I am considering the following function:
f(x) = 2x + 20*pi*sin(2*pi*x)
I know, from graphing the function, that the largest root comes at x= 30.782.
However, I am having trouble deriving this. Here is what I have done so far:
For f(x) = 0 to be true, 2x = -20*pi*sin(2*pi*x). The largest possible value that 20*pi*sin(2*pi*x) can have is when sin is at a minimum, so-20pi.
So, I believed that the last root will come when -20pi < 2x. After that, 2x will be bigger than any possible value of the sin part, so the function will always be positive and no more roots will occur. Solving for x, I get x = 10pi = ~31.4159 which is different than the value of 30.782 I got from just graphing the funciton.
What am I doing wrong? How can I prove that 30.782 is the last root, or that there will be a finite number of roots?
Thanks!
As you said, $$2x=-20\pi\sin(2\pi x)$$ Since $|20\pi\sin(2\pi x)|\leq 20\pi$, you can only have roots when $|2x|\leq 20\pi \Leftrightarrow |x|\leq 10\pi$. So $10\pi$ is the upper bound for the roots and $-10\pi$ the lower bound, but that doesn't mean $x=10\pi$.
You should plot each side of the equation separately (on the same plot). You have a root if both graphs meet. Notice that the meeting points, hence roots, are in each of the intervals $[0.5,1], [1.5,2], [2.5,3],\dots$. You obtain the next interval by adding $1$ because $\sin(2\pi x)$ has a period of $1$. So you eventually get to the intervals $[30.5, 31], [31.5,32]$. At this point you can stop because we've reached the upper bound: $31.5>10\pi$, so the greatest root is in $[30.5, 31]$.
You can also notice there are exactly $2$ roots in each of those intervals, and we have $31$ intervals. That gives us $62$ positive roots.