Hello from a math teacher in Pakistan. I am glad to see teachers taking initiatives and helping students in their problems. I am positive our videos are a great source of help for them. Good work
Haha Steve, i had exact the same "trouble" distinguishing between the 5 and the s back in the university...my workaround method was writing the s with some horns added in both its ends, you can't imagine how fancy they look...
I have just found ur channel today and hands down ur already one of my favorite teachers on youtube. I wish i knew about u earlier. Ive been studying for some hours now and this is something i didnt do in a very long time. Your videos are very informative and very entertaining.
I am taking differential equations in MIT and literally, you are saving my time with excellent exercises. Our book is just awful. Just imagine, some of your exercises appeared in my midterm exam
This expression could have been written as 1/8[1/(s^2-4 -2) -1/(s^2+4)] and then 1/16[2/(s^2-4 -2) -2/(s^2+4)]. The Laplace Transform would , therefore be 1/16[Sinh(2t) -Sin(2t)], which is what Black Pen Red Pen got but in a convoluted way.
Coming back and re-watching this video a couple years later, it occurs to me that on question 20 and other hard partial fraction decomposition problems the residue theorem from complex analysis can be used to help with it. You'd just have to calculate a couple derivatives for building up the powers of s, and the rest is fine.
19 Here we can be tricky and build difference of squares from linear factor of denominator Then we will get constant term if we combine difference of squares with the other factor of denominator We will get 13=(s^2+9)-(s+2)(s-2) If we replace numerator by 1/13((s^2+9)-(s+2)(s-2)) we will have nice cancelling 20 16=s^4-(s^2-4)(s^2+4) and we have nice cancelling If we know hyperbolic functions we dont need partial fractions
Another way to solve the convolution of multiple trig functions: Based on the degree in the denominator for (s^2 + w^2)^n, the value of (n - 1) tells you how many times you'll eventually multiply trig by t. So you form a linear combination of t^k*sin(w*t) and t^k*cos(w*t), where w is the angular frequency, and k is a power that builds from 0 to (n-1). You then find corresponding Laplace transforms to each of these terms, and add up a linear combination with unknown coefficients, to equate to the original transform. Use the function parity property of convolution, you can eliminate half of the terms, and have half as many unknowns to solve for. f_odd(t) conv g_odd(t) = odd function f_even(t) conv g_even(t) = odd function f_odd(t) conv g_even(t) = even function If expecting odd functions, this means you can eliminate all t^even * cos(w*t) terms and t^odd * sin(w*t) terms. Vice versa, if you are expecting even functions. Then you proceed with solving for the unknown coefficients.
I've just finished the Laplace Transform Ultimate Study Guide video now I'm going to start watching this one, it's going to take me a while like the other one because I have other things to do.
Hii, thank you bprp for these marathon videos. It is very helpful even after 3 years and it will stay helpful. I would like to point that i couldn't open the file, which is not a big problem because we have the functions in the video and the description, but still it would be nicer to have them printed. Thank uu again
Residues are alternative way to partial fraction decomposition In fact complex partial fractions decomposition works better Residues are more comfortable also for inverse Z transform
For question 12 you can really easily simplify the partial fractions by letting some w = s^2 and then doing the partial fractions on w, and then later substituting s back in. edit: A simular trick can be used for Q16, where you can split up (s^4-16) into (s^2+4)(s^2-4) and again let w = s^2, do the partial fraction, reverse into the s world, then you can simplify it all down into 1/16(sinh(t) - sin(t))
Number 16 no need to do that to find C and D You just have to multiply bt (s^2+4) both sides and then evaluate at s=2i It will follow -1/8 =C(2i) +D Immediately D=-1/8 and C=0 1/(s-2)(s+2)=1/(s^-4) evaluated at s=2i equals -1/8
It looks like the (6) case the cos(t-π/2) can be also sin(t)... In the (7) case cos(t)-sin(t) can be sqrt(2)*sin(π/4-t)... This should be possibly simplified in the t-domain. (10) delta(t-a) correct, but maybe then mistake before the inverse laplace if this is result. The delta is more practical in s-domain than in t-domain... (26) Try inverse laplace of s^2/(s^2+a^2)... How to make this? Why result is different for L^-1{1-a^2/(s^2+a^2)} versus L^-1{s/(s^2+i×a)} ∗ L^-1{s/(s^2-i×a)} (∗ = convolution)? Can this prove that convolution theorem and other inverse laplace can differ? Is correct answer -a×sin(a×t) ? Maybe is it possible to reconfigure the transformation to present the frequencies reference point at t=0- (zero minus) so that resulted delta(0) would be delta(0-) and then by using laplace validity for t>=0 removing this delta-function?
Thank you Mathematic Marathons GOD!!!, we appreciate your big brain, but there are a topic, Limits, can you give us a marathon about it? (I'm learn English, I speak spanish )
Dang I havent done laplace transforms since ODEs in my first year of uni. I graduated with an applied math bachelors back in 2016 but never had to do these again haha. Ive used the laplacian in PDEs many times but never these again😅. Blast from the past! Thanks for the video! Was able to do these since you mentioned LT is linear and with your note of what the laplace transform is its easy to go backwards Thank you!! Idk if Ill ever use this again (even in my masters when I start it) but it was fun to watch haha
Do you have a whiteboard at your house? Great effort, I really do think you must have several clones who you swap in every 10min. Keep up the quality content!
Whenever I got a problem that stated "s" as a variable, my first line on my answer was "Let s = t" or something like that. S looks way too similar to 5 to be used a variable name. For the same reason, I have crossed hand-written every z for a very long time so that they don't look like 2.
