I have solution with Gamma function for odd n a_{n}=(1-(-1)^n)/4*2^n*Γ(1/2n-1/4)/(Γ(1/4)Γ(n+1)) for even n a_{n}=-(1+(-1)^n)/8*2^n*Γ(1/2n-1/4)/(Γ(3/4)Γ(n+1))
you are incredibly gifted at explaining mathematics in an algorithmic way. I've watched hundreds of your videos and you are carrying me through my undergraduate math courses. Thank you for all your free videos. I will buy your merch someday i promise!
This is a great video explaining exactly what I needed in my differential equation class. I have to tell you that you explain this and other mathematical materials so well that I wish you were our class teacher! The one I currently have teaching the DE class is useless. All your videos that I watched are very helpful and you REALLY EXPLAIN THE MATERIAL VERY VERY WELL!!!
I have my ODE final exam this evening, your video has brightened my view on power series solutions which I was struggling with. You should do some videos on Laplace transform
I used to just watch your videos for fun but I never thought you'd actually teach me something I need to know for a class better than my professor does, thanks man
Watching this by recommendation of my professor for help with differential equations during the Covid-19 quarantine. Thank you so much for making this video! It helped a lot!
Excellent explanation. Working out the solution to this problem shows the kind of patience required for and rigor of this level of Mathematics. Professor Milnorski always told us to push the pen and get dirt under our finger nails.
Very beautiful indeed! What I like about your videos is that you explicitly demonstrate that mathematics is not "rocket science" so to say. It is bits and pieces of very logical building blocks and all you need to do is not to rush and be very attentive and systematic in your steps and sometimes step back and have a look at your work from a distance and you will eventually get there... Even if you find yourself having gone in the wrong direction, if you have been systematic, it will be easy to trace your steps back and seek a different path rather than getting lost... I greatly enjoy your exercises in various disciplines even though I am neither an engineer nor do I have a practical use for higher maths. I just love it for its own sake...
It's 2:30 am and I just finished watching the whole thing. Please tell me I'm not the only one who started laughing at the end and everything is okay with me. Thanks
Quite literally, you have explained so many mathematical problems in such great detail that I wonder if you are a professor. If not, you should probably look into it :)! Thanks for this video as well as all of the others!
On the series multiplied by a0, we had 1*x^0/0! + (-1)*x^2/2!. If one lets L = 1 for the expression 4L - 5, then 4L - 5 = -1, which is the numerator of the second summand in the series. Hence we could say the numerator P(k)= Π([from n = 0 to n = k], 4n - 1) If k = 0, then P(0) = -1 and P(1) = 3. Meanwhile, P(-1) is the empty product, so 1. So this accounts for the first term in the series as well. Hence the first series could be rewritten as a0*Σ([from k = 0 to infinity], P(k - 1)*x^2k/(2k)!), where P(k) is as defined above. There is no closed form for P(k) as far as I know, so it will have to do. As for the second series, one simply has in the numerator Q(k - 1), instead, where Q(k) = Π([t = 0 to t = k], 4t + 1). What I find interesting is that y = a0*Series 1 + a1*series 2. Series 1 is even and series 2 is odd. If we call them f(x) and g(x) and let a0 = A and a1 = B, then we essentially just have the linear combination form of the general solution, with f(x) and g(x) being specific solutions. Since f(x) is even, it can be written as a Fourier series of only cosines, and g(x) with only sines.
Absolutely amazing! The explanation was in the point ... I was having so much trouble with these and you literally saved me ... Thank you for your great work 😊
You're missing a factorial sign right at the end 37:46. At around 22:14, you missed an "a_1" before the "x^9". For those of you wondering if there's an elementary form for this solution, no there isn't. These are exotic things like Hermite polynomials and Kummer functions.
Thanks man. I have a final exam coming soon. You saved me from failing my exam. (I seriously thought that I'd fail for sure, even though I'm not bad at math) I had lost all hope for ever understanding how to solve a differential equation by using series but you returned it to me. I knew you wouldn't disappoint me. Thank you so much. P.S : Now I know the only difficult part of it is having the patience and concentration to do all of that. Which is clear that it was also difficult for you. I felt the relief on your face when you finished the problem😂😂😂😂😂
Great videos man, please continue doing this videos until you will teach, because there are not a lot of teachers like you who love their work, keep up your great videos and sorry for My bad english ahahah. Greets from Italy!
