Thread: 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!!!

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 and say that f(f(f(f(f(f(f(f(f(x)))))))))) yields the fixed point?

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...

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



Here is why

consider

note that

so the derivative is always negative on the above interval.

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



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

I hope this helps.

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...

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





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

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.