The uniqueness of a power series representation of a function guarantees that this power series (centered at x=0, notice each term has powers of x, and not powers of (x-c)) is the Maclaurin series. You would only have to plug in x=0 for each derivative if you were trying to find the Maclaurin series from the formula, rather than deriving it from a different power series.
@@___hwjkk I believe the radius of convergence would be 1/2 and -1/2, as inputting either of those gives you arctan(1) and arctan(-1), which are the valid limits
that's amazing, beautiful math!
You da sugar honey ice tea, Renee. I'm super thankful for your super clear explanation! I'm gonna apply your approach for arcsin(x). #Subscribed :)
thanks a lot, it's really helpful!
This was really helpful. Thankss
You’re saving me
The uniqueness of a power series representation of a function guarantees that this power series (centered at x=0, notice each term has powers of x, and not powers of (x-c)) is the Maclaurin series. You would only have to plug in x=0 for each derivative if you were trying to find the Maclaurin series from the formula, rather than deriving it from a different power series.
Thank you for the info👍 and what are we supposed to do for finding same series for tan^-1 (x^2)
that was a very good method thanks
Great video
Thank you!
Amazing thank you.
you saved me thank you
شكرا😍
Thanks
So it's almost the same as sinx's Maclaurin series, the only difference is the denominator is 2n+1 instead of (2n+1)!.
thx!
I enjoyed the video. Then, does the radius of convergence of arctan (2x) become [-2,2]?
I'm sorry for commenting on the video 5 years ago, but I hope you let me know. I don't know much about math.
@@___hwjkk I believe the radius of convergence would be 1/2 and -1/2, as inputting either of those gives you arctan(1) and arctan(-1), which are the valid limits
But the value of x can also be outside the interval[-1,1]
thankss
Tnx
Thanks mrs i need it arctan z i think is the same way i get it the idea what you mean