Pulling the sum outside the integral, the terms of the sum are half the inner products of sin(x)/x and sin(nx)/x. By plancherel's theorem, this is equal to the inner product of their Fourier transforms. Those transforms are box functions of height sqrt(pi/2) and width 1 and n, respectively. So their product is a box function of height pi/2 and width 1, easily integrated to give pi. The terms of the sum are now pi/(2 n!), which clearly sums to (pi/2)(e-1).
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
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.
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.
I've noticed that the sum under the integral is computable and converges to e^(cosx)*sin(sinx). While it seemed nightmerish, int from 0 to \infty sinx * e^(cosx) * sin(sinx) / x^2 dx is an integral of an even, complex-analytic function (one can check x=0 is also regular point). So I integrated f(z) = sinz * e^(cosz) * sin(sinz) / z^2 over a standard countour (-R -> -epsilon, small epsilon arc around z=0, epsilon -> R, large R arc). Arc integrals evaluate to zeros by direct limiting, they behave as O(epsilon), O(1/R). The remainding two integrals are the same and equal to the original one, so the countour integral is twice the original one. Yet the integrand is analytic, so they both equal to zero. That is wrong, of course. I fail to see any issues, can anyone assist?
@@maths_505 I would think the same, but the function is analytic in the neighborhood . Analyzing that circular integral manually I get O(epsilon), do you get some different expr?
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
Hey! Thats! My rule! I rule this kingdom! You do what i say! My rule is your rule... Thats the rule of me... The rules of montather muhamed... Thats canoflames rule.... I own the rule now...
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.
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.
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!
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 I found it. It's in "A treatise of the integral calculus" part II, page 278. It goes like this: The integral from 0 to pi of: [f(a+e^ix)+f(a+e^-ix)]/(1-2*b*cos(x)+b^2)dx equals: 2*pi/(1-c^2)*f(a+c) for |c|1
Misremembered badly but it still is quite general. Especially the factor of a+ Derivation really is using Fourier series for both functions and orthogonality of cosines.
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
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
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?
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?
Pulling the sum outside the integral, the terms of the sum are half the inner products of sin(x)/x and sin(nx)/x. By plancherel's theorem, this is equal to the inner product of their Fourier transforms. Those transforms are box functions of height sqrt(pi/2) and width 1 and n, respectively. So their product is a box function of height pi/2 and width 1, easily integrated to give pi. The terms of the sum are now pi/(2 n!), which clearly sums to (pi/2)(e-1).
Beautiful
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.
Hi,
6:02 : dx
"ok, cool" : 2:10 , 4:09 , 6:57 , 7:42 ,
"terribly sorry about that" : 2:55 , 5:54 , 7:34 , 9:56 , 12:48 .
I love Blackpenredpen! He is such a lovely guy and you too! Straight up you two!
Amazing generalized solution! You make it look so easy too
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
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.
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.
@@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?
I've noticed that the sum under the integral is computable and converges to e^(cosx)*sin(sinx). While it seemed nightmerish,
int from 0 to \infty sinx * e^(cosx) * sin(sinx) / x^2 dx is an integral of an even, complex-analytic function (one can check x=0 is also regular point). So I integrated f(z) = sinz * e^(cosz) * sin(sinz) / z^2 over a standard countour (-R -> -epsilon, small epsilon arc around z=0, epsilon -> R, large R arc). Arc integrals evaluate to zeros by direct limiting, they behave as O(epsilon), O(1/R). The remainding two integrals are the same and equal to the original one, so the countour integral is twice the original one. Yet the integrand is analytic, so they both equal to zero. That is wrong, of course.
I fail to see any issues, can anyone assist?
The integral over the small circle should not be zero
@@maths_505 I would think the same, but the function is analytic in the neighborhood . Analyzing that circular integral manually I get O(epsilon), do you get some different expr?
@@pavlopanasiuk7297 I have a video on a couple of similar integrals in the contour integration playlist. Perhaps they may help.
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
Hey! Thats! My rule! I rule this kingdom! You do what i say! My rule is your rule... Thats the rule of me... The rules of montather muhamed... Thats canoflames rule.... I own the rule now...
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.
Just curious, are you a student in college or school preparing for jee
@@Dharun-ge2fo Just a few credits short of getting my BA in EE
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.
@ 2:50 There shouldn't be a k in the exponent.
glad to involve in your video❤
TH-cam forgot the notification but doesn't matter cause I check on you anyways.
Now that was beautiful
Indeed
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!
Is this connected in any way to Fourier series?
Yeah, you can use Plancherel's theorem to switch to the Fourier transforms of sin(x)/x and sin(nx)/x, which results in a trivial integral.
done with sinxsinkx=1/2[cosx(k-1)-cosx(k+1)] and contour integral of complex analysis
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...
Would love to hear more about it when you remember the details.
@@maths_505 I found it. It's in "A treatise of the integral calculus" part II, page 278.
It goes like this:
The integral from 0 to pi of:
[f(a+e^ix)+f(a+e^-ix)]/(1-2*b*cos(x)+b^2)dx
equals:
2*pi/(1-c^2)*f(a+c) for |c|1
Misremembered badly but it still is quite general. Especially the factor of a+
Derivation really is using Fourier series for both functions and orthogonality of cosines.
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
I believe that the answer is ln4
Wait for me to post my process
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
@@vancedforU alright found it thank you so much🥰🥰
Heyy brother fan from Sri Lanka ❤🎉
Lovely hearing from you my friend
Very Nice. Thanks
12:01
My right ear just got tickled through my headphones.
Dude, a hearted comment. I’m honored! I love your videos man!
@@RandomLogan156thanks bro
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
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?
Lobachevsky works when f(x) is pi periodic. Here, e^(cos(x)) isn't so Lobachevsky doesn't apply.
@@maths_505 Oops, you're right 👌
Lmao this problem admits a ton of different solutions
which app do you use for notes?
samsung notes
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
Maybe take a course? There’s a whole lotta maths right there
Do a problem involving quaternions with rotations or vectors. Or mabye integrate wrt quaternions!
By the way , I love your videos
Do you like the show "the lord if the rings"?
Terribly sorry
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
I guessed it to be 3 and I'm glad I was right on the money
Right I forgot about that
This is the rule of Muntadhir al-Hasani
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?
You are a thief
nuh uh