Truncation error

 ( Numerical Analysis ) 

In numerical analysis and scientific computing, truncation error is an error caused by approximating a mathematical process.

The term truncation comes from the fact that these simplifications often involve the truncation of an infinite series expansion so as to make the computation possible and practical.

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Examples

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Infinite series A summation series for $e^x$ is given by an infinite series such as $$e^x=1+\; x+\; \backslash frac\{x^2\}\{2!\}\; +\; \backslash frac\{x^3\}\{3!\}+\; \backslash frac\{x^4\}\{4!\}+\; \backslash cdots$$

In reality, we can only use a finite number of these terms as it would take an infinite amount of computational time to make use of all of them. So let's suppose we use only three terms of the series, then $$e^x\backslash approx\; 1+x+\; \backslash frac\{x^2\}\{2!\}$$

In this case, the truncation error is $\backslash frac\{x^3\}\{3!\}+\backslash frac\{x^4\}\{4!\}+\; \backslash cdots$

Example A:

Given the following infinite series, find the truncation error for if only the first three terms of the series are used.

Solution

Using only first three terms of the series gives

The sum of an infinite geometrical series is given by $$S\; =\; \backslash frac\{a\}\{1-r\}$$

For our series, and , to give $$S=\backslash frac\{1\}\{1-0.75\}=4$$

The truncation error hence is $$\backslash mathrm\{TE\}\; =\; 4\; -\; 2.3125\; =\; 1.6875$$

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Differentiation The definition of the exact first derivative of the function is given by $$f\text{'}(x)\; =\; \backslash lim\_\{h\; \backslash to\; 0\}\; \backslash frac\{f(x+h)-f(x)\}\{h\}$$

However, if we are calculating the derivative numerically, $h$ has to be finite. The error caused by choosing $h$ to be finite is a truncation error in the mathematical process of differentiation.

Example A:

Find the truncation in calculating the first derivative of $f(x)=5x^3$ at $x=7$ using a step size of $h=0.25$

Solution:

The first derivative of $f(x)=5x^3$ is $$f\text{'}(x)\; =\; 15x^2,$$ and at $x=7$, $$f\text{'}(7)\; =\; 735.$$

The approximate value is given by $$f\text{'}(7)\; =\; \backslash frac\{f(7+0.25)-f(7)\}\{0.25\}\; =\; 761.5625$$

The truncation error hence is $$\backslash mathrm\{TE\}\; =\; 735\; -\; 761.5625\; =\; -26.5625$$

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Integration The definition of the exact integral of a function $f(x)$ from $a$ to $b$ is given as follows.

Let $f:\; a,b\; \backslash to\; \backslash Reals$ be a function defined on a closed interval $a,b$ of the real numbers, $\backslash Reals$, and $$P\; =\; \backslash left\; \backslash \{x\_0,x\_1,\; x\_1,x\_2,\; \backslash dots,x\_\{n-1\},x\_n\; \backslash right\; \backslash \},$$ be a partition of I, where $$\backslash int\_\{a\}^b\; f(x)\; \backslash ,\; dx\; =\; \backslash sum\_\{i=1\}^\{n\}\; f(x\_i^*)\backslash ,\; \backslash Delta\; x\_i$$ where $\backslash Delta\; x\_i\; =\; x\_i\; -\; x\_\{i-1\}$ and $x\_i^*\; \backslash in\; x\_\{i-1\},$.

This implies that we are finding the area under the curve using infinite rectangles. However, if we are calculating the integral numerically, we can only use a finite number of rectangles. The error caused by choosing a finite number of rectangles as opposed to an infinite number of them is a truncation error in the mathematical process of integration.

Example A.

For the integral $$\backslash int\_\{3\}^\{9\}x^\{2\}\{dx\}$$ find the truncation error if a two-segment left-hand Riemann sum is used with equal width of segments.

Solution

We have the exact value as

Using two rectangles of equal width to approximate the area (see Figure 2) under the curve, the approximate value of the integral

Occasionally, by mistake, round-off error (the consequence of using finite precision Floating point on computers), is also called truncation error, especially if the number is rounded by chopping. That is not the correct use of "truncation error"; however calling it truncating a number may be acceptable.

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Addition Truncation error can cause $(A+B)+C\; \backslash neq\; A+(B+C)$ within a computer when $A\; =\; -10^\{25\},\; B\; =\; 10^\{25\},\; C\; =\; 1$ because $(A+B)+C\; =\; (0)+C\; =\; 1$ (like it should), while $A+(B+C)\; =\; A+(B)=0$. Here, $A+(B+C)$ has a truncation error equal to 1. This truncation error occurs because computers do not store the least significant digits of an extremely large integer.

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See also

• Quantization error

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