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FP2 approximation help

Show that the equation

e^x=cot x

has a root between 0.5 and 0.55. I did this by using the change of sign rule.

I'm stuck on this next part:

Rewrite the equation as

x=F(x)

, where F(x)=

x+k(e^x-cotx)

, and find a value of k for which F'(x) is small and negative over the interval 0.5<x<0.55. Use this to find the root to 5 d.p.

So I differentiated F(x) to get:

1+k(e^x-cotx)*(e^x+cosec^2x)

. Then I'm not too sure what to do.

Bump.

Original post by Music99

So I differentiated F(x) to get:

1+k(e^x-cotx)*(e^x+cosec^2x)

. Then I'm not too sure what to do.

Huh!

You might like to rethink that - the F'(x).

Reply 3

steve10

The best I could come up with was using a simple iterative method. But it took me 14 iterations to get a root value of x = 0.53139. (used Maple)

Oh yes, as mentioned, your differentiation is wrong.