Wrong forum probably, but I'm not sure where to post...
My friend challenged me with this curious equation:
Like I know the answer is 3, but I don't know how to mathematically prove this.
Therefore by the intermediate value theorem there
So now utilizing probably three but for the sake of the absence of luck in math
say the starting point for Newton-Raphson method is 2.5
we see that as
The check step
Sorry that is bad but I cannot think of anything else as of present
it works though
It might looks simple, but it's a real challenge. In fact, if , then f has 2 zeros. It means that the solution is not singular. But to prove it... not easy.My friend challenged me with this curious equation:
2^x - 5 = x
. We want to study its sign. So we want to find when it is equal to . .
Now if . It means that is decreasing from negative infinite to . As when , , is increasing from to positive infinite.
f(0)=-4. As it will increase (strictly), it must cross the x axis on a single point when . You can find a similar conclusion calculating for example . It means that the roots of f are respectively between and . (5 because ). Now to find the roots, you can say that there is an easy one to see, when . To calculate the other root, I suggest you the bisection method which will work.