- #1
Euler2718
- 90
- 3
In my problem the linear modal is defined as the first term in the series expansion of [itex]\sin(x)[/itex] so:
[tex] \sin(x) = x - \frac{x^{3}}{3!}+\dots [/tex]
[itex]\sin(x) = x[/itex] is the linear modal.
So with this, I then have to write [itex] \frac{d^{2}x}{dt^{2}} = -\sin(x) [/itex] as a system of [itex]x^{\prime}[/itex] and [itex]y^{\prime}[/itex], so:
[tex] x^{\prime} = y [/tex]
[tex] y^{\prime} = \sin(x) [/tex]
I tried the linear modal in Euler's method, with initial conditions X(1) = 1 and V(1)=0 :
Where s is the step size. But apparently I'm supposed to get a circle when I plot V with respect to X which makes sense, but all I get is a straight line.
If I change it to:
With s=0.8 I get a spiral, which looks like a development but I'm no closer to the circular shape that I am expecting. I think I just need a fresh pair of eyes to see where perhaps an obvious error lies.
[tex] \sin(x) = x - \frac{x^{3}}{3!}+\dots [/tex]
[itex]\sin(x) = x[/itex] is the linear modal.
So with this, I then have to write [itex] \frac{d^{2}x}{dt^{2}} = -\sin(x) [/itex] as a system of [itex]x^{\prime}[/itex] and [itex]y^{\prime}[/itex], so:
[tex] x^{\prime} = y [/tex]
[tex] y^{\prime} = \sin(x) [/tex]
I tried the linear modal in Euler's method, with initial conditions X(1) = 1 and V(1)=0 :
Code:
for i = 1:1000
V(i+1) = V(i)-(1.*s) ;
X(i+1) = V(i);
end
Where s is the step size. But apparently I'm supposed to get a circle when I plot V with respect to X which makes sense, but all I get is a straight line.
If I change it to:
Code:
for i = 1:1000
V(i+1) = V(i)-(X(i).*s) ;
X(i+1) = V(i);
end
With s=0.8 I get a spiral, which looks like a development but I'm no closer to the circular shape that I am expecting. I think I just need a fresh pair of eyes to see where perhaps an obvious error lies.