Problem 4
edit
One wants to solve the equation
x
+
ln
x
=
0
{\displaystyle x+\ln x=0\!\,}
r
≈
0.5
{\displaystyle r\approx 0.5\!\,}
x
k
+
1
=
−
ln
x
k
{\displaystyle x_{k+1}=-\ln x_{k}\!\,}
x
k
+
1
=
e
−
x
k
{\displaystyle x_{k+1}=e^{-x_{k}}\!\,}
x
k
+
1
=
x
k
+
e
−
x
k
2
{\displaystyle x_{k+1}={\frac {x_{k}+e^{-x_{k}}}{2}}\!\,}
Problem 4a
edit
Which of the three methods can be used?
Solution 4a
edit
Methods {ii} and {iii} are appropriate fixed point iterations for
x
+
ln
(
x
)
=
0
{\displaystyle x+\ln(x)=0\!\,}
{i} is not a fixed point iteration
edit
To be a suitable fixed point iteration, the range of each iteration must be within the domain of the next iteration. In the case of {i},
x
k
+
1
=
−
ln
x
k
{\displaystyle x_{k+1}=-\ln x_{k}\!\,}
(
−
∞
,
∞
)
{\displaystyle (-\infty ,\infty )\!\,}
(
0
,
∞
)
{\displaystyle (0,\infty )\!\,}
{ii} and {iii} are fixed point iterations
edit
Let
f
(
x
)
:=
e
−
x
{\displaystyle f(x):=e^{-x}\!\,}
g
(
x
)
:=
x
+
e
−
x
2
{\displaystyle g(x):={\frac {x+e^{-x}}{2}}\!\,}
f
:
(
−
∞
,
∞
)
→
(
0
,
∞
)
{\displaystyle f:(-\infty ,\infty )\to (0,\infty )\!\,}
and
g
:
(
−
∞
,
∞
)
→
(
0
,
∞
)
{\displaystyle g:(-\infty ,\infty )\to (0,\infty )\!\,}
D
=
(
0
,
1
)
{\displaystyle D=(0,1)}
|
f
(
x
)
−
f
(
y
)
|
≤
|
f
′
(
η
)
|
|
x
−
y
|
{\displaystyle |f(x)-f(y)|\leq |f'(\eta )||x-y|\!\,}
η
∈
D
{\displaystyle \eta \in D\!\,}
where
max
η
∈
D
|
f
′
(
η
)
|
=
max
η
∈
D
|
e
−
x
|
≤
1
{\displaystyle \max _{\eta \in D}|f'(\eta )|=\max _{\eta \in D}|e^{-x}|\leq 1\!\,}
|
g
(
x
)
−
g
(
y
)
|
≤
|
g
′
(
η
)
|
|
x
−
y
|
{\displaystyle |g(x)-g(y)|\leq |g'(\eta )||x-y|\!\,}
η
∈
D
{\displaystyle \eta \in D\!\,}
where
max
η
∈
D
|
g
′
(
η
)
|
=
max
η
∈
D
|
1
2
(
1
−
e
−
x
)
|
≈
0.316060279
{\displaystyle \max _{\eta \in D}|g'(\eta )|=\max _{\eta \in D}|{\frac {1}{2}}(1-e^{-x})|\approx 0.316060279\!\,}
Problem 4b
edit
Which method should be used?
Solution 4b
edit
For the interval (0,1), {iii} is a better fixed point iteration than {ii} since its Lipschitz constant is smaller.
Problem 4c
edit
Give an even better iterative formula.
Solution 4c
edit
Newton's method gives an iterative formula that has quadratic convergence, compared to {iii}'s linear convergence.
