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2.03: Numerical Differentiation of Functions at Discrete Data Points

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Lesson 1: Numerical Differentiation of Functions Given as Discrete Data Points - First Derivative

Lesson Objectives

After successful completion of this lesson, you should be able to:

1) choose the most proper scheme among FDD, BDD, and CDD based on order of accuracy and data availability,

2) find numerical values of the first derivatives of functions that are given at discrete data points.

Introduction

To find the derivatives of functions that are given at discrete points, several methods are available. Although these methods are mainly used when the data is spaced unequally, they can be used for data that is spaced equally as well.

We know

f(x)=lim

We developed finite divided difference approximations for the first and second derivatives of continuous functions in a previous lesson. We can apply the same formulas for discrete functions as well. However, some limitations exist, and one should note them before applying any of the approximations.

So given n + 1 data points in ascending order of x values, \left( x_{0},y_{0} \right), \ \left( x_{1},y_{1} \right), \ \left( x_{2},y_{2} \right),\ldots,\left( x_{n},y_{n} \right), the value of f^{\prime}(x_i) by FDD for x_{i} \leq x \leq x_{i + 1}, i = 0,...,n - 1, is given by

f^{\prime}\left( x_{i} \right) \approx \frac{f\left( x_{i + 1} \right) - f\left( x_{i} \right)}{x_{i + 1} - x_{i}}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.1}) \nonumber

the value of f^{\prime}(x_i) by BDD for x_{i - 1} \leq x \leq x_{i}, i = 1,...,n, is given by

f^{\prime}\left( x_{i} \right) \approx \frac{f\left( x_{i} \right) - f\left( x_{i - 1} \right)}{x_{i} - x_{i - 1}}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.2}) \nonumber

the value of f^{\prime}(x_i) by CDD for x_{i - 1} \leq x \leq x_{i + 1}, i = 1,...,n - 1, is given by

f^{\prime}\left( x_{i} \right) \approx \frac{f\left( x_{i + 1} \right) - f\left( x_{i - 1} \right)}{x_{i + 1} - x_{i - 1}}\;\;\;\;\;\;\;\;\;\;\;\;(\PageIndex{1.3}) \nonumber

provided that x_{i + 1} - x_{i} = x_{i} - x_{i - 1}. \nonumber

Example \PageIndex{1.1}

The upward velocity of a rocket is given as a function of time in Table \PageIndex{1.1}.

Table \PageIndex{1.1}. Velocity as a function of time.
t\ \text{(s)} v(t)\ (\text{m/s})
0 0
10 227.04
15 362.78
20 517.35
22.5 602.97
30 901.67

a) Use forward divided difference, if possible, to find the acceleration of the rocket at t = 15\text{ s}.

b) Use backward divided difference, if possible, to find the acceleration of the rocket at t = 15\text{ s}.

c) Use central divided difference, if possible, to find the acceleration of the rocket at t = 15\text{ s}.

Solution

a) To find the acceleration at t = 15\text{ s} with the forward divided difference method, a data point ahead of t = 15\text{ s} should be available. All these conditions are met, and we will use velocity values at t = 15\text{ s} and t = 20\text{ s}

a\left( t_{i} \right) \approx \frac{v\left( t_{i + 1} \right) - v\left( t_{i} \right)}{t_{i+1}-t_i} \nonumber

t_{i} = 15 \nonumber

t_{i + 1} = 20 \nonumber

\begin{split} a\left( 15 \right) &\approx \frac{v\left( 20 \right) - v\left( 15 \right)}{20-15}\\\\ &= \frac{517.35 - 362.78}{5}\\\\ &=30.914\ \text{m/s}^{2}\end{split} \nonumber

b) To find the acceleration at t = 15\text{ s} with the backward divided difference method, a data point behind t = 15 \text{ s} should be available. All these conditions are met, and we will use velocity values at t = 10\text{ s} and t = 15\text{ s}

a\left( t_{i} \right) \approx \frac{v\left( t_{i} \right) - v\left( t_{i - 1} \right)}{t_i-t_{i-1}} \nonumber

t_{i - 1} = 10 \nonumber

t_{i} = 15 \nonumber

\begin{split} a\left( 15 \right) &\approx \frac{v\left( 15 \right) - v\left( 10 \right)}{15-10}\\\\ &= \frac{362.78 - 227.04}{5}\\\\ &=27.148 \text{ m/s}^{2}\end{split} \nonumber

c) To find the acceleration at t = 15\text{ s} with the central divided difference method,

  • a data point behind t = 15\text{ s} should be available,
  • a data point ahead of t = 15\text{ s} should be available, and
  • the spacing should be the same on the time domain from t = 15\text{ s}.

All these conditions are met, and we will use velocity values at t = 10\text{ s} and t = 20\text{ s}

a\left( t_{i} \right) \approx \frac{v\left( t_{i + 1} \right) - v\left( t_{i - 1} \right)}{t_{i+1}-t_{i-1}} \nonumber

t_{i - 1} = 10 \nonumber

t_{i + 1} = 20 \nonumber

\begin{split} a\left( 15 \right) &\approx \frac{v\left( 20 \right) - v\left( 10 \right)}{20-10}\\\\ &= \frac{517.35 - 227.04}{10}\\\\ &={29.031 \text{ m/s}^{2}}\end{split} \nonumber

The values of acceleration at t=15 \text{ s} obtained using the three divided difference schemes is given in Table \PageIndex{1.2}.

Table \PageIndex{1.2}. Estimated acceleration for various divided difference schemes.
Method a(15)\text{ (m/s}^{2})
Forward 30.914
Backward 27.148
Central 29.031

Lesson 2: Numerical Differentiation of Functions Given as Discrete Data Points - Second Derivative

Learning Objectives

After successful completion of this lesson, you should be able to:

1) find approximate values of the second derivatives of functions that are given at discrete data points

Introduction

To find the derivatives of functions that are given at discrete points, several methods are available. Although these methods are mainly used when the data is spaced unequally, they can be used for data spaced equally.

In a previous lesson, we developed finite divided difference approximations for the second derivatives of continuous functions. We can apply the same formulas for discrete functions as well. However, some limitations exist, and one should note them before applying any of the approximations.

So given n + 1 data points in ascending order of x values, \left( x_{0},y_{0} \right), \ \left( x_{1},y_{1} \right), \ \left( x_{2},y_{2} \right),\ldots,\left( x_{n},y_{n} \right),

the value of f^{\prime\prime}(x_{i}) by CDD for x_{i - 1} \leq x_{i} \leq x_{i + 1}, i = 1,\ldots,n - 1, is given by

f^{\prime\prime}\left( x_{i} \right) \approx \frac{f\left( x_{i + 1} \right) - 2f\left( x_{i} \right) + f\left( x_{i - 1} \right)}{(x_{i + 1} - x_{i})^{2}} \nonumber

provided that x_{i + 1} - x_{i} = x_{i} - x_{i - 1}. \nonumber

Example \PageIndex{2.1}

The upward velocity of a rocket is given as a function of time in Table \PageIndex{2.1}.

Table \PageIndex{2.1}. Velocity as a function of time.
t\ (\text{s}) v(t)\ (\text{m/s})
0 0
10 227.04
15 362.78
20 517.35
22.5 602.97
30 901.67

Use central divided difference, if possible, to find the jerk of the rocket at t = 15\ s.

Solution

To find the jerk at t = 15\text{ s} with the central divided difference method,

a data point behind t = 15\text{ s} should be available,

a data point ahead of t = 15\text{ s} should be available, and

the spacing should be the same on the time domain from t = 15\text{ s}.

All these conditions are met, and we will use velocity values at t = 10\text{ s}, t = 15\text{ s}, and t = 20\text{ s}

j\left( t_{i} \right) \approx \frac{v\left( t_{i + 1} \right) - 2v\left( t_{i} \right) + v\left( t_{i - 1} \right)}{(t_{i + 1} - t_{i})^{2}} \nonumber

t_{i - 1} = 10 \nonumber

t_{i} = 15 \nonumber

t_{i + 1} = 20 \nonumber

\begin{split} j\left( 15 \right) &\approx \frac{v\left( 20 \right) - 2v\left( 15 \right) + v\left( 10 \right)}{(20 - 15)^{2}}\\ &= \frac{517.35 - 2 \times 362.78 + 227.04}{(20 - 15)^{2}}\\ &= 0.7532 \ \text{m/s}^{3}\end{split} \nonumber

