By using the mean value therom prove that |sinx-siny|< or = |x-y| for all x and y.

Printable View

- November 18th 2006, 05:17 PMmyoplex11Mean value theorem
By using the mean value therom prove that |sinx-siny|< or = |x-y| for all x and y.

- November 19th 2006, 04:51 AMTriKri
Notice it won't make any difference if we swap x and y in both equations, so let's say . So if it isn't already, just swap x and y.

Say and .

Since , .

If sin(x) = sin(y), will be 0 and the relation will be satisfied, since .

If sin(x) > sin(y),

If sin(x) < sin(y),

I none of the cases the derivative is > 1, since x and y is real numbers (I assume).

Now the mean value theorem states that there is a point on any given section of a smooth curve that has the same derivative as the average derivative of the section. That mean the average derivative a of the section [y, x] of can't be greater than 1, since the derivative at any given point on is . Hence, can't be greater than x - y, since , and x-y is a non-negative number. And (I have to break the line here, LaTeX won't tale any longer expressions)

. Hence can't be greater than x-y, so for . And as I wrote in the begining of this post it won't make any difference if .

I don't know if that last part was completely correct. And if you wan't a more accurate explanation than mine, you'll just have to wait until any of the math gurus on this forum answers your thread. ;)

-Kristofer - November 19th 2006, 05:35 AMvio_viovalue theorem
There is x< c <y, if x<y or y<c<x if y<x such that (sinx-siny)/(x-y)= cosc

or │(sinx-siny)/(x-y)│=│cosc│

But │cosc│≤ 1

Then

│sinx-siny│/│x-y│≤1