How to compare the values $\sqrt 2$ and $\ln(3).$

Question

How to compare the values $\sqrt 2$ and $\ln(3)?$ I know only $\ln(x)<x$ and $\sqrt{x}<x$. Please help. Thanks.

Answer 1

"Argument" 1

Compute: $\sqrt{2}\approx 1.41421356237$ (source) and $\ln(3) \approx 1.09861228867$ (source)

There's no way floating point arithmetic is that inaccurate, so the desired inequality is true by direct computation. :)

But beware floating point arithmetic when you desire high levels of precision!

$python
>>> .1 + .2 == .3
False
>>> .1 + .2
0.30000000000000004

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Argument 2

Surely we can do better than a black box calculator. Motivated by how a calculator might carry out a direct computation, we argue via numerical approximation ...

As the exponential function is monotone, the desired comparison is equivalent to comparing $e^{\sqrt{2}}$ to $3$. The number $3$ is easy to compute, so we focus on the left side of the equation. From the calculator, we suspect that $e^\sqrt{2} > 3$, so we are led to truncate the exponential series: \begin{align*} e^\sqrt{2} &= 1 + \sqrt{2} + \frac{1}{2!}\sqrt{2}^2 + \frac{1}{3!}\sqrt{2}^3 + \cdots \\ &\geq 1 + \sqrt{2} + \frac{1}{2}\sqrt{2}^2 \\ &= 2 + \sqrt{2} \\ &> 2 + 1 \\ &= 3 \end{align*}

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Argument 3

Motivated by graphing the functions: The square root starts ahead of the logarithm and doesn't look back -- it not only is greater, its rate of increase is also greater, and so we can reason from their derivatives.

Let $F(x) = \sqrt{x}$ and $G(x) = \ln(1+x)$. Notice that both functions are smooth on $(0,\infty)$ and have $$F(x) = \int_0^x\frac{1}{2\sqrt{t}}\ dt$$ and $$G(x) = \int_0^x \frac{1}{t+1}\ dt $$ Now observe that $$ 2\sqrt{t} \leq t+1 $$ with equality achieved at $t=1$, as $2\sqrt{t}$ is concave down and the function $t\mapsto t+1$ is the equation of the tangent line of $2\sqrt{t}$ at $t=1$.

Therefore $\frac{1}{2\sqrt{t}} \geq \frac{1}{t+1} > 0$ for all $t>0$. As integrating positive functions preserves inequality, we have $F(x) \geq G(x)$ for all $x>0$. In particular, $F(2) \geq G(2)$, so $\sqrt{2}\geq \ln(3)$.

Answer 2

With simple numerical calculation we can show $\ln(3)< \frac75<\sqrt2$.

First calculate $e^{7/5}$ using the series expansion for exponential function.

Taking just three terms of the series: $e^{7/5}> 1+ 7/5 + 7^2/(2!5^2)=169/50>3$ This shows natural log for 3 is less than 7/5.

Now $7^2/5^2=49/25<2$ And we are done.

Answer 3

The trick is to look at $f(x) = \sqrt{x} - \ln(1+x)$.

Note that for $x = 1$,

$$1 - \ln(2) > 0$$

This is true since $\ln(x)$ is an increasing function and $1 < 2 < e$ implies

$$0 = \ln(1) < \ln(2) < \ln(e) = 1$$

Now look at the derivative

$$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{x+1}$$

We would like to show that $f'(x) \ge 0$ for $x > 1$, which means $f(x)$ is increasing and must therefore be positive for $x > 1$ (since it is already positive at $x=1$ per above).

Consider the following equivalent inequalities (when $x > 1$):

$$\begin{eqnarray} \frac{1}{2\sqrt{x}} - \frac{1}{x+1} &\ge& 0\\ \frac{1}{2\sqrt{x}} &\ge& \frac{1}{x+1}\\ x + 1 &\ge& 2\sqrt{x} \end{eqnarray}$$

For the last inequality, $1+1 = 2 = 2\sqrt{1}$ and for $x > 1$, taking the derivatives agains shows that the left side grows at a rate of $1$ and the right side grows at the rate of $\frac{2}{2\sqrt{x}} = \frac{1}{\sqrt{x}} < 1$ when $x > 1$.