Lesson 3: Direct Fit Polynomials to Find Derivatives of Functions Given as Discrete Data Points

Learning Objectives

After successful completion of this lesson, you should be able to:

1) find numerical derivatives of functions given at discrete points via the direct method of interpolation

Introduction

To find the derivatives of functions that are given at discrete points, several methods are available. Although the method of divided differences is quick to apply, the need for independent variable data to be equally spaced is a limitation for central divided difference scheme, and for higher-order derivatives. In such cases, polynomial interpolation is a method that can be used.

Vandermonde Polynomial

In this method, given n + 1 data points \left( x_{0},y_{0} \right), \ \left( x_{1},y_{1} \right), \ \left( x_{2},y_{2} \right),\ldots,\left( x_{n},y_{n} \right), one can fit a n^{th} order polynomial given by

P_{n}\left( x \right) = a_{0} + a_{1}x + \ldots\ldots + a_{n - 1}x^{n - 1} + a_{n}x^{n}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{3.1}) \nonumber

To find the first derivative,

P_{n}^{\prime}\left( x \right) = \frac{dP_{n}(x)}{dx} = a_{1} + 2a_{2}x + \ldots\ldots + \left( n - 1 \right)a_{n - 1}x^{n - 2} + na_{n}x^{n - 1}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{3.2}) \nonumber

Similarly, other derivatives such as second, third, etc., can also be found.

Example \PageIndex{3.1}

The upward velocity of a rocket is given as a function of time in Table \PageIndex{3.1}.

Table \PageIndex{3.1}. Velocity as a function of time
t\ (\text{s}) v(t)\ (\text{m/s})
0 0
10 227.04
15 362.78
20 517.35
22.5 602.97
30 901.67
Graph of velocity vs time, plotting the data points listed in Table 2.3.3.1.
Figure \PageIndex{3.1}. Velocity vs time data.

Using a second-order polynomial interpolant for velocity, find the acceleration of the rocket at t = 16\ \text{s}.

Solution

For the second-order polynomial (also called quadratic interpolation), we choose the velocity given by

v\left( t \right) = a_{0} + a_{1}t + a_{2}t^{2} \nonumber

Since we want to find the velocity at t = 16\ \text{s}, and we are using a second-order polynomial, we need to choose the three points closest to t = 16, and that also bracket t = 16, to evaluate it.

The three points are t_{0} = 10,\ t_{1} = 15, and t_{2} = 20.

t_{0} = 10,\ v\left( t_{0} \right) = 227.04 \nonumber

t_{1} = 15,\ v\left( t_{1} \right) = 362.78 \nonumber

t_{2} = 20,\ v\left( t_{2} \right) = 517.35 \nonumber

such that

\begin{split} v\left( 10 \right) &= 227.04 = a_{0} + a_{1}\left( 10 \right) + a_{2}\left( 10 \right)^{2}\\ v\left( 15 \right) &= 362.78 = a_{0} + a_{1}\left( 15 \right) + a_{2}\left( 15 \right)^{2}\\ v\left( 20 \right) &= 517.35 = a_{0} + a_{1}\left( 20 \right) + a_{2}\left( 20 \right)^{2}\end{split} \nonumber

Writing the three equations in matrix form, we have

\begin{bmatrix} 1 & 10 & 100 \\ 1 & 15 & 225 \\ 1 & 20 & 400 \\ \end{bmatrix}\begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \\ \end{bmatrix} = \begin{bmatrix} 227.04 \\ 362.78 \\ 517.35 \\ \end{bmatrix} \nonumber

Solving the above three equations gives

\begin{split}a_{0} &= 12.050 \\a_{1} &= 17.733 \\a_{2} &= 0.37660\end{split} \nonumber

Hence

\begin{split} v\left( t \right) &= a_{0} + a_{1}t + a_{2}t^{2}\\ &= 12.050 + 17.733t + 0.37660t^{2},\ 10 \leq t \leq 20 \end{split} \nonumber

The acceleration at t = 16 is given by

a\left( 16 \right) = \left.\frac{d}{dt}v \left( t \right) \right|_{t = 16} \nonumber

