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### Course: 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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# Visualizing Taylor series approximations

The larger the degree of a Taylor polynomial, the better it approximates the function. See that in action with sin(x) and its Taylor polynomials. Created by Sal Khan.

## Want to join the conversation?

- So we can approximate any function using Taylor expanison and derive it's polynomial representation.

Is it possible to approximate any function using a formula of sine curves?(40 votes)- yes. Look into the Fourier series. it has a lot of applications like electrical engineering. the wikipedia page has snapshots of how the sum of sine functions can add up to a sawtooth wave, square wave or whatever...(54 votes)

- I think I get how the Taylor/Maclaurin series works, so can somebody tell me if my thinking is correct:

The infinite series works because we build into the polynomial an infinite chain of derivatives of the original function. It gives the value of the function at a certain point, and also the rate the function is changing at that point (1st der.), and also the way that change is changing (2nd der.), and the way that the change that changes the rate of change is changing (3rd der.), ad infinitum, so that we essentially have a "seed" that gives us the entire function just from that point. Am I thinking along the right lines?(39 votes)- Yes, that's basically how the Maclaurin and Taylor series work. Good job on figuring that out!(21 votes)

- I found something quite interesting : If you take the Maclauren series of sin(x) with a finite polynomial, then whatever how small the coefficients are, for a very big x, the biggest power will overcome the others, and p(x) begin to be very big and is going really far away from sin(x). However, if we take only the first approximation, we get p(x)= 0, which is never further than 1 from sin(x) for any x. Isn't it counter-intuitive?(7 votes)
- Maybe we can think of this as the cost of being more precise in the center of the function.

For example, imagine you are trying to get a wire to fit along ripples on a beach. Each time you put a bend in the wire (add a term to the polynomial) you cause the end of the wire to move further away from the surface of the sand ...

I'll stop this analogy now, before everyone gets too bent out of shape ...(6 votes)

- Why is it that by using the values of the function and its derivatives at zero, we can obtain an approximation of the function at values other than zero? Are the other values of the function somehow "encoded" in the behavior of the curve at x = 0? I can see how that would be the case for simple functions like lines and parabolas, but does that continue to hold for more complex functions? For arbitrary functions?

In a piecewise function, I can see that this method will fail (I imagine a function that is equal to sin(x) at zero but is something entirely different at other values of x), so there are clearly restrictions on the types of functions for which Taylor/MacLaurin series can be taken. Is it simply a requirement that the function be continuous and differentiable?(6 votes)- For this to work, the function must be continuous and you must be able to differentiate it infinitely many times.

There is nothing special about x=0, it is just easier to calculate. You can use any value of x in the domain for a Taylor series.(4 votes)

- How come I cant find any videos about taylor series and taylor polynomials involving 2 variables f(x,y)? Im sure this has been covered, and some help to find these videos would be highly appreciated. If that is not the case, maybe requesting this is not such a bad idea. Thank you.(8 votes)
- Look for information involving applying "Taylor's theorem" to "multivariate functions". I don't know if Sal has done any videos there but it would involve partial derivitives and gradients to approximate the surface. I'm not familair with this, but it's not immediately clear to me that you could approximate a function of N variables using polynomials (or some other N-dimensional building block --what?) but those are some search terms for you. Hope it helps.(5 votes)

- how would you approximate piecewise functions then? like f(x) = sin(x) when x>0 and cos(x) when x<0?(4 votes)
- do i assume thst since i know Mac's series i now Taylor's series? do i just take my derivative at any point and i will arrive at the Taylor's approximate?(4 votes)
- No, you just know the Taylor series at a specific point (also the Maclaurin series) or, to be more clear, each succeeding polynomial in the series will hug more and more of the function with the specified point that x equals being the one point that every single function touches (in the video above, x equals 0). The Taylor series is generalized to x equaling every single possible point in the function's domain. You can take this to mean a Maclaurin series that is applicable to every single point; sort of like having a general derivative of a function that you can use to find the derivative of any specific point you want. Check out "Generalized Taylor Series Approximation" for a better explanation.(3 votes)

- Would this chapter (sequences,series, and function approximation specifically taylor series) be helpful or give a intuition to learn the Fourier Series & Transform which is a subject in the engineering mathmetic course..?(3 votes)
- Yes, I think they would help you as an introduction to Fourier series, since Taylor series are much simpler than Fourier series, but they have many similitudes.(3 votes)

- Is there a simple way of determining at which points the approximation to the nth degree is exactly equal to the given function? Like, at n=1, the sin(x) approximation is equal to sin(x) only at x=0, so that's once. Is there a formula that tells you how many points are exactly equal given just the nth degree you're approximating to?(2 votes)
- What's the O(x^10) at the end of the series expansion?

