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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 7

Lesson 14: Maclaurin series of sin(x), cos(x), and eˣ

# 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.

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• 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? • 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.
• 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! ?? • how does n=0 give us 1 ? • Can somebody tell me what a Power Series is? • 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) • 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 