Find the Taylor Series of f(x)=cosxf\left(x\right)=\cos x centered x=πx=\pi. Then, write the power series using summation notation.

∑ n = 0 ∞ ( − 1 ) n + 1 ( 2 n ) ! ( x − π ) 2 n \sum_{n=0}^{\infty}\frac{\left(-1\right)^{n+1}}{\left(2n\right)!}\left(x-\pi\right)^{2n}

Find the Taylor Series of f ( x ) = cos x f\left(x\right)=\cos x centered x = π x=\pi . Then, write the power series using summation notation.

Here we're asked to find the Taylor series for the function f of x equals cosine of x centered at x equals pi. Then we want to write the power series using summation notation. So since we're working with a Taylor series of a function here, that means we want to take the sum from n equals zero to infinity of the nth derivative of our function evaluated at the center, which here is pi. Then dividing that by n factorial and multiplying by x minus our center, again, pi, raised to the power of n. Now since we need those first couple of derivatives evaluated at our center, let's start there. If our function is the cosine of x, we can go ahead and evaluate that at pi. So cosine of pi is going to give us negative one. Then as we take that first derivative, that's going to be negative sine of x. So negative sine of pi evaluating that at our center will give us here zero. Then taking our second derivative, differentiating that negative sine of x will give us negative cosine of x. So negative cosine of pi here is going to end up being a positive one. Then taking our third derivative, differentiating that negative cosine is going to give us a positive sine of x. So sine of pi here is still going to be zero. Then finally our fourth derivative here is going to be cosine of x. We're right back at our original function. And the cosine of pi is going to be negative one. So now that we have those values found, let's go ahead and plug in our values of n and look at these first couple of terms of our Taylor series. So starting from n equals zero, we have the zeroth derivative of our function, which is really just the function itself evaluated at pi. So that's negative one over zero factorial times x minus pi to the power of zero. Then plugging in one for n, this is going to be the first derivative of our function evaluated at pi. That's that zero that we already found. So zero over one factorial times x minus pi to the power of one. Then for our next term here, we have that positive one for that second derivative evaluated at the center. So one over two factorial times x minus pi squared. And then moving on to my second line here, our next term up, this is n equals three. So our third derivative evaluated at pi is zero. Zero over three factorial times x minus pi cubed. One more term that we're going to look at here, we're adding this together with our fourth derivative evaluated at pi, so that's negative one, over four factorial times x minus pi to the power of four. Then this continues on. Let's do some simplification here. Looking at that first term, zero factorial is one, and anything raised to the power of zero is also one. So that just leaves me here with negative one. So that first term here, negative one. Then looking at this term here with n equals one plugged in, zero is multiplying everything here. So this entire term goes to zero. Then for my next term here, one over two factorial is just one half, so this is plus one half times x minus pi squared. Then for n equals three, we again have a zero multiplying everything, so this whole term goes to zero. And finally for this last term here, negative one over four factorial is going to be negative one over 24. Remember four factorial four times three times two times one times x minus pi to the power of four. Then of course this continues on. Now that we've simplified this, we want to go ahead and write our power series using summation notation. In order to do that, we need to recognize any patterns that are going on in our series. So writing this as the sum from n equals zero to infinity, the first thing that I notice about the terms in my series here is that I start out with a negative, then I get a positive, and then I get a negative again. So that tells me I'm dealing with an alternating series and I need to take negative one and raise it to a particular power. Now because here my first term starts out as negative and I'm having my series start at zero, that means that this negative one can't just be raised to the nth power because that would make our first term positive. Instead, we need to raise it to the power of n plus one. So that takes care of the alternating piece of our series, but let's think about what else is going on. I have this two in my denominator and I also have this 24. We know that these are the same values as two factorial and four factorial. Now if we also think about that first term having a factorial with it, that would be divided by zero factorial because zero factorial is just one. So if we think about this first term being for n equals zero, the second for n equals one, and then n equals two, how do those values in my denominator relate? Well, if I take two and multiply it by one, that would give me two. If I take two and multiply it by two, that would give me four. Two factorial and four factorial. So that means that in my denominator here, I wanna have a two n factorial, and that accounts for the coefficients that I see in my series. Then I have this piece of x minus pi raised to a certain power. So x minus pi squared and x minus pi to the power of four. Now, this is the same exact thing we saw in our denominator with that two n. So here I need to take x minus pi and raise it to that power two n. So this is the summation notation of my series and thus the Taylor series representation of the function cosine of x. Let us know if you have any questions in the comments, and I'll see you in the next one.

Find the Taylor Series of f ( x ) = cos ⁡ x f\left(x\right)=\cos x centered x = π x=\pi . Then, write the power series using summation notation.

∑ n = 0 ∞ ( − 1 ) n + 1 ( 2 n ) ! ( x − π ) 2 n \sum_{n=0}^{\infty}\frac{\left(-1\right)^{n+1}}{\left(2n\right)!}\left(x-\pi\right)^{2n}

∑ n = 0 ∞ ( 1 ) n + 1 ( 2 n ) ! ( x − π ) 2 n \sum_{n=0}^{\infty}\frac{\left(1\right)^{n+1}}{\left(2n\right)!}\left(x-\pi\right)^{2n}

∑ n = 0 ∞ ( π ) n ( 2 n ) ! ( x − π ) 2 n \sum_{n=0}^{\infty}\frac{\left(\pi\right)^{n}}{\left(2n\right)!}\left(x-\pi\right)^{2n}

∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! ( x + π ) 2 n \sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n\right)!}\left(x+\pi\right)^{2n}

15. Power Series

Taylor Series & Taylor Polynomials

Calculus

15. Power Series

Taylor Series & Taylor Polynomials

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Multiple Choice

Find the Taylor Series of $f (x) = \cos x$ centered $x = π$ . Then, write the power series using summation notation.

$\sum_{n = 0}^{\infty} \frac{{(1)}^{n + 1}}{(2 n)!} {(x - π)}^{2 n}$

$\sum_{n = 0}^{\infty} \frac{{(π)}^{n}}{(2 n)!} {(x - π)}^{2 n}$

$\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{(2 n)!} {(x - π)}^{2 n}$

$\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} {(x + π)}^{2 n}$

Verified step by step guidance

Step 1: Recall the general formula for the Taylor series of a function f(x) centered at x = a. It is given by:

f(x) = ∑

_n=0^∞ f⁽ⁿ⁾(a)/n! ⋅ (x - a)ⁿ. Here, f⁽ⁿ⁾(a) represents the nth derivative of f(x) evaluated at x = a.

Step 2: Compute the derivatives of f(x) = cos(x). The first few derivatives are:

f'(x) = -sin(x)

f''(x) = -cos(x)

f'''(x) = sin(x)

, and

f''''(x) = cos(x)

. Notice the derivatives repeat cyclically every four terms.

Step 3: Evaluate the derivatives at x = π. For

f(x) = cos(x)

, we have:

f(π) = -1

f'(π) = 0

f''(π) = 1

f'''(π) = 0

, and so on. Only even derivatives contribute to the series since odd derivatives are zero.

Step 4: Substitute the values of the derivatives into the Taylor series formula. For even derivatives, the nth term is given by:

f

⁽²ⁿ⁾(π)/(2n)! ⋅ (x - π)²ⁿ. Since

f

⁽²ⁿ⁾(π) alternates between 1 and -1, the series includes a factor of

(-1)

ⁿ.

Step 5: Write the Taylor series in summation notation. The final expression is:

∑

_n=0^∞ (-1)ⁿ/(2n)! ⋅ (x - π)²ⁿ. This represents the power series expansion of

cos(x)

centered at

x = π

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