Linear Independence of Functions & The Wronskian
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- เผยแพร่เมื่อ 8 ก.ค. 2024
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What is the analog of Linear Independence for functions? If this was just vectors linear independence would mean the only way you can take a linear combination that adds up to zero is the trivial linear combination with all the coefficients being zero. Same for functions. However, we do have one extra tool called the Wronskian, which is a nifty little determinant that we can use to create a quick test for linear dependence of a set of functions. The connection to differential equations will be that we will be looking for a fundamental set of solutions which will have to be linearly independent, more on that in the next video.
0:00 2D case
1:22 Geometric Picture
2:27 Linear Independence
5:54 The Wronskian
8:37 Example
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Dr. on 0:48 aren't you saying the opposite of what the text states? The text being the correct one in this case right?
Good catch! Yes I meant “not linearly dependent”, can I call that a “speak-o”??
@@DrTrefor Thank you for the clarification!
Sir, I had to discontinue maths in college due to financial reasons, but I find it fascinating, and I have decided to start learning maths and physics from where I left it in college. I am so thankful to you to provide such contact free of cost. a lot of people don't tell you how help full these courses, I absolutely love you and please continue to teach and spread your knowledge free of cost like this. I am here only coz I am interested in the subject and i can't even tell you how happy i am to have found your channel. Thanks a lot for everything sir.
I definitely will continue, thank you!
Please Check out the Playlist of Professor V, The Organic Chemistry Tutor, Professor Leonard and Professor RobBob. These are excellent Playlist in Mathematics , Chemistry and Physics.
I really love how you explained the Wronskian clearly.
Professor Bazett, this is an exceptional video/lecture on Linear Independence and The Wronskian. These are two powerful tools from Linear Algebra that is also used in Differential Equations.
Doing Differentials, and I got to admit you helped me out through with your videos. Sometimes, I second guess myself and often cases concept grasping is difficult for me. Now I am doing better, thank you sir!
I am doing ODE as a part of my university math course.Worked on linear algebra last year (from 3blue1brown). Seeing same thing from two different perspectives and connecting the theory behind is something I like about Math most.
Getting comfortable with ODE. Maybe will work on it even after my semester.
Thank you SIR.
I love your videos, they are easy to follow, they make sense and it all comes so nicely together!
Glad you like them!
I was struggling with my university lectures, thank you very much sir, Nice presentation
great video, really cleared things up
You"re just life saviour
Love your videos sir
Very explanatory as always :)
Great Mr
i love this man
Thanks a lot sir 🔥🔥🔥
Thanks
great video, much better than my professors indecipherable notes :)
Great video
What a great mathematician!!! U helped me and my frineds a lot...
Thank you!
I think you're erecting a large, fictional wall between functions and linear algebra. Function spaces ARE vector spaces when equipped with appropriate, obvious operations (although often infinite dimensional), and studying them this way IS linear algebra.
Good day sir
What should I study to make simulations like thoes in your videos?
The Wronskian? More like "Sharing great information is your mission!" 👍
Do you have a video on solving exact ODE's?
awesome!
great. keep it up. keep educating the students who are sleeping in their maths class.
haha yup that's like what 50% of my views are, the sleeping students who realize they have a test due the next day:D
tnxxxx
I made notes along with the videos, does these 24 videos cover the entire ODE claculus 3 course?
When will you make a video on partial differential equation . I am eager to know how will you make this concept easier 😀😀😀
Laplace vids would be rlly helpful rn my semester ends may 1
Have a whole playlist on them, check out my channels homepage!
@@DrTrefor I will thank you, have you ever tried psilocybin or mdma ?
Hi Dr Bazett thank you for your wonderful videos, they are a joy to watch and learn from. I didn't quite understand the part where you explained a1z + a2z^2 +a3z^3 = 0 must be linearly independent. How do you know that there does not exist a non-zero a1, a2 and a3 that result in the equation being equal to zero? For example if you simply set a1, a2 and a3 to 1 then the equation works (I guess you can use any 3 numbers as there will always exist three roots that satisfy the equation). So I feel like I am missing something obvious but not sure what. Is it that for any a1,a2 and a3 the equation has to equal zero for all values of z and therefore a1z+ a2z^2+a3z^3 has to be linearly independent (as max of 3 roots)?
Im not sure if this helps but dont forget that z is substitute for e^t, which is always non-zero positive so you can take any value for z actually
Love from INDIA ❤️❤️
I clicked so fast as soon as I saw the notification of your video
You're the best!
@@DrTrefor
Amazing video Sir
Thanks for the great explanation! :)
My pleasure!
What if your wronskian is some function that is equal to zero on some value of t?
I know this is an older video but I was wondering, what would w(t) look like if you had more terms w[y1,y2,y3] would it be something like {y1(y2')(y3'')}+{y2(y3'')(y1'')}+{y3(y1')(y2'')}=0
You said at the beginning of this video that 'sin x and cosine x are NOT linearly independent' probably a slip of tongue. Actually they ARE linearly independent, as you go on with the counter example.
I am watching the video now- which is your latest one. You may correct the inadvertent error. Or, correct me if I am wrong!
Quite right, good catch!
