FASCINATING MIT Integral Solved With Recursion And Sums
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- Опубліковано 31 тра 2024
- I hope everyone has found this interesting! I loved how much thinking outside of the box this integral took and so wanted to share it!!! Please comment with any questions or suggestions for new topics, and as always, subscribe to stay updated.
~ Thanks for watching all.
Thanks again to everyone for all the recent support, and @maths_505 for reposting another of my videos - he inspired me to start in the first place so am immensely grateful and a bit starstruck!
#maths #mathematics #integrals #MIT #recursion #problemsolving #whoknew #fascinating #functions #euler #funproblems #proofs #functions #physics #sums #series #limits #whiteboard #math505
I've found a very clean solution (imo) :
3f(x) = f(2x) 3 * int_0^1 f(x) dx = int_0^1 f(2x) dx
3 * 1 = int_0^1 f(2x) dx
u = 2x, du = 1/2 dx, bounds become 0 to 2
3 = 1/2 * int_0^2 f(u) du
6 = int_0^1 f(u) du + int_1^2 f(u) du
6 = 1 + I
Giving back I = 5.
That's great! Using the fact that 0 will not be scaled down at all by the u sub is so smart. Great solution.
Had the same idea lol, quite clean and elegant solution. I always love how math has several paths that can lead to the same solution.
yea this is the first thought that comes through my mind because there are 2x and x so I immidiately think about u-sub and it works
Nice solution!
Nice solution!
I think there's a much simpler way to, using King's property. We can write the given integral as int(0 to 1) f(1-x) dx and then write f(1-x) = f(2-2x)/3. Now letting u = 2-2x, and doing some arithmetic, we get int(0 to 2) f(u)du = 6, which gives that the req. integral evaluates to 5
That's so smart! Kings rule is so helpful with problems like this.
@@OscgrMaths Thanks😇😇
Got another way to do it:
Let d/dx F(x) = f(x)
Trying to differentiate F(2x) gives us 2f(2x) which is 6f(x)
And integrating back d/dx F(2x) = 6f(x) leads to F(2x) = 6F(x)
F(2*0) = F(0) = 6F(0) gives us that F(0) = 0
and the given integral gives us F(1) - F(0) = 1
which means F(1) = 1
Go back to the original integral which is equal to F(2) - F(1) = 6F(1) - F(1) = 5F(1) = 5
That's such a good approach, looking at F(x) and f(x) generally always leads to a good solution with problems like these.
@@OscgrMaths Thanks! Your solution is also excellent! It is difficult for one to think about a recursive solution.
I think that this has been most entertaining, refreshing and an original aproach. That was fun :-)
@@alphalunamare Thank you so much!! Really appreciate it.
I was about to write the same comment as mismis but I had to eat dinner. Nonetheless this method gives you more content the upload schedule is going crazy rn
GREGGGGGG
So elegant mate.
This is great, you’re an awesome teacher! Glad this was recommended to me
Thanks so much!! That means a lot, glad you enjoyed.
Would love a video on partial derivatives and what those integrals would look like.
Ive always wanted to understand maxwell's equations too.
What a clean solution, good job !
Thanks so much!
Great video!
Thanks so much! Glad you enjoyed.
amazing!
Thank you!
3:39 what you did with the u sub, really buttered my biscuit.
These videos are so good.
Thanks for the feedback!
Would be interesting to see the space of solutions of that functional equation. One could see that this equation is linear, because of that linear combination of solutions leads to another solution. You can check that if you plug f(x) = C*x^a, a = ln3/ln2 + 2pi*n/ln2 * n -> this leads to a family of solutions. One example would be f(x) = C*x^(ln3/ln2) * cos(2pi/ln2 * ln(x))
Actually, with change of variables: ln(x) = t * ln(2) and change of functions: f(2^t) = g(t) the equation would be:
3g(t) = g(t+1). It's almost like a periodic condition for Fourier series and all that stuff.
And integral constraint would be int_0^1 f(x) dx = int_{-inf}^0 g(t) 2^t ln2 dt = 1
That's such an interesting idea wow - thanks so much for the comment!
We can then further state that for any f(x) satisfying f(2x)=3f(x), we have a function g(x) with period of 1 and a constant A such that f(x)=Ax^(log_2(3)) * g(log_2(x)). So long as the integral doesn’t evaluate to 0, we can always re-choose an appropriate constant A so that our integral constraint is met.
I have another way of solving that:
int_0^1 f(x) dx = 1
F(1) - F(0) = 1
Then by according to some formula I found that states that:
int f(mx+b) dx = (1/m)*F(mx+b) + C
We can say that:
int_0^1 f(2x)/3 dx = 1
int_0^1 f(2x) dx = 3
(F(2*1)/2) - (F(2*0)/2) = 3
F(2) - F(0) = 6
So we have:
F(1) - F(0) = 1
F(2) - F(0) = 6
And by subtracting the first equation from the second we get
F(2) - F(1) = 5
So
int_1^2 f(x) dx = 5, or I=5
That's a really nice method thanks for sharing! That formula makes sense since all they've done is divide by the derivative of the inside of f(mx+b) (which is just m, and you'd always do that when making a u-sub) and then change f to F to show the function has been integrated.
WOW
I enjoy your videos greatly, wish I had been half as talented as you are when I was in university! Although please do write bigger if you can... I have trouble following some of your formulas when reading on my phone.
Thanks so much for the comment! I'll definitely try and write bigger in the future - I'm planning to move to a bigger whiteboard at my college soon so that should help. Really glad you're enjoying the videos.
appreciate you homie.
thanks so much, really appreciate it ☺️
f(x) will be something like Ae^(kx) or something ,lol...e^{2x} = 3e^x [B], B a constant equal to Ae^2 - A e^1 or something, etc...
Yr videos are helping me a lot
❤🇮🇳
So glad to hear that! Thanks so much for watching.
I've never seen a problem quite like this!
Recursion integrals! I have another one ready if you'd like to see more...
@@OscgrMaths I'd love it! Are there any applications for such things?
@@worldnotworld Can be used in all sorts of places! - stats, computer science, physics etc...
@@OscgrMaths Looking forward to learning more! (Maybe mention that what you're getting into "gets used" in various domains?)