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show that the equation has a root in the interval $(0, \pi)$ $$e^{x/\pi} + \sin(x) = x^2$$

I have solved it with the Intermediate Value Theory (as below) but I don’t think that my justification is enough, and I don’t know how to do it in a different way?

My answer:

let: $f(x) = e^{x/\pi} + \sin(x) - x^2$

Note that: $f(0) = 1+0-0 = 1$

Note that: $f(\pi) = e+0-\pi^2 =$ a negative number (how can I justify this w/o a calculator)

So, by the intermediate value theory, all numbers between $0$ and (that negative number) is defined.

Also, later I noticed that the interval in question is an open interval and it didn’t say that this equation is continuous. So, back to square one?

Michael Hardy
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Abdulrahman Alattas
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    Presumably you know that $e<3<\pi<\pi^2$ which is enough to show that $f(\pi)<0$. And the function is a sum of continuous functions, hence continuous. – lulu Feb 05 '18 at 20:18
  • you can also Show that the function $$f(x)=e^{x/ \pi}+\sin(x)-x^2$$ is strictly monotonously decreasing in the interval $(\frac{\pi}{2};\pi)$ – Dr. Sonnhard Graubner Feb 05 '18 at 20:23
  • The open interval is not a problem, the conclusion of the intermediate value theorem works for the interior of an interval (in this case). – Michael Burr Feb 05 '18 at 20:24
  • Weird how in math hours of work can lead you to nothing then a simple sentence can lead to everything. Thank you, both (edit: all). – Abdulrahman Alattas Feb 05 '18 at 20:25

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If you know $0<e<3<\pi$ that's enough to conclude $e-\pi^2<0.$

Let's look at how to show $3<\pi.$

Inscribe a regular hexagon in a circle of diameter $1.$ A bit of elementary geometry shows the perimeter of the hexagon is $3.$ Therefore the circumference of the circle is $>3,$ so you have $\pi>3.$

Now let's look at how to show $e<3.$

Suppose you know $\displaystyle e^x = \sum_{n=0}^\infty \frac {x^n}{n!}.$ Then $$ e^{-1} = 1 - 1 + \frac 1 2 - \frac 1 6 + \frac 1 {24} - \cdots. $$ Since the terms alternate in signs while decreasing, you must have $$ \frac 1 2 - \frac 1 6 < e^{-1} < \frac 1 2 - \frac 1 6 + \frac 1 {24}, $$ and therefore $e^{-1}>\dfrac 1 3,$ so $e<3.$

Michael Hardy
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  • Incidentally, knowing $0<e<\pi$ is not enough to conclude $e-\pi^2<0$. Just as knowing $0<\frac13<\frac12$ is not enough to conclude $\frac13-\frac1{2^2}<0$ – Hagen von Eitzen Feb 05 '18 at 20:52
  • @HagenvonEitzen : Sorry -- should should have said $0<e<3<\pi.$ That's what is shown in this answer. Except that I just took it for granted that $0<e. \qquad$ – Michael Hardy Feb 05 '18 at 22:19