I have a doubt regarding the linear independence of terms e^rt and t.e^rt. Here in this video in order to find the wronskian of these two functions you have differentiated t.e^rt as e^rt+t.e^rt by considering t and e^rt as two different functions of that is the case shouldn't we add a dt every time we differentiate the term e^rt. This is stuck in my mind since you have stated that e^rt and t.e^rt are linearly independent in your previous videos. Btw thank you for the marvelous explanation
In this case, differentiating removes the dt differential, since our operator is d/dt, rather than just d in general.
Well l literally wanna learn about the importance of liner independence (in Quantum mechanics) Damn I am so early I need to wait for the next video😅
I just started reading Quantum mech (from Griffiths) please let me know If anyone have any suggestions on that
It's been 15 years since I had to read Griffiths, even the name makes me shiver lol!
@@DrTrefor 😂😂
I would like to say, why are we going through so much trouble to find linear dependency. Isnt it easy to divide both the functions and If we get a constant K, it means they are linearly dependent but if we get a function f(x) it means they are linearly independent.
This is only true for 2 functions, but for 3 or more it is much more complicated
please see if there is a way to make english subtitles available...
i am seeing vietnamese auto generated subtitiles
TH-cam is so weird. Every once in a while it makes subtitles automatically for the wrong language despite me setting defaults to English and I have no idea how to fix it other than to say MOST of my videos don’t have this problem
@@DrTrefor yes... thank you for response
6:47 Are y1 y2 and y3 functions of t?
Yup
Dear professor, at 7.43 u said LI iff W(t) is non zero.
Example:
f(x)=x^2, g(x) =x|x| on R are LI, but their wronskian is 0 everywhere. Please clarify
g(x) = x |x| is not differentiable everywhere, so the Wronskian isn't technically zero "everywhere". The Wronskian is zero at the well-behaved domains of this function where it is differentiable, but due to the problem point at x=0, where there is a sudden change in its derivative, that is where it gets its linear independence from f(x) = x^2.
The Wronskain needs to be zero everywhere to prove that the functions are linearly dependent. But one counterexample where the Wronskian is either non-zero, or undefined, means that the functions can be linearly independent. It doesn't necessarily prove that they are linearly independent until you have a confirmed non-zero Wronskian at at least one point, where the Wronskian is defined. So this is a function pair where the Wronskian test is inconclusive, at proving linear independence.
@@carultch x|x| is differentiable every where . Its derivative is 0 at x=0,
for x>0, its 2x,
for x
@@sripad72You're right, it is differentiable. However, there is still a way to use this procedure to prove that they are linearly independent.
Suppose we introduce a third function, h(x) = x^3. If f(x) and g(x) were linearly dependent, then the 3rd order Wronskian of f(x), g(x), and h(x) should be zero everywhere. We run into a problem, when g"(x) is undefined at x=0. The Wronskian is also undefined at x=0. It is zero everywhere else. Wolfram Alpha produced the Wronskian of -x^3 (x (x Abs''(x) - 2 Abs'(x)) + 2 abs(x)), and for real x-values, -x^5 Abs''(x). Abs"(0) is a spike to infinity, of 2*delta(x), so we have an indeterminant form when we multiply it by 0^5.
f(x) and g(x) ultimately are linearly independent, because there is no non-trivial linear combination of the both of them, that equals zero everywhere. You'd require at least one of your coefficients in front of one of these functions to have a jump discontinuity at x=0, to add up to zero, in which case it is by definition, not a constant, since it depends on x.
@@carultch To prove x^2, x|x| are linearly independent on R, consider ax^2+bx|x|=0. Take x=1, x=-1 we get a+b=0, a-b=0 .Solving we get a=b=0. Hence they are LI on R.
7:33 why the wronskian = 0 somewhere and not everywhere?
the size of wonkskian matrix gives us info about the order of differential equation true or false
Sir I have one question how i will connect you??
My main "public" channels are here on TH-cam or on my Twitter @treforbazett:)
@@DrTrefor okk sir Thanks
I will send message on your Twitter account
Second 😊😎
hahah that's still pretty good:D
Sir, the subtitles that TH-cam has automatically generated are in Vietnamese language. Could you please either upload english one or change the auto translate from Vietnamese to English.
I think this occurred because of your accent.
it's called wronskian, NOT ronskian
In English, the W is silent in the WR digraph, like in write and wrench. You can't really mix W's sound with R's sound anyway. Yes, the W is supposed to sound like V in that word, such that it would sound like Vronskian, but it's common to not bother with this when anglicizing W-words from Germanic and Slavic languages.
@@carultch OK cool, but Józef Hoene-Wroński was a Polishman, not an Englishman, so his name should be pronounced according to Polish rules, shouldn't? And in Polish, the W is always pronounced like English V, never being silent.
@@filoreykjavik It's a topic for a math class, not a Polish language class. The details of exactly how to say all of its contributors' names isn't really the main point.
I wouldn't expect people to say my name exactly as I say it, if the letters L and R don't mix in their language at the end of a word. I'll settle for being called "Car" or "Call" out of linguistic necessity.
Hell, if Spanish speakers drop the H in my last name, because the H is mute in their language, I'll understand and I'm fine with it. Even though I know they can say it, since my H sounds like the J in Jalapeño.
Thanks