Math / Calculus

QuestionTaylor Series Formula $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)(x-a)^{n}}{n!}$
Find the Taylor Series for the following functions at the given value by using the definition and finding the derivatives at the given value.
1. $f(x)=e^{x}$ at $a=e$
4. $f(x)=\cos (2 x)$ at $a=\frac{\pi}{4}$
2. $f(x)=e^{2 x}$ at $a=3$
5. $f(x)=\frac{1}{x} \quad$ at $\quad a=7$
3. $f(x)=\cos (x)$ at $a=-\pi$
6. $f(x)=\ln (x)$ at $a=e$

Studdy Solution

STEP 1

What is this asking? We need to build Taylor series for a bunch of functions, centered around different values, using the Taylor series formula. Watch out! Don't forget the factorial in the Taylor series formula, and make sure to evaluate the derivatives at the correct center point!
Also, remember that $\cos(x)$ has a repeating pattern in its derivatives.

STEP 2

1. $e^x$ at $a=e$
2. $e^{2x}$ at $a=3$
3. $\cos(x)$ at $a = -\pi$
4. $\cos(2x)$ at $a = \frac{\pi}{4}$
5. $\frac{1}{x}$ at $a=7$
6. $\ln(x)$ at $a=e$

STEP 3

Define the function and the center point: $f(x) = e^x$ and $a = e$ .

STEP 4

Calculate the derivatives at $a = e$ .
Since the derivative of $e^x$ is always $e^x$ , all derivatives are simply $e^e$ ! $f(e) = f'(e) = f''(e) = \dots = e^e$ .

STEP 5

Plug into the Taylor series formula: $\sum_{n=0}^{\infty} \frac{f^{(n)}(e)(x-e)^n}{n!} = \sum_{n=0}^{\infty} \frac{e^e(x-e)^n}{n!}$

STEP 6

Define the function and the center point: $f(x) = e^{2x}$ and $a = 3$ .

STEP 7

Calculate the derivatives at $a = 3$ .
The $n$ -th derivative is $2^n e^{2x}$ , so $f^{(n)}(3) = 2^n e^6$ .

STEP 8

Plug into the Taylor series formula: $\sum_{n=0}^{\infty} \frac{2^n e^6 (x-3)^n}{n!} = \sum_{n=0}^{\infty} \frac{e^6 2^n (x-3)^n}{n!}$

STEP 9

Define the function and the center point: $f(x) = \cos(x)$ and $a = -\pi$ .

STEP 10

Calculate the derivatives at $a = -\pi$ .
We know $f(-\pi) = \cos(-\pi) = -1$ , $f'(-\pi) = -\sin(-\pi) = 0$ , $f''(-\pi) = -\cos(-\pi) = 1$ , and so on.
The pattern repeats every four derivatives.

STEP 11

Plug into the Taylor series formula: $\sum_{n=0}^{\infty} \frac{f^{(n)}(-\pi)(x+\pi)^n}{n!} = -1 + \frac{(x+\pi)^2}{2!} - \frac{(x+\pi)^4}{4!} + \frac{(x+\pi)^6}{6!} - \dots = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}(x+\pi)^{2n}}{(2n)!}$

STEP 12

Define the function and the center point: $f(x) = \cos(2x)$ and $a = \frac{\pi}{4}$ .

STEP 13

Calculate the derivatives at $a = \frac{\pi}{4}$ .
The $n$ -th derivative is given by $(-1)^{\lfloor\frac{n}{2}\rfloor + 1} 2^n \left(\frac{d}{dx}\right)^n \cos(2x)$ .
So, $f(\frac{\pi}{4}) = 0$ , $f'(\frac{\pi}{4}) = -2$ , $f''(\frac{\pi}{4}) = 0$ , $f'''(\frac{\pi}{4}) = 8$ , and so on.

STEP 14

Plug into the Taylor series formula: $\sum_{n=0}^{\infty} \frac{f^{(n)}(\frac{\pi}{4})(x-\frac{\pi}{4})^n}{n!} = -2(x-\frac{\pi}{4}) + \frac{8(x-\frac{\pi}{4})^3}{3!} - \frac{32(x-\frac{\pi}{4})^5}{5!} + \dots = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}2^{2n+1}(x-\frac{\pi}{4})^{2n+1}}{(2n+1)!}$

STEP 15

Define the function and the center point: $f(x) = \frac{1}{x}$ and $a = 7$ .

