Prove that $2^n > 10n^2$ for sufficiently large $n$ using calculus

Question

My approach to doing these types of questions is finding the derivative of the LHS and the RHS and finding the limit of that derivative as $n$ approaches infinity. So, for this question, it would be:

Let $f(x) = 2^n$ --> $f'(x) = 2^nlog(2)$

Let $g(x) = 10n^2$ --> $g'(x) = 20n$

But I don't know how to proceed from here using limits as $n$ goes to infinity. Any help would be appreciated!

Answer 1

It's true for all $n\geq10$.

Indeed, $$f''(x)=2^x\ln^22-20>0$$ $$f'(x)=2^x\ln2-20x>0$$ and from here $f(x)>0$ for all $x>0$.

Answer 2

define $$f(n)=2^n-10n^2$$ then we get $$f'(n)=2^n\ln(2)-20n$$ $$f''(n)=2^n(\ln(2))^2-20$$ can you proceed?

Answer 3

$$2^n=\exp(n\log(2))=\sum_{k\geq0}\frac{n^k\log^k(2)}{k!}$$ $$2^n-10n^2=1+n\log(2)+n^2\left(\frac{\log^2(2)}{2}-10\right)+n^3\frac{\log^3(2)}{6}+\sum_{k\geq4}\frac{n^k\log^k(2)}{k!}$$ So: $$2^n-10n^2\geq1+n\log(2)+n^2\left(\frac{\log^2(2)}{2}-10\right)+n^3\frac{\log^3(2)}{6}=:f(n)$$ The coefficient of $n^2$ is less than $0$, but the coefficient of $n^3$ is greater than $0$.
For $a$,$b$,$c$,$d \geq0$: $\lim\limits_{x \to \infty}ax^3-bx^2+cx+d=\infty$, so $f(n)$ will be positive for large enough $n$. And since $2^n-10n^2\geq f(n)$, $2^n-10n^2$ will be also greater than $0$.

Answer 4

$$ 2^n > 10 n^2 \iff n\times log(2) > 1+2\times log(n)\iff $$

$$ n \times log(2) -2\times log(n)> 1\iff $$

$$n\ge10$$

Answer 5

My way of doing it would be

Let $f(x) = 2^x - 10x^2$ (which is continuous) then

$f'(x) = 2^x*\ln 2 - 20x$.

$f''(x) = 2^x*(\ln 2)^2 - 20$

$f'''(x) = 2^x *(\ln 2)^2 > 0;$ for all $x$.

So $f''(x)$ is strictly increasing. If there is any $x_2$ where $f''(x_2) \ge 0$ then $f''(x) > 0$ for all $x > x_2$.

Which means $f'(x)$ is strictly increasing for all $x > x_2$. If there is an $x_1$ where $x_1 \ge x_2$ and $f'(x_1) > 0$ then $f'(x) > 0$ for all $x >x_1$.

Which means $f(x)$ is strictly increasing for all $x > x_1$. So if there is an $x_0$ and $x_0 \ge x_1$ and $f(x_0) > 0$ then $f(n)>0$ and $2^n > 10n^2$ for all $n > x_0$. Hence our result.

Now to find these $x_0,x_1,x_2$.... well, there is no need for accuracy.... we can be as crude as we like.

We want $2^x_2*(\ln 2)^2 > 20$, just let $x_2 > \log_2 (\frac {20}{(\ln 2)^2})$. No need for accuracy: $\frac 12 < \ln 2 < 1$; $\frac 14(\ln 2)^2<1$ so $20< \frac {20}{(\ln 2)^2}<80$ so $4=\log_2 16 < \log_2 20< \log_2 \frac {20}{(\ln 2)^2} < \log_2 80 < \log_2 128 = 7$ so let $x_2 = 7$.

We want $2^{x_1}*\ln 2 - 20x$ so... I dunno... $\ln 2 > \frac 23$ so $2^{8}\ln 2 - 160 > 256*2/3 - 160 > 170 - 160 = 10 > 0$ so $x_1 = 8$ will do.

Just need to find any $x_0 \ge 8$ so that $2^{x_0} \ge 10x_0^2$. $x_0 = 10$ will do as $2^{10} = 2014 > 1000 = 10*10^2$.

So that's it. For any $n \ge 10$ we have proven $2^n > 10*n^2$.