Question:

Find the 3rd Taylor Polynomial of
𝑓
(
𝑥
)
=
cos
𝑥
at
𝑥
=
𝜋
4
.

Solution steps:

Write a table similar to this on a sheet of paper. Fill it out with information regarding the function in the question.

𝑛
:
𝑛
th derivative: Evaluated at
𝑥
=
𝑎
: Term in Taylor Polynomial:
0
𝑓
(
𝑥
)
=

𝑓
(
𝑎
)
=

𝑓
(
𝑎
)
=

1
𝑓
′
(
𝑥
)
=

𝑓
′
(
𝑎
)
=

𝑓
′
(
𝑎
)
(
𝑥
−
𝑎
)
=

2
𝑓
″
(
𝑥
)
=

𝑓
″
(
𝑎
)
=

𝑓
″
(
𝑎
)
2
!
(
𝑥
−
𝑎
)
2
=

3
𝑓
‴
(
𝑥
)
=

𝑓
‴
(
𝑎
)
=

𝑓
‴
(
𝑎
)
3
!
(
𝑥
−
𝑎
)
3
=

4
𝑓
(
4
)
(
𝑥
)
=

𝑓
(
4
)
(
𝑎
)
=

𝑓
(
4
)
(
𝑎
)
4
!
(
𝑥
−
𝑎
)
4
=


Add the terms in the last column to find the Taylor Polynomial.

Which of the following options is the correct 3rd Taylor Polynomial of
𝑓
(
𝑥
)
=
cos
𝑥
at
𝑥
=
𝜋
4
?

Question

Question:

Find the 3rd Taylor Polynomial of 
𝑓
(
𝑥
)
=
cos
𝑥
  at 
𝑥
=
𝜋
4
.

Solution steps:

Write a table similar to this on a sheet of paper. Fill it out with information regarding the function in the question.

𝑛
: 	
𝑛
th derivative: 	Evaluated at 
𝑥
=
𝑎
: 	Term in Taylor Polynomial:
0	
𝑓
(
𝑥
)
=
 	
𝑓
(
𝑎
)
=
   	
𝑓
(
𝑎
)
=
   
1	
𝑓
′
(
𝑥
)
=
  	
𝑓
′
(
𝑎
)
=
   	
𝑓
′
(
𝑎
)
(
𝑥
−
𝑎
)
=
     
2	
𝑓
″
(
𝑥
)
=
  	
𝑓
″
(
𝑎
)
=
    	
𝑓
″
(
𝑎
)
2
!
(
𝑥
−
𝑎
)
2
=
     
3	
𝑓
‴
(
𝑥
)
=
  	
𝑓
‴
(
𝑎
)
=
     	
𝑓
‴
(
𝑎
)
3
!
(
𝑥
−
𝑎
)
3
=
     
4	
𝑓
(
4
)
(
𝑥
)
=
  	
𝑓
(
4
)
(
𝑎
)
=
    	
𝑓
(
4
)
(
𝑎
)
4
!
(
𝑥
−
𝑎
)
4
=

Add the terms in the last column to find the Taylor Polynomial.

Which of the following options is the correct 3rd Taylor Polynomial of 
𝑓
(
𝑥
)
=
cos
𝑥
  at 
𝑥
=
𝜋
4
?

BlackTom AI · Accepted Answer

We need to find the 3rd Taylor polynomial of f(x) = cos x expanded about a = π/4.

First, identify the Taylor series around a: P3(x) = f(a) + f'(a)(x−a) + f''(a)/2! (x−a)^2 + f'''(a)/3! (x−a)^3, since we want up to the 3rd degree.

Compute the necessary derivatives and evaluate at a = π/4:
- f(x) = cos x, so f(a) = cos(π/4) = √2/2.
- f'(x) = −sin x, so f'(a) = −sin(π/4) = −√2/2.
- f''(x) = −cos x, so f''(a) = −cos(π/4) = −√2/2.
- f'''(x) = sin x, so f'''(a) = sin(π/4) = √2/2.

Substitute these into the Taylor formula with h = x − a:
P3(x) = (√2/2) + (−√2/2) h + (−√2/2)/2 · h^2 + (√2/2)/6 · h^3.

Simplify the coefficients:
- (−√2/2) h = −(√2/2) h.
- (−√2/2)/2 = −√2/4, so the h^2 term is −(√2/4) h^2.
- (√2/2)/6 = √2/12, so the h^3 term is (√2/12) h^3.

Thus, the 3rd Taylor polynomial is
P3(x) = √2/2 − (√2/2)(x − π/4) − (√2/4)(x − π/4)^2 + (√2/12)(x − π/4)^3, with h = x − π/4.

If you compare to any given multiple-choice options, you would expect coefficients that reflect these exact constants multiplied by the appropriate powers of (x − π/4). Any option that uses 1/2, 1/6, or other coefficients not matching the above derivation (and factoring out √2 appropriately) would be inconsistent with the correct Taylor expansion shown here.

Note: In the provided data, the answer choices field appears empty, so there is no selectable option to verify against. The derivation above stands as the correct form for P3 around a = π/4.

类似问题

Find the Taylor polynomial of order 2 for f(x)=−8x3−x2+x−9 around x=0.

Find the Taylor polynomial of order 2 for 𝑓 ( 𝑥 ) = − 7 𝑥 3 + 9 𝑥 2 − 2 𝑥 − 8 around 𝑥 = 0 .

Let [math: P(x)] be the second-order Taylor polynomial of [math: sin⁡x]\sin {x} at the reference point [math: x=π6]x=\dfrac {\pi }{6}. Compute [math: P(1)]. (Correct the answer to 2 decimal places.)

Let [math: P(x)] be the second-order Taylor polynomial of [math: sin⁡x]\sin {x} at the reference point [math: x=π4]x=\dfrac {\pi }{4}. Compute [math: P(2)]. (Correct the answer to 2 decimal places.)

Let 𝑃 2 ( 𝑥 ) be the quadratic Taylor approximation of the function 𝑓 ( 𝑥 ) = 𝑥 𝑒 2 𝑥 about x=0. What is 𝑃 2 ( 1 ) ?

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