- #1
chwala
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- 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: