a nice non-linear recursion problem 

I know it doesn't matter for the point of the problem, but shouldn't it have been 505 and not 550?

6:59 NOT 550 but 505

14:06 Homework 14:43 Good Place To Stop

I don't feel that the "continuous system" is the same; it is the equivalent to (x_(n+1)-x_n)=(x_n-1)/(x_n+1) since y' corresponds to the difference between consecutive terms. I think the corresponding continuous system would be y'=(y-1)/(y+1)-y=-(y^2+1)/(y+1). An alternative would be for y to be a solution to the functional equation f(x+1)=(f(x)-1)/(f(x)+1)

0:00 Thumbnail is wrong. Sum starts at n=1 not n=0.

I am very impressed with your brilliance and I really enjoy your videos, Michael. You narrate very well your thought process and your delivery is clear and well timed plus, I might add, you come across as having a very pleasant disposition.

Thank you, Michael! I saw a new type of problems in your video. It's pretty interesting.

The second quadratic equation you get can even more easily be solved than with the quadratic formula by noticing that replacing x0 with -x0 gives the other equation.

There is a group type structure where the four solutions are related by the transformation, and the transformation cycles through the roots. That is, no matter what solution one picks for x0, the unordered set {x0,x1,x2,x3} is the same. This is a particular version of the Mobius transformations. Maybe MP can comment. Best

Great vid as always :)

14:06 that looks like a separable DE and once you have the family of solutions you can get the antiderivative and substitute.

Putting the basic recursion in a spreadsheet, it was immediately clear that there was a cycle of 4 values. Summing 4 values & trying a few numbers for x0, I quickly narrowed in to near 2.4. On a guess, I went to 1+√2, and all the other values fell out.

The problem in the thumbnail is not the problem in the video! Summing from 0 to 2020 gives a much less simple answer than 1 to 2020...

Fun, thanks.

Since I spent quite some time on the supplementary problem, here is my effort for the differential equation. (a) the equation is separable and the general solution is y+2Ln(y-1)=x+C b)Defining a=y(1) and b=y(2020) (which will appear in the limits of the condition) we can eliminate the constant C and get the first equation for a,b which is a-b+2Ln((a-1)/(b-1)=2019. c) Next we work with the integral condition. We change from the x to the y variable. Using the differential equation we solve for dx and we get dx=(y+1)*dy/(y-1). Substituting we get the integral y*(y+1)*dy/(y-1) with limits from a to b for the y variable. Since this integral is 0 we get a second equation for a and b which after some manipulations is (b-a)(4+a+b)=4Ln((b-1)/(a-1). d) Defining a temporary parameter K=2Ln((b-1)/(a-1) we get a simpler system b-a+K=2019,(b-a)(4+a+b)=2K which allows to find a and b as rational functions of K. These expressions are substituted in the definition of K and a non linear equation has to be solved numerically for K. I wish someone else has a simpler approach to this or even to prove me wrong in the above. The requirement for a solution to exist is (a>1 ,b>1) or (a

6:00 "attack" :)

great i solved in the same way. Calculate x[n+2]=-1/xn you will see that the sequence is a loop of size 4

Subscribed! Reaching pi*10^5 is a worthy goal

I tried the thumbnail version where the sum starts at 0 💀

X0=tan(x) X1=tan(x-pi/4).. We can do this It is a periodic sequence.

If the nr. Of terms is mult. Of 4, when x0 is solution, x1, x2 and x3 also are. If you invert x0 or change sign of x0, the sum changes sign, so, if x0 is solution also are 1/x0 and - x0 (and - 1/x0, that is x2). So solving x1=-x0, you have 2 solutions, and with x1=1/x0 have the other 2. Given the recursive func. Is ratio al with num. and den. 1st. Degree, and Given periodicity of. 4, we cannot have more than 4th Degree (in num. when sum x0 to x3 in funct of x0) , so no more than 4 solutions. So the previous are all sulutions.

Very nice

Set 1/x_n = tan(a_n). I believe we get a_{n+1} = a_n + pi/4.

Coming back to continuous version and worked out a numerical solution, but I'm not sure whether it is correct or not. By direct integration of the differential equation y' = (y-1)/(y+1), we get y + 2ln|y-1| = x+C (eqn 1). Rearrange, we also have (y+1)y' = y-1 or y = (y^2/2 + y + x)', integrating over (1, 2020), we have [y^2/2 + y + x] = 0 or [(y+1)^2 + 2x -1] = 0 for x = 2020 and 1. [(y(1)+1)^2 + 2(1) -1] = [(y(2020)+1)^2 + 2(2020) -1], using eqn 1, C can be solved out numerically as C = -998.7 (approximately). Back substitute into eqn 1, y(0) = -1012.5 (approx.) When we plot x = g(y) = y + 2ln|y-1| + 999 (y as a variable), it is asymptotic at large absolute y values (x = y + 999, a straight line at the two ends) with a singularity (cusp) at y = 1. BTW, y(1) = -1009.5 and y(2020) = 1007.5 (approx.)

14:05 Time to do some homework! Let's try to prove no solution exists. For simplicity let *N := 2020* , define *z(x) := 1 - y(x)* and only consider *y, z* on the interval *z: D := [1; N] -> ℝ with initial condition z(1) := c* As long as *z ≠ 0* we move all terms to one side: *1 = -z' * (2 - z) / (-z) = z' * ( 2/z - 1 ) = d/dx ( 2 * ln |z| - z ) =: d/dx f(z)* Integrate both sides from *1* to *x* and get *x - 1 = f(z(x)) - f(c) => x - k = f(z(x)) | k := 1 - f(c)* We found *z(x)* must lie in the pre-image of *{x - k}* under *f* (we can't simply take the inverse as *f* is not bijective). To get an idea what that actually looks like, do the following: 1.) Sketch the graph of *f(z)* 2.) Swap axes, thus flipping the graph 3.) Shift the flipped graph by *k* to the right (results in a left-shift if *k < 0* ) Notice the resulting graph has (at most) three different branches we can choose *z(x)* from: - First branch: *z(x) ≥ 2* - Second branch: *z(x) ∈ (0; 2]* - Third branch: *z(x) < 0* The graph of the continuous function *z* must stay on one of these branches. To check which one actually may lead to a solution, consider the integral condition (transformed to *z* ): *∫_1^N z dx = ∫_1^N (1 - y) dx = N - 1 > 0 ( * )* The first and third branch result in integral values *> 2N - 2* or *< 0* , respectively, so they cannot lead to a solution. If *z* lies on the second branch, the integral *( * )* represents a flipped and shifted sub-area of *∫_0^2 |f(z)| dz = ∫_0^2 -f(z) dz = [ z^2 / 2 - 2z * ( ln |z| - 1) ]_0^2 = 6 - 4 * ln(2)* As a sub-area the value of the integral *( * )* will be smaller than *6 - 4 ln(2) < N - 1* , regardless of the shift *k* we may introduce: Even on the second branch we can never satisfy the integral condition!

In fact you should mention that x_0 cannot be -1, 0 or 1 as in those cases the reduction mod 4 of the indices doesn't hold

y+2ln(y-1)=x+c or e^y(y-1)^2=Ae^x, subject to B.C. int(1,2020)[y]=0.

Its pretty easy to just guess one solution, you can do xn+1=-xn --> -x=(x-1)/(x+1) and solve that directly and get -1+-root(2) very easily. Ofc there could be more but getting one is pretty easy

I vaguely remember this is cyclic except for some initial values

another way to finish up the problem and solve the last equation is to make the substitution u=x-1/x, then (x-1)/(x+1)-(x+1)/(x-1)=-4/u

I wonder if the "continuous" version even has a solution. y(x) has a perpendicular tangent (y' becomes infinite) w in hen y=-1, y'>=0 when y>=1, so for y(x) to cross the x-axis 0

If professor Penn were to sing Do-Re-Mi from The Sound of Music, he would start it like this: "Let's stop at the very ending, a very good place to stop..."

Please , Could u solve the last ex that you've mentioned in the end, i tried to simplify the y' = y(x+h) - y(x) / h i don't know if it's the good way to achieve the solution or not !!!

