My pleasure, and thanks for the comment. Yes, the power reducing formula can also be used, though it may be better for larger powers, such as the discussion here: HOW TO DERIVE AND USE REDUCTION FORMULA FOR INTEGRATING SINE N-TH POWER - A WORKED EXAMPLE INCLUDED th-cam.com/video/2k-8uV3Ur4A/w-d-xo.html
Thanks for introducing this approach. However, it is not a fast method and we still have to use the trigonometric formulas twice. Once for Sin(pi/2-x) and next for Sin^2 x+Cos^2 x.
You’re welcome and thanks for the comment. You have a valid point. However, the Pythagorean identity is a primary identity (the most primary identity I would say), whereas the double angle identities are secondary to that. Co-functions are introduced even before Pythagorean. Because of that I thought the method in the video is somewhat faster especially for a student in a lower level calculus class, given their more familiarity with Pythagorean identity compared to the double angle identities. Also, in my experience, I’ve seen people often tend to forget cos (pi), and also cos (0), and that can be avoided using the method discussed. Secondly, if Wallis product is used, this may be even faster, but it’s sometimes not discussed even so some calculus 2 classes. Good thoughts…! Thank you.
First, yeah, that's wicked clever, so thanks! I hadn't seen that convolution before and I teach this stuff. But it doesn't look to be faster than the standard trig sub and the application appears restricted to just this function across these specific bounds. Net net, I'm not seeing a reason to introduce any of this to my students. Still, I do like it!
Thank you so much for your comment. I agree with you, I also do not discuss this in my class, those types of stuff, better to be figured out by themselves, also we have to fight time. I am glad you still liked it...!
Do you plan course like videos from beginner to advance For analysis it would be Precalculus, Single variable calculus Complex analysis, Multivariable calculus Differential equations (ordinary and later partial) Each video 45min-90min long (not shorter than 45min and not longer than 90min)
Thank you for your comment. I will keep that in mind. At this point, the prime target for this channel is to feature videos for essential basic math for undergrad non-math majors of US universities in STE(M) fields, since I think the videos of those areas will have a high demand. Depending on how the channel grows, I will consider pros/cons of including more advanced subject areas. Several years ago, I put some measure theory videos (during pandemic), nobody watched other than myself, my wife, and 4 students, who were required to watch. I heard 2 of 4 dogs at my student Jonathan's house watched them. Then I kind of stopped posting any advanced videos, but these entry level under grad areas of math seem to be somewhat demanding.
Good question. With the negative symbol, you get the same integral, true, however, you can flip the limits and make it positive, integral stays the same (just the limits are flipped).
Thanks for your question. For this example, it may be easier this way, or cosine double angle. Of course for higher orders we can use reduction, however, some instructors don’t like it unless it is explicitly proven before applying. For specific exams, the instructions may vary.
sin is the symmetric of cos by axis X=π/4 because sinx=cos(π/2-x)=cos(2*π/4-x) (No need to sée graph) And also for sin² Than Area under Curve sin²x on [π/4;π/2] is equal to Area under Curve cos²x on [0;π/4] by symmetry Finally All Area J= area under(sin²x+cos²x) on [0;π/4] to say integral of 1 between 0 & π/4 Wich is π/4
Good question. It can be seen as a special case of King’s Property. For the full asymmetric form of the theorem, please watch: CAN YOU EVALUATE THIS DEFINITE INTEGRAL? th-cam.com/video/Kpstn9E1Ses/w-d-xo.html
@@calbernandhowbe integration by parts is easy and allow to write recursive formula for sin^n(x) Integration by parts \int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = \int\limits_{0}^{\frac{\pi}{2}}\sin{x}\cdot\sin^{n-1}{x}dx = -\cos{x}\sin^{n-1}{x}\biggl_{0}^{\frac{\pi}{2}} - \int\limits_{0}^{\frac{\pi}{2}}{(-\cos{x}\cdot (n - 1)\sin^{n-2}{x}\cos{x})dx} \int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)\cos^{2}(x)dx \int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)(1-\sin^2(x))dx \int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)dx - (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx n\int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)dx \int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = \frac{n-1}{n}\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)dx So we have recursion I_{0} = \frac{\pi}{2} I_{1} = 1 I_{n} = \frac{n-2}{n}I_{n} Now we break this into even and odd cases expand each case as product and finally merge these two cases into one
@@holyshit922 In fact even better would be to define I(n) = ∫(0,π/2) sin^n(x) dx and then you can easily establish a recurrence relation between I(n) and I(n+2).
