Picard Iteration/ index
i have the following System of Differential Equations
$$ begin{pmatrix}
x'(t) \
y'(t)
end{pmatrix} = begin{pmatrix}
0 && 1\
-1 && 0
end{pmatrix} begin{pmatrix}
x(t) \
y(t)
end{pmatrix} and begin{pmatrix}
x(0) \
y(0)
end{pmatrix}= begin{pmatrix}
2 \
0
end{pmatrix} $$
When i use the Picard-Iteration, i get ($ s:= begin{pmatrix}
x(t) \
y(t)end{pmatrix} $ )
$$ s_1 = begin{pmatrix}
2 \
-2tend{pmatrix} $$
$$ s_2 = begin{pmatrix}
2-t^2 \
-2tend{pmatrix} $$
$$ s_3 = begin{pmatrix}
2-t^2 \
frac{1}{3} t^3-2tend{pmatrix} $$
I assume that $ s_{infty} = begin{pmatrix}
2cos t \
-2sint end{pmatrix} $
But when i consider the series for sin and cos, for example for the index 1, $ s_1$ will be $ s_1 = begin{pmatrix}
2-t^2 \
frac{1}{3} t^3-2tend{pmatrix} $
How do i get this right?
differential-equations
asked Nov 24 at 15:24
Steven33
144
add a comment |
i have the following System of Differential Equations
$$ begin{pmatrix}
x'(t) \
y'(t)
end{pmatrix} = begin{pmatrix}
0 && 1\
-1 && 0
end{pmatrix} begin{pmatrix}
x(t) \
y(t)
end{pmatrix} and begin{pmatrix}
x(0) \
y(0)
end{pmatrix}= begin{pmatrix}
2 \
0
end{pmatrix} $$
When i use the Picard-Iteration, i get ($ s:= begin{pmatrix}
x(t) \
y(t)end{pmatrix} $ )
$$ s_1 = begin{pmatrix}
2 \
-2tend{pmatrix} $$
$$ s_2 = begin{pmatrix}
2-t^2 \
-2tend{pmatrix} $$
$$ s_3 = begin{pmatrix}
2-t^2 \
frac{1}{3} t^3-2tend{pmatrix} $$
I assume that $ s_{infty} = begin{pmatrix}
2cos t \
-2sint end{pmatrix} $
But when i consider the series for sin and cos, for example for the index 1, $ s_1$ will be $ s_1 = begin{pmatrix}
2-t^2 \
frac{1}{3} t^3-2tend{pmatrix} $
How do i get this right?
differential-equations
asked Nov 24 at 15:24
Steven33
144
2
Writing $x(x)$ is confusing (even mildly infuriating), so please use $x(t)$ or something similar.
– MisterRiemann
Nov 24 at 15:26
I changed it. I am sorry.
– Steven33
Nov 24 at 16:53
You get the linear terms correct in the first integration, the quadratic terms in the second, and the cubic terms in the third integration. Why do you think you could get (correct or at all) cubic terms in the first integration?
– LutzL
Nov 24 at 17:21
I mean it differently. When i consider, $ 2 cos t = sum_{k=0}^{infty} frac{(-1)^k t^{2k} }{(2k)!} $, then i get for k=1: $ 2- t^2 ne s_{11}=2$ Dont i have to get the indices synchronised?
– Steven33
Nov 24 at 17:34
add a comment |
i have the following System of Differential Equations
$$ begin{pmatrix}
x'(t) \
y'(t)
end{pmatrix} = begin{pmatrix}
0 && 1\
-1 && 0
end{pmatrix} begin{pmatrix}
x(t) \
y(t)
end{pmatrix} and begin{pmatrix}
x(0) \
y(0)
end{pmatrix}= begin{pmatrix}
2 \
0
end{pmatrix} $$
When i use the Picard-Iteration, i get ($ s:= begin{pmatrix}
x(t) \
y(t)end{pmatrix} $ )
$$ s_1 = begin{pmatrix}
2 \
-2tend{pmatrix} $$
$$ s_2 = begin{pmatrix}
2-t^2 \
-2tend{pmatrix} $$
$$ s_3 = begin{pmatrix}
2-t^2 \
frac{1}{3} t^3-2tend{pmatrix} $$
I assume that $ s_{infty} = begin{pmatrix}
2cos t \
-2sint end{pmatrix} $
But when i consider the series for sin and cos, for example for the index 1, $ s_1$ will be $ s_1 = begin{pmatrix}
2-t^2 \
frac{1}{3} t^3-2tend{pmatrix} $
How do i get this right?
