nonsensical.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#27 2013-12-11 10:24:04

evinda
Member
Registered: 2013-04-13
Posts: 105

Re: Wronskian use identities !

bobbym wrote:

That makes v1(x) = v2(x) which makes the determinant 0 that means they are linearly dependent.

Seems nonsensical.

So,what have I done wrong?? hmm Is there an other way to do this?

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#28 2013-12-11 10:24:41

anonimnystefy
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From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Wronskian use identities !

Hi evinde

For the first part, use the fact that the Wronskian is non-zero and the fact that the determinant of the system of equations is the Wronskian at x=0.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#29 2013-12-11 10:25:52

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 86,449

Re: Wronskian use identities !

Why is the Wronskian non - zero?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#30 2013-12-11 10:26:41

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Wronskian use identities !

Because the exponential function is always non-zero.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#31 2013-12-11 10:28:27

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 86,449

Re: Wronskian use identities !

The determinant is 0 because v1(x) = v2(x), therefore the Wronskian is 0.

That is why I say something is very wrong.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#32 2013-12-11 10:28:46

evinda
Member
Registered: 2013-04-13
Posts: 105

Re: Wronskian use identities !

bobbym wrote:

Why is the Wronskian non - zero?

Also,because of the fact that the solutions are linearly independent!!!

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#33 2013-12-11 10:29:37

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 86,449

Re: Wronskian use identities !

Isn't that what are you trying to prove? Using it in the proof would be circular reasoning?!


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#34 2013-12-11 10:29:57

evinda
Member
Registered: 2013-04-13
Posts: 105

Re: Wronskian use identities !

anonimnystefy wrote:

Hi evinde

For the first part, use the fact that the Wronskian is non-zero and the fact that the determinant of the system of equations is the Wronskian at x=0.

How can I use these facts??I haven't understood yet... roll

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#35 2013-12-11 10:32:01

anonimnystefy
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From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Wronskian use identities !

Find the determinant of the system of equations given in the second to last line of your post 1. You will see that it actually represents the Wronskian at x=0.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#36 2013-12-11 10:37:32

evinda
Member
Registered: 2013-04-13
Posts: 105

Re: Wronskian use identities !

anonimnystefy wrote:

Find the determinant of the system of equations given in the second to last line of your post 1. You will see that it actually represents the Wronskian at x=0.

Don't you mean these two:

, d_{1}{v_{1}}'(0)+d_{2}{v_{2}}'(0)={f}'(0) ?
Can I  find the determinant,although I haven't shown that there
and
that satisfy these conditions??

Last edited by evinda (2013-12-11 10:41:16)

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#37 2013-12-11 10:40:43

anonimnystefy
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From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Wronskian use identities !

Well, the point of finding the determinant is to show that there are solutions, so, yes, you can. Of course, you should treat d1 and d2 as the unknowns when finding the determinant.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#38 2013-12-11 10:43:29

evinda
Member
Registered: 2013-04-13
Posts: 105

Re: Wronskian use identities !

anonimnystefy wrote:

Well, the point of finding the determinant is to show that there are solutions, so, yes, you can. Of course, you should treat d1 and d2 as the unknowns when finding the determinant.

So,I find the determinant and notice that it equals to the Wronskian,so it is

.And then?

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#39 2013-12-11 10:50:29

anonimnystefy
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From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Wronskian use identities !

Yes, but first you need to find the Wronskian. You can do this using Abel's identity.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#40 2013-12-11 11:21:16

evinda
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Registered: 2013-04-13
Posts: 105

Re: Wronskian use identities !

anonimnystefy wrote:

Yes, but first you need to find the Wronskian. You can do this using Abel's identity.

Do we have a

we could use?

Last edited by evinda (2013-12-11 11:21:58)

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#41 2013-12-12 01:23:26

evinda
Member
Registered: 2013-04-13
Posts: 105

Re: Wronskian use identities !

Do you mean that I have to use

? I tried and got W(x)=(v1(0)v2'(0)-v2(0)v1'(0))*e^(-ax).Is it right so far?and how can I continue?

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#42 2013-12-12 01:39:40

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 86,449

Re: Wronskian use identities !

What did you get for

(v1(0)v2'(0)-v2(0)v1'(0))?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#43 2013-12-12 01:45:13

evinda
Member
Registered: 2013-04-13
Posts: 105

Re: Wronskian use identities !

bobbym wrote:

What did you get for

(v1(0)v2'(0)-v2(0)v1'(0))?

How can I find this value?

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#44 2013-12-12 01:49:14

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,449

Re: Wronskian use identities !

What do you get for v1(x) and v2(x)?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#45 2013-12-12 02:00:19

evinda
Member
Registered: 2013-04-13
Posts: 105

Re: Wronskian use identities !

bobbym wrote:

What do you get for v1(x) and v2(x)?

I haven't found them..how could I do this?

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#46 2013-12-12 02:06:41

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,449

Re: Wronskian use identities !

It has been stated the v1 and v2 are the 2 solutions to the DE. Do you agree?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#47 2013-12-12 04:33:18

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Wronskian use identities !

There is an infinite nunber of solutions of the DE.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#48 2013-12-12 06:24:52

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,449

Re: Wronskian use identities !

Without fixing c1 and c2, that determinant will always be zero. That means the Wronskian will be 0. Do you agree?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#49 2013-12-12 06:26:54

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Wronskian use identities !

Huh?


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#50 2013-12-12 06:28:36

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,449

Re: Wronskian use identities !

(v1(0)v2'(0)-v2(0)v1'(0))

Without getting the 2 constants v1 = v2, therefore the above expression will always be 0.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

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