Lagrange Error Bound - AP Calculus BC

$R_1(0.1) = \frac{M(0.1)^2}{2}$. The second derivative of $\ln(1+x)$ gives the maximum $M$.

$M = e^2$. The function $e^x$ is increasing, so maximum occurs at $x = 2$.

Approaches zero, improving approximation. Higher-degree polynomials provide increasingly accurate approximations.

$R_2(0.5) = \frac{e^{0.5} (0.5)^3}{3!}$. Since $f'''(x) = e^x$, the maximum on $[0, 0.5]$ is $e^{0.5}$.

It must be continuous on the interval. Continuity ensures the maximum value $M$ exists.

$R_1(0.5) = \frac{M(0.1)^2}{2}$. Distance is $|0.5 - 0.4| = 0.1$ from center $a = 0.4$.

$R_1(0.1) = \frac{M(0.1)^2}{2}$. The second derivative of $x^2$ is constant 2, so $M = 2$.

Maximum value of $|f^{(n+1)}(c)|$ on the interval. This is the maximum absolute value of the $(n+1)$-th derivative.

Estimates the error of a Taylor polynomial approximation. This provides an upper bound on the approximation error.

$R_2(0.1) = \frac{M(0.1)^3}{3!}$. The third derivative of $\sin(x)$ is $-\cos(x)$, with maximum 1.

$M = \frac{1}{1}$. For $\ln(x)$, the second derivative $-\frac{1}{x^2}$ has maximum at $x = 1$.

$M = 1$. All derivatives of $\cos(x)$ have absolute value at most 1.

Differentiable function. The function must have enough derivatives for the error bound.

$R_n(x) = \frac{M|x-a|^{n+1}}{(n+1)!}$. This bounds the error between the function and its Taylor polynomial.

The derivative $f^{(n+1)}$ over the interval. We need the maximum of $|f^{(n+1)}(x)|$ over the interval.

$c$ is some value in the interval between $a$ and $x$. This is where the $(n+1)$-th derivative is evaluated in the error formula.

The $(n+1)$-th derivative of the function. This derivative determines the maximum value $M$ in the error bound.

Degree of the Taylor polynomial. This determines how many terms are in the Taylor polynomial.

$R_3(1.5) = \frac{M(0.5)^4}{4!}$. Distance is $|1.5 - 1| = 0.5$ from the center $a = 1$.

Error term in Taylor approximation. This measures how much the polynomial differs from the function.

$x$ must be close to $a$ for best approximation. Closer values to $a$ give smaller error bounds.

$R_3(0.1) = \frac{M(0.1)^4}{4!}$. The fourth derivative of $\sin(x)$ is $\sin(x)$, with maximum 1.

Denominator scaling factor in error term. The factorial grows rapidly, making higher-order errors much smaller.

$R_3(0.1) = \frac{M(0.1)^4}{4!}$. The fourth derivative of $x^4$ is 24, which is constant.

$R_2(0.1) = \frac{M(0.1)^3}{3!}$. The third derivative of $x^3$ is 6, which is constant.

The remainder or error term of Taylor polynomial. This represents the difference between $f(x)$ and its polynomial approximation.

The error of a Taylor polynomial approximation. This gives the maximum possible difference from the true function value.

$x$ must be within the interval where $M$ is maximum. This ensures $M$ is well-defined on the interval containing $x$ and $a$.

Factorial of $(n+1)$ in the denominator of the error term. This factorial makes the error bound decrease rapidly as $n$ increases.