differential-equations
asked Nov 24 at 15:24
Steven33
144
i have the following System of Differential Equations
$$ begin{pmatrix}
x'(t) \
y'(t)
end{pmatrix} = begin{pmatrix}
0 && 1\
-1 && 0
end{pmatrix} begin{pmatrix}
x(t) \
y(t)
end{pmatrix} and begin{pmatrix}
x(0) \
y(0)
end{pmatrix}= begin{pmatrix}
2 \
0
end{pmatrix} $$
When i use the Picard-Iteration, i get ($ s:= begin{pmatrix}
x(t) \
y(t)end{pmatrix} $ )
$$ s_1 = begin{pmatrix}
2 \
-2tend{pmatrix} $$
$$ s_2 = begin{pmatrix}
2-t^2 \
-2tend{pmatrix} $$
$$ s_3 = begin{pmatrix}
2-t^2 \
frac{1}{3} t^3-2tend{pmatrix} $$
I assume that $ s_{infty} = begin{pmatrix}
2cos t \
-2sint end{pmatrix} $
But when i consider the series for sin and cos, for example for the index 1, $ s_1$ will be $ s_1 = begin{pmatrix}
2-t^2 \
frac{1}{3} t^3-2tend{pmatrix} $
How do i get this right?
differential-equations
differential-equations
asked Nov 24 at 15:24
Steven33
144
asked Nov 24 at 15:24
Steven33
144
edited Nov 24 at 15:33
asked Nov 24 at 15:24
Steven33
144
asked Nov 24 at 15:24
Steven33
144
asked Nov 24 at 15:24
Steven33
144
144
2
Writing $x(x)$ is confusing (even mildly infuriating), so please use $x(t)$ or something similar.
– MisterRiemann
Nov 24 at 15:26
I changed it. I am sorry.
– Steven33
Nov 24 at 16:53
You get the linear terms correct in the first integration, the quadratic terms in the second, and the cubic terms in the third integration. Why do you think you could get (correct or at all) cubic terms in the first integration?
– LutzL
Nov 24 at 17:21
I mean it differently. When i consider, $ 2 cos t = sum_{k=0}^{infty} frac{(-1)^k t^{2k} }{(2k)!} $, then i get for k=1: $ 2- t^2 ne s_{11}=2$ Dont i have to get the indices synchronised?
– Steven33
Nov 24 at 17:34
add a comment |
2
Writing $x(x)$ is confusing (even mildly infuriating), so please use $x(t)$ or something similar.
– MisterRiemann
Nov 24 at 15:26
I changed it. I am sorry.
– Steven33
Nov 24 at 16:53
You get the linear terms correct in the first integration, the quadratic terms in the second, and the cubic terms in the third integration. Why do you think you could get (correct or at all) cubic terms in the first integration?
– LutzL
Nov 24 at 17:21
I mean it differently. When i consider, $ 2 cos t = sum_{k=0}^{infty} frac{(-1)^k t^{2k} }{(2k)!} $, then i get for k=1: $ 2- t^2 ne s_{11}=2$ Dont i have to get the indices synchronised?
– Steven33
Nov 24 at 17:34
2
2
Writing $x(x)$ is confusing (even mildly infuriating), so please use $x(t)$ or something similar.
– MisterRiemann
Nov 24 at 15:26
Writing $x(x)$ is confusing (even mildly infuriating), so please use $x(t)$ or something similar.
– MisterRiemann
Nov 24 at 15:26
I changed it. I am sorry.
– Steven33
Nov 24 at 16:53
I changed it. I am sorry.
– Steven33
Nov 24 at 16:53
You get the linear terms correct in the first integration, the quadratic terms in the second, and the cubic terms in the third integration. Why do you think you could get (correct or at all) cubic terms in the first integration?