We found

v\left( t \right) = 12.050 + 17.733t + 0.37660t^{2},\ 10 \leq t \leq 20, \nonumber

and hence

\begin{split} a\left( t \right)\ &= \frac{d}{dt}v\left( t \right)\\ &= \frac{d}{dt}\left( 12.050 + 17.733t + 0.37660t^{2} \right)\\& = 17.733 + 0.75320t,\ 10 \leq t \leq 20\end{split} \nonumber \begin{split} a\left( 16 \right)\ &= 17.733 + 0.75320(16)\\ &= 29.784 \text{ m/s}^{2} \end{split} \nonumber

Example \PageIndex{3.2}

The upward velocity of a rocket is given as a function of time in Table \PageIndex{3.2}.

Table \PageIndex{3.2}. Velocity as a function of time.
t\ (\text{s}) v(t)\ (\text{m/s})
0 0
10 227.04
15 362.78
20 517.35
22.5 602.97
30 901.67

Using a third-order polynomial interpolant for velocity, find the acceleration and jerk (it is a measure of discomfort caused to passengers in a vehicle) of the rocket at t = 16\ \text{s}.

Solution

For the third-order polynomial (also called cubic interpolation), we choose the velocity given by

v\left( t \right) = a_{0} + a_{1}t + a_{2}t^{2} + a_{3}t^{3} \nonumber

Since we want to find the velocity at t = 16\ \text{s}, and we are using a third-order polynomial, we need to choose the four points closest to t = 16, and that also bracket t = 16 to evaluate it.

The four points are t_{0} = 10,\ t_{1} = 15,t_{2} = 20 and t_{3} = 22.5.

t_{0} = 10,\ v\left( t_{0} \right) = 227.04 \nonumber

t_{1} = 15,\ v\left( t_{1} \right) = 362.78 \nonumber

t_{2} = 20,\ v\left( t_{2} \right) = 517.35 \nonumber

t_{3} = 22.5,\ v\left( t_{3} \right) = 602.97 \nonumber

such that

\begin{split} v\left( 10 \right) &= 227.04 = a_{0} + a_{1}\left( 10 \right) + a_{2}\left( 10 \right)^{2} + a_{3}\left( 10 \right)^{3}\\ v\left( 15 \right) &= 362.78 = a_{0} + a_{1}\left( 15 \right) + a_{2}\left( 15 \right)^{2} + a_{3}\left( 15 \right)^{3}\\ v\left( 20 \right) &= 517.35 = a_{0} + a_{1}\left( 20 \right) + a_{2}\left( 20 \right)^{2} + a_{3}\left( 20 \right)^{3}\\ v\left( 22.5 \right) &= 602.97 = a_{0} + a_{1}\left( 22.5 \right) + a_{2}\left( 22.5 \right)^{2} + a_{3}\left( 22.5 \right)^{3} \end{split} \nonumber

Writing the four equations in matrix form, we have

\displaystyle \begin{bmatrix} 1 & 10 & 100 & 1000 \\ 1 & 15 & 225 & 3375 \\ 1 & 20 & 400 & 8000 \\ 1 & 22.5 & 506.25 & 11391 \\ \end{bmatrix}\begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \\ \end{bmatrix} = \begin{bmatrix} 227.04 \\ 362.78 \\ 517.35 \\ 602.97 \\ \end{bmatrix} \nonumber

Solving the above four equations gives

\begin{split} a_{0} &= - 4.3810 \\ a_{1} &= 21.289 \\a_{2} &= 0.13065 \\a_{3} &= 0.0054606 \end{split} \nonumber

Hence

\begin{split} v\left( t \right) &= a_{0} + a_{1}t + a_{2}t^{2} + a_{3}t^{3}\\ &= - 4.3810 + 21.289t + 0.13065t^{2} + 0.0054606t^{3},\ 10 \leq t \leq 22.5 \end{split} \nonumber

The acceleration at t = 16 is given by

a\left( 16 \right) = \left.\frac{d}{dt}v \left( t \right) \right|_{t = 16} \nonumber

