MATH 417 Lecture 8

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Quiz

Find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_4(x)} interpolating polynomial of $2\cos {(\pi \,x)}+6x+14$ for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \left\{ 0, 1, 2, 3 \right\}}

The Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6x + 14} can be omitted for the computation of the and added at the end

Chapter 4: Numerical Integration and Differentiation

Suppose we wanted to find the length of the curve Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \cos{x}} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_a^b \sqrt{1 + \sin^2{x}} \,\mathrm{d}x} is an elliptic integral that is impossible to find.

We could use an interpolating polynomial Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_n(x)} such that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = p_n(x) + \frac{f^{(n+1)}(\xi(x))}{(n+1)!} \, \prod_{i=0}^n \left( x-x_i \right)}

In which case we can approximate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x_j) = p_n'(x_j) + \frac{f^{(n+1)}(\xi)}{(n+1)!} \prod_{i \ne j} \left( x_j - x_i \right)}

We find that the error between two points is approximately linear in regards to the distance between them:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x_0) - \frac{f(x_0 + h) - f(x_0)}{h} = \frac{h}{2} \, f''(\xi) \in O(h)}

Example

Let's differentiate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_2(x) = f(x_0) \, \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} + f(x_1) \, \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x-x_2)} + f(x_2) \, \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}} :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_2'(x) = \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} \, \left( x-x_2 + x-x_1 \right) + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)} \, \left( x - x_2 + x-x_0 \right) + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)} \, \left( x - x_1 + x-x_0 \right)}

If the points are equidistant at distance Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_2'(x) = \frac{-3f(x_0) + 4f(x_0+h) - f(x_0+2h)}{2h}}

To estimate the error, we use

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x_0) - \frac{-3f_0 + 4f_1 - f_2}{2h} = \frac{f'''(\xi)}{6} h \cdot 2h = \frac{h^2}{3} \, f'''(\xi) = O(h^2)}

The moral of the story: For 3 equidistant points on a parabola, the slope of the tangent line at the middle point is equal to the slope of the secant line of the first and last points.