prove that tan z = z has one real roots
Printable View
prove that tan z = z has one real roots
If by "root" you mean "solution", it doesn't?
Counterexamples: Let z -> 0, and z -> between 4.493 and 4.494.
The tangent function goes from negative infinity to positive infinity as z goes through one period of length pi. That is, in each period, tan(z) takes on every value from -infinity to +infinity [not including the infinities]. In each period, tan(z) must take on one value of z in the range of that period--that is, in each period, tan(z) = z once. (Graph tan(z) - z and look for zeros if you want to see this graphically)
So... there are an infinite number of solutions...
Am I missing something?
Edit: I finally thought of a good argument as to why tan(z) must take on one value of z in the range of each period.
Let the period be the interval (a, b).
Since tan(z) takes on an arbitrarily small value (close to -infinity) near a, there must exist some sub range (a, a1) where tan(z) < z [where z is in the sub range].
Similarly, tan(z) takes on arbitrarily large values (close to +infinity) near b, so there must exist some sub range (b1, b) where tan(z) > z [where z is in this new sub range].
In the middle range [a1, b1], there must be some point z* such that tan(z*) = z*, since both z and tan(z) are smooth in this period's range. There is exactly one point like this in each period since the slopes of the two sides of the equation are equal only instantaneously, though I won't get into that.
There are an infinite number of distinct period ranges to the tan(z) function, so there are an infinite number of corresponding z* points where tan(z*) = z*. Thus, tan(z) = z has an infinite number of solutions, and even has approximately (d-c)/pi solutions for any given range [c, d] of values for z.