I love Blackpenredpen! He is such a lovely guy and you too! Straight up you two!
@joshuaiosevich3727
18 сағат бұрын
I think drunkenness helped me solve this. I proved that the integral sin(nx)sin(x)/x^2 is constant, it’s an instant solution.
@venkatamarutiramtarigoppul2078
23 сағат бұрын
1/2 cos (k-1)x-1/2 cos k+1x /x2 now use feynmens trick ( both the integrals are convergent ) so it would be integral from (k-1,0) to (k+1,infinty) sinxy/y dy dx
@adityagoel5746
Күн бұрын
Let a denote the positive real root of the polynomial x ^ 2 - 3x - 2 . Compute the remainder when [a ^ 1000] is divided by the prime number 997. [.] denotes the greatest integer function. Pls solve this too.
@sbares
19 сағат бұрын
Fun little problem. Letting b be the other root, one easily sees that -1 < b < 0, so 0 < b^1000 < 1. Therefore [a^1000] = a^1000 + b^1000 - 1 (EDIT: since a^1000 + b^1000, being a symmetric polynomial in the roots, is an integer, which must then be the next integer up from a^1000). It turns out that the polynomial x^2 - 3x - 2 does not have a root in F997 as can be seen by applying quadratic reciprocity to its discriminant 17. However, we can extend F997 to a larger field F997[a] = F997[X]/(X^2 - 3X - 2) in which it does have two roots (abusively also denoted by a and b here). Since there exists a homomorphism from Z[a] to F997[a] mapping the integers to themselves (modulo 997), a to a and b to b, we may do our calculations in this field. Now for the fun part. The Frobenius map x |-> x^997 generates the Galois group Gal(F997[a]/F997) (this is a general fact about finite fields) and hence exchanges the two roots. In F997[a] we thus have a^1000 + b^1000 - 1 = a^997 a^3 + b^997 b^3 - 1 = b a^3 + a b^3 - 1 = ab(a^2 + b^2) - 1 = ab((a+b)^2 - 2ab) - 1 From the coefficients of the polynomial, we can read off ab= -2, a+b = 3, so ab((a+b)^2 - 2ab) - 1 = (-2)(3^2 - 2*(-2)) - 1 = -27 = 970. Back in Z, we therefore have [a^1000] == 970 mod 997.
@adityagoel5746
19 сағат бұрын
@@sbaresbro what kind of god are you? And do you use whatsapp and can you send me a photo of the solution using pen and paper?
@ericknutson8310
13 сағат бұрын
I was thinking you were gonna use a Lebachebsky ID for pi periodic functions to reduce from 0 to infinity to 0 to pi/2 and also remove a sinx/x. I haven’t checked to see if that sum is pi periodic.
@Greg852I
Күн бұрын
Cool Solution! I found a path to the same solution representing the sin(x)sin(kx)/x product as a int(-1,1) sin(kx+ax). By use of a nice substitution t=(k+a)x in the (0,inf) integral we seperate the sum and the two integrals into a Dirichlet integral, a sum giving basically e-1 and an integral over the number one from -1 to 1. Giving the same as your Answer! I was so happy that this worked
@mcalkis5771
21 сағат бұрын
KZitem forgot the notification but doesn't matter cause I check on you anyways.
@MrWael1970
9 сағат бұрын
Very Nice. Thanks
@leeshaocheng239
19 сағат бұрын
glad to involve in your video❤
@DiffgaFysgie
23 сағат бұрын
Umm... I'm sorry to bother you, but I wish to let you know that purple text on black background is somewhat hard to read. Could you possibly use some other colours to increase contrast and readability of your notes? Thank you!
@krisbrandenberger544
20 сағат бұрын
@ 2:50 There shouldn't be a k in the exponent.
@ericthegreat7805
16 сағат бұрын
Is this connected in any way to Fourier series?
