# Find the number of real solutions of $x^2+e^x=\sin x$

Find the number of real solutions of: $e^x+x^2= \sin x$

Graphically, I am unable to derive any solutions. Also, how could I estimate the real solutions?

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Which set does $x$ belong to ? – The very fluffy Panda May 9 '14 at 9:27
@PandaBear OP is aksing for "real solutions". – 5xum May 9 '14 at 9:29
I am more interested in the number of solutions than the exact solutions. – Aditya Parson May 9 '14 at 10:09

There are no solutions. Intuitively, this graph shows you why there are no solutions.

For the analytic proof there are no solutions, Prove that for $x\notin[-2,0]$, the function $e^x+x^2$ is larger than $1$, while on $[-2,0]$, it is larger than $0$.

More hints: Showing that $e^x+x^2 > 0$ should be trivial. The fact that $e^x + x^2>x^2$ should prove that $e^x+x^2>1$ for $x<-2$, and a similar expression will take care of $x>0$.

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Why do you plot $e^2+x^2$ and not $e^x+x^2$? – gammatester May 9 '14 at 9:32
@gammatester I fixed it, it was my sloppy writing. – 5xum May 9 '14 at 9:33
How can I show this rigorously? – Aditya Parson May 9 '14 at 10:29
@AdityaParson I added one more hint. – 5xum May 9 '14 at 11:46

It's actually quite simple.

Clearly $x^2+e^x>1$ when $x>0$ so there are no positive solutions

Also, clearly that for $x<-1$, $x^2+e^x>1$ so there is no solution below -1.

For $-1\leq x<0$ you have $\sin x<0$ but $x^2+e^x>0$ you still get no solution.

This leaves you with only $x=0$. However, $0^2+e^0=1$ while $\sin 0=0$ so it isn't a solution either.

Thus there are no real solutions.

If you wanted complex solutions instead, there are two answers: $x\approx0.07315\pm0.80702i$

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