I really like your teaching style :) One small comment though: When taking the derivative of your infinite series, you have to keep in mind that your 0-th term (i.e. for n=0) is a constant. So when differentiating your series will start at n=1, instead. Because the first term vanishes. Then you can perform an index shift, letting your series start at n=0 again. And since you have infinitely many terms, this will not change the end of summation (it is still infinity). This will slightly change your result to: f(x) = /sum_{n=0}^/infty /frac{ (-1)^n (n+1) x^(n+3) }{ 3^(n+2) } .... using LaTeX notation. But nevertheless keep up the great work ^^
I’m glad you mentioned that the series only converges for |x| < 3, which is what many people forget as the geometric approximation only gives us a power series which is accurate locally near x = 0
I studied this subject a long time ago, 20 years ago to be a little more precise. At that time I didn't understand any of that and mechanically I did the exercises, but this exercise clarified everything for me. Congratulations!
I found a method to the solution, but if you actually want to compute that integral it will take a while. Start by using u substitution of x=tanθ, this will mean that dx=sec^2(θ)dθ, and (1+x^2)^2023=1/(sec^4046(θ)), to simplify you will have to integrate: (tan^2024(θ)/(sec^4044(θ))) dθ , to simplify more with trig identities this will equal to integral: (sin^2024(θ)*cos^2020(θ)) dθ . Now we can do integration by parts, I recommend using the DI table method since it is easier to follow. What you want to do is derive the sin part and integrate the cosin part, however you want to keep donating a sin to the cosine so that when you use substitution again dθ=du/(-sinx), keep going and i believe the last term you have should be cos^4044(x) with some coefficient, call that co, c ,and add it to the other side for (1/sec^4044(θ)) *(tan^2024(θ) + c), now sub it back to the θ world to get Integral: of x²⁰²⁴/(1+x²)²⁰²³ + c/(1+x²)²⁰²³dx = (a bunch of sines θ and cosines θ you should sub back via tan^-1(x)=θ) . Minus the integral of c/(1+x²)²⁰²³dx. With this do u sub again u=1+x^2, du/2x=dx, x= (u-1)^0.5 . Do the DI method integrate 1/u^2023 til 0, and diff (u-1)^0.5 . Now this is when you realize I have no clue what I'm talking about because a solution shouldn't take that long to compute and there's probably some faster pattern that i haven't identified.
Your teaching method is understandable and powerful. You make mathematics some easy subject. Thank you so much!
I really like your teaching style :)
One small comment though: When taking the derivative of your infinite series, you have to keep in mind that your 0-th term (i.e. for n=0) is a constant. So when differentiating your series will start at n=1, instead. Because the first term vanishes. Then you can perform an index shift, letting your series start at n=0 again. And since you have infinitely many terms, this will not change the end of summation (it is still infinity). This will slightly change your result to: f(x) = /sum_{n=0}^/infty /frac{ (-1)^n (n+1) x^(n+3) }{ 3^(n+2) } .... using LaTeX notation.
But nevertheless keep up the great work ^^
Yes yes yes. Thank you!
Been wondering about these for a long time - that was a great overview in the first minute. Thank you!
I’m glad you mentioned that the series only converges for |x| < 3, which is what many people forget as the geometric approximation only gives us a power series which is accurate locally near x = 0
It's really helpful the way you teaching, especially that you explain everything slowly
I studied this subject a long time ago, 20 years ago to be a little more precise. At that time I didn't understand any of that and mechanically I did the exercises, but this exercise clarified everything for me. Congratulations!
Great Mathematician. Regards
Thank you for everything, I really benefit from your videos to refresh my memory in one side, and to upgrade my maths knowledge in the other side.
Another great video. Thanks for sharing your awesome knowledge to us.
Fantastic,I got it. Thanks sir.
Butifull
VIDEO IDEA: Proof of the Quadratic Formula
If r= -x/3, then |r| = |x|/3, not just x/3. Otherwise, this was a very clear video.
Wow!
This guy loves math!
Sir please show e^x power series
Sir! Please help me solve this : Integral of x²⁰²⁴/(1+x²)²⁰²³ dx
Start with the Integration by Parts formula
I found a method to the solution, but if you actually want to compute that integral it will take a while. Start by using u substitution of x=tanθ, this will mean that dx=sec^2(θ)dθ, and (1+x^2)^2023=1/(sec^4046(θ)), to simplify you will have to integrate: (tan^2024(θ)/(sec^4044(θ))) dθ , to simplify more with trig identities this will equal to integral: (sin^2024(θ)*cos^2020(θ)) dθ . Now we can do integration by parts, I recommend using the DI table method since it is easier to follow. What you want to do is derive the sin part and integrate the cosin part, however you want to keep donating a sin to the cosine so that when you use substitution again dθ=du/(-sinx), keep going and i believe the last term you have should be cos^4044(x) with some coefficient, call that co, c ,and add it to the other side for (1/sec^4044(θ)) *(tan^2024(θ) + c), now sub it back to the θ world to get Integral: of x²⁰²⁴/(1+x²)²⁰²³ + c/(1+x²)²⁰²³dx = (a bunch of sines θ and cosines θ you should sub back via tan^-1(x)=θ) . Minus the integral of c/(1+x²)²⁰²³dx. With this do u sub again u=1+x^2, du/2x=dx, x= (u-1)^0.5 . Do the DI method integrate 1/u^2023 til 0, and diff (u-1)^0.5 . Now this is when you realize I have no clue what I'm talking about because a solution shouldn't take that long to compute and there's probably some faster pattern that i haven't identified.
Hi, thx for this video. May be another one to explain what's the practical point of all of this with an example ?
Just replace x+3 by y and live a beautiful life.
Thanks, brother
Are you a golfer by any chance? Your thumbnails have golf ball background and your hats are very golfy.
Not a golfer. Maybe someday