Tip for people new to the Taylor series world Remember your expansion for e^x sin(x) is the imaginary part of e^ix Substitute ix into expansion for e^x. Write out a few terms and you’ll be able to derive the summation expression for sin(x). The pattern that emerges will jog your memory and you’ll recognise it quickly. Take the derivative and you get cos(x). Bit less that you have to memorise 👍👍
You what confounds me? Let's look at the terms of the Taylor series for sin (2*pi). All those terms - each of them (2*pi) raised to an integer exponent - somehow add up to exactly zero. It should not be possible, and yet it happens.
@@kingbeauregard okay, so your comment was to say you’re in awe that infinitely many terms add up to exactly zero? I can see that. It’s sorta demystified for me when I realise that every term is just 0^n, but I get the confoundedness:)
Now that I've seen the solution, it's obvious in retrospect. Well done!
this explanation made me smile. brilliant, thank you!
Tip for people new to the Taylor series world
Remember your expansion for e^x
sin(x) is the imaginary part of e^ix
Substitute ix into expansion for e^x. Write out a few terms and you’ll be able to derive the summation expression for sin(x). The pattern that emerges will jog your memory and you’ll recognise it quickly.
Take the derivative and you get cos(x).
Bit less that you have to memorise 👍👍
Thank you so much you explain very well I finally understand 😊
Glad it helped!
You are a genius. THANK YOU!
Thank you Sir for this, very helpful.
Amazing explanation! Thank you!!
Nice explanation
Thanks and welcome
very nice question, thank you for posting!
thank you sooo much :D
This reminds me series expansion of sine
pi-pi^3/(4^2•3!)+… = 4 (pi/4 - (pi/4)^3 / 3! + (pi/4)^5 / 5! - …) = 4sin(pi/4) = 4 sqrt(2)/2 = 2sqrt(2)
How to find summation of serries
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😂😂 Amazing
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You what confounds me? Let's look at the terms of the Taylor series for sin (2*pi). All those terms - each of them (2*pi) raised to an integer exponent - somehow add up to exactly zero. It should not be possible, and yet it happens.
It’s because sin(2pi)=0
@@Tristanlj-555 Yes, I know. Not at all my point.
@@kingbeauregard okay, so your comment was to say you’re in awe that infinitely many terms add up to exactly zero? I can see that. It’s sorta demystified for me when I realise that every term is just 0^n, but I get the confoundedness:)