I love mathematicians 😭❤️💯🙌🏽 That was mesmerizing to watch, mate. Thank you! I’m looking for as many examples of contour integration as possible, and your playlist is a massive help. Greetings from the US 🎊🥳
A more generalised version for the integral 1/(x^n+1) for x>0 will be Θ/sinΘ. Where Θ=π/n. For even n we can extended our result to the entire real axis and the result becomes 2Θ/sinΘ, for odd n however, the integral makes sense only for x>0 as there is a pole at x=-1.
I was trying to do this by factoring the denominator and finding residues at simple poles for z=i(i )**1/2 and z= i,but the solution is coming to be a complex 😥. Why we have to put it in exponential form?
I think you might have make a slight mistake in your problem. When you rotated by multiplying and dividing by "i" (pi/2), the "i" in the denominator should cancel the "i" in the (2pi)i part in the numerator. You should get cos(pi/4), not sin(pi/4). Since cos(pi/4) = sin(pi/4), you got the correct answer. Of course I could be all wrong, but it's worth checking out.
@@qncubed3 At about 9:10. Looks like you should have multiplied top and bottom by "i". e^(pi/2)i. You would end up with pi times cos(pi/4) and that gives pi/sqrt(2).
@@qncubed3 I ended up with e^ (-3/4 ipi) and e^(-9/4 ipi). -9pi/4 = -8pi/4 - pi/4. -8pi/4 = -2pi, a complete rotation, so we are left with -pi/4 and -3pi/4. Ok, now multiply top and bottom by e^(ipi/2). We get pi/2 - pi/4 = pi/4, and pi/2 - 3pi/4 = - pi/4. So we get e^(ipi/4) and e^(-ipi/4). So we get (2pi i / 4i) ( e^(i pi/4) + e^(-i pi/4) ) This simplifies down to cos(pi/4).
@@OleJoe Yes, that is an equivalent way to calculate it. You simply used an angle of pi/4, whereas I used 3pi/4. For your angle, you would indeed end up with a cosine function. Recall the trig identity sin(x)=cos(x-pi/2) which is basically what's being used here since you multiplied by e^(i*pi/2). Hence my answer = sin(3pi/4) = cos(3pi/4-pi/2) = cos(pi/4) = your answer.
A more efficient method is integrating over a sector since you'll only need to evaluate one residue. I use this in my video evaluating the integral of 1/(x^n+1) from 0 to infinity
I don't see that you need a long calculation involving the reverse triangle inequality. If you know that |z| ≤ R on \Gamma, then just use the triangle inequality + this assumption in the denominator to deduce directly that |1 / (z^4 + 1)| ≤ 1 / ( |z|^4 + 1) ≤ 1 / (R^4 + 1).
You should be careful when you apply the triangle inequality: since |z^4 + 1| ≤ R^4 + 1, you have that 1 / |z^4 + 1| ≥ 1 / (R^4 + 1) which doesn't tell you much when R tends to infinity. The idea is that you need a lower bound on |z^4 + 1| that depends on R. In that case, be considering the inequality with the reciprocals, you get 1 / |z^4 + 1| ≤ (something that goes to 0 as R approaches infinity).
What you can do is factor out a^4 on the denominator, this will leave you with a^4((x/a)^4+1) Taking 1/a^4 out from the integral as a constant, you will be left with the integral of 1/((x/a)^4+1) dx From here, you can do a simple substitution: u=x/a => du=dx/a => dx=a*du Note that bounds will not be changed. Substituting back in you will get 1/a^3*int(1/(u^4+1))dx, the same integral in the video. This evaluates to 1/a^3 * pi*sqrt(2)/2 = pi*sqrt(2)/(2a^3) Hope that helps :)
@qncubed3 , you helped me a lot ! I actually got a problem such as x^3 * sinx / x^4 + a^4 , following your explanation I hope I got the right solution, it's pi * sqr(2) / 2i Thank you very much for your additional help, kindness and time.
@@dewman7477 I used L'hopital's rule as directly subbing in the pole would result in 0/0, however we would like some value for the limit. To do this, we differentiate the top and bottom separately with respects to z. As the numerator is a linear function with a leading coefficient of 1, the derivative becomes 1.
