Respective Initial Value Solutions for Nonlinear Equations Using Newton's Method

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Homework Statement


Find the zero for the following equation with the respective initial value...
a)cos(x)=(e^x)-1,x0=0
b)ln(x)=x^2-1,x0=1
c)x^2=2x+2,x0=2.6
the answers i get are...
a)0.6931
b)1
c)2.7321
but here comes my question...
i'm attempting a class teach us to use the software matlab...
but 2 out of 3 answers that i get from using the software is different from what i count myself...
the software MATLAB give me the answers are
a)0.6013
b)1
c)error
can someone help me??
please and thank you...

Homework Equations





The Attempt at a Solution

 
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I assume you are using some kind of iterative method to numerically solve these equations- but what method? Secant method? Newton's method?
 
HallsofIvy said:
I assume you are using some kind of iterative method to numerically solve these equations- but what method? Secant method? Newton's method?

i'm using Newton's method...
is that effect the answers??
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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