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.
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.
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.
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.
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.
Awesome explanation. Thank you (from Brazil ) for the video. I have a question. Wronskian = 0 doesn't imply necessarily that the funcions are linearly dependent, right? But Wronskian = 0 imply two functions are linearly dependent only if these functions are solutions to the same differential equation?? Thanks in advance, professor.
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
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!
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
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)?
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
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.
@@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.
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.
UA-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
Sir, the subtitles that UA-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.
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.
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.
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!
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.
I was struggling with my university lectures, thank you very much sir, Nice presentation
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
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.
great video, really cleared things up
Wonderful video. You are a great teacher.
Thanks sir, lots of things are cleared now😊
great video, much better than my professors indecipherable notes :)
7:33 why the wronskian = 0 somewhere and not everywhere?
You"re just life saviour
Thanks
What a great mathematician!!! U helped me and my frineds a lot...
Thank you!
Love your videos sir
Awesome explanation. Thank you (from Brazil ) for the video.
I have a question. Wronskian = 0 doesn't imply necessarily that the funcions are linearly dependent, right?
But Wronskian = 0 imply two functions are linearly dependent only if these functions are solutions to the same differential equation??
Thanks in advance, professor.
Great Mr
i love this man
Thanks a lot sir 🔥🔥🔥
The Wronskian? More like "Sharing great information is your mission!" 👍
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 ?
I clicked so fast as soon as I saw the notification of your video
You're the best!
@@DrTrefor
Amazing video Sir
When will you make a video on partial differential equation . I am eager to know how will you make this concept easier 😀😀😀
Love from INDIA ❤️❤️
Thanks!
Very explanatory as always :)
I made notes along with the videos, does these 24 videos cover the entire ODE claculus 3 course?
Great video
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.
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!
Good day sir
What should I study to make simulations like thoes in your videos?
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 😂😂
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
the size of wonkskian matrix gives us info about the order of differential equation true or false
6:47 Are y1 y2 and y3 functions of t?
Yup
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
What if your wronskian is some function that is equal to zero on some value of t?
tnxxxx
My book diff. eqs. for dummies says the wronskian is a determinant that gives you the constants of the functions.
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.
Do you have a video on solving exact ODE's?
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
awesome!
please see if there is a way to make english subtitles available...
i am seeing vietnamese auto generated subtitiles
UA-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
Sir I have one question how i will connect you??
My main "public" channels are here on UA-cam or on my Twitter @treforbazett:)
@@DrTrefor okk sir Thanks
I will send message on your Twitter account
Thanks for the great explanation! :)
My pleasure!
Second 😊😎
hahah that's still pretty good:D
Sir, the subtitles that UA-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