Sequences, series, and function approximation

# Maclaurin series and Euler's identity

## Cosine Taylor series at 0 (Maclaurin)

Approximating f(x)=cos x using a Maclauren Series (special case of a Taylor series at x=0)
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## Cosine Taylor series at 0 (Maclaurin)

Discussion and questions for this video
I could understand the steps, end result, but a question as to how some function determined at 0 is valid over the full range. What I feel is since cosine is periodic function this holds true. Someone correct me if I am wrong.
Two points.
1. If you look at the pattern of your derivatives, you'll see that after 4 derivatives, it goes back to it's original derivative, which means that it will just continue to repeat this pattern no matter how far you go out.
2. It then has to do with the series representation. If you look at the series representation, you'll see that it's E ((-1)^k)*x^2k)/(2k!). This is for all k's. You'll notice that the x is just x^2k. If it were (x^2k)-3, then it would be centered at 3. The formula for "power" series is E an(x-c)^n. If there is no "c" in the series representation, then the function is centered at 0.

1 Comment
So... the Taylor Series is a way to represent a function? :S
With Taylor and Maclaurin series you can approximate a function with a polynomial. This is useful because you can turn a complicated function (defined by a limit, for example) into simple multiplication and exponentiation of numbers. That's how your calculator gets the sine and cosine of angles almost immediately: It doesn't have a table of sines and cosines for all possible values, it approximates them using Taylor/Maclaurin series
This is supposed to approximate the function cos(x). Does it approximate the entire function all the way to infinity, or only near x=0?
There are methods for determining the maximum size of the error produced by an approximation like this, so you can figure out how many terms you need to have in order to get an approximation within the tolerance you desire. Hopefully Sal will cover those in an upcoming video. Also, cos x is periodic, so if you can get within your desired tolerance on, say, the interval from 0 <= x <= 2 pi, then you can just shift over any multiple of 2 pi to get your approximation if your x lies outside that int
How is Maclaurin and Taylor Series related to Complex Number's Polar Form, given that they share the same playlist?

I'm honestly not getting what the relation between these two might be.
Is it actually proven that the Maclaurin/Taylor-representations of these functions are EQUIVALENT to the corresponding functions if the number of terms in the representation goes to infinity?
Lot's of it! And it can be done in various ways, via L'Hopital, via induction, via application of the Mean Value Theorem:
Mean Value: http://www.math.harvard.edu/~pflueger/math1b/Lecture14.pdf
Induction: http://www.math.csusb.edu/faculty/pmclough/SPTT.pdf
"L'Hopital: http://en.wikipedia.org/wiki/Taylor%27s_theorem#Proof_for_Taylor.27s_theorem_in_one_real_variable
Very cool concept but I'm wondering if you can do it in reverse, take an infinite polynomial and turn it into a simple function?
I transformed that last polynomial equation ( produced after the Mac Lauren series was applied ) and transformed it into a power series summation ( like Sal taught us on his last playlist), and I got :

Summation from n=0 to infinity of - ( x^2 / 2n! )^n , which is 1 / [ 1 + ( x^2/2n! )] with radius of convergence x < sqrt( 2n! ).

So my question is, is that correct? Can we represent cos(x) three different ways (MacLaurin polynomial, power series(sigma notation), power series summation after applying the formula 1 / 1-r ) ??
Why exactly does taking derivatives at a point give you the function of the polynomial? I don't get the part of how the 3rd,4th,5th etc. derivative will attempt to match the slope at any point other than at zero, is it because the slope of the slope at zero approximates the points around it?
Here's the basic idea: Hm, I wonder if I can make a polynomial that has all of the same derivatives at a point as a particular function? Look at that, now we can.
And here's a bonus: it even looks like the function! In fact, for certain functions, I can "tailor" (pun intended) an infinite polynomial that exactly equals the function at all values of x just by this process of creating a polynomial with the same derivatives as my function.
Why is -sin(x) the derivative of cos(x)? I mean, where does this information comes from? How can I calculate this?

at 1:18
There are a variety of ways to prove this. I prefer direct computation using the definitions of the sine and cosine functions:
Note that:
sin x ≡ ½*i*e^(-x*i) - ½*i*e^(x*i)
cos x ≡ ½ e^(-x*i) + ½e^(x*i)
Thus:
d/dx cos x = d/dx { ½ e^(-x*i) + ½e^(x*i)}
d/dx cos x = d/dx {½ e^(-x*i)} + d/dx {½e^(x*i)}
d/dx cos x = ½ d/dx e^(-x*i) + ½ d/dx e^(x*i)
d/dx cos x = ½ e^(-x*i) d/dx(-x*i) + ½ e^(x*i) d/dx(x*i)
d/dx cos x = ½ e^(-x*i)(-i) + ½ e^(x*i)(i)
d/dx cos x = − [½ i e^(-x*i) - ½ i e^(x*i)]
d/dx cos x = − sin x
If we define the MacLaurin series for cos x as ∑ from n = 0 to infinity of (-1)^n * x^(2n)/(2n!) wouldn't we artificially take x = 0 out of the domain because 0^0 is not defined? I know this notation isn't used in this video but I've seen it used before, including in Khan Academy exercises.
Most mathematicians assert that 0⁰ = 1, though this has not been established by a formal proof. And this is in fact used in the Maclaurin Series for cos x, such that the first term is 1 no matter what x is equal to.
At around 2:15 he states that the steps work whether you are using radians or degrees, but it doesn't seem reasonable that the polynomial will give you the same answer whether you use x = pi/2 or x=90. Is there some reason that you makes this approximation work specifically with radians?
That is only because you are dealing with an angle of 0, which is the same in radians and degrees. For any other angle, you need to use radians. At this level of study, degrees should not be used at all.
If calculators use a MacLauren series to approximate trig functions, how come when I graph cos x it takes a couple seconds, but when I graph the first 10 terms of it's MacLauren series it takes around half a minute? Does it have to do with the fact that trig functions are periodic, and hence the series only needs to be accurate between -pi and pi and then can be horizontally translated to find the cosine of angles outside that domain?
Calculators use a method called CORDIC to compute the trig functions. While this method is quite good for the way that computers process numbers, it isn't really very good for the way that humans do computations by hand. So, that is why it is not routinely taught in math classes.

So, no, calculators do not use Maclaurin series for the trig functions.