§8.2 Taylor polynomials (continued)

Taylor polynomial approximations centered at an arbitrary value

Summary

For $f(x) = \ln(x)$, the derivatives at $a = 1$ begin $f'(1) = 1$, $f''(1) = -1$, and $f'''(1)$ equals

For $f(x) = \ln(x)$ centered at $a = 1$, the derivatives give $f'(1) = 1$ and $f''(1) = -1$. The first two coefficients of $T_4$ are

For $T_4(x) = (x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4$ approximating $\ln(x)$, the condition $|f-T_4| \lt 0.1$ holds approximately on

For $\ln(x)$ centered at $1$, moving from $T_5$ to $T_6$ shifts the sub-tenth window from about $(0.24,1.999)$ to $(0.21,2)$. Unlike cosine, the window

Centering the $\ln(x)$ approximation at $a = 2$ instead of $1$ produces $T_4$ beginning

For the degree $4$ polynomial of $\ln(x)$ centered at $a = 2$, the tolerance $|f-T_4| \lt 0.1$ holds roughly on

Centering the sine expansion at $a = \frac{\pi}{2}$ produces

For the quadratic $f(x) = \frac{1}{2}x^2-2x+5$ centered at $a = 2$, the degree $2$ Taylor polynomial reads $T_2(x) = 3+\frac{1}{2}(x-2)^2$. This form reveals

A function satisfies $f(0) = 2$, $f'(0) = -3$, $f''(0) = -1$, $f'''(0) = 0$, $f^{(4)}(0) = -3$, giving $T_4(x) = 2-3x-\frac{1}{2}x^2-\frac{3}{24}x^4$. Estimate $f(0.5)$ from $T_4(0.5)$, rounded to three decimals.

For the quadratic $f(x) = \frac{1}{2}x^2-2x+5$ centered at $a = 2$, we have $T_2(x) = 3+\frac{1}{2}(x-2)^2$. Compute $T_2(4)$.

For $f(x) = \ln(x)$, the first derivative at the center $a = 2$ serves as the linear coefficient source. Compute $f'(2)$ as a decimal.

The Taylor series of $\ln(x)$ centered at $1$ is obtained from the series of $\ln(1+x)$ centered at $0$ by replacing $x$ with $x-1$.

Whether centered at $1$ or at $2$, the degree $4$ polynomial approximates $\ln(x)$ far more accurately, and on a wider interval, than the tangent line $T_1$.