Thanks for your comment. This channel simply discusses random math problems, not targeting any exams. This channel will be evolved as a basic math channel for non-math majors, i.e. to have just college algebra, trig, calc 1, 2 and 3, as per the US undergrad math syllabi for non-math STE(M).
Good to hear you remember that….! In fact, most of the calculus 1 classes discuss it, usually not discussed as a part of calculus 2 classes, since calculus 2 classes discuss the integration of general n-th power of trig functions. Even though it is discussed in calculus 1, many people don’t use the property anywhere. Almost 90+% of the students use the cosine double angle identities to convert it to linear, which is correct, but slightly inefficient. I’ll have a better one in the next video…! 👌😄.
@@gayansamarasekara I feel like double angle formula would be better though because you will not always have a case where the lower bound for the integral is 0
not really? the equality of the integrals of f(x) and f(a-x), sort of, its a very simple consequence of variable substitution that you never think about until you see it, but this trick for sine squared ive never seen... it's not very complicated, but it could be a nice trick to introduce the gaussian integral! (assuming students know change of coordinates, of course)
@@geekjokes8458 Yeah I agree that its a nice trick but I'd prefer the double angle method, this method requires you to be very analytical. I haven't learned about the gaussian integral yet as I'm taking Calculus 3 at the moment but I know what it is just not how to compute it without elementary integration techniques.
Do you want propsition for video ? Derive formula for coefficients of Chebyshev polynomials by deriving and solving ordinary differential equation T_{n}(x) = cos(n*arccos(x)) Let t = arccos(x) y(t) = cos(n*t) Let's differentiate it twice y'(t) = -n*sin(n*t) y''(t) = -n^2cos(n*t) y''(t) = -n^2y(t) y''(t)+n^2y(t) = 0 Let's change independent variable t = arccos(x) t = arccos(x) dt/dx = -1/sqrt(1-x^2) dy/dt = dy/dx * dx/dt dy/dt = dy/dx * (-sqrt(1-x^2)) d^2y/dt^2 = d/dt(dy/dt) d^2y/dt^2 = d/dx(dy/dx*dx/dt)*dx/dt d^2y/dt^2 = d/dx(dy/dx*(-sqrt(1-x^2)))*(-sqrt(1-x^2)) d^2y/dt^2 = sqrt(1-x^2)*d/dx(dy/dx*sqrt(1-x^2)) d^2y/dt^2 = sqrt(1-x^2)*(d^2y/dx^2*sqrt(1-x^2) - x/sqrt(1-x^2)*dy/dx) d^2y/dt^2 = (1-x^2) d^2y/dx^2 - x*dy/dx d^2y/dt^2 + n^2y(t) = (1-x^2) d^2y/dx^2 - x*dy/dx + n^2y(x) y''(t)+n^2y(t) = 0 (1-x^2) d^2y/dx^2 - x*dy/dx + n^2y(x) Lets solve this equation with power series y(x)=\sum\limits_{m=0}^{\infty}c_{m}x^{m} (1-x^2)(\sum\limits_{m=0}^{\infty}m(m-1)c_{m}x^{m-2}) -x(\sum\limits_{m=0}^{\infty}mc_{m}x^{m-1})+n^2(\sum\limits_{m=0}^{\infty}c_{m}x^{m}) = 0 \sum\limits_{m=0}^{\infty}m(m-1)c_{m}x^{m-2} - \sum\limits_{m=0}^{\infty}m(m-1)c_{m}x^{m} - \sum\limits_{m=0}^{\infty}mc_{m}x^{m} + n^2(\sum\limits_{m=0}^{\infty}c_{m}x^{m}) = 0 \sum\limits_{m=2}^{\infty}m(m-1)c_{m}x^{m-2} - \sum\limits_{m=0}^{\infty}(m(m-1)+m - n^2)c_{m}x^{m} = 0 \sum\limits_{m=2}^{\infty}m(m-1)c_{m}x^{m-2} - \sum\limits_{m=0}^{\infty}(m^2 - n^2)c_{m}x^{m} = 0 Here as you can see it is not a good idea to immediately start indexing from 2 after differentiating twice , start indexing from 1 after differentianing once etc After solving this equation we will get (m+2)(m+1)c_{m+2} - (m-n)(m+n)c_{m} = 0 Now lets solve it for c_{m} c_{m} = \frac{(m+2)(m+1)}{(m-n)(m+n)}c_{m+2} Let m = m - 2 c_{m-2} = \frac{m(m-1)}{(m-2-n)(m-2+n)}c_{m} Finally we will get y_{1}(x) = c_{n}(\sum\limits_{k=0}^{\lfloor\frac{n}{2} floor}\frac{(-1)^k}{2^n}\cdot\frac{n}{n-k}\cdot {n-k \choose k} \cdot (2x)^{n-2k}) Now if we want to calculate leading coefficient we use that y(1) = 1 c_{n}(\sum\limits_{k=0}^{\lfloor\frac{n}{2} floor} \frac{(-1)^k}{4^k}\cdot\frac{n}{n-k}\cdot {n-k \choose k} ) = 1 So we need sum \sum\limits_{k=0}^{\lfloor\frac{n}{2} floor} \frac{(-1)^k}{4^k}\cdot\frac{n}{n-k}\cdot {n-k \choose k} to calculate leading coefficient but this sum is difficult for me I can make a conjecture about it but i cant prove it That is my proposition for video for you