In finals , you are kindly requested to do this in a quarter or 20 mins tops... in the meanwhile , professors are so bored to do this properly on board , instead they leave the solution as a homework ... from a fed up engineer
I dont know if its too "hard" but you can put the (-1/2)x^2 in to a0 sum because m=1 4m-5= -1. (modular arithmetic). And i dont know if this is helpfull at all ^^
I think it would you dont subtrakt the sum you "add" it and with the -1 do the rest of the job ^^ doesn,t it? so it would a0(1 + sum from k=1 to inf. (-1*3*7...(4k-5)) and so on. it is 3 o'clock in the morning so it could be wrong... :). if it's ture does it "help" to simplyfy it more?
4:50 Why can't we reduce from n=1 to n=0? It seems that if "n" were 0 then we have 2*0*a0*X^0 = 0, so we don't affect total sum if we write sum from n=0
Hello! Please let us know why you choose zero and one at the time of 11:10 min for the arbitrary values. Thank you for being so gentle and generous with your time !!!
If you mean 11:49 it's because we don't know their value. They can be anything. That's why they're arbitrary so we write them as they are. They act like a constant in our general answer for a differential equation.
You could have just started the sum from n=0 in the middle term - it's zero at n=0 due to the factor of n. No need to fiddle with each term to start at n=1.
Not gonna lie, was totally trying to do y''+xy=0 and was confused because making X^n constant through each term ended up making one of the starting numbers negative, which maybe wouldn't be bad but surely it can't be the same as just keeping n equaling 0
@@angelmendez-rivera351 No much easier, the only difference are the steps to find the recurrence equation. lowering it will provide (n+2)(n+1)a_{n+2} - (1-2n)a_{n} = 0
@@angelmendez-rivera351 How isn't it? it gives you immediately the equation for a_n+2, the only thing that you need to do is calculate a2 in the equation instead of from what you have taken out. But that does not take long
After so much work, it is worth to add one more step and write the product as a Pochhammer symbol or directly as Gamma functions... Product[a m - b,{m,1,k}] = a^k (1-b/a)_k = a^k Gamma[k+1-b/a]/Gamma[1-b/a]
Also worth to mention that this is the Hermite's Equation (for lambda=1) and the series solutions end up being a nice closed form of the 1F1 (Confluent Hypergeometric Function) mathworld.wolfram.com/HermiteDifferentialEquation.html
00:01 Solving differential equation using power series 02:19 Explanation on adjusting power series solution 07:15 Finding the power series solution to a differential equation. 09:44 Solving for the conditions a and n when a is equal to 0 14:41 Illustration of power series solution 17:16 Deriving the power series solution for a differential equation 21:48 Power series expansion simplifies differential equation solutions. 24:02 Developing a formula for the coefficients in the expanded form of the series. 28:46 Deriving a formula for incrementing numbers by 4 30:44 Formula for arithmetic sequence with common difference 4 35:01 Introduction to the capital PI notation for product. 37:09 Explanation of calculating terms in a power series solution
The moment you start giggling when writing the a0 and a1 terms separetly in colors you had me laughing so bad 🤣🤣🤣 and the moment when the capital pi was obviously coming I was like "come on!!!!! not a pi in sigma!!! 🤣🤣🤣🤣
There is no radius of convergence because function is analytic for all x in R, and he didn't define what should be regular point, but e.g if we want to examine if x0=1 is regular point: a2(x)=1 ; a1(x)=-2x ; a0(x)=2; a2(1)=1 and isnt 0; then, check if p(x) and q(x) don't have snigularities (if they have singularity then radius of convergence is length between x0 and nearest singularity) --> p(x)= a1/a2= -2x and q(x)= a0/a2=2 ---> p(x) and q(x) dont have singularities that mean that there don't exist lenght to nearest sing. and there is no radius of convergence and that mean that function y''-2xy'+y=0 is analytic everywhere. so the first power serie would be SUM(from k=0 to inf)= Cn * (x-x0)^n ---> SUM(from k=0 to inf)= Cn * (x-1)^n because he didnt define regular point then write just SUM(from k=0 to inf)= Cn * x^n (because x0=0) hahaha and now i realise that his regular point is x0=0 sorry, just do the whole procedure for x0=0; You do this to EVERY POWER SERIES: check if a2(x0)isn't 0 then check if p(x) and q(x) don't have singularities, and p(x) and q(x) ARE ALWAYS p(x)= a1/a2 and q(x)= a0/a2 if p and q have singularity radius of convergence is length from x0 to neares singularity
We did an example question for each type of solution / method but this type of question was not set in our exams 😁😅 c1976 so I saved all the notes for retirement. Here I am now trying not to make mistakes, its been so long!
My MCQ question for you for my information. Solution of power series in ODE is (A) liner independent (B) liner dependent (C) both A and B solution (liner dependent and independent) Which option is correct A or B or C
That was a long ride.. but thanks for clarifying the context of Stewart's example. You got this from his book right? If you can't do the IVP version, it's alright. I'll do it myself. Thanks again!
You're welcome. Yes, this is an example from the book. I really didn't want to do the IVP version anymore after 3+ hrs of recording + editing + other things....
blackpenredpen Power Series is sure really frustrating but still enjoyable to do so! I've got one more Bizarre request: Can you do the Snowflake Curve problem for series? It's hard for me to do so.
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This is not a full lecture. This is just one problem...
blackpenredpen
so where is the other vidios ??
@@gatocorniossatanicos6542 Just give up if it's really impossible
Thank you very much. This video showed I was doing some things wrong.
I have solution with Gamma function
for odd n
a_{n}=(1-(-1)^n)/4*2^n*Γ(1/2n-1/4)/(Γ(1/4)Γ(n+1))
for even n
a_{n}=-(1+(-1)^n)/8*2^n*Γ(1/2n-1/4)/(Γ(3/4)Γ(n+1))
@@remagairsoftllc mine dumped me directly into exams 😅
Absolutely phenomenal explanation. I swear, I pay my teachers for nothing...
you are incredibly gifted at explaining mathematics in an algorithmic way. I've watched hundreds of your videos and you are carrying me through my undergraduate math courses. Thank you for all your free videos. I will buy your merch someday i promise!
I remember doing these problems in my DiffEq class. I love math, but this was pure hell
was that the longest in the class u took? and were there harder questions than that? im taking it next semester and im worried
@@farooquenadeem5300power series and laplace transforms usually took my professor almost half the class to solve 1 or 2 so yes its long
Power series was the biggest part I was like whaaaaat tf
This is a great video explaining exactly what I needed in my differential equation class. I have to tell you that you explain this and other mathematical materials so well that I wish you were our class teacher! The one I currently have teaching the DE class is useless. All your videos that I watched are very helpful and you REALLY EXPLAIN THE MATERIAL VERY VERY WELL!!!
The satisfaction you get at the end solving one question after working for 40 minutes cannot be explained in words😭
Omg, if I had a teacher like this I would have straight A’s, just solved my entire chapter with absolute ease after watching ur videos. Thank you
It's my pleasure to help! Thanks for leaving a comment!
+blackpenredpen
You are not only great at teaching, but really cute as well.
@@Peter_1986 dammnnnn laurel get it lmao
I have my ODE final exam this evening, your video has brightened my view on power series solutions which I was struggling with. You should do some videos on Laplace transform
Waycee Harvey I have a ton of videos on that already. You can check out my playlist.
ODE prof never taught this, now my grad PDE prof expects us to know it. You sir are a godsend
I don't know what i would do with my Engineering Mathematics without you sir..i really appreciate your help..Thank you.
who else has finals comin up
facts tho
Me
Me in like 3hrs time, and I don't grab shit about power series
Mid Semester
@@hamedurrahman3408 i forgot i posted this LOL, it ended up being top comment:D
I used to just watch your videos for fun but I never thought you'd actually teach me something I need to know for a class better than my professor does, thanks man
you watch math videos for fun?
Same, used to watch him in high school. Now I actually need to learn this for my final.
Watching this by recommendation of my professor for help with differential equations during the Covid-19 quarantine. Thank you so much for making this video! It helped a lot!
Someone get him a wireless microphone
for real!!
This is by far the most organized solution on this topic, Thanks!
Excellent explanation. Working out the solution to this problem shows the kind of patience required for and rigor of this level of Mathematics. Professor Milnorski always told us to push the pen and get dirt under our finger nails.
17:12 "This is...a lot of fun....i guess"
Mad color pen juggling skills with to-the-point explanations!
Doug Wallace thanks!
You explained in 11 minutes a topic I had been struggling on for hours. Thank you so much!
It's refreshing to have him wear a Berkeley shirt while I do the university's homework.
I've stopped going to classes. This is so much better than listening to my professor.
Very beautiful indeed!
What I like about your videos is that you explicitly demonstrate that mathematics is not "rocket science" so to say. It is bits and pieces of very logical building blocks and all you need to do is not to rush and be very attentive and systematic in your steps and sometimes step back and have a look at your work from a distance and you will eventually get there... Even if you find yourself having gone in the wrong direction, if you have been systematic, it will be easy to trace your steps back and seek a different path rather than getting lost...
I greatly enjoy your exercises in various disciplines even though I am neither an engineer nor do I have a practical use for higher maths. I just love it for its own sake...
It's 2:30 am and I just finished watching the whole thing. Please tell me I'm not the only one who started laughing at the end and everything is okay with me. Thanks
13:55 2+2 is 4 minus 1 thats 3 quick maths
I laughed also XDXD
2 plus 2 is four.. minus free that's one. QUICK MAFFS
Finding for this exact comment
Thanks for your help! I love that you go through every step and are very thorough!
Quite literally, you have explained so many mathematical problems in such great detail that I wonder if you are a professor. If not, you should probably look into it :)! Thanks for this video as well as all of the others!
he is
On the series multiplied by a0, we had 1*x^0/0! + (-1)*x^2/2!. If one lets L = 1 for the expression 4L - 5, then 4L - 5 = -1, which is the numerator of the second summand in the series. Hence we could say the numerator P(k)= Π([from n = 0 to n = k], 4n - 1) If k = 0, then P(0) = -1 and P(1) = 3. Meanwhile, P(-1) is the empty product, so 1. So this accounts for the first term in the series as well. Hence the first series could be rewritten as a0*Σ([from k = 0 to infinity], P(k - 1)*x^2k/(2k)!), where P(k) is as defined above. There is no closed form for P(k) as far as I know, so it will have to do.
As for the second series, one simply has in the numerator Q(k - 1), instead, where Q(k) = Π([t = 0 to t = k], 4t + 1).
What I find interesting is that y = a0*Series 1 + a1*series 2. Series 1 is even and series 2 is odd. If we call them f(x) and g(x) and let a0 = A and a1 = B, then we essentially just have the linear combination form of the general solution, with f(x) and g(x) being specific solutions. Since f(x) is even, it can be written as a Fourier series of only cosines, and g(x) with only sines.
Absolutely amazing! The explanation was in the point ... I was having so much trouble with these and you literally saved me ... Thank you for your great work 😊
THIS WAS GOOOOOOOOOD🤩 !!!!! why cant i have a teacher like this in my uni??🤧
I have an Exam tomorrow... and this video help me understand power series solutions better.. thanks for the help man 😃💯💯
you my friend saved me from failing my finals horribly. thanks a lot
This guy is definitely about his business!
Best Explanation, even my professor could not explain like this, thank you so much!!
Glad to help 😃
You're missing a factorial sign right at the end 37:46. At around 22:14, you missed an "a_1" before the "x^9".
For those of you wondering if there's an elementary form for this solution, no there isn't. These are exotic things like Hermite polynomials and Kummer functions.
Thanks man. I have a final exam coming soon. You saved me from failing my exam. (I seriously thought that I'd fail for sure, even though I'm not bad at math)
I had lost all hope for ever understanding how to solve a differential equation by using series but you returned it to me. I knew you wouldn't disappoint me. Thank you so much.
P.S : Now I know the only difficult part of it is having the patience and concentration to do all of that. Which is clear that it was also difficult for you. I felt the relief on your face when you finished the problem😂😂😂😂😂
This video was extremely helpful ! You've saved my grade! Thanks again
This lecture is masterpiece. Can't wait to practice! Thanks for your passionate and brilliant explanation. 👍👍
Thank you!
Great videos man, please continue doing this videos until you will teach, because there are not a lot of teachers like you who love their work, keep up your great videos and sorry for My bad english ahahah.
Greets from Italy!
Trybal Clips thank you. I will keep making videos.
Before i watch the video, i give it like and this is comment. Now i will start enjoining.
Lol
If you begin the a0 product at -1 and go up to (4k-5), you can incorporate the negative sign and the second (x^2) term into the sum.
Thank you so much! I have my final for Calc 2 today and this saved my life
mudhara makaipa its really of great help, ndabatsirikana
Thanks so much for taking the time to do all this work for us. I appreciate it!
In finals , you are kindly requested to do this in a quarter or 20 mins tops... in the meanwhile , professors are so bored to do this properly on board , instead they leave the solution as a homework ... from a fed up engineer
wow.. watched an entire 38 min video without getting bored !!!
IDK 🤷♂️ I have solved a similar problem in my final exam 🙃 using your method of solving. Remarkable!
