Question

Order the following function in descending order:

2^2^n, n^2, lg^2(n), 1, 2^n, n!, (5/3) ^n, e^n​

Answer 1

Answer:

2^2^n,e^n,2^n,n^2,(5/3)^n, n! , 1, lg^2(n)

Explanation:

assign the value for n

ex: n=2 , you will get two terms(n^2, 2^n) as same value

then

assign value n=3, you can get the answer

Answer 2

In descending order, the functions are: n! > 2^n > e^n > (5/3)^n > 2^(2^n) > lg^2(n) > n^2 > 1

The explanation for the above descending order,

• The factorial function n! grows faster than any exponential function.
• As n increases, n! multiplies larger and larger numbers, while exponential functions only involve repeated multiplication of a fixed base.
• Among the exponential functions, 2^n grows faster than e^n and (5/3)^n because the base of the exponent is larger.
• e^n grows faster than (5/3)^n because e is approximately 2.718, which is larger than 5/3.

• 2^(2^n) > lg^2(n) > n^2 > 1:
• The function 2^(2^n) grows faster than any of the other functions because it is an exponential function of an exponential function.
• As n increases, the inner exponent 2^n increases exponentially, causing the outer exponent 2^(2^n) to grow even faster.
• lg^2(n) grows slower than n^2 because the square of the logarithm is a sublinear function of n.
• This means that lg^2(n) increases at a slower rate than the square of n, which is a polynomial function of degree 2.
• n^2 grows slower than 1 because it is a polynomial function of degree 2, while 1 is a constant function.
• As n increases, the growth rate of n^2 becomes increasingly negligible compared to the constant growth rate of 1.

For further reference about descending order visit,

https://brainly.in/question/16388599?referrer=searchResults

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