Solved on Feb 28, 2024

Calculate the 2nd and 3rd order Taylor polynomials for $f(x) = 8 \tan(x)$ centered at $a = 0$ . Express the polynomials using symbolic notation and fractions.

STEP 1

Assumptions
1. The function given is $f(x) = 8 \tan(x)$ .
2. The Taylor polynomials $T_{2}$ and $T_{3}$ are to be centered at $a = 0$ .
3. The Taylor polynomial of degree $n$ for a function $f(x)$ centered at $a$ is given by: $T_{n}(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k$
4. We need to calculate the first three derivatives of $f(x)$ to find $T_{2}$ and $T_{3}$ .

STEP 2

First, we need to find the value of the function $f(x)$ at $a = 0$ .
$f(0) = 8 \tan(0)$

STEP 3

Calculate $f(0)$ .
$f(0) = 8 \cdot 0 = 0$

STEP 4

Next, we need to find the first derivative of $f(x)$ , $f'(x)$ .
$f'(x) = \frac{d}{dx}(8 \tan(x))$

STEP 5

Use the derivative rule for $\tan(x)$ , which is $\frac{d}{dx}(\tan(x)) = \sec^2(x)$ , to find $f'(x)$ .
$f'(x) = 8 \sec^2(x)$

STEP 6

Now, find the value of $f'(x)$ at $a = 0$ .
$f'(0) = 8 \sec^2(0)$

STEP 7

Calculate $f'(0)$ .
$f'(0) = 8 \cdot 1 = 8$

STEP 8

Next, we need to find the second derivative of $f(x)$ , $f''(x)$ .
$f''(x) = \frac{d}{dx}(8 \sec^2(x))$

STEP 9

Use the derivative rule for $\sec^2(x)$ , which is $\frac{d}{dx}(\sec^2(x)) = 2\sec^2(x)\tan(x)$ , to find $f''(x)$ .
$f''(x) = 16 \sec^2(x) \tan(x)$

STEP 10

Now, find the value of $f''(x)$ at $a = 0$ .
$f''(0) = 16 \sec^2(0) \tan(0)$

STEP 11

Calculate $f''(0)$ .
$f''(0) = 16 \cdot 1 \cdot 0 = 0$

STEP 12

Next, we need to find the third derivative of $f(x)$ , $f'''(x)$ .
$f'''(x) = \frac{d}{dx}(16 \sec^2(x) \tan(x))$

STEP 13

Use the product rule and the derivative rules for $\sec^2(x)$ and $\tan(x)$ to find $f'''(x)$ .
$f'''(x) = 16 \left( \frac{d}{dx}(\sec^2(x)) \tan(x) + \sec^2(x) \frac{d}{dx}(\tan(x)) \right)$

STEP 14

Apply the derivative rules from previous steps.
$f'''(x) = 16 \left( 2\sec^2(x)\tan(x) \tan(x) + \sec^2(x) \sec^2(x) \right)$

STEP 15

Simplify the expression for $f'''(x)$ .
$f'''(x) = 16 \left( 2\sec^2(x)\tan^2(x) + \sec^4(x) \right)$

STEP 16

Now, find the value of $f'''(x)$ at $a = 0$ .
$f'''(0) = 16 \left( 2\sec^2(0)\tan^2(0) + \sec^4(0) \right)$

STEP 17

Calculate $f'''(0)$ .
$f'''(0) = 16 \left( 2 \cdot 1 \cdot 0 + 1 \right) = 16$

STEP 18

Now, we can construct the Taylor polynomial $T_{2}(x)$ using the values we found for $f(0)$ , $f'(0)$ , and $f''(0)$ .
$T_{2}(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2$

STEP 19

Plug in the values into $T_{2}(x)$ .
$T_{2}(x) = 0 + \frac{8}{1}x + \frac{0}{2}x^2$

STEP 20

Simplify $T_{2}(x)$ .
$T_{2}(x) = 8x$

STEP 21

Next, we can construct the Taylor polynomial $T_{3}(x)$ using the values we found for $f(0)$ , $f'(0)$ , $f''(0)$ , and $f'''(0)$ .
$T_{3}(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3$

STEP 22

Plug in the values into $T_{3}(x)$ .
$T_{3}(x) = 0 + \frac{8}{1}x + \frac{0}{2}x^2 + \frac{16}{6}x^3$