I dont know if its too "hard" but you can put the (-1/2)x^2 in to a0 sum because m=1 4m-5= -1. (modular arithmetic).
And i dont know if this is helpfull at all ^^
gaier19 but the signs wouldn't work out
I think it would you dont subtrakt the sum you "add" it and with the -1 do the rest of the job ^^ doesn,t it?
so it would a0(1 + sum from k=1 to inf. (-1*3*7...(4k-5)) and so on. it is 3 o'clock in the morning so it could be wrong... :). if it's ture does it "help" to simplyfy it more?
gaier19 ah I see! Yea, if I had not put the neg in the front, then yes. -1,3,7,11 etc, which is a sequence that's increasing by 4
you are my real university teacher
I'm surprised I still know this and its been 6 years since I took a D.E. course. I am happy that I do remember it though!
this is literally the same exact question in my homework haha thnx!! Your explanation was great as well! The pen color switching was super helpful
oh snap GO BEARS!! haha currently a freshman @ uc berkeley studying for math 1b
you are great at explaining math👍 your videos have helped me out so many times
I wish I have a professor like you.
4:50 Why can't we reduce from n=1 to n=0? It seems that if "n" were 0 then we have 2*0*a0*X^0 = 0, so we don't affect total sum if we write sum from n=0
Exactly.
You can do it, but you will have to go another way to solve the recursive formula and produce the series
Explaining style to make it understand is excellent sir
Hello! Please let us know why you choose zero and one at the time of 11:10 min for the arbitrary values. Thank you for being so gentle and generous with your time !!!
If you mean 11:49 it's because we don't know their value. They can be anything. That's why they're arbitrary so we write them as they are. They act like a constant in our general answer for a differential equation.
Ahh, Power Series Diff EQ...I remember those torturous days! Good luck to all of you currently studying this topic lol
I'm currently studying that. It's very painful😑😑
Indeed
Another excellent video.
trialen thank you.
trialen yes all bprp videos are excellent you dont say
The best video on this topic. Thank you so much!
Great explanation brother loved your video very much
Thanks for making this very much understandable.
😂😂 agba
Could you hear the sound ?
Thank you and thank the double tapping option on the right of the screen (over half an hour!!)
You could have just started the sum from n=0 in the middle term - it's zero at n=0 due to the factor of n. No need to fiddle with each term to start at n=1.
Not gonna lie, was totally trying to do y''+xy=0 and was confused because making X^n constant through each term ended up making one of the starting numbers negative, which maybe wouldn't be bad but surely it can't be the same as just keeping n equaling 0
no, that would have made solving the difference equations much harder.
@@angelmendez-rivera351 No much easier, the only difference are the steps to find the recurrence equation. lowering it will provide (n+2)(n+1)a_{n+2} - (1-2n)a_{n} = 0
Rudy Niet That isn't any easier, lol, are you insane or what?
@@angelmendez-rivera351 How isn't it? it gives you immediately the equation for a_n+2, the only thing that you need to do is calculate a2 in the equation instead of from what you have taken out. But that does not take long
Thank you for this!!!!! This has helped me so so much! Keep doing these they are fantastic yayayaya
4:45, i think you actually can start at n=0 for the second sum. That sum, at n=0, evaluates to 0--and so does not change the result.
22:15... a1 missing... You´re awesome... Thanks for sharing
Thanks, I learned a lot from this video. P.S. you're missing a factorial (denominator of odd coefficient series) in your final answer! 37:50
After so much work, it is worth to add one more step and write the product as a Pochhammer symbol or directly as Gamma functions...
Product[a m - b,{m,1,k}] = a^k (1-b/a)_k = a^k Gamma[k+1-b/a]/Gamma[1-b/a]
Also worth to mention that this is the Hermite's Equation (for lambda=1) and the series solutions end up being a nice closed form of the 1F1 (Confluent Hypergeometric Function)
mathworld.wolfram.com/HermiteDifferentialEquation.html
I understand this more better than my prof in Advance Mathematicsheck
00:01 Solving differential equation using power series
02:19 Explanation on adjusting power series solution
07:15 Finding the power series solution to a differential equation.
09:44 Solving for the conditions a and n when a is equal to 0
14:41 Illustration of power series solution
17:16 Deriving the power series solution for a differential equation
21:48 Power series expansion simplifies differential equation solutions.
24:02 Developing a formula for the coefficients in the expanded form of the series.
