- #1

chwala

Gold Member

- 2,728

- 382

- Homework Statement
- see attached.

- Relevant Equations
- iterative techniques

part (a)

Asymptote at ##x=0.5##

part (b)

##\dfrac{e^x}{4x^2-1}= -2##

##e^x=2-8x^2##

##2e^x= 4-16x^2##

##16x^2=4-2e^x##

##x^2= \dfrac{4-2e^x}{16}##

##x=\dfrac{\sqrt{4-2e^x}}{4}##

part (c)

...

##x_{2}=0.2851##

##x_{3}=0.2894##

##x_{4}=0.2881##

##x_{5}=0.2885##

##x_{6}=0.2884##

##x_{7}=0.2884##

##α=0.2884##

For part(d)

not sure here, but i checked directly with

##F(x_n) = \ln (2-8x^2_n)## ...and noted that after a few iterations we were ending up with natural log of negative numbers thus ##α## cannot be found.

on a different approach, using second derivative,

##F^{'}(x) = \left[\dfrac{1}{2-8x^2} × -16x\right]##

##F^{'}(x) = \dfrac{8x}{4x^2-1}##

##F^{'}(0.3) = -3.75##

##F^{''}(x)=- \left[\dfrac{8+32x^2}{(4x^2-1)^2}\right]##

This will always be negative irrespective of ##x## value thus ##α## cannot be found.

Last edited: