Indicate whether the first function of each of the following pairs has a smaller, same or larger order of growth (to within a constant multiple) than the second function. Justify your answer
1. n(n+1) and 2000 n
2 + 34 n 2. In n and Ign
3. 2n-1 and 2n
4. 2 n² and 0.001 n3 - 2 n

Respuesta :

Answer:

1. The first function [tex]n^2 + n[/tex] has the same order of growth as the second function [tex]2000n^2 + 34n[/tex] within a constant multiple.

2. The first [tex]ln(n) \\[/tex]  and the second [tex]log(n)[/tex] logarithmic functions have the same order of growth within a constant multiple.

3. The first function [tex]\frac{1}{2}2^{n}[/tex] has the same order of growth as the second function [tex]2^n[/tex] within a constant multiple.

4. The first function [tex]2n^2\\[/tex] has a smaller order of growth as the second function [tex]0.001n^3 - 2n[/tex] within a constant multiple.

Explanation:

The given functions are

1. [tex]n(n +1 )[/tex] and [tex]2000n^2 + 34n[/tex]

2. [tex]ln(n) \\[/tex] and [tex]log(n)[/tex]

3. [tex]2^{n-1}[/tex] and [tex]2^n[/tex]

4. [tex]2n^2\\[/tex] and [tex]0.001n^3 - 2n[/tex]

The First pair:

[tex]n(n +1 )[/tex] and [tex]2000n^2 + 34n[/tex]

The first function can be simplified to

[tex]n(n +1 ) \\\\(n \times n) + (n\times1)\\\\n^2 + n[/tex]

Therefore, the first function [tex]n^2 + n[/tex] has the same order of growth as the second function [tex]2000n^2 + 34n[/tex] within a constant multiple.

The Second pair:

[tex]ln(n) \\[/tex] and [tex]log(n)[/tex]

As you can notice the difference between these two functions is of logarithm base which is given by

[tex]log_a \: n = log_a \: b\: log_b \: n[/tex]

Therefore, the first [tex]ln(n) \\[/tex]  and the second [tex]log(n)[/tex] logarithmic functions have the same order of growth within a constant multiple.

The Third pair:

[tex]2^{n-1}[/tex] and [tex]2^n[/tex]

The first function can be simplified to

[tex]2^{n-1} \\\\\frac{2^{n}}{2} \\\\\frac{1}{2}2^{n} \\\\[/tex]

Therefore, the first function [tex]\frac{1}{2}2^{n}[/tex] has the same order of growth as the second function [tex]2^n[/tex] within a constant multiple.

The Fourth pair:

[tex]2n^2\\[/tex] and [tex]0.001n^3 - 2n[/tex]

As you can notice the first function is quadratic and the second function is cubic.

Therefore, the first function [tex]2n^2\\[/tex] has a smaller order of growth as the second function [tex]0.001n^3 - 2n[/tex] within a constant multiple.