Excellent question and barely no one asked. My first thing was NOT to have “math” in my channel name. Then I wanted to say something to intrigue other. Then I also wanted to point out the obvious thing that made me stand out. I think these were the things I was thinking about 10 years ago.
Hi Bprp, as always enjoying your infinite amount of ideas. Now for q8 , I did the convolution and got a result like you had on paper with a t^2 • cos t term, but i cannot see how it matches your result (and the result from partial fractions). Could there be more than 1 solution?
Since convolution is ultimately a definite integral, just with a variable upper limit of integration, there is only one solution to each stage, and one solution tot he final stage. There could be multiple ways of expressing it that ultimately are equivalent, but there still is only one solution. Here's my partial fractions solution to that one: 1/(s^3*(s^2 + 1)) = A/s^3 + B/s^2 + C/s + (D*s + E)/(s^2 + 1) Using the function parity property of convolution, we see that we have an even function (t^2, corresponding to 1/s^3) that is convolved with an odd function, (sin(t) corresponding to 1/(s^2 +1)). A mixture of both function parities, means that we should expect an even function. If we had equal function parities, such as odd conv odd, or even conv even, then we'd expect an odd function. This means, by inspection, we can know in advance that B and E are zero. 1/(s^3*(s^2 + 1)) = A/s^3 + C/s + D*s/(s^2 + 1) Heaviside coverup finds A, at s=0: A = 1/(covered*(0^2 + 1)) = 1 Let s= 1, and s=2, to find the remaining constants: s=1: 1/2 = 1 + C + D/2 -1 = 2*C + D s = 2: 1/(8*(4 + 1)) = 1/8 + C/2 + D*2/(4 + 1) -1 = 5*C + 4*D Solutions: C = -1, D=1 Partial fraction result: 1/s^3 + -1/s + s/(s^2 + 1) Inverse Laplace solution: t^2 - 1 + cos(t)
Non-integer powers of t have a Laplace transform that uses a function that interpolates the factorials, called the Gamma function. So where L{t^n} = n!/s^(n + 1), the corresponding Laplace transform of non-integer powers of t is: L{t^p} = Gamma(p + 1)/s^(p +1) For reasons I can't explain, the Gamma function is offset from the factorials by a shift of 1. Your given transform is: sqrt(pi/(4*s^3)) which we can rewrite as sqrt(pi/4) * 1/s^(3/2) This means (p+1) = 3/2, implying that the power p = 1/2. We want Gamma(3/2) to appear upstairs, so we multiply by 1 in a fancy way to achieve this: sqrt(pi/4)/Gamma(3/2) * Gamma(3/2)/s^(3/2) Now we can take the inverse Laplace and get: sqrt(pi/4)/Gamma(3/2) * t^(1/2) Special cases of the Gamma function occur half way between integers, where: Gamma(n + 1/2) = (product of odds up to 2*n-1)/2^n * sqrt(pi) Gamma(3/2) = (product of odds up to 1)/2^1 * sqrt(pi) = 1/2*sqrt(pi) Thus: sqrt(pi/4)/(1/2*sqrt(pi)) = 1 And our result is: sqrt(t)
Sir, I have a question that little bothering me for such a long time. We know that 1+(1/2)+(1/3)+(1/4)+..... which goes infinity. And 1^2+(1/2)^2+(1/3)^2+(1/4)^2+....which sums up pi^2/6 right? So I wonder what is the "number of power" that makes this series goes converge with an exact number and once the number of power gets smaller than the specific number of power, it goes diverge. I know that number must be between 1 and 2, I am just really curious. I will appreciate if I get the solution. Thank you.
@@ken52037g I would highly recommend you to check out the riemann zeta function. It is a generalisation of this series that has a lot of special properties (there is a 1 million dollar price for the person that solves riemann's hypothesis).
@@ken52037g Interestingly enough, the cutoff point for divergence the series of 1/n^p, and the integral of 1/x^p from x=1 to infinity, occurs at p=1 for both of them. 1/n^1 and 1/x diverges, while just adding any number no matter how small to the exponent of 1, will make both of these converge. It's an easy Calc 1 problem to find the any solution to integral 1/x^p dx from 1 to infinity, or to any upper limit of integration in between, but it is a very complicated problem that is still unsolved, to find the exact solution to the series of 1/n^p, from n=1 to infinity or any upper limit of the sum. Where p>1, so that it converges, of course. These two problems are very closely related to each other, but the difference between discrete and continuous makes all the difference.
Question: If there is 7.8 billion people on earth and we are all suppose to keep a 6 foot distance between ourselves to prevent the spread Cov-19, how much land do we need? It seems if we were to use a simple square grid we would need trillions of square miles. What would be the most efficient layout and is there enough land?
The most efficient layout would be a hexagonal close packing. Each person gets a hexagon to themselves, with a cross-flats distance (like the way you measure a wrench size) of 6 ft. To calculate the cross-points distance on a hexagon, given the cross-flats distance, you multiply by 2/sqrt(3). Let F equal the cross-flats distance, and P equal the cross-points distance. Such a hexagon is equal in area to 6 equilateral triangles of side length P/2, which each have an altitude of F/2. So the area of each triangle is: 1/8*P*F, and the hexagon area is 3/4*P*F. Plug in P=F*2/sqrt(3), and simplify, and we get: A_1hex = sqrt(3)/2*F^2 This means we'd need 31.177 square feet for each person. Multiply by 7.8 billion, and we get 2.43*10^11 ft^2 Divide by (5280 ft/mi)^2, and we get 8773 square miles required.