STEP 23

Simplify $T_{3}(x)$ .
$T_{3}(x) = 8x + \frac{16}{6}x^3$

STEP 24

Further simplify $T_{3}(x)$ by reducing the fraction.
$T_{3}(x) = 8x + \frac{8}{3}x^3$
The Taylor polynomials centered at $a=0$ for the function $f(x)=8 \tan (x)$ are: $T_{2}(x)= 8x$ $T_{3}(x)= 8x + \frac{8}{3}x^3$

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Calculate the 2nd and 3rd order Taylor polynomials for f(x)=8tan⁡(x)f(x) = 8 \tan(x)f(x)=8tan(x) centered at a=0a = 0a=0. Express the polynomials using symbolic notation and fractions.

STEP 1

STEP 2

First, we need to find the value of the function f(x)f(x)f(x) at a=0a = 0a=0.f(0)=8tan⁡(0) f(0) = 8 \tan(0) f(0)=8tan(0)

STEP 3

Calculate f(0)f(0)f(0).f(0)=8⋅0=0 f(0) = 8 \cdot 0 = 0 f(0)=8⋅0=0

STEP 4

Next, we need to find the first derivative of f(x)f(x)f(x), f′(x)f'(x)f′(x).f′(x)=ddx(8tan⁡(x)) f'(x) = \frac{d}{dx}(8 \tan(x)) f′(x)=dxd​(8tan(x))

STEP 5

Use the derivative rule for tan⁡(x)\tan(x)tan(x), which is ddx(tan⁡(x))=sec⁡2(x)\frac{d}{dx}(\tan(x)) = \sec^2(x)dxd​(tan(x))=sec2(x), to find f′(x)f'(x)f′(x).f′(x)=8sec⁡2(x) f'(x) = 8 \sec^2(x) f′(x)=8sec2(x)

STEP 6

Now, find the value of f′(x)f'(x)f′(x) at a=0a = 0a=0.f′(0)=8sec⁡2(0) f'(0) = 8 \sec^2(0) f′(0)=8sec2(0)

STEP 7

Calculate f′(0)f'(0)f′(0).f′(0)=8⋅1=8 f'(0) = 8 \cdot 1 = 8 f′(0)=8⋅1=8

STEP 8

Next, we need to find the second derivative of f(x)f(x)f(x), f′′(x)f''(x)f′′(x).f′′(x)=ddx(8sec⁡2(x)) f''(x) = \frac{d}{dx}(8 \sec^2(x)) f′′(x)=dxd​(8sec2(x))

STEP 9

STEP 10

Now, find the value of f′′(x)f''(x)f′′(x) at a=0a = 0a=0.f′′(0)=16sec⁡2(0)tan⁡(0) f''(0) = 16 \sec^2(0) \tan(0) f′′(0)=16sec2(0)tan(0)

STEP 11

Calculate f′′(0)f''(0)f′′(0).f′′(0)=16⋅1⋅0=0 f''(0) = 16 \cdot 1 \cdot 0 = 0 f′′(0)=16⋅1⋅0=0

STEP 12

Next, we need to find the third derivative of f(x)f(x)f(x), f′′′(x)f'''(x)f′′′(x).f′′′(x)=ddx(16sec⁡2(x)tan⁡(x)) f'''(x) = \frac{d}{dx}(16 \sec^2(x) \tan(x)) f′′′(x)=dxd​(16sec2(x)tan(x))

STEP 13

STEP 14

Apply the derivative rules from previous steps.f′′′(x)=16(2sec⁡2(x)tan⁡(x)tan⁡(x)+sec⁡2(x)sec⁡2(x)) f'''(x) = 16 \left( 2\sec^2(x)\tan(x) \tan(x) + \sec^2(x) \sec^2(x) \right) f′′′(x)=16(2sec2(x)tan(x)tan(x)+sec2(x)sec2(x))

STEP 15

Simplify the expression for f′′′(x)f'''(x)f′′′(x).f′′′(x)=16(2sec⁡2(x)tan⁡2(x)+sec⁡4(x)) f'''(x) = 16 \left( 2\sec^2(x)\tan^2(x) + \sec^4(x) \right) f′′′(x)=16(2sec2(x)tan2(x)+sec4(x))