STEP 16

Calculate the derivatives at $a = 7$ .
The $n$ -th derivative is $\frac{(-1)^n n!}{x^{n+1}}$ , so $f^{(n)}(7) = \frac{(-1)^n n!}{7^{n+1}}$ .

STEP 17

Plug into the Taylor series formula: $\sum_{n=0}^{\infty} \frac{\frac{(-1)^n n!}{7^{n+1}}(x-7)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n (x-7)^n}{7^{n+1}}$

STEP 18

Define the function and the center point: $f(x) = \ln(x)$ and $a = e$ .

STEP 19

Calculate the derivatives at $a = e$ . $f(e) = 1$ , $f'(e) = \frac{1}{e}$ , $f''(e) = -\frac{1}{e^2}$ , $f'''(e) = \frac{2}{e^3}$ , and so on.

STEP 20

Plug into the Taylor series formula: $\sum_{n=0}^{\infty} \frac{f^{(n)}(e)(x-e)^n}{n!} = 1 + \frac{x-e}{e} - \frac{(x-e)^2}{2e^2} + \frac{(x-e)^3}{3e^3} - \dots = 1 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(x-e)^n}{ne^n}$

STEP 21

We found the Taylor series for all six functions!
They are listed above in their respective sections.

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Studdy Solution

STEP 1

STEP 2

1. exe^xex at a=ea=ea=e2. e2xe^{2x}e2x at a=3a=3a=33. cos⁡(x)\cos(x)cos(x) at a=−πa = -\pia=−π4. cos⁡(2x)\cos(2x)cos(2x) at a=π4a = \frac{\pi}{4}a=4π​5. 1x\frac{1}{x}x1​ at a=7a=7a=76. ln⁡(x)\ln(x)ln(x) at a=ea=ea=e

STEP 3

**Define the function** and the **center point**: f(x)=exf(x) = e^xf(x)=ex and a=ea = ea=e.

STEP 4

**Calculate the derivatives** at a=ea = ea=e.Since the derivative of exe^xex is always exe^xex, all derivatives are simply eee^eee! f(e)=f′(e)=f′′(e)=⋯=eef(e) = f'(e) = f''(e) = \dots = e^ef(e)=f′(e)=f′′(e)=⋯=ee.

STEP 5

**Plug into the Taylor series formula**: ∑n=0∞f(n)(e)(x−e)nn!=∑n=0∞ee(x−e)nn! \sum_{n=0}^{\infty} \frac{f^{(n)}(e)(x-e)^n}{n!} = \sum_{n=0}^{\infty} \frac{e^e(x-e)^n}{n!} n=0∑∞​n!f(n)(e)(x−e)n​=n=0∑∞​n!ee(x−e)n​

STEP 6

**Define the function** and the **center point**: f(x)=e2xf(x) = e^{2x}f(x)=e2x and a=3a = 3a=3.

STEP 7

**Calculate the derivatives** at a=3a = 3a=3.The nnn-th derivative is 2ne2x2^n e^{2x}2ne2x, so f(n)(3)=2ne6f^{(n)}(3) = 2^n e^6f(n)(3)=2ne6.

STEP 8

**Plug into the Taylor series formula**: ∑n=0∞2ne6(x−3)nn!=∑n=0∞e62n(x−3)nn! \sum_{n=0}^{\infty} \frac{2^n e^6 (x-3)^n}{n!} = \sum_{n=0}^{\infty} \frac{e^6 2^n (x-3)^n}{n!} n=0∑∞​n!2ne6(x−3)n​=n=0∑∞​n!e62n(x−3)n​

STEP 9

**Define the function** and the **center point**: f(x)=cos⁡(x)f(x) = \cos(x)f(x)=cos(x) and a=−πa = -\pia=−π.

STEP 10

STEP 11

STEP 12

**Define the function** and the **center point**: f(x)=cos⁡(2x)f(x) = \cos(2x)f(x)=cos(2x) and a=π4a = \frac{\pi}{4}a=4π​.