@@dewman7477 Our initial interval of integration was from -infinity to infinity on the real axis. Hence, we choose a contour which also passes through the real axis. Enclosing all 4 poles would result in more residues to be calculated, and would be quite challenging to construct a suitable contour.
Yes, you would still get the same result. Polar form is easier to work with when you raise complex numbers to a power. (Otherwise the algebra can get messy)
Sorry i dont get the mathematics behind the arc length, integral of dz =piR, can you explain it mathematically, because dz =Ri(e^it)dt, and when i integrated it from pi to zero equals R
I made an error, there should have been an absolute value around the dz when I first applied the triangle inequality. So int_Gamma |dz| = int_[0,pi] R dt = pi*R
For sir Math "qncubed 3": 1. You are very smart 2. Do you explain this problem only to Euler,...Gauss, Green? etc., or NOT for all people who wish to learn and understand your solution? In this case, I have to avoid you, or 3. Maybe you MUST change your CARIOCAS with a PEN black color to understand your writing, which is horrible. Take my idea as something friendly and useful for both. Sorry about that.
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I love mathematicians 😭❤️💯🙌🏽 That was mesmerizing to watch, mate. Thank you! I’m looking for as many examples of contour integration as possible, and your playlist is a massive help. Greetings from the US 🎊🥳
Thanks! Glad you're finding the videos helpful 👍
Thank you so much! Your explanation is so clear! I love your one-by-one steps!
that was like magic! thanks for your clear explanation sir...!!!
Wow what an amazing explanation 😍😍😍😍😍😍😍😍😍😍😍
It's answer just remembered me the Euler Reflection Formula!
Thank you so much dear *QN³*
A more generalised version for the integral 1/(x^n+1) for x>0 will be Θ/sinΘ. Where Θ=π/n. For even n we can extended our result to the entire real axis and the result becomes 2Θ/sinΘ, for odd n however, the integral makes sense only for x>0 as there is a pole at x=-1.
Cool!
I was trying to do this by factoring the denominator and finding residues at simple poles for z=i(i )**1/2 and z= i,but the solution is coming to be a complex 😥. Why we have to put it in exponential form?
thanks, I discovered with you a simpler method
Thanks a lot. I enjoyed the video - great👍
I think you might have make a slight mistake in your problem.
When you rotated by multiplying and dividing by "i" (pi/2), the "i" in the denominator should cancel the "i" in the (2pi)i part in the numerator. You should get cos(pi/4), not sin(pi/4).
Since cos(pi/4) = sin(pi/4), you got the correct answer.
Of course I could be all wrong, but it's worth checking out.
Not sure if I can spot the error you mentioned... Do you mind giving me an exact timestamp?
@@qncubed3 At about 9:10. Looks like you should have multiplied top and bottom by "i". e^(pi/2)i. You would end up with pi times cos(pi/4) and that gives pi/sqrt(2).
@@OleJoe We cannot end up with cosine, as one of the exponentials will need to have a negative. So we can only produce the sine function.
@@qncubed3 I ended up with e^ (-3/4 ipi) and e^(-9/4 ipi). -9pi/4 = -8pi/4 - pi/4. -8pi/4 = -2pi, a complete rotation, so we are left with -pi/4 and -3pi/4. Ok, now multiply top and bottom by e^(ipi/2). We get pi/2 - pi/4 = pi/4, and pi/2 - 3pi/4 = - pi/4. So we get e^(ipi/4) and e^(-ipi/4).
So we get (2pi i / 4i) ( e^(i pi/4) + e^(-i pi/4) )
This simplifies down to cos(pi/4).
@@OleJoe Yes, that is an equivalent way to calculate it. You simply used an angle of pi/4, whereas I used 3pi/4. For your angle, you would indeed end up with a cosine function. Recall the trig identity sin(x)=cos(x-pi/2) which is basically what's being used here since you multiplied by e^(i*pi/2).
Hence my answer = sin(3pi/4) = cos(3pi/4-pi/2) = cos(pi/4) = your answer.
Super Like 👍
Thank you dear *QN³* 💓
Good refresher.
