# Thread: real analysis question

1. ## real analysis question

i need to prove that sin(x)-x=0 has a unique solution in the open interval (0,pi/2). indicate a computational procedure for finding that solution approximately. on the previous parts of the problem i proved that every contraction mapping is continuous and that every contraction on a complete metric space has a unique fixed pt, so we can use these facts. my intuition is that we'll need to set f(x) = sin x and use the contraction mapping idea, but it's hard to prove that sin x is a contraction mapping. help!!!

2. Originally Posted by squarerootof2
i need to prove that sin(x)-x=0 has a unique solution in the open interval (0,pi/2). indicate a computational procedure for finding that solution approximately. on the previous parts of the problem i proved that every contraction mapping is continuous and that every contraction on a complete metric space has a unique fixed pt, so we can use these facts. my intuition is that we'll need to set f(x) = sin x and use the contraction mapping idea, but it's hard to prove that sin x is a contraction mapping. help!!!
This is not a proof but could you state that sin(x) has a fixed point on $(0,2\pi)$ and say that f(f(f(f(f(f(f(f(f(x)))))))))) yields the fixed point?

3. hmm do you think that i'm right in saying that proving sin (x) is a contraction mapping would do the job to show this given the previous parts that i've proven? thanks...

4. Originally Posted by squarerootof2
i need to prove that sin(x)-x=0 has a unique solution in the open interval (0,pi/2). indicate a computational procedure for finding that solution approximately. on the previous parts of the problem i proved that every contraction mapping is continuous and that every contraction on a complete metric space has a unique fixed pt, so we can use these facts. my intuition is that we'll need to set f(x) = sin x and use the contraction mapping idea, but it's hard to prove that sin x is a contraction mapping. help!!!
since you said the open interval the statement is not true becuase

$0 \notin (0,\frac{\pi}{2})$

Here is why

consider $f:[0,\pi/2] \to \mathbb{R} f(x)=\sin(x)-x \mbox{ then } f'(x)=\cos(x)-1$

note that $\cos(x) < 1 \mbox{ } \forall x \in (0,\pi/2)$

so the derivative is always negative on the above interval.

Now by the Mean Value Theorem we have
let a=0 and b=x

$f'(c)(b-a)=f(b)-f(a) \iff f'(c)(x)=f(x)$

since the derivative is always negative and x is positive we know that f(x) is negative on the interval i.e f(x) < 0

so finally we get

$f(x)=\sin(x)-x \iff \sin(x)-x < 0 \iff \sin(x) < x$

by the last they they are never equal, so no solution exits.

I hope this helps.

5. isnt there a way to apply the intermediate theorem to show that this solution exists? because the way the instructor asked the question, he implied that there is a solution... might be a mistake on the instructor's part...

6. Originally Posted by squarerootof2
isnt there a way to apply the intermediate theorem to show that this solution exists? because the way the instructor asked the question, he implied that there is a solution... might be a mistake on the instructor's part...
Here is another way to think about it

$\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...=x+\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$

$x-\sin(x)=x-\left( x+\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} \right)=-\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$

The only way that sum will equal zero is if x is zero.

Here is a graph as well

The only solution is x=0

7. Originally Posted by squarerootof2
isnt there a way to apply the intermediate theorem to show that this solution exists? because the way the instructor asked the question, he implied that there is a solution... might be a mistake on the instructor's part...
If it was on a different interval yes, but not on the one given above.

Any open interval containing the origin would have exactly one solution.