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Number of real solution of the equation $4\cos(e^x) = 2^x+2^{-x}$

$\bf{My\; Try::}$ . Now for $|x|\geq \pi>3$

No real solution exists, So we will check for $|x|<\pi$

Now i did not understand how can i calculate umber of real solution in that interval

Help me, Thanks

juantheron
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  • Is there a specific method you need? I would just subtract the LHS from the RHS to get a single function $f(x)$ and analyze it with basic calculus and Rolle's theorem... For example, $f(-1)>0$ and $f(-2)<0$ so there must be a zero there. Do the same for the other $3$ points using easy values to calculate. Then just prove that for all $|x|<2$ $f(x)$ is decreasing (proving the first derivative is always negative here, using Taylor series and showing the dominance of $-2^x-2^{-x} $ over any similar technique) – Brevan Ellefsen Mar 05 '16 at 21:41
  • Proving continuity should be trivial (I wouldn't be surprised if the function is smooth) – Brevan Ellefsen Mar 05 '16 at 21:43

2 Answers2

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Answer - 4

If you draw a rough schedule, it turns 4 root. To prove the existence of enough to show the values of different signs. Here, one of the gaps is quite narrow, where the function has a small positive "hump" above the x-axis. It is clear that there is a point with a positive value, that is, it is possible to count on a calculator.

No more solutions. have estimates using the second derivative. Technically, all this can be done

Roman83
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    Three things 1) You may have noticed that your equation is $2 \cos(e^x)=cosh( x \ln(2))$ 2) Were you supposed to give a graphical proof (which by no means is a proof...) ? 3) $x=0$ is not a good answer. – Jean Marie Mar 05 '16 at 14:24
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$3π/2>e^x>π/2$

$0.196<x<0.673$, $cose^x<0$

$cose^{1.8}=3.9$

$2^{1.8}+2^{-1.8}=3.73$

By$|4cose^x|≦4$ and $2^x+2^{-x}>4$, $x>2$ There are 2 roots around $x=2$, and $x≒0, x≒-2$

Takahiro Waki
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