4
$\begingroup$

$$\tan\left(x+\frac{\pi}8\right)>e^x+\ln x,\ x\in\left(0,\frac{3\pi}8\right)$$

By plotting the graph I found that this is indeed true (in fact I found this inequality through plotting), but how can I prove this? There are various functions in this inequality and I don't know how to start. Any hints will be appreciated.

Edit: I was suggested to post the graph (from WolframAlpha). This is the graph of $\tan(x+\frac{\pi}8)-e^x-\ln x$.

graph

$\endgroup$
  • 2
    $\begingroup$ The left-hand side is undefined for $x=\frac38\pi$, and the right-hand side is undefined for $x=0$. Should the range for $x$ have been an open interval instead? $\endgroup$ – Henning Makholm Sep 25 '16 at 4:35
  • $\begingroup$ @HenningMakholm Oh yes, my mistake, sorry. I'll fix it right away $\endgroup$ – Colescu Sep 25 '16 at 4:37
  • $\begingroup$ post the graph? likely just take a derivative and confirm the function is increasing or decreasing. $\endgroup$ – djechlin Sep 25 '16 at 4:47
  • 1
    $\begingroup$ @djechlin Have you taken the derivative to discern anything useful here? The function $\tan(x+\pi/8)-e^x-\log(x)$ is neither everywhere increasing nor everywhere decreasing. $\endgroup$ – Mark Viola Sep 25 '16 at 5:00
  • $\begingroup$ Well I need to clarify that this is not a calculus problem. I don't know what methods or techniques are to be used, but I think calculus might help so I hashtagged calculus. $\endgroup$ – Colescu Sep 25 '16 at 5:24
3
$\begingroup$

To prove that $$\tan(x+\frac{\pi}8)>e^x+\ln x \qquad\ x\in(0,\frac{3\pi}8)$$ I suppose that it is sufficient to show that function $$F(x)=\tan(x+\frac{\pi}8)-e^x-\ln x $$ is always positive in the given range.

The function is positive infinite at both ends. So, let us show that $F(x)$ goes through aminimum value which is positive. Compute the derivatives $$F'(x)=\sec ^2\left(x+\frac{\pi }{8}\right)-e^x-\frac{1}{x}$$ $$F''(x)=2 \tan \left(x+\frac{\pi }{8}\right) \sec ^2\left(x+\frac{\pi }{8}\right)-e^x+\frac{1}{x^2}$$ The first derivative is $-\infty$ at the left bound and $\infty$ at the upper bound. If you plot $F'(x)$, you will notice that $F'(x)=0$ has a single root.

Since no explicit solution can be obtained for such as a transcendental equation, we need to use some numerical method. So, let us use Newton method will will give for the iterates $$x_{n+1}=x_n-\frac{F'(x_n)}{F''(x_n)}$$ and start iterating at the midpoint of the interval $(x_0=\frac{3 \pi }{16})$. This will give the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.5890486225 \\ 1 & 0.6131825159 \\ 2 & 0.6121561606 \\ 3 & 0.6121539948 \end{array} \right)$$

Now, using your pocket calculator, $$F(0.6121539948)\approx 0.22053$$ $$F''(0.6121539948)\approx 11.7739$$ The second derivative test confirms that the point is a minimum.

$\endgroup$
2
$\begingroup$

We show that there exists $m\in \mathbb{R}^+$ such that for all $x\in \left(0,\frac{3\pi }{8}\right)$ $$\tan \left(x+\frac{\pi }{8}\right)>m\left(x-\frac{\pi }{16}\right)>e^x+\ln x$$ $m$ is the slope of the tangent line to $$f(x)=\tan \left(x+\frac{\pi }{8}\right)$$ at $x=a$ and passing through the point $\left(\frac{\pi }{16},0\right)$

Since $f(x)$ is concave up in the interval we have $$f(x)>m\left(x-\frac{\pi }{16}\right)$$ for $x\in \left(0,\frac{3\pi }{8}\right)$

Now, $a$ satisfies the equation $$\sec ^2\left(a+\frac{\pi }{8}\right)=\frac{\tan \left(a+\frac{\pi }{8}\right)}{a-\frac{\pi }{16}}$$ which simplified gives $$2a-\frac{\pi }{8}=\sin \left(2a+\frac{\pi }{4}\right)$$ From this we get $a\approx 0.637575$ and $$m=\sec ^2\left(a+\frac{\pi }{8}\right)\approx 3.77648$$

Since $g(x)=e^x+\ln x$ is strictly increasing over the interval, to show that $$g(x)<m\left(x-\frac{\pi }{16}\right)$$ it suffices to verify the inequality at $x=\frac{3\pi }{8}$

Check $$\frac{g\left(\frac{3\pi }{8}\right)}{\frac{5\pi }{16}}\approx 3.47552 <3.77648\approx m$$

Therefore, $g(x)<f(x)$ over the given interval

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.