Thanks for your comment. This channel will feature the most basic required math for non-math undergraduate majors of STE(M) fields of the US universities, since it has a wider reach. The targeted areas covered in this channel are: college algebra, trig, calculus 1,2,3 and into statistics. There are some occasional travel videos posted for fun, the others are focused. So even the Chebyshev’s inequality is a bit too heavy and may not reach a reasonable crowd on TH-cam, so those are not planned at this point since the channel is just taking off only. But your comment is interesting, and I’ll take a closer look and try to work on it when possible.
@@gayansamarasekara I have problem with this sum \sum\limits_{k=0}^{\lfloor\frac{n}{2} floor}\frac{(-1)^k}{4^{k}}\cdot\frac{n}{n-k}\cdot{n-k \choose k} I made conjecture that \sum\limits_{k=0}^{\lfloor\frac{n}{2} floor}\frac{(-1)^k}{4^{k}}\cdot\frac{n}{n-k}\cdot{n-k \choose k} = \frac{1}{2^{n-1}} , for n > 0 but i dont know how to prove it and thats why I would be intertested in that video The same problem I have with Legendre polynomials when I want to get it via ordinary differential equation but for Legendre polynomials it is quite easy to expand generating function 1/sqrt(1-2xt+t^2) = \sum\limits_{n=0}^{\infty}P_{n}(x)t^{n} So I am interested more in video about coefficients of Chebyshev polynomials than Legendre polynomials Also once we know coefficients of Chebyshev polynomial we can easily expand cos(nx) which can be useful for your approach to the integral of powers of cosine and sine
Thanks for your comment. I agree, in fact once the concept is well practiced, it should just take 5 seconds or less to write the answer and go on to the next question.
This is good. We were taught this in coaching way back when i was a jee aspirant. 👍
Thank you. I believe I also saw this in a JEE-IIT prep book, some 25+ years ago.
Thanks! They usually have us use the power-reducing formula for sin^2…
My pleasure, and thanks for the comment. Yes, the power reducing formula can also be used, though it may be better for larger powers, such as the discussion here:
HOW TO DERIVE AND USE REDUCTION FORMULA FOR INTEGRATING SINE N-TH POWER - A WORKED EXAMPLE INCLUDED
th-cam.com/video/2k-8uV3Ur4A/w-d-xo.html
I love your handwriting, it is a pleasure to read and enhances the presentation.
Thank you so much. Your comment gives me confidence.
Thanks for introducing this approach. However, it is not a fast method and we still have to use the trigonometric formulas twice. Once for Sin(pi/2-x) and next for Sin^2 x+Cos^2 x.
You’re welcome and thanks for the comment. You have a valid point. However, the Pythagorean identity is a primary identity (the most primary identity I would say), whereas the double angle identities are secondary to that. Co-functions are introduced even before Pythagorean. Because of that I thought the method in the video is somewhat faster especially for a student in a lower level calculus class, given their more familiarity with Pythagorean identity compared to the double angle identities. Also, in my experience, I’ve seen people often tend to forget cos (pi), and also cos (0), and that can be avoided using the method discussed. Secondly, if Wallis product is used, this may be even faster, but it’s sometimes not discussed even so some calculus 2 classes. Good thoughts…! Thank you.