Thanks for the explanation.
hi, how can i solve x^8+1 with the same method? i have to do 4 residues or can samplify something? thanks
A more efficient method is integrating over a sector since you'll only need to evaluate one residue. I use this in my video evaluating the integral of 1/(x^n+1) from 0 to infinity
Explanation is so nice
I don't see that you need a long calculation involving the reverse triangle inequality. If you know that |z| ≤ R on \Gamma, then just use the triangle inequality + this assumption in the denominator to deduce directly that |1 / (z^4 + 1)| ≤ 1 / ( |z|^4 + 1) ≤ 1 / (R^4 + 1).
You should be careful when you apply the triangle inequality: since |z^4 + 1| ≤ R^4 + 1, you have that 1 / |z^4 + 1| ≥ 1 / (R^4 + 1) which doesn't tell you much when R tends to infinity. The idea is that you need a lower bound on |z^4 + 1| that depends on R. In that case, be considering the inequality with the reciprocals, you get 1 / |z^4 + 1| ≤ (something that goes to 0 as R approaches infinity).
What if we had x^4 + a^4 instead of x^4 + 1 ?
Could you help me about it? Thank you
What you can do is factor out a^4 on the denominator, this will leave you with a^4((x/a)^4+1)
Taking 1/a^4 out from the integral as a constant, you will be left with the integral of 1/((x/a)^4+1) dx
From here, you can do a simple substitution: u=x/a => du=dx/a => dx=a*du
Note that bounds will not be changed.
Substituting back in you will get 1/a^3*int(1/(u^4+1))dx, the same integral in the video.
This evaluates to 1/a^3 * pi*sqrt(2)/2 = pi*sqrt(2)/(2a^3)
Hope that helps :)
@qncubed3 , you helped me a lot !
I actually got a problem such as x^3 * sinx / x^4 + a^4 , following your explanation I hope I got the right solution, it's pi * sqr(2) / 2i
Thank you very much for your additional help, kindness and time.
@@dewman7477 I used L'hopital's rule as directly subbing in the pole would result in 0/0, however we would like some value for the limit. To do this, we differentiate the top and bottom separately with respects to z. As the numerator is a linear function with a leading coefficient of 1, the derivative becomes 1.
@@dewman7477 Our initial interval of integration was from -infinity to infinity on the real axis. Hence, we choose a contour which also passes through the real axis. Enclosing all 4 poles would result in more residues to be calculated, and would be quite challenging to construct a suitable contour.
You're the best
Can't you say that 1/abs(z^4+1)
This only works if z^4 is positive. If z^4=-2 for example then you would get 1
Why was it necessary to convert the singularities to polar form? Would this still work if you hadn't?
Yes, you would still get the same result. Polar form is easier to work with when you raise complex numbers to a power. (Otherwise the algebra can get messy)
@@qncubed3 thank you! Could you possibly do a video on branch cuts and the complex log?
Currently planning videos on integration with branch cuts :)
@@qncubed3 i would love that- i have an exam on Saturday and im extremely worried. I'm just glad I found you channel.
Sorry i dont get the mathematics behind the arc length, integral of dz =piR, can you explain it mathematically, because dz =Ri(e^it)dt, and when i integrated it from pi to zero equals R
I made an error, there should have been an absolute value around the dz when I first applied the triangle inequality. So int_Gamma |dz| = int_[0,pi] R dt = pi*R
@@qncubed3 ok thanks that makes sense
Muchas Gracias ( Thank's a lot)
8:33
Super class
Thanks
Should have used Jordan’s lemma
thank you
Can I say that; I love you? ❤
Intoxicating mathematics there is no better intoxicant maths, sincerely yrs
mit Partialbruchzerlegung oder Betafunktion ruclips.net/video/fTT_jFy0Tew/видео.html
Bêta fonction mieux
For sir Math "qncubed 3": 1. You are very smart 2. Do you explain this problem only to Euler,...Gauss, Green? etc., or NOT for all people who wish to learn and understand your solution? In this case, I have to avoid you, or 3. Maybe you MUST change your CARIOCAS with a PEN black color to understand your writing, which is horrible. Take my idea as something friendly and useful for both. Sorry about that.