@WhatToSayWTS
2 сағат бұрын
How are you so good at math im only familiar with symbols can you suggest ways for me to understand a tiny proportion if not all what you explain
@192chickenking
Күн бұрын
done with sinxsinkx=1/2[cosx(k-1)-cosx(k+1)] and contour integral of complex analysis
@Noam_.Menashe
Күн бұрын
This reminds me that sometime ago I found the integral from 0 to pi of f(e^ix)/(x^2+1-2xcos(a))dx or something like that. It's not that hard using the fourier series but it was nearly two years ago and I forgot the details. I probably have the notes kept somewhere...
@maths_505
Күн бұрын
Would love to hear more about it when you remember the details.
@vancedforU
Күн бұрын
Nice I solved it by pulling out the summation operator, then integrating by parts (integrate 1/x^2, differentiate sinx sin(nx)) Then you can simplify it down to dirichlet integral
@joshuaiosevich3727
18 сағат бұрын
Lmao this problem admits a ton of different solutions
@RandomLogan156
Күн бұрын
12:01 My right ear just got tickled through my headphones.
@RandomLogan156
Күн бұрын
Dude, a hearted comment. I’m honored! I love your videos man!
@maths_505
Күн бұрын
@@RandomLogan156thanks bro
@Sugarman96
Күн бұрын
6:09 Papa Fourier comes in handy with those. You can extend the integral over from -inf to inf, then simply replace the cosines with e^j(k±1)x (the imaginary part of the integral is 0). At which point electrical engineers will recognize the denominator as looking like the Fourier transform of e^-a|t|, which makes the integrals the inverse fourier transforms (constants not withstanding), at which point you just plug in (k±1) into the original function.
@Dharun-ge2fo
Күн бұрын
Just curious, are you a student in college or school preparing for jee
@Sugarman96
Күн бұрын
@@Dharun-ge2fo Just a few credits short of getting my BA in EE
@lokithe.godofmischief
Күн бұрын
Ques i found, which seemed impossible Let f(x)=(x⁴)+(x³)+(x²) Let y depict the inverse of the function f(x) Then, evaluate the integral from (lower bound)2A to (upper bound)8A of (1/(y²+y⁴)) with the limit A tends to ♾️infinity The variable of integration is obviously dx, as y will be a function of x only
@vancedforU
Күн бұрын
I believe that the answer is ln4 Wait for me to post my process
@vancedforU
Күн бұрын
You can look at my post I think I spent more time writing the details in a formal way than actually solving the problem haha
@lokithe.godofmischief
16 сағат бұрын
@@vancedforU alright found it thank you so much🥰🥰
@Mario_Altare
8 сағат бұрын
Hi 🙂Could we use Lobachevsky's integral to solve this one? I.e. sin(nx) = Im e^(inx)/n! So: I = Im ∫_0^∞ sinx/x^2 ∑_(n=1)^∞ (e^ix)^n/n! dx] → I=Im∫_0^∞〖sinx/x^2 (e^cosx e^(i sin x)-1)dx〗 Now Loba comes in: I = Im ∫_0^(π/2)〖(e^cosx e^(i sin x)-1)/sinx dx〗 At this point, I'd like to use Feynman here, but how? Or is it just a wrong way to try to solve it?
@maths_505
8 сағат бұрын
Lobachevsky works when f(x) is pi periodic. Here, e^(cos(x)) isn't so Lobachevsky doesn't apply.
@Mario_Altare
8 сағат бұрын
@@maths_505 Oops, you're right 👌
@AndyBaiduc-iloveu
Күн бұрын
Do a problem involving quaternions with rotations or vectors. Or mabye integrate wrt quaternions!
@AndyBaiduc-iloveu
Күн бұрын
By the way , I love your videos
@senpai12349
Күн бұрын
which app do you use for notes?
@sundaresanabishek5127
Күн бұрын
Heyy brother fan from Sri Lanka ❤🎉
@maths_505
Күн бұрын
Lovely hearing from you my friend
@anuragkr3026
8 сағат бұрын
I am scrolling through math channel and commnets to find people who would like to help in my math idea its related to 3d geometry Any math guy here who thinks he is able to discuss?
@Tosi31415
Күн бұрын
what the hell i did it in my head in like 5 seconds, using lobachevski's rule, and then with some rough guesses in my head
@maths_505
Күн бұрын
I guessed it to be 3 and I'm glad I was right on the money
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