Given that

v\left( t \right) = - 4.3810 + 21.289t + 0.13065t^{2} + 0.0054606t^{3},\ 10 \leq t \leq 22.5, \nonumber

\begin{split} a\left( t \right) &= \frac{d}{dt}v\left( t \right)\\ &= \frac{d}{dt}\left( - 4.3810 + 21.289t + 0.13065t^{2} + 0.0054606t^{3} \right)\\ &= 21.289 + 0.26130t + 0.016382t^{2},\quad 10 \leq t \leq 22.5 \end{split} \nonumber

\begin{split} a\left( 16 \right)\ &= 21.289 + 0.26130\left( 16 \right) + 0.016382\left( 16 \right)^{2}\\ &= 29.664 \text{ m/s}{^2} \end{split} \nonumber

The jerk at t = 16\text{ s} is given by

j\left( 16 \right) = \left. \frac{d}{dt}a \left( t \right) \right|_{t = 16} \nonumber

Given that a\left( t \right) = 21.289 + 0.26130t + 0.016382t^{2},\quad 10 \leq t \leq 22.5, \nonumber

\begin{split} j\left( t \right)\ &= \frac{d}{dt}a\left( t \right)\\ &= \frac{d}{dt}\left( 21.289 + 0.26130t + 0.016382t^{2} \right)\\ &= 0.26130 + 0.032764t,\quad 10 \leq t \leq 22.5 \end{split} \nonumber

\begin{split} j\left( 16 \right)\ &= 0.26130 + 0.032764(16)\\ &= 0.785524\text{ m/s}^{3} \end{split} \nonumber

Multiple Choice Test

(1). The definition of the first derivative of a function f(x) is

(A) \displaystyle f^{\prime}(x) = \frac{f(x + \Delta x) + f(x)}{\Delta x}

(B) \displaystyle f^{\prime}(x) = \frac{f(x + \Delta x) - f(x)}{\Delta x}

(C) \displaystyle f^{\prime}(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) + f(x)}{\Delta x}

(D) \displaystyle f^{\prime}(x) = \lim_{\Delta x \rightarrow 0}\frac{f(x + \Delta x) - f(x)}{\Delta x}

(2). Using the forward divided difference approximation with a step size of 0.2, the derivative of the function at x = 2 is given as

x 1.8 2.0 2.2 2.4 2.6
f\left( x \right) 6.0496 7.3890 9.0250 11.023 13.464

(A) 6.697

(B) 7.389

(C) 7.438

(D) 8.180

(3). A student finds the numerical value of f^{^{\prime}}(x) = 20.220 at x = 3 using a step size of 0.2. Which of the following methods did the student use to conduct the differentiation if f\left( x \right) is given in the table below?

x 2.6 2.8 3.0 3.2 3.4 3.6
f\left( x \right) e^{2.6} e^{2.8} e^{3} e^{3.2} e^{3.4} e^{3.6}

(A) Backward divided difference

(B) Calculus, that is, exact

(C) Central divided difference

(D) Forward divided difference

(4). The upward velocity of a body is given as a function of time as

t,\ \text{s} 10 15 20 22
v,\ \text{m/s} 22 36 57 10

To find the acceleration at t = 17\text{ s}, a scientist finds a second-order polynomial approximation for the velocity, and then differentiates it to find the acceleration. The estimate of the acceleration in \text{m/s}^{2} at t = 17\text{ s} is most nearly

(A) 4.060

(B) 4.200

(C) 8.157

(D) 8.498

(5). The velocity of a rocket is given as a function of time as

t,\text{ s} 0 0.5 1.2 1.5 1.8
v,\text{ m/s} 0 213 223 275 300

Allowed to use the forward divided difference, backward divided difference or central divided difference approximation of the first derivative, the best estimate for the acceleration \displaystyle \left( a = \frac{dv}{dt} \right) in \text{ m/s}^{2} of the rocket at t = 1.5 \ \text{s} is

(A) 83.33

(B) 128.33

(C) 173.33

(D) 183.33

(6). In a circuit with an inductor of inductance L, a resistor with resistance R, and a variable voltage source E(t),

\displaystyle E(t) = L\frac{di}{dt} + Ri. The current, i, is measured at several values of time as