STEP 13

STEP 14

STEP 15

**Define the function** and the **center point**: f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ and a=7a = 7a=7.

STEP 16

**Calculate the derivatives** at a=7a = 7a=7.The nnn-th derivative is (−1)nn!xn+1\frac{(-1)^n n!}{x^{n+1}}xn+1(−1)nn!​, so f(n)(7)=(−1)nn!7n+1f^{(n)}(7) = \frac{(-1)^n n!}{7^{n+1}}f(n)(7)=7n+1(−1)nn!​.

STEP 17

**Plug into the Taylor series formula**: ∑n=0∞(−1)nn!7n+1(x−7)nn!=∑n=0∞(−1)n(x−7)n7n+1 \sum_{n=0}^{\infty} \frac{\frac{(-1)^n n!}{7^{n+1}}(x-7)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n (x-7)^n}{7^{n+1}} n=0∑∞​n!7n+1(−1)nn!​(x−7)n​=n=0∑∞​7n+1(−1)n(x−7)n​

STEP 18

**Define the function** and the **center point**: f(x)=ln⁡(x)f(x) = \ln(x)f(x)=ln(x) and a=ea = ea=e.

STEP 19

**Calculate the derivatives** at a=ea = ea=e. f(e)=1f(e) = 1f(e)=1, f′(e)=1ef'(e) = \frac{1}{e}f′(e)=e1​, f′′(e)=−1e2f''(e) = -\frac{1}{e^2}f′′(e)=−e21​, f′′′(e)=2e3f'''(e) = \frac{2}{e^3}f′′′(e)=e32​, and so on.

STEP 20

STEP 21

We found the Taylor series for all six functions!They are listed above in their respective sections.

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1. $e^x$ at $a=e$
2. $e^{2x}$ at $a=3$
3. $\cos(x)$ at $a = -\pi$
4. $\cos(2x)$ at $a = \frac{\pi}{4}$
5. $\frac{1}{x}$ at $a=7$
6. $\ln(x)$ at $a=e$

Define the function and the center point: $f(x) = e^x$ and $a = e$ .

Calculate the derivatives at $a = e$ .
Since the derivative of $e^x$ is always $e^x$ , all derivatives are simply $e^e$ ! $f(e) = f'(e) = f''(e) = \dots = e^e$ .

Plug into the Taylor series formula: $\sum_{n=0}^{\infty} \frac{f^{(n)}(e)(x-e)^n}{n!} = \sum_{n=0}^{\infty} \frac{e^e(x-e)^n}{n!}$

Define the function and the center point: $f(x) = e^{2x}$ and $a = 3$ .

Calculate the derivatives at $a = 3$ .
The $n$ -th derivative is $2^n e^{2x}$ , so $f^{(n)}(3) = 2^n e^6$ .

Plug into the Taylor series formula: $\sum_{n=0}^{\infty} \frac{2^n e^6 (x-3)^n}{n!} = \sum_{n=0}^{\infty} \frac{e^6 2^n (x-3)^n}{n!}$

Define the function and the center point: $f(x) = \cos(x)$ and $a = -\pi$ .

Define the function and the center point: $f(x) = \cos(2x)$ and $a = \frac{\pi}{4}$ .

Define the function and the center point: $f(x) = \frac{1}{x}$ and $a = 7$ .

Calculate the derivatives at $a = 7$ .
The $n$ -th derivative is $\frac{(-1)^n n!}{x^{n+1}}$ , so $f^{(n)}(7) = \frac{(-1)^n n!}{7^{n+1}}$ .

Plug into the Taylor series formula: $\sum_{n=0}^{\infty} \frac{\frac{(-1)^n n!}{7^{n+1}}(x-7)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n (x-7)^n}{7^{n+1}}$

Define the function and the center point: $f(x) = \ln(x)$ and $a = e$ .

Calculate the derivatives at $a = e$ . $f(e) = 1$ , $f'(e) = \frac{1}{e}$ , $f''(e) = -\frac{1}{e^2}$ , $f'''(e) = \frac{2}{e^3}$ , and so on.

We found the Taylor series for all six functions!
They are listed above in their respective sections.