Polynomials #4 - Questions and Answers

Question 1 Suppose you want to estimate 26 26 ​ using a fourth-order Taylor polynomial centered at 𝑥 = 𝑎 x=a for 𝑓 ( 𝑥 ) = 𝑥 f(x)= x ​ . Choose an appropriate value for the center 𝑎 a. Answer: Given Function: 𝑓 ( 𝑥 ) = 𝑥 . f(x)= x ​ . Taylor Polynomial Formula (nth-order): 𝑃 𝑛 ( 𝑥 ) = 𝑓 ( 𝑎 ) + 𝑓 ′ ( 𝑎 ) ( 𝑥 − 𝑎 ) + 𝑓 ′ ′ ( 𝑎 ) 2 ! ( 𝑥 − 𝑎 ) 2 + 𝑓 ( 3 ) ( 𝑎 ) 3 ! ( 𝑥 − 𝑎 ) 3 + 𝑓 ( 4 ) ( 𝑎 ) 4 ! ( 𝑥 − 𝑎 ) 4 . P n ​ (x)=f(a)+f ′ (a)(x−a)+ 2! f ′′ (a) ​ (x−a) 2 + 3! f (3) (a) ​ (x−a) 3 + 4! f (4) (a) ​ (x−a) 4 . Center Choice: Choose 𝑎 = 25 a=25 because it is a perfect square close to 26 26. **Compute 𝑓 ( 𝑎 ) f(a), 𝑓 ′ ( 𝑎 ) f ′ (a), 𝑓 ′ ′ ( 𝑎 ) f ′′ (a), 𝑓 ( 3 ) ( 𝑎 ) f (3) (a), and 𝑓 ( 4 ) ( 𝑎 ) f (4) (a): 𝑓 ( 𝑎 ) = 25 = 5. f(a)= 25 ​ =5. 𝑓 ′ ( 𝑥 ) = 1 2 𝑥 ,   𝑓 ′ ( 25 ) = 1 2 × 5 = 1 10 . f ′ (x)= 2 x ​ 1 ​ ,f ′ (25)= 2×5 1 ​ = 10 1 ​ . 𝑓 ′ ′ ( 𝑥 ) = − 1 4 𝑥 3 / 2 ,   𝑓 ′ ′ ( 25 ) = − 1 4 × 125 = − 1 500 . f ′′ (x)=− 4x 3/2 1 ​ ,f ′′ (25)=− 4×125 1 ​ =− 500 1 ​ . 𝑓 ( 3 ) ( 𝑥 ) = 3 8 𝑥 5 / 2 ,   𝑓 ( 3 ) ( 25 ) = 3 8 × 3125 = 3 25000 . f (3) (x)= 8x 5/2 3 ​ ,f (3) (25)= 8×3125 3 ​ = 25000 3 ​ . 𝑓 ( 4 ) ( 𝑥 ) = − 15 16 𝑥 7 / 2 ,   𝑓 ( 4 ) ( 25 ) = − 15 16 × 78125 = − 15 1250000 = − 1 83333.33 . f (4) (x)=− 16x 7/2 15 ​ ,f (4) (25)=− 16×78125 15 ​ =− 1250000 15 ​ =− 83333.33 1 ​ . Substitute into the Taylor Polynomial: 𝑃 4 ( 𝑥 ) = 𝑓 ( 25 ) + 𝑓 ′ ( 25 ) ( 𝑥 − 25 ) + 𝑓 ′ ′ ( 25 ) 2 ! ( 𝑥 − 25 ) 2 + 𝑓 ( 3 ) ( 25 ) 3 ! ( 𝑥 − 25 ) 3 + 𝑓 ( 4 ) ( 25 ) 4 ! ( 𝑥 − 25 ) 4 . P 4 ​ (x)=f(25)+f ′ (25)(x−25)+ 2! f ′′ (25) ​ (x−25) 2 + 3! f (3) (25) ​ (x−25) 3 + 4! f (4) (25) ​ (x−25) 4 . Substitute the values: 𝑃 4 ( 𝑥 ) = 5 + 1 10 ( 𝑥 − 25 ) − 1 1000 ( 𝑥 − 25 ) 2 + 1 75000 ( 𝑥 − 25 ) 3 − 1 