Rates in algorithm analysis? [closed]

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WHY do logarithms grow slower than any polynomial? What is the (understandable) proof for this?

Similarly,

WHY do exponentials always grow faster than any polynomial?

Answer 1

EDIT: This answer is essentially doing what PengOne said.

We take the limit of

log_2(x) / x^p

for constant p > 0 and show that the limit is zero. Since both log_2(x) and x^p go to infinity as x grows without bound, we apply l'Hopital's rule. This means our limit is the same as the limit of

1/(x*ln2) / p*x^(p-1)

Using simple rules of fractions, we reduce this to

1 / (p * x^p * ln2)

Since the denominator goes to infinity while the numerator is constant, we can evaluate the limit - it's zero, which means that log_2(x) grows asymptotically more slowly than x^p, regardless of the (positive) value of p.

EDIT2:

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Answer 2

Given two (nonnegative) real-valued functions f and g, you want to compute

lim_{x -> infinity} f(x) / g(x)

This limit is:

• 0 if and only if f grows slower than g
• infinity if and only if f grows faster than g
• c for some constant 0 < c < infinity if and only if f and g grow at the same rate

Now you can take any examples you like and compute the limits to see which grows faster.

Answer 3

You could consider the derivatives.

d(x^n)/dx = nx^(n-1)

d(ln x)/dx = 1/x

for n >= 1 nx^(n-1) increases with x or stays the same, whereas 1/x decreases with x, so the polynomial grows quicker.

The logarithm of e^x is x, whereas the logarithm of n^x is n ln x, so using the above argument to compare the logarithm of e^x and the logarithm of x^n, e^x grows quicker.