@@gayansamarasekara Thanks for explanation. Keep up the good work.
Thank you…!
Hi ! Nice video !
Which virtual whiteboard are you using in the video ?
its just apple notes on an ipad
Just the Notes on iPad.
First, yeah, that's wicked clever, so thanks! I hadn't seen that convolution before and I teach this stuff.
But it doesn't look to be faster than the standard trig sub and the application appears restricted to just this function across these specific bounds. Net net, I'm not seeing a reason to introduce any of this to my students.
Still, I do like it!
Thank you so much for your comment. I agree with you, I also do not discuss this in my class, those types of stuff, better to be figured out by themselves, also we have to fight time. I am glad you still liked it...!
Do you plan course like videos from beginner to advance
For analysis it would be
Precalculus, Single variable calculus
Complex analysis, Multivariable calculus
Differential equations (ordinary and later partial)
Each video 45min-90min long (not shorter than 45min and not longer than 90min)
Thank you for your comment. I will keep that in mind. At this point, the prime target for this channel is to feature videos for essential basic math for undergrad non-math majors of US universities in STE(M) fields, since I think the videos of those areas will have a high demand. Depending on how the channel grows, I will consider pros/cons of including more advanced subject areas. Several years ago, I put some measure theory videos (during pandemic), nobody watched other than myself, my wife, and 4 students, who were required to watch. I heard 2 of 4 dogs at my student Jonathan's house watched them. Then I kind of stopped posting any advanced videos, but these entry level under grad areas of math seem to be somewhat demanding.
I am kind of confused. lets say at 3:41 you substitute the value of u back into the equation, wont you get the same integral but with a minus sign.
Good question. With the negative symbol, you get the same integral, true, however, you can flip the limits and make it positive, integral stays the same (just the limits are flipped).
Sir, can we use reduction formulae of sine cosine.
Thanks for your question. For this example, it may be easier this way, or cosine double angle. Of course for higher orders we can use reduction, however, some instructors don’t like it unless it is explicitly proven before applying. For specific exams, the instructions may vary.
interesting, thanks for the insight :)
My pleasure. Thanks for your comment.
From Indonesia, good job
Thank you so much.
sin is the symmetric of cos by
axis X=π/4
because
sinx=cos(π/2-x)=cos(2*π/4-x)
(No need to sée graph)
And also for sin²
Than
Area under Curve sin²x on
[π/4;π/2] is equal to Area under Curve cos²x on [0;π/4] by symmetry
Finally
All Area J= area under(sin²x+cos²x) on [0;π/4] to say integral of 1 between 0 & π/4
Wich is π/4
Nice….!
You are a lovely person 🎉❤
Thank you for such a lovely comment 🎉 ❤️
These tricks are insaneee
Thank you, I am glad you enjoyed.
nice presentation, i like that you quickly proved the property
Thank you so much...! I am glad you liked it.
Really cool method! There is also Walley’s product!
Or Wallis
Thank you. Yes, and that’ll be a good future video for this channel….! 👍🏻.
I didn’t learned about this theorem, where is it called.
Good question. It can be seen as a special case of King’s Property. For the full asymmetric form of the theorem, please watch: CAN YOU EVALUATE THIS DEFINITE INTEGRAL?
th-cam.com/video/Kpstn9E1Ses/w-d-xo.html
Interesting
Thank you.
Oh king? Also that proof can be done just by shifting graph of f(x) @one-liner
That’s correct….! 👍🏻
That's a nice method :)
Thank you.
Integration by parts is better
Works for all integer n >= 2
Or De Moivre’s theorem, integration by parts will be difficult for high powers of sin^n(x).