I just imagine that the secret problem suggester is just Michael with a disguise on.

great

Z tranforms would also solve problems similar to this. Could you perhaps do some problems using the engineer's favourite discrete tool?

Once again mixed feelings between the interest of the problem + an overall nice way to solve it on one side and some random complexities + the horror of the mistake when dividing by 4 on the other...

6:59 where do you get the 550 from?

In the picture of the video the sum is from 0 to 2020 (2021 terms). Almost the same, but the expression of the solutions are not so nice. El

I think i proved that the continous problem has no solution, that, to me, it's pretty strange given that the discrete analogous has solutions

What is this problem suggestor that Michael mentions so often?

asnwer=1 . 0 isit 🤣🤣🤣🤣🤣

in fact, if f(x) = 1-2/(x+1) and if we draw the graph of the function and the element of the sequences, after 4 steps we meet the first element; the sum is zero if the first element is +/-1+/-2^0.5. thanks!!

very good but it is 505 instead of 550

8:46 Sub-Zero Wins!

could you make a video about Stirling's approximation/formula

What a about a non-nice linear problem?

My solution for the homework: y=y’(y+1)+1=y’y+y’+1 A primitive of y is therefore : (1/2).y^2 + y + x So the integral equals : (1/2).y(2020)^2+y(2020)+2020-(1/2).y(0)^2-y(0) If it’s equal to 0 we have y(0)=??? Maybe I missed something.. let me know!

505 not 550 2020/4=505

Ok, so for homework, I think I almost got it, but the part that I didn't get I was exploring numerically, and it seemed that the answer diverged, but anyway, let's get to it. I'll present my solution, hoping that someone can continue it, or at least give me a hint to how to finish it. I'll also show what I think we were supposed to do. y' = (y-1)/(y+1) Which is a separable differential equation ∫((y+1)/(y-1)) dy = ∫dx u = y-1, du = dy x = ∫((u+2)/u) du x = ∫(1+2/u) du x = u + 2 ln u + C' e^x = C⋅u²⋅e^u e^x/C = u²⋅e^u √(e^x/C)/2 = (u/2)⋅e^(u/2) W(√(e^x/C)/2) = (u/2) y = 1 + 2W(√(e^x/C)/2) And we got y(x) with an integration constant C. Now we can perform ∫y(x) dx = 0, and from now on, all integrals are evaluated from 1 to 2020. 0 = ∫y(x) dx 0 = ∫[1 + 2W(√(e^x/C)/2)] dx 0 = 2019 + 2∫W(√(e^x/C)/2) dx v = W(√(e^x/C)/2) v⋅e^v = √(e^x/C)/2 4C⋅v²⋅e^(2v) = e^x x = ln[4C⋅v²⋅e^(2v)] x = ln(4C)+2ln(v)+2v dx = 2(1/v+1)dv Substituting v and dv in the integral, and changing the integration limits to v(1) to v(2020), meaning W(√(e/C)/2) to W(√(e^2020/C)/2). 0 = 2019 + 4∫v⋅(1/v+1)dv 0 = 2019 + 4v(2020) - 4v(1) + 2v²(2020) - 2v²(1) 0 = 2019 + 4W(√(e^2020/C)/2) - 4W(√(e/C)/2) + 2W²(√(e^2020/C)/2) - 2W²(√(e/C)/2) Now we have an equation with only C as variable, but I couldn't figure out a way to isolate C in this equation. I was going to try to do it numerically, but as you can see, the e^2020 part explodes. Anyway, I tried to substitute the number 2020 in the equation by a lower number like 5, but even then, it seemed that only an infinite C would make the right-hand side of the equation equal to 0. So I probably got something wrong. I was also thinking that we probably were supposed to start the solution similarly to the video, so one thing we could is to differentiate the equation y' = (y-1)/(y+1) y'' = 2y' / (y+1)² And substitute y' into y'', just like he did y'' = 2(y-1)/(y+1)³ Doing the same thing again y''' = 4(2-y)y' / (y+1)⁴ y''' = -8(y-2)(y-1) / (y+1)⁷ But that doesn't seem to get us anywhere...