-TX520(2 votes)

## Video transcript

I've talked a lot
about using polynomials to approximate functions, but
what I want to do in this video is actually show you that
the approximation is actually happening. So right over here-- and I'm
using WolframAlpha for this. It's a very cool website. You can do all sorts of crazy
mathematical things on it. So WolframAlpha.com-- I got this
copied and pasted from them. I met Steven Wolfram at a
conference not too long ago. He said yes, you
should definitely use WolframAlpha in your videos. And I said, great. I will. And so that's what
I'm doing right here. And it's super useful,
because what it does is-- and we could
have calculated a lot of this on
our own, or even done it on a graphic calculator. But you usually can do it just
with one step on WolframAlpha-- is see how well we can
approximate sine of x using-- you could call it a
Maclaurin series expansion, or you could call it a
Taylor series expansion-- at x is equal to 0 using
more and more terms. And having a good
feel for the fact that the more terms we add, the
better it hugs the sine curve. So this over here in
orange is sine of x. That should hopefully look
fairly familiar to you. And in previous
videos, we figured out what that Maclaurin
expansion for sine of x is. And WolframAlpha does
it for us as well. They actually calculate
the factorials. 3 factorial is 6, 5 factorial
is 120, so on and so forth. But what's interesting
here is you can pick how many of
the approximations you want to graph. And so what they did
is if you want just one term of the
approximation-- so if we didn't have this whole thing. If we just said that our
polynomial is equal to x, what does that look like? Well, that's going to be
this graph right over here. They tell us which
term-- how many terms we used by how many dots there are
right over here, which I think is pretty clever. So this right here, that is the
function p of x is equal to x. And so it's a very
rough approximation, although for sine of x,
it doesn't do a bad job. It hugs the sine curve
right over there. And then it starts to veer
away from the sine curve again. You add another term. So if you have the x minus
x to the third over 6. So now you have two
terms in the expansion. Or I guess we should say we
were up to the third-order term, because that's how their
numbering the dots. Because they're not talking
about the number of terms. They're talking about
the order of the terms. So they have one
dot here, because we have only one first-degree term. When we have two
terms here, since we-- when you do the
expansion for sine of x, it doesn't have a
second-degree term. We now have a third-degree
polynomial approximation. And so let's look
at the third-degree. We should look for three dots. That's this curve
right over here. So if you just have
that first term, you just get that straight line. You add the negative x to
the third over 6 to that x. You now get a curve
that looks like this. And notice it starts hugging
sine a little bit earlier. And it keeps hugging
it a little bit later. So once again, just
adding that second term does a pretty good job. It hugs the sine
curve pretty well, especially around
smaller numbers. You add another term. And now we're at an order five
polynomial, right over here. So x minus x to the third over
6 plus x to the fifth over 120. So let's look for the five dots. So that's this one right over
here-- one, two, three, four, five. So that's this curve
right over here. And notice it begins hugging
the line a little bit earlier than the magenta
version, and it keeps hugging it a little bit longer. Then it flips back up like this. So it hugged it a
little bit longer. And you can see I'll keep going. If you have all these
first four terms, it gives us a seventh
degree polynomial. So let's look for the
seven dots over here. So they come in just like this. And then once again, it hugs the
curve sooner than when we just had the first three terms. And it keeps hugging the
curve all the way until here. And then the last one. If you have all of these
terms up to x to the ninth, it does it even more. You start here. It hugs the curve
longer than the others. And goes out. And if you think about
it, it makes sense, because what's happening here is
each successive term that we're adding to the expansion,
they have a higher degree of x over a much, much,
much, much larger number. So for small x
value-- so when you're close to the origin
for small x values, this denominator is
going to overpower the numerator, especially
when you're below 1. Because when you
take something that has absolute value
less than 1 to a power, you're actually
shrinking it down. So when you're
close to the origin, these latter terms
don't matter much. So you're kind of
not losing some of the precision of some
of the earlier terms. When these tweaking
terms come in, these come in when
the numerator can start to overpower
the denominator. So this last term, it starts
to become relevant out here, where all of a sudden x to the
ninth can overpower 362,880. And the same thing
on the negative side. So hopefully this
gives you a sense. We only have one, two,
three, four, five terms here. Imagine what would
happen if we had an infinite number of terms. I think you get a
pretty good sense that it would kind of hug the
sine curve out to infinity. So hopefully that makes you feel
a little bit better about this. And for fun, you
might want to go type in-- you can type in
Taylor expansion at 0 and sine of x, or Maclaurin
expansion or Maclaurin series for sine of x,
cosine of x, e to the x, at WolframAlpha.com. And try it out for a bunch
of different functions. And you can keep adding
or taking away terms to see how well
it hugs the curve.