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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 10

Lesson 14: Finding Taylor or Maclaurin series for a function- Function as a geometric series
- Geometric series as a function
- Power series of arctan(2x)
- Power series of ln(1+x³)
- Function as a geometric series
- 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
- Geometric series interval of convergence

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# Worked example: cosine function from power series

Given a power series, we recognize it as the Maclaurin series of cos(x³) and evaluate it at a given x-value.

## Want to join the conversation?

- when i tried it first, i just factored out the x^3, so it looked like 1+ x^3{ (-x^2/2!) +(x^4/4!)+.........} , then i added 1 inside and subtracted it so it looked like 1+ x^3{-1+1 - (x^2/2!)+.......} which then equals to 1+x^3{-1 + cos(x)}. Can anyone tell me why this way doesnt work?(3 votes)
- You did not factor correctly.

If you factor x³ from x⁶ + x¹² ....

you get x³ ( x³ + x⁹ ....), not what you computed.

You should also be aware that the commutative and associative properties do not necessarily hold for infinite series. Therefore, you cannot use either property without first establishing that the property really does hold for that particular infinite series.(13 votes)

- if i have to find the function of 1 + x + x^2/2! + x^4/4! + x^5/5! + ......

what would be the function of this?

is it (exp^x ) - (x^3)/3! ??(3 votes)- No. It is much simpler. That is the Taylor series for e^x.(0 votes)

- how does n=0 give us 1 ?(2 votes)
- Can somebody tell me what a Power Series is?(2 votes)
- A power series is the summation of a sequence of numbers where each following number in the series is raised to a higher and higher power.

This wouldn't converge and would never be used but just illustrates what a power series is.

'''

Sum from n = 0 to infinity, of x^n, then the function is whatever value is 1 + x + x^2 + x^3 + ...

'''(1 vote)

- Just out of curiosity, since we are given a number for X, does that mean that it is is the point in which the series converges to? or is that simply a point within the series?(1 vote)
- No, in this case the series is a function (in terms of x) which converges to (spoiler!) cos(x^(1/3)) for all x. It has varying values depending on the specific value of x. In other words, since the series is in terms of x, we must be given a number for x to evaluate it.(3 votes)

- I thought I had a pretty good handle on this. Until I got this problem in the mastery challenge.

Sum to Infinity of (-1) to the (n+1) * x to the (2n + 3) / (2n!) is the taylor series of which function. I had no problem with working this series out.

The solution was -x to the (3rd) cos (x)

My understanding was that the approximation is: f(x) + f'(x)*x to the (2nd) over two factorial and so on.

Using the product rule, the second derivative, the x to the 2nd over 2n Factorial term is

-6x(cos(x)) + 3x(to the 2nd)sin(x) + 3x(to the 2nd) and so on.

each of these terms has an x. And evaluated at x = 0 all of these terms will be 0.

They cannot equal x to the 5th over 2 Factorial.

The first term with a non zero coefficient would be the 4th (3rd derivative term). I feel like I am missing something fundamental here.

Do we only use Maclaurin for the cos(x) function and do not take the derivatives of x to the 3rd?

thanks so much for anyone who will take the trouble to reading and answering this.(1 vote)- Notice that the expansion of a Macluarin series is defined as follows

T(x) = f(0) + f'(0)x + f''(0)x^2 / 2! + f'''(0) x^3 / 3! + ...... so on

Indeed it is true that the second derivative equals zero therefore the term containing x^2 is cancelled and the same thing goes for the term containing x and the constant term they are all equal to zero you will notice we only get a nonzero value at the 3rd derivative that means the coef. of x^3 in out expansion is non zero therefor we have a x^3 term in our expansion.

if you look at the series given to you, it agrees with our results when n=0 it starts with our x^3 term.

if you set n=1 it tells you that the coef of the x^5 term in our expansion equals 1/2 that means that our fifth derivative at x=0 equals to 60 because when you divide 60 by 5! you get 1/2.

I think you are confused and I tried to explain things clearly but I may have ended up confusing you more maybe watch the videos again and I am sorry if I was confusing.

