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## Integral Calculus

### Unit 5: Lesson 15

Maclaurin series of eˣ, sin(x), and cos(x)- Maclaurin series of cos(x)
- Maclaurin series of sin(x)
- Maclaurin series of eˣ
- Worked example: power series from cos(x)
- Worked example: cosine function from power series
- Worked example: recognizing function from Taylor series
- Maclaurin series of sin(x), cos(x), and eˣ
- Visualizing Taylor series approximations
- Euler's formula & Euler's identity

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# Maclaurin series of sin(x)

AP.CALC:

LIM‑8 (EU)

, LIM‑8.E (LO)

, LIM‑8.E.1 (EK)

, LIM‑8.F (LO)

, LIM‑8.F.2 (EK)

Approximating sin(x) with a Maclaurin series (which is like a Taylor polynomial centered at x=0 with infinitely many terms). It turns out that this series is exactly the same as the function itself! Created by Sal Khan.

## Want to join the conversation?

- How come he plugged in zero into all of the derivatives but not into the x's with exponents?(3 votes)
- The polynomial p(X) is a representation of a funtion f(x). SO if you wanted to find the value of cos(0.1) it would be almost impossible without a calculator to use f(0,1).

So instead they found a way to manipulate f(x) into a p(x) polynomial form. that is to say that the f(x) and p(x) graphs ( where p(x) is to the nth approximation) would look the same.

And if the graphs look the same, they will give you the same corresponding y values for whatever value of x you you need to find the answer for.

so you would be able to plug that (0.1) value into the p(x) polynomial ie: p(0.1) and find a value with relatively simple arithmetic.

eg: for f(x)= cos(x), what would the 4th approximation of cos(0.1) be?

p(x)= 1-x^2/2!+(0.1)^4/4! ( I just plugged 0.1 wherever there is an x )

p(0.1)=1-(0.1)^2/2+x^4/24 = 0.995....

the actual value of cos(0.1)=.999998476

which is pretty close, the acuracy will increase the higher the degree of your polynomial.

I hope that helped you Greg.(18 votes)

- 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.(9 votes)- Ooooold question, but my guess is that :

1. complex numbers are deeply connected to trigonometry (as you can see when studying the polar form)

2. these series include the approximation for cos(x) and sin(x).

I'm doing a class on Complex Calculus and series are a part of the program.(0 votes)

- Is the maclaurin series related someway to the parity of mathematical functions?(5 votes)
- That's a valuable observation. If a function is even, then its Maclaurin series contains only terms with even exponents, and the same for odd functions.(5 votes)

- What are the applications of studying this in real life?(1 vote)
- I've created a function on python that sums fifty terms of this Maclaurin series and returns the number rounded to 14 digits, this is my own home made sine calculator, I also did this for the cosine function.

If you don't know about python programming, See the computer science playlist, here is a link.

http://www.khanacademy.org/science/computer-science(9 votes)

- What is the application of series? Can you use it to approximate the equation of an unknown function? Sorry if my question is stupid, I'm a bit lost; I'll go back and review the previous videos. :/(3 votes)
- This process can only be done on a known function, and it is very difficult unless the derivatives can be calculated easily or repeat. It is very useful for things like sine and cosine functions, where it is impossible to calculate things like sin(0.237) without using the taylor or maclaurin series or actually drawing a circle and taking measurements. This also has many applications in sound processing.(5 votes)

- Maclaurin Taylor Series at 0 for sinx - odd powers over odd factorials

Maclaurin Taylor Series at 0 for cosx - even powers over even factorials

Is this is any way related to the fact that sinx is an odd function while cosx is an even function?

Or are the two ideas completely irrelevant?(5 votes)- Yeah. Odd power functions have odd symmetry and even power functions have even symmetry. So a polynomial comprised of only odd power functions still has odd symmetry and likewise for even. Since the cosine function has even symmetry its polynomial representation cannot have any odd powers and likewise for sine.(2 votes)

- So I'm wondering what presuppositions must be made in order to come up with these trigonometric equations in terms of only x. All I can see is that we have to know what the derivatives of these trig functions are. So could somebody tell me what presuppositions must be made and perhaps how to prove that they are true? I just never understood how we can know what a sine or cosine of an angle is without like measuring lengths of a triangle or something (which probably wouldn't be too accurate).(4 votes)
- Really, were not actually "knowing" the sine or cosine, we are getting a infinitely close polynomial approximation, that happens to have the same value of sine or cosine. It's like saying 1.999999999999999...=2, just a little more complicated.(1 vote)

- Can you see the property of sin(x)^2 + cos(x)^2 = 1 using Taylor series?(3 votes)
- Great question!

