4:32 For this, I used different function to solve. First, let I(a) = integral(0 to inf) (e^(-2x) * sin(ax) / x) dx. (The integral we want to compute will be I(3).) Observe that I(0) = 0. Compute I'(a) by integration by part to reach I'(a) = 2 / (a^2 + 4). So I(a) = arctan(a / 2), then the answer we want is integral(0 to inf) (e^(-2x) * sin(3x) / x) dx = I(3) = arctan(3 / 2).
Isn't the first problem super easy if you apply the sine and cosine sum formulae to (sin 20x + sin 22x) and (cos 20x + cos 22x) which nicely reduces the expression under the root to (2 cos x + 3 )² (sin²21x + cos²21x) which immediately reduces to ±(2cos x + 3) from which the result follows immediately.
Here's a slicker solution to Problem 1 using complex numbers. Let z = e^{ix}. Then the integrand is the modulus of z^20 + 3z^21 + z^22. But | z^20 + 3z^21 + z^22 | = |z|^21 * | z + 1/z + 3 | = 1 * | 3 + 2 cos x | = 3 + 2 cos x. Hence the integral is just 3x + 2 sin x.
Great Video! I enjoyed seeing your solutions to these problem. I have a much nicer solution to the last problem. We notice that the floor of the log10 of a number is literally just the number of digits (or the exponent when represented in scientific notation). So as long as our asnwer is accurate to the nearest power of 10, we are fine. We also note that 10^(-x^3) at x = 2022 drops fast. Really fast. Like really really fast. Simply going from x=2022 to x=2023 drops the output by 10 million orders of magnitude. Only a very, very small section of the integral is actually contributing to the relvant part of the integral (again we only care about anything that can contribute to anywhere near the leading digit). In fact, we can stop evaluating the integral when we reach x such that x^3=1+2022^3, as after this point, the rest of the integral will be too small to contribute. Using the linear approximation for x^3 (the one you used in your video), we find that our upper bound of integration is x=2022 + 1/(3*2022^2), or around 2022+8*10^-8. Evaluating the integral using the trapezoid method gets us I = (Left + Right)*length/2 = (10^(-2022^3)+10^(-2022^3-1))*(8*10^-8)/2 = 4.4*10^(-2022^3-8). The trapezoid is a good (over)estimate, and this sliver of the integral is literally contributing orders of magnitude more than the rest of the integral, so we can be confident that this answer is accurate to the power of 10. (To get an appropriate underestimate, I believe that we can use the right rectangle estimate with 3 partitions, but I could be wrong). Now we take the log and floor it, to get an answer of -2022^3 - 8.
For problem 4 I thought a u =1/x sub would be a good start. You follow that and quickly find you have to calc integral of 3x^4/(x^3 + 1)^3 in bounds of zero to infinity...how hard could that be? Kept making errors in partial fraction decomp, the third order pole in contour integration is a bitch. I had to dig out my notes on Ostrogradsky method. Several pages later I can confirm you're right! I need a beer.
@ 39:58 The denominator should be 3*4*10^6×2, not 3*8*10^6×2. The 2000=2*10^3 is being squared, not cubed. However, I did enjoy the video! Keep up the good work! 👍
se partecipassi alla finale, avrei ottenuto questo giusto insieme ai primi 2. Nessuna possibilità diavolo avrei mai ottenuto gli altri 2 giusti in una competizione.
@@maths_505... Il 1 é - arctg(2/3)+pi/2...non ho capito cosa volevi dire.. Non sono esercizi semplicissimi e io li svolgo coi miei tempi.. A volte sono bravo, altre volte sono meno bravo
(2cos(x)+3) is always positive, so you don't get +/-! You just get the positive root.
bro solves the MIT integrals just because he's bored? What a beast!
4:34 can’t we use Laplace here ?
To evaluate the integral from the table
Just asking
4:32 For this, I used different function to solve.
First, let I(a) = integral(0 to inf) (e^(-2x) * sin(ax) / x) dx. (The integral we want to compute will be I(3).)
Observe that I(0) = 0.
Compute I'(a) by integration by part to reach I'(a) = 2 / (a^2 + 4).
So I(a) = arctan(a / 2), then the answer we want is integral(0 to inf) (e^(-2x) * sin(3x) / x) dx = I(3) = arctan(3 / 2).
