Solved on Feb 28, 2024

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

STEP 1

Assumptions
1. The function given is f(x)=8tan(x)f(x) = 8 \tan(x).
2. The Taylor polynomials T2T_{2} and T3T_{3} are to be centered at a=0a = 0.
3. The Taylor polynomial of degree nn for a function f(x)f(x) centered at aa is given by: Tn(x)=k=0nf(k)(a)k!(xa)k 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)f(x) to find T2T_{2} and T3T_{3}.

STEP 2

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

STEP 3

Calculate f(0)f(0).
f(0)=80=0 f(0) = 8 \cdot 0 = 0

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

Calculate f(0)f'(0).
f(0)=81=8 f'(0) = 8 \cdot 1 = 8

STEP 8

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

STEP 9

Use the derivative rule for sec2(x)\sec^2(x), which is ddx(sec2(x))=2sec2(x)tan(x)\frac{d}{dx}(\sec^2(x)) = 2\sec^2(x)\tan(x), to find f(x)f''(x).
f(x)=16sec2(x)tan(x) f''(x) = 16 \sec^2(x) \tan(x)

STEP 10

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

STEP 11

Calculate f(0)f''(0).
f(0)=1610=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)f'''(x).
f(x)=ddx(16sec2(x)tan(x)) f'''(x) = \frac{d}{dx}(16 \sec^2(x) \tan(x))

STEP 13

Use the product rule and the derivative rules for sec2(x)\sec^2(x) and tan(x)\tan(x) to find f(x)f'''(x).
f(x)=16(ddx(sec2(x))tan(x)+sec2(x)ddx(tan(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(2sec2(x)tan(x)tan(x)+sec2(x)sec2(x)) 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).
f(x)=16(2sec2(x)tan2(x)+sec4(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)f'''(x) at a=0a = 0.
f(0)=16(2sec2(0)tan2(0)+sec4(0)) f'''(0) = 16 \left( 2\sec^2(0)\tan^2(0) + \sec^4(0) \right)

STEP 17

Calculate f(0)f'''(0).
f(0)=16(210+1)=16 f'''(0) = 16 \left( 2 \cdot 1 \cdot 0 + 1 \right) = 16

STEP 18

Now, we can construct the Taylor polynomial T2(x)T_{2}(x) using the values we found for f(0)f(0), f(0)f'(0), and f(0)f''(0).
T2(x)=f(0)+f(0)1!x+f(0)2!x2 T_{2}(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2

STEP 19

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

STEP 20

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

STEP 21

Next, we can construct the Taylor polynomial T3(x)T_{3}(x) using the values we found for f(0)f(0), f(0)f'(0), f(0)f''(0), and f(0)f'''(0).
T3(x)=f(0)+f(0)1!x+f(0)2!x2+f(0)3!x3 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 T3(x)T_{3}(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

STEP 23

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

STEP 24

Further simplify T3(x)T_{3}(x) by reducing the fraction.
T3(x)=8x+83x3 T_{3}(x) = 8x + \frac{8}{3}x^3
The Taylor polynomials centered at a=0a=0 for the function f(x)=8tan(x)f(x)=8 \tan (x) are: T2(x)=8x T_{2}(x)= 8x T3(x)=8x+83x3 T_{3}(x)= 8x + \frac{8}{3}x^3

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