Math / Calculus

QuestionFind the Taylor series for $f(x)$ centered at the given value of $a$ . [Assume that $f$ has a power series expansion. Do not show that $R_{n}(x) \rightarrow 0$ .] $\begin{array}{r} f(x)=\sin (x), \quad a=\pi \\ f(x)=\sum_{n=0}^{\infty}(\square) \end{array}$
Find the associated radius of convergence, $R$ . $\square$ $R=$

Studdy Solution

STEP 1

What is this asking? We need to find the Taylor series of $\sin(x)$ centered at $a = \pi$ and its radius of convergence. Watch out! Remember that the Taylor series centered at a is not the same as the Maclaurin series, which is centered at 0!

STEP 2

1. Derivatives of sine
2. Evaluate at pi
3. Assemble the Taylor series
4. Radius of convergence

STEP 3

Let's start by finding the derivatives of $f(x) = \sin(x)$ .
We'll need these for the Taylor series formula.

STEP 4

$f(x) = \sin(x)$ $f'(x) = \cos(x)$ $f''(x) = -\sin(x)$ $f'''(x) = -\cos(x)$ $f^{(4)}(x) = \sin(x)$ And the pattern repeats!

STEP 5

Now, we evaluate these derivatives at $x = a = \pi$ .

STEP 6

$f(\pi) = \sin(\pi) = 0$ $f'(\pi) = \cos(\pi) = -1$ $f''(\pi) = -\sin(\pi) = 0$ $f'''(\pi) = -\cos(\pi) = 1$ $f^{(4)}(\pi) = \sin(\pi) = 0$ The zero and one pattern also repeats, but with alternating signs!

STEP 7

The Taylor series formula is: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$

STEP 8

Let's plug in our values, where $a = \pi$ : $f(x) = \frac{f(\pi)}{0!}(x-\pi)^0 + \frac{f'(\pi)}{1!}(x-\pi)^1 + \frac{f''(\pi)}{2!}(x-\pi)^2 + \frac{f'''(\pi)}{3!}(x-\pi)^3 + \dots$ $f(x) = \frac{0}{1} \cdot 1 + \frac{-1}{1}(x-\pi) + \frac{0}{2}(x-\pi)^2 + \frac{1}{6}(x-\pi)^3 + \frac{0}{24}(x-\pi)^4 + \frac{-1}{120}(x-\pi)^5 + \dots$ $f(x) = -(x-\pi) + \frac{1}{6}(x-\pi)^3 - \frac{1}{120}(x-\pi)^5 + \dots$

STEP 9

We can rewrite this using summation notation.
Notice only the odd powers are present, and the signs alternate. $f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} (x-\pi)^{2n+1}$

STEP 10

The radius of convergence for the Taylor series of $\sin(x)$ is infinity, regardless of the center.

STEP 11

This is because the sine function is analytic everywhere, meaning its Taylor series converges to the function's value for all $x$ .

STEP 12

The Taylor series for $f(x) = \sin(x)$ centered at $a = \pi$ is: $\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} (x-\pi)^{2n+1}$ The radius of convergence is $R = \infty$ .

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Studdy Solution

STEP 1

What is this asking? We need to find the Taylor series of $\sin(x)$ centered at $a = \pi$ and its radius of convergence. Watch out! Remember that the Taylor series centered at a is not the same as the Maclaurin series, which is centered at 0!

STEP 2

1. Derivatives of sine
2. Evaluate at pi
3. Assemble the Taylor series
4. Radius of convergence

STEP 3

Let's start by finding the derivatives of $f(x) = \sin(x)$ .
We'll need these for the Taylor series formula.

STEP 4

$f(x) = \sin(x)$ $f'(x) = \cos(x)$ $f''(x) = -\sin(x)$ $f'''(x) = -\cos(x)$ $f^{(4)}(x) = \sin(x)$ And the pattern repeats!

STEP 5

Now, we evaluate these derivatives at $x = a = \pi$ .

STEP 6

$f(\pi) = \sin(\pi) = 0$ $f'(\pi) = \cos(\pi) = -1$ $f''(\pi) = -\sin(\pi) = 0$ $f'''(\pi) = -\cos(\pi) = 1$ $f^{(4)}(\pi) = \sin(\pi) = 0$ The zero and one pattern also repeats, but with alternating signs!

