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I am stuck the following problem :

Solve the following problem:

$(1-\tan^2x)\sec^2x+2^{\tan^2x}=0$

Taking $p=\tan^2x,$ we get $(1-p^2)+2^p=0$. Now,I do not know how to go from here.

Simple observation says $x=2n\pi \pm \frac{\pi}{3}$,satisfies the equation and so it will be the solution but I do not know how to get there. Can someone help?

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learner
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  • @qaphla That's what the OP has used. –  Apr 04 '14 at 04:37
  • Yes ,I have noted that and used it to reach the result $(1-p^2)+2^p=0,,$but I still do not get your hint. Can you explain more? – learner Apr 04 '14 at 04:38
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    @learner qaphla didn't read your solution probably. Slight correction though, $-\pi/3$ isn't a solution. $+\pi/3$ is. Also it is quite elementary to show that $x^2-1=2^x$ has two solutions in $(-\infty,\infty)$ therefore this equation will have two solutions in $(-\pi/2,\pi/2)$. The problem is finding these solutions. – Guy Apr 04 '14 at 04:39
  • @Sabyasachi Be careful with slopes... the graph of 2^x will take revenge when x^2-1 goes above it. 3 solutions. – evil999man Apr 04 '14 at 04:46
  • Also, we need only positive solutions. – evil999man Apr 04 '14 at 04:48
  • I think your teacher made the same mistake Sabyasachi did and mistook that 3 was the only solution. wolfram alpha return no closed form for other solution – evil999man Apr 04 '14 at 04:50
  • http://www.wolframalpha.com/input/?i=2%5Ex%3Dx%5E2-1 – evil999man Apr 04 '14 at 04:51
  • @Awesome I am careful. Two solutions. See for yourself (take pain to type it in yourself, sorry. ) – Guy Apr 04 '14 at 04:52
  • @Sabyasachi You mean to say wolfram alpha is wrong? – evil999man Apr 04 '14 at 04:55
  • @Awesome I've seen it go crazy before, but no not this time. My bad. But it did go crazy. $x\approx 3$ wtf! – Guy Apr 04 '14 at 04:57
  • @Sabyasachi yeah... but below that in integer solution it says x=3 – evil999man Apr 04 '14 at 04:58
  • @learner Calculating the other closed forms is hard, calculating $3$ is simple enough, you observe that, $2^p=(p+1)(p-1)$ and therefore $p$ is odd and $p\gt1$. The first odd number you try is a solution. – Guy Apr 04 '14 at 05:00
  • @Awesome all the more reason to call it crazy. Is $x\approx3$ and $x=3$ two solutions. :p – Guy Apr 04 '14 at 05:00
  • @Sabyasachi Click on exact forms. – evil999man Apr 04 '14 at 05:01
  • @Awesome I know. that is beside the point. Let's focus on the question, ok? I'm just saying WA is still a machine, and I've seen it screw some integrals before. – Guy Apr 04 '14 at 05:02
  • @learner you can get bounds on the other 2 solutions. If you take $f(x)=2^x-x^2+1$, It has a root $x=3$, and is decreasing near $x=3$. If you try $x=4$, $f(4)$ is positive, therefore it must have another root in $(3,4)$. You can also see $f(-1)\gt0$ and $f(-2)\lt0$. Therefore another root in $(-2,-1)$ – Guy Apr 04 '14 at 05:06
  • @Sabyasachi thanks a lot for your time and effort. It surely helped me a lot. – learner Apr 04 '14 at 05:06
  • @learner you're welcome. – Guy Apr 04 '14 at 05:07

1 Answers1

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If you look at the equation $$f(p)=2^p-p^2+1$$ there are three solutions, one of them being exactly $p=3$. The other ones are irrational and can only be found using numerical methods such as Newton. These roots are $p=3.40745$ and $p=-1.19825$. The negative root can be discarded since $p=\tan^2(x)$.

To $p=3.40745$ corresponds $x=1.07432$ which is another solution very close to $x=\frac {\pi} {3}=1.04720$

Edit

It is interesting to notice that, if we develop the function as a Taylor series at $x=3$, we obtain $$f(p)= (8\log (2)-6)(p-3)+ \left(4 \log ^2(2)-1\right)(p-3)^2+O\left((p-3)^3\right)$$ which cancels at $$p=\frac{3+12 \log ^2(2)-8 \log (2)}{4 \log ^2(2)-1}\approx 3.49340$$ which is very close to the second solution. Expanding for one more order and solving the quadratic would give a solution $\approx 3.41174$.

It is also interesting to notice that the roots of the first derivatives are explicit in terms of Lambert function $$\left\{p= -\frac{W\left(-\frac{1}{2} \log ^2(2)\right)}{\log (2)}\right\},\left\{p= -\frac{W_{-1}\left(-\frac{1}{2} \log ^2(2)\right)}{\log (2)}\right\}$$ Developing $f(p)$ as a Taylor series to second order at the second (the largest root) allows to estimate the roots at $3.00933$ and $3.41554$.

Claude Leibovici
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