Question

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

Answer 1

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

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

3. The first function
`(1)/(2)2^(n)` has the same order of growth as the second function
`2^n` within a constant multiple.

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

Step-by-step explanation:

The given functions are

1.
`n(n +1 )` and
`2000n^2 + 34n`

2.
`ln(n) \\` and
`log(n)`

3.
`2^(n-1)` and
`2^n`

4.
`2n^2\\` and
`0.001n^3 - 2n`

The First pair:

`n(n +1 )` and
`2000n^2 + 34n`

The first function can be simplified to

`n(n +1 ) \\\\(n * n) + (n*1)\\\\n^2 + n`

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

The Second pair:

`ln(n) \\` and
`log(n)`

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

`log_a \: n = log_a \: b\: log_b \: n`

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

The Third pair:

`2^(n-1)` and
`2^n`

The first function can be simplified to

`2^(n-1) \\\\(2^(n))/(2) \\\\(1)/(2)2^(n) \\\\`

Therefore, the first function
`(1)/(2)2^(n)` has the same order of growth as the second function
`2^n` within a constant multiple.

The Fourth pair:

`2n^2\\` and
`0.001n^3 - 2n`

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

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