\text{Time},\ t\ (\text{s}) 1.00 1.01 1.03 1.1
\text {Current},\ i\ (\text{A}) 3.10 3.12 3.18 3.24

If L = 0.98 Henries and R = 0.142 ohms, the most accurate expression for E(1.00) is

(A) 0.98\left( \displaystyle \frac{3.24 - 3.10}{0.1} \right) + (0.142)(3.10)

(B) (0.142)(3.10)

(C) 0.98\left( \displaystyle \frac{3.12 - 3.10}{0.01} \right) + (0.142)(3.10)

(D) 0.98\left( \displaystyle \frac{3.12 - 3.10}{0.01} \right)

For complete solution, go to

http://nm.mathforcollege.com/mcquizzes/02dif/quiz_02dif_discrete_solution.pdf

Problem Set

(1). Given below is the value of a function at discrete values of x

x 0 0.4 0.9 1.8 2.5
f(x) 20 75 121 126 171

Estimate f^{\prime}(0.9) using

a) Forward divided difference scheme,

b) Backward divided difference scheme, and

c) Central divided difference scheme.

Answer

a)\ 5.5556\ \ b)\ 92\ \ c)\ 58.89

(2). The upward velocity of a rocket is given as a function of time in the table below

t,\ \text{s} 0 10 15 20 22.5 30
v(t),\ \text{m/s} 0 227.04 362.78 517.35 602.97 901.67

Use the first, second and third-order polynomial direct method (Vandermonde) interpolation to find the acceleration at t = 9.5 \ \text{s}. What is the true error if I told you that the data given in the table was derived from the formula

v\left( t \right) = 2000\ln\left\lbrack \frac{14 \times 10^{4}}{14 \times 10^{4} - 2100t} \right\rbrack - 9.8t,\ 0 \leq t \leq 30? \nonumber

Answer

\text{In m/s}^2:\ 22.704,\ 25.370,\ 25.152

\text{True errors}:\ 2.4814,\ -0.18498,\ 0.032927

(3). The upward velocity of a rocket is given as a function of time in the table below.

t,\ \text{s} 0 10 15 20 22.5 30
v(t),\ \text{m/s} 0 227.04 362.78 517.35 602.97 901.67

From the data given, use the first, second and third-order Lagrangian method interpolation to find the acceleration at t = 9.5\ \text{s}. What is the true error, if I told you that the data given in the table was derived from the formula

v\left( t \right) = 2000\ln\left\lbrack \frac{14 \times 10^{4}}{14 \times 10^{4} - 2100t} \right\rbrack - 9.8t,\quad 0 \leq t \leq 30? \nonumber

Answer

\text{In m/s}^2:\ 22.704,\ 25.370,\ 25.152

\text{True errors}:\ 2.4814,\ -0.18498,\ 0.032927

(4). An aircraft position during an emergency landing exercise on a runway was timed as follows

t,\ s 0 0.4 1.00 1.75 2.5
x,\ m 20 71 110 161 178

What is your best estimate for the velocity of the aircraft at t = 1.75 \ \text{s}?

Answer

45.33\ \text{m/s} from CDD, 54.024 \ \text{m/s} if using third order polynomial interpolation

(5). An aircraft position during an emergency landing exercise on a runway was timed as follows

t,\ \text{s} 0 0.4 1.00 1.75 2.5
x,\ \text{m} 20 71 110 161 178

Estimate the acceleration of the jet fighter at t = 1.75 \ \text{s} by any method.

Answer

CDD gives -60.444 \ \text{m/s}^2; 3rd order polynomial for location gives -60.442\ \text{m/s}^2

(6). The location of a futuristic car is given as a function of time

t,\ \text{s} 0 0.5 1.2 1.5 1.8
x,\ \text{m} 0 213 223 275 300

What is your best estimate for the momentum M = mv (m = mass, v = velocity) of the car at t = 1.5 seconds? Assume the mass of the car is 2006\ \text{kg}.

Answer

CDD gives 257437\ \text{kg} \cdot \text{m/s}; using 3rd order polynomial gives 300354\ \text{kg} \cdot \text{m/s}


This page titled 2.03: Numerical Differentiation of Functions at Discrete Data Points is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Autar Kaw via source content that was edited to the style and standards of the LibreTexts platform.

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