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 $R_{n}(x) \rightarrow 0$ .] $\begin{array}{c} f(x)=\sin (x), \quad a=\pi \\ f(x)=\sum_{n=0}^{\infty}(-1)^{n} \cdot \frac{x^{2 n}}{(2 n)!} \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to build a Taylor series for $\sin(x)$ , but instead of building it around $x = 0$ , we're building it around $x = \pi$ ! Watch out! Remember, the Taylor series lets us approximate a function using its derivatives at a specific point.
Don't mix up the center point!

STEP 2

1. Find Derivatives
2. Evaluate Derivatives
3. Build the Series

STEP 3

Let's start by finding the derivatives of $f(x) = \sin(x)$ .
This is the exciting part!

STEP 4

* $f(x) = \sin(x)$ * $f'(x) = \cos(x)$ * $f''(x) = -\sin(x)$ * $f'''(x) = -\cos(x)$ * $f^{(4)}(x) = \sin(x)$

STEP 5

Notice how the derivatives repeat every four derivatives!
This pattern will be super helpful later.

STEP 6

Now, let's evaluate these derivatives at our center point, $x = \pi$ .

STEP 7

* $f(\pi) = \sin(\pi) = 0$ * $f'(\pi) = \cos(\pi) = -1$ * $f''(\pi) = -\sin(\pi) = 0$ * $f'''(\pi) = -\cos(\pi) = 1$ * $f^{(4)}(\pi) = \sin(\pi) = 0$

STEP 8

See how the pattern of zeros and ones (and negative ones!) emerges?

STEP 9

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

STEP 10

Let's plug in our values, remembering that $a = \pi$ :
$f(x) = \frac{f(\pi)}{0!} \cdot (x-\pi)^0 + \frac{f'(\pi)}{1!} \cdot (x-\pi)^1 + \frac{f''(\pi)}{2!} \cdot (x-\pi)^2 + \frac{f'''(\pi)}{3!} \cdot (x-\pi)^3 + ...$

STEP 11

Substituting the evaluated derivatives:
$f(x) = \frac{0}{1} \cdot 1 + \frac{-1}{1} \cdot (x-\pi) + \frac{0}{2} \cdot (x-\pi)^2 + \frac{1}{6} \cdot (x-\pi)^3 + \frac{0}{24} \cdot (x-\pi)^4 + \frac{-1}{120} \cdot (x-\pi)^5 + ...$

STEP 12

Simplifying gives us:
$f(x) = -(x-\pi) + \frac{1}{6} \cdot (x-\pi)^3 - \frac{1}{120} \cdot (x-\pi)^5 + ...$

STEP 13

We can rewrite this using summation notation, noticing the pattern of alternating signs and odd powers:
$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} \cdot (x-\pi)^{2n+1}$

STEP 14

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

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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 $R_{n}(x) \rightarrow 0$ .] $\begin{array}{c} f(x)=\sin (x), \quad a=\pi \\ f(x)=\sum_{n=0}^{\infty}(-1)^{n} \cdot \frac{x^{2 n}}{(2 n)!} \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to build a Taylor series for $\sin(x)$ , but instead of building it around $x = 0$ , we're building it around $x = \pi$ ! Watch out! Remember, the Taylor series lets us approximate a function using its derivatives at a specific point.
Don't mix up the center point!

STEP 2

1. Find Derivatives
2. Evaluate Derivatives
3. Build the Series

STEP 3

Let's start by finding the derivatives of $f(x) = \sin(x)$ .
This is the exciting part!

STEP 4

* $f(x) = \sin(x)$ * $f'(x) = \cos(x)$ * $f''(x) = -\sin(x)$ * $f'''(x) = -\cos(x)$ * $f^{(4)}(x) = \sin(x)$

STEP 5

Notice how the derivatives repeat every four derivatives!
This pattern will be super helpful later.

STEP 6

Now, let's evaluate these derivatives at our center point, $x = \pi$ .

STEP 7

* $f(\pi) = \sin(\pi) = 0$ * $f'(\pi) = \cos(\pi) = -1$ * $f''(\pi) = -\sin(\pi) = 0$ * $f'''(\pi) = -\cos(\pi) = 1$ * $f^{(4)}(\pi) = \sin(\pi) = 0$

STEP 8

See how the pattern of zeros and ones (and negative ones!) emerges?

