Question

How many roots are there for
cos x -logx=0

Answer 1

it would be easiest to think about cos x = log x instead, then you can compare the graphs and look for intersections... these will correspond to the roots of the original equation.

Answer 2

finding
the roots by several method like newton-rapson, comparing graph ...! but without such methods is there any other one

Answer 3

it's very likely that there isn't a more explicit method... maybe infinite series expansion, but I don't know of anything with higher analytic firepower.. :)

Answer 4

I suspect series method is not a fair method here

Answer 5

you might be able to get a bound on the number of solutions with an argument about the constantly positive derivative of log(x), but not sure if that would give you the precise number.

Answer 6

Of course, once log(x) exceeds 1 there are no more roots possible.

Answer 7

Is it log as in ln?

Answer 8

But log x can take values
-1<x<1

Answer 9

I mean the actual value of log(x) as in log(x) = 1 is the upper bound, which occurs when x = 10. so the only x values we care about are those from 0 < x <= 10. This is the region of possible roots.

Answer 10

now at about x = 6.28 cos(x) hits a maximum and comes back down, this will form two intersections with log(x) since it doesn't reach a height of 1 until x=10.

Answer 11

and the third root comes from x = 0 implies cos(x) = 1 diving to x = 3.14 implies cos(x) = -1 which creates an intersection with log(x) at some point in between.

Answer 12

this is all very qualitative and "wishy-washy," but we are dealing with transcendental functions which are always annoying. :)

Answer 13

Well said brother.!

Answer 14

lol, but yeah it is a very intriguing question nonetheless, partly because it won't submit to most methods.