Thus the last equivalent inequality holds for $x > 1$, which implies $f'(x) \ge 0$ for $x > 1$ which implies $f(x)$ is positive for $x > 1$. In particular, $f(2) = \sqrt{2} - \ln(3) > 0$. Therefore $\sqrt{2} > \ln(3)$.

Answer 4

By the Cauchy-Schwarz inequality

$$ \log(3) = \int_{0}^{2}\frac{dx}{1+x}\leq \sqrt{\int_{0}^{2}\frac{dx}{(1+x)^2}\int_{0}^{2}dx}=\frac{2}{\sqrt{3}}\tag{1}$$ and trivially $\frac{2}{\sqrt{3}}<\sqrt{2}$, hence $\color{red}{\log(3)<\sqrt{2}}$.
The same approch shows that for any $t>0$ we have $$ \log(t+1) < \frac{t}{\sqrt{t+1}}\tag{2} $$ $$ \log(t+1) < \frac{2\sqrt{t+1}-2}{(t+1)^{1/4}}\tag{3} $$ and the inequalities are pretty tight if $t$ is close to zero.
For instance, $(3)$ immediately proves $\color{red}{\log(4)<\sqrt{2}}$.

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Addendum. A nice alternative is provided by Beuker-type integrals. We have $\log(3)=\int_{0}^{1}\frac{4\,dx}{4-x^2}$ and $$ \int_{0}^{1}\frac{4x^4(1-x^2)^4}{4-x^2}\,dx = 1296\log(3)-\frac{109632}{77}\tag{4}$$ is bounded between $0$ and $\frac{1}{40}$. It follows that $\log(3)\approx \frac{2284}{2079}=[1; 10, 7, 14, 2]$ is clearly less than $\sqrt{2}=[1;2,2,2,2,\ldots]$, by using the typical notation for continued fractions.

Answer 5

Here's another series argument:

Use the fact that $\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} -\cdots$ and $\ln(2+x) = \ln 2 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{24} \cdots$. The first expansion is obtained by successfully differentiating $(1+x)^n$ and substituting $0$ with $n = \frac{1}{2}$.

Then $\sqrt{2} - \ln 3$ equals:

$$\left(1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} \cdots \right) - \left(\ln2 + \frac{1}{2} - \frac{1}{8} + \frac{1}{24} \right)$$ $$=\frac{49}{48} - \ln 2$$

and approximating $\ln 2$ as $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12}$, we have:

$$\frac{49}{48} - \frac{7}{12} = \frac{7}{16} $$

The Lagrange error bound for $\sqrt{1+x}$ is:

$$|R_n| ≤ \frac{|1/16|(1-0)^4}{4!} \Rightarrow |R_n| ≤ \frac{1}{384} (\approx 0.0026)$$

and the Lagrange error bound for $\ln(1+x)$ is:

$$|R_n| ≤ \frac{|1/24|(1-0)^4}{4!} \Rightarrow |R_n| ≤ \frac{1}{576} (\approx 0.0017)$$

Meanwhile, the maximum error for approximating $\ln 2$ to order $x^4$ is:

$$|R_n| ≤ \frac{|-1/4|(1-0)^5}{5!} \Rightarrow |R_n| ≤ \frac{1}{480} (\approx 0.0021)$$

In the worst possible scenario, adding up all these errors gets nowhere near $\frac{7}{16}$. Therefore $\sqrt{2} - \ln 3 > 0$ and $\sqrt{2} > \ln 3$.

Answer 6

We have $\exp(\sqrt 2)= \sum\limits_{n=0}^{+\infty} \frac{(\sqrt 2)^n}{n!}$, then, $\exp(\sqrt 2) \geq 1+\sqrt 2 + \frac{(\sqrt{2})^2}{2}=2+\sqrt 2 > 3$, then $\sqrt 2 > \ln 3$.