In theory, you could if you find the taylor polynomial if "Sin(x)^2" and the Taylor polynomial of "Cos(x)^2", and if you add them all the terms except for the number "1" would cancel. So, this would verify that sin(x)^2+cos(x)^2 = 1(2 votes)

- I understand that you can essentially rewrite the functions by using this method but I don't understand why it has to be zero. From what I know a MacLaurin series is when x=0 but a Taylor series can be any other value or am I incorrect? What does that value represent? Also is there a proof somewhere of this method?(2 votes)
- You're right; the center doesn't
**have**to be 0, it's just often very convenient to use 0 because it reduces the number of terms we have to handle. And yes, a Maclaurin series is just a particular kind of Taylor series that is centered at 0 (it's the same theorem).

That number, 0 or whatever you choose, represents the "center" of the series; it's the point around which we're building our successive approximations of the function. Taylor's formula is about creating better and better polynomial approximations of a function based on its derivative behavior at a given point; so we choose the point in which we're interested and construct a polynomial approximation around that point. When we make an approximation, we also have to consider what values of x allow that approximation to actually equal the desired value whenever we sum the infinite series. Sometimes the approximation will converge for all values of x, and sometimes it will only converge in a finite interval around the center that we choose; it depends on the function. But for the approximations that don't converge for all x, it's useful to be able to move our "window" of convergence so that it encompasses the value that we want to approximate.

Say we're approximating`ln(e + 0.1)`

. For one thing, we can't use a Maclaurin series because the function isn't even defined at 0. We might choose a Taylor series centered at x = e rather than at x = 1 because at x = 1, the approximation will only converge on the interval (0, 2), which doesn't include our value (about 2.8).

As far as proofs go, I'd check the Wikipedia page for Taylor's Theorem or http://math.stackexchange.com/questions/481661/simplest-proof-of-taylors-theorem .

I hope that helps.(3 votes)

- how can i graph all the derivatives of sinx?(1 vote)
- you would graph sin x, - sin x, cos x, and - cos x...that represents all the derivatives. Beyond that they keep repeating(2 votes)

## Video transcript

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 a Taylor series, where we are centering
our approximation around x is equal to 0. So it's just a special
case of a Taylor series. So let's take f of
x in this situation to be equal to sine of x. And let'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 the sine of x is the derivative of cosine of
x, which is negative sine of x. The third derivative is going
to be the derivative of this. So I'll just write
a 3 in parentheses there, instead of doing
prime, prime, prime. So the third derivative
is the derivative of this, which is
negative cosine of x. 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 times. And we care-- in order to do
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. I'll do it in this purple color. So f-- that's hard
to see, I think So let's do this
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 then the fourth
derivative evaluated at 0 is going to be 0 again. And we could keep going,
but once again, it seems like there's a pattern. 0, 1, 0, negative
1, 0, then you're going to go back to positive 1. So on and so forth. So let's find its
polynomial representation using the Maclaurin series. And just a reminder,
this one up here, this was approximately
cosine of x. And you'll get closer and
closer to cosine of x. I'm not rigorously
showing you how close, in that it's definitely
the exact 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 go to
infinity, you're 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're not even going
to need to include that. The next term is
going to be f prime of 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're 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 to the third
over 3 factorial. And then the next
term is going to be 0, because that's the
fourth derivative. The fourth derivative evaluated
at 0 is the next coefficient. We see that that is going to be
0, so it's going to drop off. And what you're
going to see here-- and actually maybe I haven't
done enough 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 the fifth
derivative of x is going to be cosine of x again. The fifth derivative-- we'll
do it in that same color, just so it's consistent-- the
fifth derivative evaluated at 0 is going to be 1. So the fourth derivative
evaluated at 0 is 0, then you go to the fifth
derivative evaluated at 0, it's going to be positive 1. And if I kept doing this,
it would be positive 1-- I have to write the 1
as the coefficient-- times x to the fifth over 5 factorial. So there's something
interesting going on here. For cosine of x, I had 1,
essentially 1 times x to the 0. Then I don't have x
to the first power. I don't have x to the
odd powers, actually. And then I just essentially have
x to all of the even powers. And whatever power it is, I'm
dividing it by that factorial. And then the signs
keep switching. 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, because-- well 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, especially 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. This is x to the
third over 3 factorial plus x to the fifth
over 5 factorial. Yeah, 0 would be an even number. Anyway, 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 sines. X to the seventh over
7 factorial plus x to the ninth over 9 factorial. So there's something
interesting here. You once again see this
kind of complimentary nature between sine and cosine here. You see almost
this-- they're kind of 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 its
polynomial representation, is all of the odd powers of
x divided by its factorial, and you switch sines. 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 really, really mind blowing.