28:46 Deriving a formula for incrementing numbers by 4
30:44 Formula for arithmetic sequence with common difference 4
35:01 Introduction to the capital PI notation for product.
37:09 Explanation of calculating terms in a power series solution
Sir your smile is a very special one and also a very sweet one, keep smiling, 12:47 :)
Great video!
This helped me understand the process of solving these problems,
Thank you
Your lectures are so clear! Thank you so much
The moment you start giggling when writing the a0 and a1 terms separetly in colors you had me laughing so bad 🤣🤣🤣 and the moment when the capital pi was obviously coming I was like "come on!!!!! not a pi in sigma!!! 🤣🤣🤣🤣
thank you for the great explanation by the way
Beautifully explained. Excellent video!
Wait... what's the radius of convergence for this series?
There is no radius of convergence because function is analytic for all x in R, and he didn't define what should be regular point, but e.g if we want to examine if x0=1 is regular point:
a2(x)=1 ; a1(x)=-2x ; a0(x)=2; a2(1)=1 and isnt 0;
then, check if p(x) and q(x) don't have snigularities (if they have singularity then radius of convergence is length between x0 and nearest singularity) --> p(x)= a1/a2= -2x and q(x)= a0/a2=2 ---> p(x) and q(x) dont have singularities that mean that there don't exist lenght to nearest sing. and there is no radius of convergence and that mean that function y''-2xy'+y=0 is analytic everywhere.
so the first power serie would be SUM(from k=0 to inf)= Cn * (x-x0)^n ---> SUM(from k=0 to inf)= Cn * (x-1)^n
because he didnt define regular point then write just SUM(from k=0 to inf)= Cn * x^n (because x0=0) hahaha and now i realise that his regular point is x0=0 sorry, just do the whole procedure for x0=0;
You do this to EVERY POWER SERIES:
check if a2(x0)isn't 0 then check if p(x) and q(x) don't have singularities, and p(x) and q(x) ARE ALWAYS p(x)= a1/a2 and q(x)= a0/a2 if p and q have singularity radius of convergence is length from x0 to neares singularity
Kni Flic I give up, time to be a male stripper at my local bar
@@kniflapunxFucknazzi a0(x)=1
watching this vid gives me a slight feeling what infinity really means ...
I felt the same too
LOL
What a Marathon! Really nice video!
13:50 QUICK MAFS
wish you were my teacher.. very good explanation!
That must’ve been exhausting, especially holding that microphone the whole time
Wow this is gonna be a long one to watch!
Thanks a lot for your videos sir!
It was a LONG one for me to make as well. Thanks for watching!
Thanks you helped me get through a problem I was stuck on
Yes i wanted the variable "k"
You're videos are really great thanks! Please make more on Differential Equations!
Do we have to go that far with terms ?
If you start the capital PI notation (37:00) with i=1, so that the product becomes (-1×3×7×11...) then you get a shorter factor for a0 ...
The homogenous ones with ordinary points are pretty easy.
The ones near a Regular Singular Point suck a little more.
Please start doing these differential equations again, units 4,6 and onwards. Need help studying for upcoming test🙏🏼🙌🏼
at about 27:00 it doesn‘t go from 1 to 3, it goes from negative 1 to 3, which is also a difference of 4
We did an example question for each type of solution / method but this type of question was not set in our exams 😁😅 c1976 so I saved all the notes for retirement. Here I am now trying not to make mistakes, its been so long!
thanks for putting valuable content , it really helps me!!!
Who else is here in 2023😢
Me! 😆
2024
2025
tank you sir...
hope continue like this
My MCQ question for you for my information.
Solution of power series in ODE is
(A) liner independent
(B) liner dependent
(C) both A and B solution (liner dependent and independent)
Which option is correct A or B or C
That was pretty tutorable, thanks for the toil.
Very amazingly explained. Thank you so much.
thank you very much bro, your explanation is very clearly. Keep it up 🔥
That was a long ride.. but thanks for clarifying the context of Stewart's example. You got this from his book right? If you can't do the IVP version, it's alright. I'll do it myself. Thanks again!
You're welcome. Yes, this is an example from the book. I really didn't want to do the IVP version anymore after 3+ hrs of recording + editing + other things....
blackpenredpen Power Series is sure really frustrating but still enjoyable to do so!
I've got one more Bizarre request: Can you do the Snowflake Curve problem for series? It's hard for me to do so.
Hmm, my semester is ending on Monday and I have tons to do for my students. Maybe in the future but I can't say.
This one sparks joy