Isn't the first problem super easy if you apply the sine and cosine sum formulae to (sin 20x + sin 22x) and (cos 20x + cos 22x) which nicely reduces the expression under the root to (2 cos x + 3 )² (sin²21x + cos²21x) which immediately reduces to ±(2cos x + 3) from which the result follows immediately.
Here's a slicker solution to Problem 1 using complex numbers. Let z = e^{ix}. Then the integrand is the modulus of z^20 + 3z^21 + z^22. But | z^20 + 3z^21 + z^22 | = |z|^21 * | z + 1/z + 3 | = 1 * | 3 + 2 cos x | = 3 + 2 cos x. Hence the integral is just 3x + 2 sin x.
Noice
Great Video! I enjoyed seeing your solutions to these problem.
I have a much nicer solution to the last problem. We notice that the floor of the log10 of a number is literally just the number of digits (or the exponent when represented in scientific notation). So as long as our asnwer is accurate to the nearest power of 10, we are fine.
We also note that 10^(-x^3) at x = 2022 drops fast. Really fast. Like really really fast. Simply going from x=2022 to x=2023 drops the output by 10 million orders of magnitude. Only a very, very small section of the integral is actually contributing to the relvant part of the integral (again we only care about anything that can contribute to anywhere near the leading digit). In fact, we can stop evaluating the integral when we reach x such that x^3=1+2022^3, as after this point, the rest of the integral will be too small to contribute.
Using the linear approximation for x^3 (the one you used in your video), we find that our upper bound of integration is x=2022 + 1/(3*2022^2), or around 2022+8*10^-8. Evaluating the integral using the trapezoid method gets us I = (Left + Right)*length/2 = (10^(-2022^3)+10^(-2022^3-1))*(8*10^-8)/2 = 4.4*10^(-2022^3-8). The trapezoid is a good (over)estimate, and this sliver of the integral is literally contributing orders of magnitude more than the rest of the integral, so we can be confident that this answer is accurate to the power of 10. (To get an appropriate underestimate, I believe that we can use the right rectangle estimate with 3 partitions, but I could be wrong).
Now we take the log and floor it, to get an answer of -2022^3 - 8.
Brilliant!
Great job. Thank you for your amazing effort. With my best wishes.
screw the last problem, I missed like half of your solution 😭😭😭
Floor integrals are cruel bro
s a m e
You can also use the Laplace transform of sin(3x)/x for the second integral.
They solve it in 3 minutes??
For problem 4 I thought a u =1/x sub would be a good start. You follow that and quickly find you have to calc integral of 3x^4/(x^3 + 1)^3 in bounds of zero to infinity...how hard could that be? Kept making errors in partial fraction decomp, the third order pole in contour integration is a bitch. I had to dig out my notes on Ostrogradsky method. Several pages later I can confirm you're right! I need a beer.
Or 2😂
Hey, Professor! The product of two cosines results in a sum of two cosines, not a difference.
cos x*cos y=1/2(cos (x+y)+cos (x-y))
Oh yeah I mixed that up
Doesn't matter though
That floor function floored me.
Same here bro😭
First suiiii
@ 39:58 The denominator should be 3*4*10^6×2, not 3*8*10^6×2. The 2000=2*10^3 is being squared, not cubed. However, I did enjoy the video! Keep up the good work! 👍
There is another way of doing the first integral. You can sum sin20x with sin22x and cos20x with cos22x, then more easier to simplify
Just wondering what’s the app you’re using in this video?
What software do you use to make these videos?
Gamma(1/3 +1)*gamma(2/3+1)/gamma((1/3 +1)(2/3+1))
4 integrale... B(5/3,4/3)
se partecipassi alla finale, avrei ottenuto questo giusto insieme ai primi 2. Nessuna possibilità diavolo avrei mai ottenuto gli altri 2 giusti in una competizione.
@@maths_505... Il 1 é - arctg(2/3)+pi/2...non ho capito cosa volevi dire.. Non sono esercizi semplicissimi e io li svolgo coi miei tempi.. A volte sono bravo, altre volte sono meno bravo
Il 1 integrale... La funzione integrand risulta 2cosx+3...da cui... 2sinx+3x
Tomorrow is the big game 505. I'm putting down 5 euro on Man City, will you bet against? :)
Nah mate I don't bet
Thanks for the offer though
I'm always with the underdog so Inter for me
@@maths_505 Inter for me too, but I will put my money on the Juggernaut
I hate the last one bro
Me too
What qualifications are required to understand the last question ?