The Logistic Equation

[@javgrzg]

OMG! I'm a big fan of Gilbert Strang! I discovered him at college reading Linear Algebra and Its Applicatios. Immediately I knew I was in front of something different. The way he explains is of such a high quality, so natural. Here, in Argentina, the young ones would say "tiene flow!" haha. Now I just discovered this in KZbin and is the first time I actually see him "in person". Thank you VERY much professor Gilbert Strang for your amazing books 🙏❤️. And thank you for this video also!

[@PeteZam]

Great video again. I love how you break down and dissect everything, and you do an amazing job of making it easy to follow. These videos have been extremely helpful.

[@rafaelsouza4575]

Coming from the ML area, it's nice to hear about the regression equation and the sigmoid function in another context.

[@cassioalles5116]

Gilbert Strang , The friendly professor!

[@Kori114]

Awesome lecture. I really liked the discussion of the solution.

[@xiaominsong]

This logistic equation is a great approximation of a start-up company's value's growth starting from joining a market to maturity or saturation state.

[@Palisade5810]

This looks like the fermi-dirac distribution. And the number of fermions that can occupy a space is dampered by the pauli exclusion principle which makes perfect sense since no two particle being able to occupy the same spin state puts them in competition with eachother for the availiable states.

[@victorburnett6329]

Nothing more comfortable than watching this stuff.

[@abdrrahimkhouya6660]

Thank you sir for this lecture

[@njabulomahlalela2912]

that was brilliant, thank you MIT

[@payammahbobi8029]

Wow. Clear, Precise and Beautiful. Thank you Professor Strang!

[@oneforallah]

Thanks for all the lectures MIT.

[@jaeimp]

Derivation of the solution from @2:37:user-images.githubusercontent.com/9312897/89488137-39421900-d775-11ea-9fe8-8c7b953e3f1e.png

[@nafizabdoulcarime5082]

awesome, many thanks

[@suharsh96]

This was really good! Thanks sir.

[@Tony855C]

Where did you get that burgundy sweater. Must know.

[@l1mmg0t]

always enjoy his class

[@hamadalazmi4489]

Amazing Prof. ❤️❤️❤️

[@wifrigrosario5059]

my question is the logistic expression for general supply chain, thanks

[@zeyneptufan5789]

Thank you so much, it is great and really helpful

[@omarsanhaj6904]

I would thank you professor Gilbert for you wonderful class, otherwise i still don't understand where the "a" in the numerator of the solution y(t) come from ? Thank you in advance for your clarfication

[@stephenbeck7222]

Let dz/dt = -az + b. Then separate variables to get dz/(-az+b) = dt, and integrate both sides and solve for z with normal calc 1 methods. You get z = (de^(-at)-b)/-a, where d = e^c, the constant of integration. Then you substitute back into y = 1/z to get the final solution. It's a neat trick. Normally in a calc 2 class I think you'd see it solved without the z substitution by using partial fractions.

[@christianjimenez1877]

Excelente clase. ¡Gracias!

[@PremiDhruv]

-by2 term will make it more and more negative and that increases forever. Unable to understand how does it reach a steady state. there will be a time where the derivative is zero but then it will become negative and negative. Right ?

[@geraldocarvalhobritojunior4793]

Very nice! Thanks for the lecture...

[@chinmoypal3397]

Is there any we can listen to all of lectures

[@brendawilliams8062]

Thankyou

[@feyisayoadediwura1896]

Thank you.

[@surendrakverma555]

Very good 🙏🙏🙏🙏🙏

[@gregmcn11]

What are the steps to the solution at 4.30 ?

[@xayanwelsh]

You can use e^-at as your integrating factor and solve from there to get the solution e^[ integral ( -a ) dt ] constant coefficient

[@tenzin9327]

dz /(az-b) = -dt integrate both sides substitute az-b = u adz =dr , dz=1/a *dr 1/a integrand 1/r dr = -t + c1 ln r + c2 = -at +ac1 ln r = -at + ac1 -c2 a,c1 and c2 are all constants let ac1 -c2 = c3 ln r = -at + c3 r = e^ -at +c3 r = e^-at . e^ c3 r= e^ -at .d az -b = e^ at .d az= e^at *d +b y=a/(e^at *d +b )

[@jaeimp]

Here is the derivation: user-images.githubusercontent.com/9312897/89488137-39421900-d775-11ea-9fe8-8c7b953e3f1e.png

[@phillipechavda3654]

@@tenzin9327 you did it the hard way, dz /(az-b) = -dt -> integrate bouth sides ln(az-b)/a = -t + C -> put an "e" under bouth sides az-b = e^(-at+aC) az - b = e^(-at)*d (if e^(aC) = d) z = (b+d*e^(-at))/a y = a/(b+d*e^(-at))

[@awesomecat2293]

T=0 is probably y=2020

[@andykim9536]

where does that b comes from in the equation?

[@tenzin9327]

dz /(az-b) = -dt integrate both sides substitute az-b = u adz =dr , dz=1/a *dr 1/a integrand 1/r dr = -t + c1 ln r + c2 = -at +ac1 ln r = -at + ac1 -c2 a,c1 and c2 are all constants let ac1 -c2 = c3 ln r = -at + c3 r = e^ -at +c3 r = e^-at . e^ c3 r= e^ -at .d az -b = e^ at .d az= e^at *d +b y=a/(e^at *d +b )

[@Wondermass]

We did a video (featuring the logistic equation) about where equations come from in social science: kzbin.info/www/bejne/sIm8Z397rK2al6s

[@Tony855C]

Huge respect for being 40+ older than anyone else teaching math on KZbin. He's pwning newbs on a website he ain't even know how to use.

[@oldPrince22]

Anyone considered the case that the population is 1, not 0 neither a/b? But one person cannot give birth to babies...