Homework 4.9 (Homework)

Q1.

Find the Taylor polynomial Tn(x)T_n(x) for the given function at the number aa

f(x)=x2−e−4xf(x) = x² -e^{-4x}    a=0a=0        n=3n=3

T_3(x) = ___________

Taylor polynomial formula:

  • Tn(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f(3)(a)3!(x−a)3T_n(x) = f(a) + f^{\prime}(a)(x-a) + \frac {f^{\prime\prime}(a)}{2!}(x-a)²+\frac {f^{(3)}(a)} {3!} (x-a)³

When a=0a=0, this simplifies to:

  • T3(x)=f(0)+f′(0)x+f′′(0)2!x2+f(3)(0)3!x3T_3(x) = f(0) + f^\prime(0)x + \frac {f^{\prime\prime}(0)}{2!}x²+\frac {f^{(3)}(0)} {3!}x³

f(x)=x2−e−4xf(x) = x²-e^{-4x}
f′(x)=2x+4e−4xf^\prime(x) = 2x +4e^{-4x}

f′′(x)=2−16e−4xf^{\prime\prime} (x) = 2 - 16e^{-4x}

f(3)(x)=64e−4xf^{(3)}(x) = 64e^{-4x}

f(0)=02−e−4(0)=−1f(0) = 0² - e^{-4(0)} = -1

f′(0)=2(0)+4e0=4f^\prime(0) = 2(0) + 4e^0 = 4

f′′(0)=2−16e0=−14f^{\prime\prime} (0) = 2 - 16e^0 = -14

f(3)=64e0=64f^{(3)} = 64e^0 = 64

T3(x)=(−1)+(4)x+−142x2+646x3T_3(x) = (-1) + (4)x + \frac {-14}2x² + \frac {64}6 x³

T3(x)=−1+4x−7x2+323x3T_3(x) = -1 + 4x - 7x²+\frac {32}3x³

Q2.

Find the Taylor polynomial Tn(x)T_n(x) for the given function at the number aa.

f(x)=sin⁡(2x)f(x) = \sin(2x)        a=π4a=\frac {\pi}4        n=4n=4

T4(x)T_4(x)=___________

  • T_{n}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^{2}+\frac{f^{(3)}(a)} {3!}(x-a)^3+ \frac {f^{(4)}(a)}{4!}(x-a)^4$

When a=\frac \pi4</p><p></p><p>f(\frac \pi 4) + f^\prime(\frac \pi 4)(x-\frac \pi 4) + \frac {f^{\prime\prime}(\frac \pi 4)} {2!} (x-\frac \pi 4)² + \frac {f^{(3)}(\frac \pi 4)}{3!} (x-\frac \pi 4)³ + \frac {f^{(4)}(\frac \pi 4)}{4!}(x-\frac \pi 4)^4</p><p></p><p></p><p></p><p>f(x) = \sin (2x)</p><p></p><p>f^\prime (x)= 2\cos(2x)</p><p></p><p>f^{\prime\prime} (x) = -4\sin(2x)</p><p></p><p>f^{(3)} (x) = -8\cos(2x)</p><p></p><p>f^{(4)} (x) = 16\sin(2x)</p><p></p><p></p><p></p><p>f(\frac \pi 4) = \sin(\frac \pi 2)= 1</p><p></p><p>f^\prime (\frac \pi 4) = 2\cos(\frac \pi 2) = 0</p><p></p><p>f^{\prime\prime} (\frac \pi 4) = -4 \sin (\frac \pi 2) = -4</p><p></p><p>f^{(3)} (\frac \pi 4) = -8\cos(\frac \pi 2) = 0</p><p></p><p>f^{(4)}(\frac \pi 4) = 16\sin(\frac \pi 2) = 16</p><p></p><p></p><p></p><p>T_4(x) = 1 + \frac {-4}{2!}(x-\frac \pi 4)²+\frac{16}{4!}(x-\frac \pi 4)^4</p><p></p><p>T_4(x) = 1 -2(x-\frac \pi 4)² + \frac 23 (x-\frac \pi 4)^4</p><divdata−type="horizontalRule"><hr></div><p>Q3.</p><p>FindtheTaylorpolynomial</p><div data-type="horizontalRule"><hr></div><p>Q3.</p><p>Find the Taylor polynomialT_n(x) for the given function at the number a.</p><p></p><p>.</p><p></p><p>f(x) = \tan ^{-1}(2x)        a=\frac 12        n=3</p><p></p><p></p><p></p><p>f(x) = \tan^{-1}(2x)</p><p></p><p>f^\prime (x) = \frac 2{1+4x²}</p><p></p><p>f^{\prime\prime}(x)= -\frac {16x}{(1+4x²)²}</p><p></p><p>f^{(3)}(x) = -\frac {16}{(1+4x²)²} + \frac {(-16x)²}{(1+4x²)³}</p><p><br></p><p></p><p><br></p><p>f(\frac 12) = \tan^{-1} (2 \cdot \frac 12) = \frac \pi4</p><p></p><p>f^\prime (\frac 12) = \frac 2{1 + 4(\frac 12)²} = 1</p><p></p><p>f^{\prime\prime}(x) = -\frac {16(\frac 12)}{(1+4(\frac 12)²)²}= -2</p><p></p><p>f^{(3)}(x) = -\frac {16}{(1+4(\frac 12)²)²} + \frac {(-16(\frac 12))²}{(1+4(\frac 12)²)³}= 4</p><p></p><p></p><p></p><p>T_3(x) = \frac \pi 4+ (x-\frac 12) - \frac 2{2!}(x - \frac 12)² +\frac 4{3!}(x-\frac 12)³</p><divdata−type="horizontalRule"><hr></div><p>Find</p><div data-type="horizontalRule"><hr></div><p>FindT_n(x) for the given function at the number a.

f(x) = e^{4x}        a=0        n=3        0\le x\le 0.3</p><p></p><p></p><p></p><p>f(x) = e^{4x}</p><p></p><p>f^\prime (x) =4e^{4x}</p><p></p><p>f^{\prime\prime} (x) = 16e^{4x}</p><p></p><p>f^{(3)} (x)= 64e^{4x}</p><p></p><p></p><p></p><p>f(0) = e^{4(0)} = 1</p><p></p><p>f^\prime (0) = 4</p><p></p><p>f^{\prime\prime} (0) = 16</p><p></p><p>f^{(3)} (0)= 64</p><p></p><p></p><p></p><p>T_3 (x) = 1 + 4x + \frac {16}{2!} x² + \frac {64}{3!} x³</p><p></p><p>T_3 (x) = 1 + 4x + 8x² + \frac {32}3 x³$$