@@calbernandhowbe integration by parts is easy and allow to write recursive formula for sin^n(x)
Integration by parts
\int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx =
\int\limits_{0}^{\frac{\pi}{2}}\sin{x}\cdot\sin^{n-1}{x}dx = -\cos{x}\sin^{n-1}{x}\biggl_{0}^{\frac{\pi}{2}} - \int\limits_{0}^{\frac{\pi}{2}}{(-\cos{x}\cdot (n - 1)\sin^{n-2}{x}\cos{x})dx}
\int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)\cos^{2}(x)dx
\int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)(1-\sin^2(x))dx
\int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)dx - (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx
n\int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = (n-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)dx
\int\limits_{0}^{\frac{\pi}{2}}\sin^{n}(x)dx = \frac{n-1}{n}\int\limits_{0}^{\frac{\pi}{2}}\sin^{n-2}(x)dx
So we have recursion
I_{0} = \frac{\pi}{2}
I_{1} = 1
I_{n} = \frac{n-2}{n}I_{n}
Now we break this into even and odd cases
expand each case as product
and finally merge these two cases into one
@@holyshit922 In fact even better would be to define I(n) = ∫(0,π/2) sin^n(x) dx and then you can easily establish a recurrence relation between I(n) and I(n+2).
π\4 is the right answer
Of course…!
can you give me the name of the theorem in text?
I saw it in a really old book, some 25 years ago. I can’t remember any name associated with the theorem.
This is a specific case of the King’s rule of integration
Correct....! King's Rule, came to my mind. This is a special case, where the lower limit is zero.
These are intro questions fir jee mains aspirants as well jer adv aspirant well for him it would a revision tool 😂😂😂😂😂
Thanks for your comment. This channel simply discusses random math problems, not targeting any exams. This channel will be evolved as a basic math channel for non-math majors, i.e. to have just college algebra, trig, calc 1, 2 and 3, as per the US undergrad math syllabi for non-math STE(M).
Fil à couper le beurre 😂
what are you talking about? this substitution is taught in every calc 2 class ever.
Good to hear you remember that….! In fact, most of the calculus 1 classes discuss it, usually not discussed as a part of calculus 2 classes, since calculus 2 classes discuss the integration of general n-th power of trig functions. Even though it is discussed in calculus 1, many people don’t use the property anywhere. Almost 90+% of the students use the cosine double angle identities to convert it to linear, which is correct, but slightly inefficient. I’ll have a better one in the next video…! 👌😄.
@@gayansamarasekara I feel like double angle formula would be better though because you will not always have a case where the lower bound for the integral is 0
not really?
the equality of the integrals of f(x) and f(a-x), sort of, its a very simple consequence of variable substitution that you never think about until you see it, but this trick for sine squared ive never seen...
it's not very complicated, but it could be a nice trick to introduce the gaussian integral! (assuming students know change of coordinates, of course)
@@geekjokes8458 Yeah I agree that its a nice trick but I'd prefer the double angle method, this method requires you to be very analytical. I haven't learned about the gaussian integral yet as I'm taking Calculus 3 at the moment but I know what it is just not how to compute it without elementary integration techniques.
@@suhail6239 youll learn it pretty soon then
Do you want propsition for video ?
Derive formula for coefficients of Chebyshev polynomials
by deriving and solving ordinary differential equation
T_{n}(x) = cos(n*arccos(x))
Let t = arccos(x)
y(t) = cos(n*t)
Let's differentiate it twice
y'(t) = -n*sin(n*t)
y''(t) = -n^2cos(n*t)
y''(t) = -n^2y(t)
y''(t)+n^2y(t) = 0
Let's change independent variable t = arccos(x)
t = arccos(x)
dt/dx = -1/sqrt(1-x^2)
dy/dt = dy/dx * dx/dt
dy/dt = dy/dx * (-sqrt(1-x^2))
d^2y/dt^2 = d/dt(dy/dt)
d^2y/dt^2 = d/dx(dy/dx*dx/dt)*dx/dt
d^2y/dt^2 = d/dx(dy/dx*(-sqrt(1-x^2)))*(-sqrt(1-x^2))
d^2y/dt^2 = sqrt(1-x^2)*d/dx(dy/dx*sqrt(1-x^2))
d^2y/dt^2 = sqrt(1-x^2)*(d^2y/dx^2*sqrt(1-x^2) - x/sqrt(1-x^2)*dy/dx)
d^2y/dt^2 = (1-x^2) d^2y/dx^2 - x*dy/dx
d^2y/dt^2 + n^2y(t) = (1-x^2) d^2y/dx^2 - x*dy/dx + n^2y(x)
y''(t)+n^2y(t) = 0
(1-x^2) d^2y/dx^2 - x*dy/dx + n^2y(x)
Lets solve this equation with power series
y(x)=\sum\limits_{m=0}^{\infty}c_{m}x^{m}
(1-x^2)(\sum\limits_{m=0}^{\infty}m(m-1)c_{m}x^{m-2}) -x(\sum\limits_{m=0}^{\infty}mc_{m}x^{m-1})+n^2(\sum\limits_{m=0}^{\infty}c_{m}x^{m}) = 0
\sum\limits_{m=0}^{\infty}m(m-1)c_{m}x^{m-2} - \sum\limits_{m=0}^{\infty}m(m-1)c_{m}x^{m} - \sum\limits_{m=0}^{\infty}mc_{m}x^{m} + n^2(\sum\limits_{m=0}^{\infty}c_{m}x^{m}) = 0
\sum\limits_{m=2}^{\infty}m(m-1)c_{m}x^{m-2} - \sum\limits_{m=0}^{\infty}(m(m-1)+m - n^2)c_{m}x^{m} = 0
\sum\limits_{m=2}^{\infty}m(m-1)c_{m}x^{m-2} - \sum\limits_{m=0}^{\infty}(m^2 - n^2)c_{m}x^{m} = 0
Here as you can see it is not a good idea to immediately start indexing from 2 after differentiating twice , start indexing from 1 after differentianing once etc
After solving this equation we will get
(m+2)(m+1)c_{m+2} - (m-n)(m+n)c_{m} = 0
Now lets solve it for c_{m}
c_{m} = \frac{(m+2)(m+1)}{(m-n)(m+n)}c_{m+2}
Let m = m - 2
c_{m-2} = \frac{m(m-1)}{(m-2-n)(m-2+n)}c_{m}
Finally we will get y_{1}(x) = c_{n}(\sum\limits_{k=0}^{\lfloor\frac{n}{2}
floor}\frac{(-1)^k}{2^n}\cdot\frac{n}{n-k}\cdot {n-k \choose k} \cdot (2x)^{n-2k})
Now if we want to calculate leading coefficient we use that
y(1) = 1
c_{n}(\sum\limits_{k=0}^{\lfloor\frac{n}{2}
floor} \frac{(-1)^k}{4^k}\cdot\frac{n}{n-k}\cdot {n-k \choose k} ) = 1
So we need sum
\sum\limits_{k=0}^{\lfloor\frac{n}{2}
floor} \frac{(-1)^k}{4^k}\cdot\frac{n}{n-k}\cdot {n-k \choose k}
to calculate leading coefficient
but this sum is difficult for me
I can make a conjecture about it but i cant prove it
That is my proposition for video for you
Thanks for your comment. This channel will feature the most basic required math for non-math undergraduate majors of STE(M) fields of the US universities, since it has a wider reach. The targeted areas covered in this channel are: college algebra, trig, calculus 1,2,3 and into statistics. There are some occasional travel videos posted for fun, the others are focused. So even the Chebyshev’s inequality is a bit too heavy and may not reach a reasonable crowd on TH-cam, so those are not planned at this point since the channel is just taking off only. But your comment is interesting, and I’ll take a closer look and try to work on it when possible.
@@gayansamarasekara
I have problem with this sum
\sum\limits_{k=0}^{\lfloor\frac{n}{2}
floor}\frac{(-1)^k}{4^{k}}\cdot\frac{n}{n-k}\cdot{n-k \choose k}
I made conjecture that
\sum\limits_{k=0}^{\lfloor\frac{n}{2}
floor}\frac{(-1)^k}{4^{k}}\cdot\frac{n}{n-k}\cdot{n-k \choose k} = \frac{1}{2^{n-1}} , for n > 0
but i dont know how to prove it and thats why I would be intertested in that video
The same problem I have with Legendre polynomials when I want to get it via ordinary differential equation but for Legendre polynomials it is quite easy to expand generating function 1/sqrt(1-2xt+t^2) = \sum\limits_{n=0}^{\infty}P_{n}(x)t^{n}
So I am interested more in video about coefficients of Chebyshev polynomials than Legendre polynomials
Also once we know coefficients of Chebyshev polynomial we can easily expand cos(nx) which can be useful for your approach to the integral of powers of cosine and sine
7mn vid for 30s mind calculation
Thanks for your comment. I agree, in fact once the concept is well practiced, it should just take 5 seconds or less to write the answer and go on to the next question.
🤣 If you take 30s to solve this question better invest 7 min in this video and learn to do it in 5 sec