STEP 9

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

STEP 10

Let's plug in our values, remembering that $a = \pi$ :
$f(x) = \frac{f(\pi)}{0!} \cdot (x-\pi)^0 + \frac{f'(\pi)}{1!} \cdot (x-\pi)^1 + \frac{f''(\pi)}{2!} \cdot (x-\pi)^2 + \frac{f'''(\pi)}{3!} \cdot (x-\pi)^3 + ...$

STEP 11

Substituting the evaluated derivatives:
$f(x) = \frac{0}{1} \cdot 1 + \frac{-1}{1} \cdot (x-\pi) + \frac{0}{2} \cdot (x-\pi)^2 + \frac{1}{6} \cdot (x-\pi)^3 + \frac{0}{24} \cdot (x-\pi)^4 + \frac{-1}{120} \cdot (x-\pi)^5 + ...$

STEP 12

Simplifying gives us:
$f(x) = -(x-\pi) + \frac{1}{6} \cdot (x-\pi)^3 - \frac{1}{120} \cdot (x-\pi)^5 + ...$

STEP 13

We can rewrite this using summation notation, noticing the pattern of alternating signs and odd powers:
$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} \cdot (x-\pi)^{2n+1}$

STEP 14

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

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Studdy Solution

STEP 1

STEP 2

1. Find Derivatives2. Evaluate Derivatives3. Build the Series

STEP 3

Let's **start** by finding the derivatives of f(x)=sin⁡(x)f(x) = \sin(x)f(x)=sin(x).This is the exciting part!

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)

STEP 5

Notice how the derivatives **repeat** every four derivatives!This pattern will be super helpful later.

STEP 6

Now, let's **evaluate** these derivatives at our center point, x=πx = \pix=π.

STEP 7

STEP 8

See how the **pattern** of zeros and ones (and negative ones!) emerges?

STEP 9

The **general formula** for a Taylor series centered at aaa is:f(x)=∑n=0∞f(n)(a)n!⋅(x−a)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} \cdot (x-a)^nf(x)=n=0∑∞​n!f(n)(a)​⋅(x−a)n

STEP 10

STEP 11

STEP 12

**Simplifying** gives us:f(x)=−(x−π)+16⋅(x−π)3−1120⋅(x−π)5+...f(x) = -(x-\pi) + \frac{1}{6} \cdot (x-\pi)^3 - \frac{1}{120} \cdot (x-\pi)^5 + ...f(x)=−(x−π)+61​⋅(x−π)3−1201​⋅(x−π)5+...

STEP 13

We can **rewrite** this using summation notation, noticing the pattern of alternating signs and odd powers:f(x)=∑n=0∞(−1)n+1(2n+1)!⋅(x−π)2n+1f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} \cdot (x-\pi)^{2n+1}f(x)=n=0∑∞​(2n+1)!(−1)n+1​⋅(x−π)2n+1

STEP 14

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)!} \cdot (x-\pi)^{2n+1}n=0∑∞​(2n+1)!(−1)n+1​⋅(x−π)2n+1

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Find Derivatives
2. Evaluate Derivatives
3. Build the Series

Let's start by finding the derivatives of $f(x) = \sin(x)$ .
This is the exciting part!

* $f(x) = \sin(x)$ * $f'(x) = \cos(x)$ * $f''(x) = -\sin(x)$ * $f'''(x) = -\cos(x)$ * $f^{(4)}(x) = \sin(x)$

Notice how the derivatives repeat every four derivatives!
This pattern will be super helpful later.

Now, let's evaluate these derivatives at our center point, $x = \pi$ .

See how the pattern of zeros and ones (and negative ones!) emerges?

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

Simplifying gives us:
$f(x) = -(x-\pi) + \frac{1}{6} \cdot (x-\pi)^3 - \frac{1}{120} \cdot (x-\pi)^5 + ...$

We can rewrite this using summation notation, noticing the pattern of alternating signs and odd powers:
$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} \cdot (x-\pi)^{2n+1}$

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