Is $\sin (1+ \frac{x}{n})$ decreasing function , x any element in bounded interval of real numbers?
My attempt: Let the interval in which x belongs be bounded by m(lower bound),M(upper bound)
$1+ \frac{x}{n}$ –> decreasing function for all $n>x > m$
Sin is increasing function in interval $[0, \pi/2]$
composition of increasing and decreasing function –> decreasing function.
So $\sin (1+ \frac{x}{n})$ decreasing function for $n>m$
But in wolfram plotting (by putting values x = 1, 2 ,3 –> sin (1+1/n), sin(1+2/n),…), i am not seeing decreasing function but some sin like oscillating function. Seems i did some mistake. Pls correct me…
http://www.wolframalpha.com/widgets/view.jsp?id=74c4d9c719a83b7ef727a22ad471f80d
Best Answer
You've proved that when you fix $x$ and increase $n$ the function decreases (for $n$ sufficiently large in terms of $x$). But then in Wolfram you fixed $n$ and increased $x$, which is the wrong way round. You should be plotting $\sin(1+x), \sin(1+x/2),\sin(1+x/3),...$