3333333.33 ( 𝑥 − 25 ) 4 . P 4 ​ (x)=5+ 10 1 ​ (x−25)− 1000 1 ​ (x−25) 2 + 75000 1 ​ (x−25) 3 − 3333333.33 1 ​ (x−25) 4 . Estimate 26 26 ​ : Substitute 𝑥 = 26 x=26: 𝑃 4 ( 26 ) = 5 + 1 10 ( 26 − 25 ) − 1 1000 ( 26 − 25 ) 2 + 1 75000 ( 26 − 25 ) 3 − 1 3333333.33 ( 26 − 25 ) 4 . P 4 ​ (26)=5+ 10 1 ​ (26−25)− 1000 1 ​ (26−25) 2 + 75000 1 ​ (26−25) 3 − 3333333.33 1 ​ (26−25) 4 . 𝑃 4 ( 26 ) = 5 + 0.1 − 0.001 + 0.0000133 − 0.0000003. P 4 ​ (26)=5+0.1−0.001+0.0000133−0.0000003. 𝑃 4 ( 26 ) ≈ 5.099. P 4 ​ (26)≈5.099. Question 2 One of the zeros is given for each of the following polynomials. Find the other zeros in the field of complex numbers. 𝑥 3 − 4 𝑥 2 + 6 𝑥 − 4 , zero:  1 − 𝑖 . x 3 −4x 2 +6x−4,zero: 1−i. Answer: Given zero: 1 − 𝑖 . 1−i. Using the Complex Conjugate Theorem: If 1 − 𝑖 1−i is a zero, then 1 + 𝑖 1+i is also a zero. Find the third zero: Use synthetic division to divide 𝑥 3 − 4 𝑥 2 + 6 𝑥 − 4 x 3 −4x 2 +6x−4 by ( 𝑥 − ( 1 − 𝑖 ) ) ( 𝑥 − ( 1 + 𝑖 ) ) = ( 𝑥 − 1 ) 2 + 1 (x−(1−i))(x−(1+i))=(x−1) 2 +1: ( 𝑥 − 1 ) 2 + 1 = 𝑥 2 − 2 𝑥 + 2. (x−1) 2 +1=x 2 −2x+2. Perform polynomial division: 𝑥 3 − 4 𝑥 2 + 6 𝑥 − 4 𝑥 2 − 2 𝑥 + 2 . x 2 −2x+2 x 3 −4x 2 +6x−4 ​ . Using synthetic or long division: Quotient = 𝑥 − 2. x−2. The third zero is: 𝑥 = 2. x=2. Final Answer: The zeros are: 1 − 𝑖 ,   1 + 𝑖 ,   2. 1−i,1+i,2. Taylor Approximation Evaluation (from the provided image): For 𝑇 3 ( 𝑥 ) = 1 + 𝑥 + 𝑥 2 2 ! + 𝑥 3 3 ! T 3 ​ (x)=1+x+ 2! x 2 ​ + 3! x 3 ​ , substitute 𝑥 = 0.12 x=0.12: Expand and Evaluate: 𝑇 3 ( 0.12 ) = 1 + 0.12 + ( 0.12 ) 2 2 + ( 0.12 ) 3 6 . T 3 ​ (0.12)=1+0.12+ 2 (0.12) 2 ​ + 6 (0.12) 3 ​ . 𝑇 3 ( 0.12 ) = 1 + 0.12 + 0.0072 + 0.000288. T 3 ​ (0.12)=1+0.12+0.0072+0.000288. 𝑇 3 ( 0.12 ) ≈ 1.127488. T 3 ​ (0.12)≈1.127488.