STEP 16

Now, find the value of f′′′(x)f'''(x)f′′′(x) at a=0a = 0a=0.f′′′(0)=16(2sec⁡2(0)tan⁡2(0)+sec⁡4(0)) f'''(0) = 16 \left( 2\sec^2(0)\tan^2(0) + \sec^4(0) \right) f′′′(0)=16(2sec2(0)tan2(0)+sec4(0))

STEP 17

Calculate f′′′(0)f'''(0)f′′′(0).f′′′(0)=16(2⋅1⋅0+1)=16 f'''(0) = 16 \left( 2 \cdot 1 \cdot 0 + 1 \right) = 16 f′′′(0)=16(2⋅1⋅0+1)=16

STEP 18

STEP 19

Plug in the values into T2(x)T_{2}(x)T2​(x).T2(x)=0+81x+02x2 T_{2}(x) = 0 + \frac{8}{1}x + \frac{0}{2}x^2 T2​(x)=0+18​x+20​x2

STEP 20

Simplify T2(x)T_{2}(x)T2​(x).T2(x)=8x T_{2}(x) = 8x T2​(x)=8x

STEP 21

STEP 22

Plug in the values into T3(x)T_{3}(x)T3​(x).T3(x)=0+81x+02x2+166x3 T_{3}(x) = 0 + \frac{8}{1}x + \frac{0}{2}x^2 + \frac{16}{6}x^3 T3​(x)=0+18​x+20​x2+616​x3

STEP 23

Simplify T3(x)T_{3}(x)T3​(x).T3(x)=8x+166x3 T_{3}(x) = 8x + \frac{16}{6}x^3 T3​(x)=8x+616​x3

STEP 24

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Calculate the 2nd and 3rd order Taylor polynomials for $f(x) = 8 \tan(x)$ centered at $a = 0$ . Express the polynomials using symbolic notation and fractions.

First, we need to find the value of the function $f(x)$ at $a = 0$ .
$f(0) = 8 \tan(0)$

Calculate $f(0)$ .
$f(0) = 8 \cdot 0 = 0$

Next, we need to find the first derivative of $f(x)$ , $f'(x)$ .
$f'(x) = \frac{d}{dx}(8 \tan(x))$

Use the derivative rule for $\tan(x)$ , which is $\frac{d}{dx}(\tan(x)) = \sec^2(x)$ , to find $f'(x)$ .
$f'(x) = 8 \sec^2(x)$

Now, find the value of $f'(x)$ at $a = 0$ .
$f'(0) = 8 \sec^2(0)$

Calculate $f'(0)$ .
$f'(0) = 8 \cdot 1 = 8$

Next, we need to find the second derivative of $f(x)$ , $f''(x)$ .
$f''(x) = \frac{d}{dx}(8 \sec^2(x))$

Now, find the value of $f''(x)$ at $a = 0$ .
$f''(0) = 16 \sec^2(0) \tan(0)$

Calculate $f''(0)$ .
$f''(0) = 16 \cdot 1 \cdot 0 = 0$

Next, we need to find the third derivative of $f(x)$ , $f'''(x)$ .
$f'''(x) = \frac{d}{dx}(16 \sec^2(x) \tan(x))$

Apply the derivative rules from previous steps.
$f'''(x) = 16 \left( 2\sec^2(x)\tan(x) \tan(x) + \sec^2(x) \sec^2(x) \right)$

Simplify the expression for $f'''(x)$ .
$f'''(x) = 16 \left( 2\sec^2(x)\tan^2(x) + \sec^4(x) \right)$

Now, find the value of $f'''(x)$ at $a = 0$ .
$f'''(0) = 16 \left( 2\sec^2(0)\tan^2(0) + \sec^4(0) \right)$

Calculate $f'''(0)$ .
$f'''(0) = 16 \left( 2 \cdot 1 \cdot 0 + 1 \right) = 16$

Plug in the values into $T_{2}(x)$ .
$T_{2}(x) = 0 + \frac{8}{1}x + \frac{0}{2}x^2$

Simplify $T_{2}(x)$ .
$T_{2}(x) = 8x$

Plug in the values into $T_{3}(x)$ .
$T_{3}(x) = 0 + \frac{8}{1}x + \frac{0}{2}x^2 + \frac{16}{6}x^3$

Simplify $T_{3}(x)$ .
$T_{3}(x) = 8x + \frac{16}{6}x^3$