– LutzL
Nov 24 at 17:21
You get the linear terms correct in the first integration, the quadratic terms in the second, and the cubic terms in the third integration. Why do you think you could get (correct or at all) cubic terms in the first integration?
– LutzL
Nov 24 at 17:21
I mean it differently. When i consider, $ 2 cos t = sum_{k=0}^{infty} frac{(-1)^k t^{2k} }{(2k)!} $, then i get for k=1: $ 2- t^2 ne s_{11}=2$ Dont i have to get the indices synchronised?
– Steven33
Nov 24 at 17:34
I mean it differently. When i consider, $ 2 cos t = sum_{k=0}^{infty} frac{(-1)^k t^{2k} }{(2k)!} $, then i get for k=1: $ 2- t^2 ne s_{11}=2$ Dont i have to get the indices synchronised?
– Steven33
Nov 24 at 17:34
add a comment |
1 Answer
1
active
oldest
votes
There is no guarantee that taking finitely many steps of Picard iteration will result in successively matched Taylor polynomials between the approximant and the solution. This does happen to occur for $y'=y,y(0)=1$ but this is in some sense a coincidence.
answered Nov 25 at 1:18
Ian
67.3k25285
In this case, my solutions work. Its easily to proof. But do you understand my index problem?
– Steven33
Nov 25 at 10:38
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is no guarantee that taking finitely many steps of Picard iteration will result in successively matched Taylor polynomials between the approximant and the solution. This does happen to occur for $y'=y,y(0)=1$ but this is in some sense a coincidence.
answered Nov 25 at 1:18
Ian
67.3k25285
In this case, my solutions work. Its easily to proof. But do you understand my index problem?
– Steven33
Nov 25 at 10:38
add a comment |
There is no guarantee that taking finitely many steps of Picard iteration will result in successively matched Taylor polynomials between the approximant and the solution. This does happen to occur for $y'=y,y(0)=1$ but this is in some sense a coincidence.
answered Nov 25 at 1:18
Ian
67.3k25285
In this case, my solutions work. Its easily to proof. But do you understand my index problem?
– Steven33
Nov 25 at 10:38
add a comment |
There is no guarantee that taking finitely many steps of Picard iteration will result in successively matched Taylor polynomials between the approximant and the solution. This does happen to occur for $y'=y,y(0)=1$ but this is in some sense a coincidence.
answered Nov 25 at 1:18
Ian
67.3k25285
There is no guarantee that taking finitely many steps of Picard iteration will result in successively matched Taylor polynomials between the approximant and the solution. This does happen to occur for $y'=y,y(0)=1$ but this is in some sense a coincidence.
answered Nov 25 at 1:18
Ian
67.3k25285
edited Nov 25 at 1:44
answered Nov 25 at 1:18
Ian
67.3k25285
answered Nov 25 at 1:18
Ian
67.3k25285
answered Nov 25 at 1:18
Ian
67.3k25285
67.3k25285
In this case, my solutions work. Its easily to proof. But do you understand my index problem?
– Steven33
Nov 25 at 10:38
add a comment |
In this case, my solutions work. Its easily to proof. But do you understand my index problem?
– Steven33
Nov 25 at 10:38
In this case, my solutions work. Its easily to proof. But do you understand my index problem?
– Steven33
Nov 25 at 10:38
In this case, my solutions work. Its easily to proof. But do you understand my index problem?
– Steven33
Nov 25 at 10:38
add a comment |
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2
Writing $x(x)$ is confusing (even mildly infuriating), so please use $x(t)$ or something similar.
– MisterRiemann
Nov 24 at 15:26
I changed it. I am sorry.
– Steven33
Nov 24 at 16:53
You get the linear terms correct in the first integration, the quadratic terms in the second, and the cubic terms in the third integration. Why do you think you could get (correct or at all) cubic terms in the first integration?
– LutzL
Nov 24 at 17:21
I mean it differently. When i consider, $ 2 cos t = sum_{k=0}^{infty} frac{(-1)^k t^{2k} }{(2k)!} $, then i get for k=1: $ 2- t^2 ne s_{11}=2$ Dont i have to get the indices synchronised?
– Steven33
Nov 24 at 17:34