STEP 7

The Taylor series formula is: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$

STEP 8

STEP 9

We can rewrite this using summation notation.
Notice only the odd powers are present, and the signs alternate. $f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} (x-\pi)^{2n+1}$

STEP 10

The radius of convergence for the Taylor series of $\sin(x)$ is infinity, regardless of the center.

STEP 11

This is because the sine function is analytic everywhere, meaning its Taylor series converges to the function's value for all $x$ .

STEP 12

The Taylor series for $f(x) = \sin(x)$ centered at $a = \pi$ is: $\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} (x-\pi)^{2n+1}$ The radius of convergence is $R = \infty$ .

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Studdy Solution

STEP 1

What is this asking? We need to find the Taylor series of sin⁡(x)\sin(x)sin(x) centered at a=πa = \pia=π and its radius of convergence. Watch out! Remember that the Taylor series centered at *a* is not the same as the Maclaurin series, which is centered at 0!

STEP 2

1. Derivatives of sine2. Evaluate at pi3. Assemble the Taylor series4. Radius of convergence

STEP 3

Let's **start** by finding the derivatives of f(x)=sin⁡(x)f(x) = \sin(x)f(x)=sin(x).We'll need these for the Taylor series formula.

STEP 4

f(x)=sin⁡(x) f(x) = \sin(x) f(x)=sin(x) f′(x)=cos⁡(x) f'(x) = \cos(x) f′(x)=cos(x) f′′(x)=−sin⁡(x) f''(x) = -\sin(x) f′′(x)=−sin(x)f′′′(x)=−cos⁡(x) f'''(x) = -\cos(x) f′′′(x)=−cos(x)f(4)(x)=sin⁡(x) f^{(4)}(x) = \sin(x) f(4)(x)=sin(x)And the pattern repeats!

STEP 5

Now, we **evaluate** these derivatives at x=a=πx = a = \pix=a=π.

STEP 6

STEP 7

The **Taylor series formula** is: f(x)=∑n=0∞f(n)(a)n!(x−a)n f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n f(x)=n=0∑∞​n!f(n)(a)​(x−a)n

STEP 8

STEP 9

We can **rewrite** this using summation notation.Notice only the odd powers are present, and the signs alternate. f(x)=∑n=0∞(−1)n+1(2n+1)!(x−π)2n+1 f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} (x-\pi)^{2n+1} f(x)=n=0∑∞​(2n+1)!(−1)n+1​(x−π)2n+1

STEP 10

The **radius of convergence** for the Taylor series of sin⁡(x)\sin(x)sin(x) is **infinity**, regardless of the center.

STEP 11

This is because the sine function is **analytic** everywhere, meaning its Taylor series converges to the function's value for all xxx.

STEP 12

The Taylor series for f(x)=sin⁡(x)f(x) = \sin(x)f(x)=sin(x) centered at a=πa = \pia=π is: ∑n=0∞(−1)n+1(2n+1)!(x−π)2n+1 \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} (x-\pi)^{2n+1} n=0∑∞​(2n+1)!(−1)n+1​(x−π)2n+1 The radius of convergence is R=∞R = \inftyR=∞.

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What is this asking? We need to find the Taylor series of $\sin(x)$ centered at $a = \pi$ and its radius of convergence. Watch out! Remember that the Taylor series centered at a is not the same as the Maclaurin series, which is centered at 0!

1. Derivatives of sine
2. Evaluate at pi
3. Assemble the Taylor series
4. Radius of convergence

Let's start by finding the derivatives of $f(x) = \sin(x)$ .
We'll need these for the Taylor series formula.

$f(x) = \sin(x)$ $f'(x) = \cos(x)$ $f''(x) = -\sin(x)$ $f'''(x) = -\cos(x)$ $f^{(4)}(x) = \sin(x)$ And the pattern repeats!

Now, we evaluate these derivatives at $x = a = \pi$ .

The Taylor series formula is: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$

We can rewrite this using summation notation.
Notice only the odd powers are present, and the signs alternate. $f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} (x-\pi)^{2n+1}$

The radius of convergence for the Taylor series of $\sin(x)$ is infinity, regardless of the center.

This is because the sine function is analytic everywhere, meaning its Taylor series converges to the function's value for all $x$ .

The Taylor series for $f(x) = \sin(x)$ centered at $a = \pi$ is: $\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} (x-\pi)^{2n+1}$ The radius of convergence is $R = \infty$ .