Sine Taylor Series at 0 (Maclaurin) Sine Taylor Series at 0 (Maclaurin)
Sine Taylor Series at 0 (Maclaurin)
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- In the last video we took the Maclaurin Series of Cosine of x
- we approximated it using this polynomial
- and we saw this pretty interesting pattern.
- Let's see if we can find a similar pattern if we try
- to approximate sine of x using a Maclaurin series.
- And once again a Maclaurin series is really the same thing
- as Taylor Series where we are centering our approximation
- around x is equal to 0 .so this is a special case of a Taylor Series
- so let's take f of x in this sitiuation to be equal to sine of x
- f of x is now equal to sine of x and lets's do the same thing
- that we did with cosine of x . Let's just take the different
- derivatives of sine of x really fast. so if you have the first
- derivative of sine of x is just cosine of x . The second derivative
- of sine of x is derivative of cos x which is negative sine of x
- the third derivative is going to be the derivative of this
- so I will just write 3 in parentheses there in stead of doing
- all the prime prime prime. so the third derivative is
- derivative of this which is negative cosine of x . the fourth
- derivative the fourth derivative is the derivative of this
- which is positive sine of x again. so you see just like cosine of x
- it kind of cycles after you take the derivative enough time.
- and we care in order to the Maclaurin series we care
- about evaluating the function and each of these derivatives at x is equal to 0 .
- so let's do that. so for this let me do this in a different
- color not that same blue. so i'll do it in this purple color.
- so f that's hard to see I think. so let's do this in the
- other blue color. so f of 0 in this situation is 0 and f the
- first derivative evaluated at 0 is 1. cosine of 0 is 1
- negative sine of 0 is going to be 0. so f prime prime
- the second derivative evaluated at 0 is 0.
- the third derivative evaluated at 0 is negative 1.
- cosine of 0 is 1 you have a negative out there it is
- negative 1 and the fourth derivative evaluated at 0 is
- going to be 0 again. we could keep going once again seems like
- there is a pattern 0 1 -1 0 then you are going to go back
- to positive 1 so on and so forth . so let's find it's
- polynomial representation using Maclaurin Series.
- just a reminder this one up here this was approximately
- cosine of x and you will get closer and closer
- to cosine of x I am not rigorously showing you how close
- and that is definitely the exactly the same thing as cosine of x
- but you get closer and closer and closer to cosine of x
- as you keep adding terms here and if you to infinity
- you are going to be pretty much at cosine of x
- now let's do the same thing for sine of x . so i'll pick
- a new color. this green should be nice. so this is
- our new p of x. so this is approximately going to be
- sine of x as we add more and more terms
- and so the first term here f of 0 that's just going to be 0
- so we are not even going to need to include that. the next term
- is going to f prime 0 which is 1 times x . so it's going to be x
- then the next term is f prime the second derivative at 0
- which we see here is 0. let me scroll down a little bit
- it is 0 so we won't have the second term
- this third term right here the third derivative of sine of x
- evaluated at 0 is negative 1 so we are now going to have
- a negative 1 . let me scroll down so you can see this
- negative 1 this is negative 1 in this case times x the third
- over 3 factorial. so negative x the third over 3 factorial
- and then the next term is going to be 0 because that's
- the fourth derivative . that's the fourth derivative evaluated
- at 0 is the next coefficient. we see that that's going to be 0
- so it's going to drop off and what you are going to see here
- and actually maybe I haven't done enough pep terms
- for you. for you to feel good about this let me do
- one more term right over here just so it becomes clear
- f of fifth derivative of x is going to be cosine of x
- again. the fifth derivative let me do that in the same color
- just so that it's consistent. the fifth derivative
- the fifth derivative evaluated at 0 is going to be 1
- so the fourth derivatives evaluated at 0 is 0. then you
- go to the fifth derivative evaluated at 0 is going to be
- positive 1 and if I kept doing this it would be positive 1 times
- I would have to write 1 as a coefficient times x to the fifth
- over 5 factorial so there is something interesting going
- on here and for cosine of x I had 1 essentially 1 times
- x to the zero then I don't have x to the first power
- I don't have x to the odd powers actually then I just
- have x to all the even powers and whatever power it is
- I am dividing it by that factorial and then the sign
- keeps switching and this is ,I shouldn't say this is an
- even power because 0 really isn't , well I guess you
- can view it as an even number cuz it.. I won't go into
- all of that but it's essentially 0 2 4 6 so on and so forth
- so this is interesting specially when you compare to
- this . this is all of the odd powers this is x to the first
- over 1 factorial I didn't write it here there's x to
- the third over 3 factorial plus x to the fifth over
- 5 factorial. ya zero would be an even number . anyway
- I don't.almost. my brain is in a different place right now
- and you could keep going if we kept this process up
- you would then keep switching signs x to the seventh
- over 7 factorial plus x to the ninth over 9 factorial
- so there is something interesting here you once
- again see this kind of complimentary nature between
- sine and cosine here. you see almost this..they kind of
- they are filling each other's gaps over here cosine of x
- is all of the even powers of x divided by that power's factorial
- sine of x when you take it's polynomial representation
- is all of the odd powers of x divided by it's factorial
- and you switch signs. In the next video I'll do e to the x
- and what's really fascinating is that e to the x starts
- to look like a little bit of a combination here. but not quite
- and you really do get the combination when you involve
- imaginary numbers and that's when it starts to get
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