Good luck and never give up(3 votes)

- Does this mean that cosx is equivalent to sum from 0 to infinity of ((-1)^n)(x^2n)/(2n)!(1 vote)

## Video transcript

- So I have an infinite series, here. The sum from n equals zero
to infinity of negative one to the n power times x to the sixth n over two n, the whole two n, factorial. And my goal in this video is
to evaluate this power series when x is equal to the
cubed root of pi over two. And I encourage you to pause this video and give it a go on your own. And I will give you a hint. The key to this is to figure out, well what function is
this the power series for? and then use that
function to evaluate this. And there's another clue here, hey, this is kind of a
mysterious or a suspicious looking number, here, pi over two. That looks like something I would use a trig function to evaluate. That might be a little
bit more straightforward. So I'll let you have a go at it. So I'm assuming you have tried, so let's try to work
through this together. And in any of these types of problems I like to at least expand
out this power series so I get a better sense of what it's like. So this right over here,
if I were to expand it out, this is going to be
equal to when n is zero, this is one, actually,
all of these are one, so it's just going to be one. When n is one, this is going to be negative one, x to the sixth, x to the sixth over two factorial. when n is two, it's going to be positive. negative one squared is positive one, times x to the 12th over four factorial. and then let's just do one
more: when x is equal to three, it's going to be negative x to the 18th. x to the 18th over six factorial. And you just keep going on and on forever. Now, offhand, I don't know a function, especially a trigonometric function, because that was kind of our clue here, this pi over two makes me feel like this might be a trigonometric
function right over here Nothing jumps out at me offhand, but this does look suspiciously familiar. This looks awfully close
to the power series, or the Maclaurin Series for cosine of x. Which we have seen multiple times. Let's just remind
ourselves of what that is. And if this doesn't look familiar, the previous video where
I do the Maclaurin Series for cosine of x goes into
detail on how I get this. The Maclaurin Series for
cosine of x is equal to... So I'll just write a few terms, so I'll write approximately equal to one minus x squared over two factorial, plus x to the fourth over four factorial, minus x to the sixth over six factorial, and just like that you're
probably seeing the similarities. Well the first term is the same, the sign negative, positive,
negative, positive, negative, positive, negative, positive. two factorial, four
factorial, six factorial. The difference is the powers, the exponents on the x. This is x squared, this is x to the sixth. This is x to the fourth,
that's x to the 12th. This is x to the sixth,
that's x to the 18th. Well, what if we... I guess, something for
you to think about is, well, how can we replace
x with something here? Because anything that I change, if I take cosine of... if I change x to, I don't know, a plus b, everywhere we see an x, you'd
replace it with an a plus b. Can we put a power of x here, so that these things end up like that? well, this x to the
sixth is the same thing, as x to the third squared. That's x to the third squared. This right over here, x to the 12th, is the same thing as x to the
third to the fourth power. This right over here is the same thing as x to the third to the sixth power. So if we could replace each of
these xs with x to the thirds we will get this power series up here. Well, how do we do that? Well, we would just say the
cosine of x to the third and actually let me do
that in a different color. So the cosine, and that's
not a different color. So the cosine of x to the
third is going to be equal to, and once again, everywhere we see an x, we replace it with x to the third. So it's one minus, and actually, I'm just going to put in
parenthesis squared, two factorial. I wanted to do that in the green, let me do all this in the green. Alright, so it's gonna be equal to: one minus parenthesis
squared over two factorial plus parenthesis to the fourth
power over four factorial. minus parenthesis to the
sixth power over 6 factorial. And now let me change
back to that mauve color. And since I'm taking the
cosine of x to the third, well this is gonna be
x to the third squared. this is gonna be x to the
third to the fourth power. this is gonna be x to the
third to the sixth power. Which is exactly what
I have right over here. So this right over here
is the power series for cosine of x to the third. So evaluating this when x is equal to the cubed root of pi over two, is the same thing as evaluating this when x is equal to the
cubed root of pi over two. Let me write that down,
because this is interesting. So this, so I'll just rewrite
it, from n equals zero to infinity of negative one to the n of x to the sixth n over two n factorial. This is equal to, this is the
power series representation, of cosine of x to the third power. So if you want to evaluate this when x is the cubed root of pi over two, we just have to evaluate this when x is the cubed root of pi over two. And this does suspiciously
work out nicely, because if you take the
cube of the cubed root, well, good things happen. So the cosine of the
cubed root of pi over two to the third power, well
that's just the same thing as cosine of pi over two, which, of course, is equal to zero. And we are done.