Truncation Errors

Truncation errors are the difference between the actual value of the function and the truncated value of the given function. The truncated value of the functions is the approximated value up to a given number of digits. For example, the speed of light in vacuum is 2.99792458 × 10⁸ ms^-1. The truncated value up to two decimal places is 2.99 × 10⁸. Hence the truncation error is the difference between 2.99792458 × 10⁸ and 2.99 × 10⁸, which is 0.00792458 × 10^8, or in scientific notation, it is 7.92458 × 10⁵.

Learn about Types of Errors.

Truncation errors mainly arise when we approximate functions represented by an infinite series instead of using the actual value. We shall use Taylor’s and Maclaurin’s series to calculate the truncation error.

Taylor’s Series to Approximate Truncation Errors

Taylor’s Theorem: If f is a continuously differentiable function on an interval containing the points a and x, then the value of function f at x is given by

where R_n is the remainder, defined as

c is any quantity between a and x. It is Lagrange’s form of remainder.

Taylor’s series is very useful for representing any continuously differentiable function as a polynomial function of infinite order.

To understand the approximation of truncation error by Taylor’s series, let us take an example. Let f(x) = e^x, clearly which is continuously differentiable function. Taking a = 0, we have f(0) = 1 and f’(0) = 1 = f^(k) (0) such that

Finally, we have

If we approximate the function up to (n + 1) terms, then truncation error is the remaining terms from (n + 1) onwards.

The remainder of Taylor’s series calculates the value of the truncation error. It is to be noted that more the number of terms in the approximation smaller the truncation error.

Estimation of Truncation Error for Geometric Series

Let S be an infinite geometric series if its terms are such that |t_{j + 1}| ≤ k|t_j| where 0 ≤ k ≤ 1 for every j ≥ n, then while approximating the series up to n terms, the truncation error R_n is given by

|R_n | = t _{n +1} + t _{n + 2} + t _{n + 3} + …

≤ t _{n +1} + k |t _{n + 1} | + k² |t _{n + 1} | + k³ |t _{n + 1} | + …

= |t _{n + 1} | ( 1 + k² + k³ + …)

= |t _{n + 1} |/ (1 – k)

⇒ |R_n | ≤ [k |t _n | ]/ (1 – k)

For example, we have to calculate the truncation error |R₆| of the infinite geometric series

Clearly,

Now, we to find k such that |t_{j + 1}| ≤ k|t_j| where 0 ≤ k ≤ 1 for every j ≥ n (n = 6)

|t_{j + 1}| ≤ k|t_j| = |t_{j + 1}|/ |t_j| ≤ k = √(1 + 1/j) 𝜋 ^–2

⇒ |t_{j + 1}|/ |t_j| ≤ √(1 + 1/6) 𝜋 ^–2 < 0.11 as j ≥ 6

Therefore, by |R_n | ≤ [k |t _n | ]/ (1 – k) and t₆ < 3 × 10^{– 6}

|R₆ | ≤ [k |t _n | ]/ (1 – k) < [0.11 × 3 × 10^{– 6}]/ (1 – 0.11)

Related Articles

• Error Measurement
• Differential Calculus Approximation
• Estimation
• Taylor’s Series
• Maclaurin’s Series

Solved Examples on Truncation Errors

Example 1:

Calculate the truncation error for approximation of the up to 5 terms of the infinite series

1 + x² + x⁴ + x⁶ + x⁸ + …

Solution:

Given infinite geometric series,

1 + x² + x⁴ + x⁶ + x⁸ + …

Now, |R₅| = t₅ + t₆ + t₇ + … = x⁸ + x¹⁰ + x¹² + …

Which is again a infinite geometric series with first term x⁸ and common ratio x², then,

|R₅| = x⁸/(1 – x²).

Example 2:

Calculate the truncation error if the function e is calculated as 1 + 1/1! + 1/2! + 1/3! + …+1/7! .

Solution:

The given function is the Maclaurin’s Series for e^x at x = 1. The truncation error |R_n| for n = 7

|R_n| ≤ |(e^x. x^{7 + 1})/ (7 + 1)!| = |(e¹ . 1⁸)/8!| = |e/8!| ≈ 0.2786 × 10^{– 4}.

The truncation error is about = 0.2786 × 10^{– 4}.