Problem 5
edit
Consider the boundary value problem
{
−
u
″
+
e
u
=
0
0
<
x
<
1
u
(
0
)
=
u
(
1
)
=
0
{\displaystyle {\begin{cases}-u''+e^{u}=0&0<x<1\\u(0)=u(1)=0&\end{cases}}\!\,}
A
U
h
+
F
h
(
U
h
)
=
0
{\displaystyle AU_{h}+F_{h}(U_{h})=0\!\,}
U
h
∈
R
m
{\displaystyle U_{h}\in \mathbb {R} ^{m}\!\,}
A
{\displaystyle A\!\,}
m
×
m
{\displaystyle m\times m\!\,}
h
{\displaystyle h\!\,}
F
h
:
R
m
→
R
m
{\displaystyle F_{h}:\mathbb {R} ^{m}\rightarrow \mathbb {R} ^{m}\!\,}
Solution 5
edit
Finding the weak form
edit
Let
V
=
H
0
1
(
0
,
1
)
:=
{
v
∈
H
1
:
v
(
0
)
=
v
(
1
)
=
0
}
{\displaystyle V=H_{0}^{1}(0,1):=\{v\in H^{1}:v(0)=v(1)=0\}\!\,}
Multiplying by a test function
v
∈
V
{\displaystyle v\in V\!\,}
∫
0
1
(
−
u
″
v
)
d
x
=
∫
0
1
u
′
v
′
d
x
−
u
′
v
|
0
1
=
∫
0
1
u
′
v
′
d
x
{\displaystyle \int _{0}^{1}(-u''v)dx=\int _{0}^{1}u'v'dx-\left.u'v\right|_{0}^{1}=\int _{0}^{1}u'v'dx\!\,}
Then the Variational formulation is: Find
u
∈
V
{\displaystyle u\in V\!\,}
a
(
u
,
v
)
=
∫
0
1
u
′
v
′
d
x
+
∫
0
1
e
u
v
d
x
=
l
(
v
)
=
0
{\displaystyle a(u,v)=\int _{0}^{1}u'v'dx+\int _{0}^{1}e^{u}vdx=l(v)=0\!\,}
v
∈
V
{\displaystyle v\in V\!\,}
Define piecewise linear basis functions
edit
Let's consider a partition of the interval
[
0
,
1
]
{\displaystyle [0,1]\!\,}
τ
h
:
0
=
x
0
<
x
1
<
⋯
<
x
m
+
1
=
1
{\displaystyle {\mathcal {\tau }}_{h}:0=x_{0}<x_{1}<\dots <x_{m+1}=1\!\,}
V
h
=
{
v
∈
V
:
v
|
[
x
i
−
1
,
x
i
]
∈
P
1
(
[
x
i
−
1
,
x
i
]
)
,
i
=
1
,
…
,
m
+
1
,
v
(
0
)
=
v
(
1
)
=
0
}
{\displaystyle V_{h}=\{v\in V:\left.v\right|_{[x_{i-1},x_{i}]}\in {\mathcal {P}}_{1}([x_{i-1},x_{i}]),i=1,\dots ,m+1,v(0)=v(1)=0\}\!\,}
V
h
{\displaystyle V_{h}\!\,}
{
ϕ
j
}
j
=
1
m
{\displaystyle \{\phi _{j}\}_{j=1}^{m}\!\,}
j
=
1
,
2
,
…
,
m
{\displaystyle j=1,2,\ldots ,m\!\,}
ϕ
j
(
x
)
=
{
x
−
x
j
−
1
h
for
x
∈
[
x
j
−
1
,
x
j
]
x
j
+
1
−
x
h
for
x
∈
[
x
j
,
x
j
+
1
]
0
otherwise
{\displaystyle \phi _{j}(x)={\begin{cases}{\frac {x-x_{j-1}}{h}}&{\mbox{for }}x\in [x_{j-1},x_{j}]\\{\frac {x_{j+1}-x}{h}}&{\mbox{for }}x\in [x_{j},x_{j+1}]\\0&{\mbox{otherwise}}\end{cases}}\!\,}
Therefore,
ϕ
j
′
(
x
)
=
{
1
h
for
x
∈
[
x
j
−
1
,
x
j
]
−
1
h
for
x
∈
[
x
j
,
x
j
+
1
]
0
otherwise
{\displaystyle \phi _{j}'(x)={\begin{cases}{\frac {1}{h}}&{\mbox{for }}x\in [x_{j-1},x_{j}]\\-{\frac {1}{h}}&{\mbox{for }}x\in [x_{j},x_{j+1}]\\0&{\mbox{otherwise}}\end{cases}}\!\,}
u
h
∈
V
h
{\displaystyle u_{h}\in V_{h}\!\,}
a
(
u
h
,
v
h
)
=
∫
0
1
u
h
′
v
h
′
d
x
+
∫
0
1
e
u
h
v
h
d
x
=
0
,
∀
v
h
∈
V
h
{\displaystyle a(u_{h},v_{h})=\int _{0}^{1}u_{h}'v_{h}'dx+\int _{0}^{1}e^{u_{h}}v_{h}dx=0,\forall v_{h}\in V_{h}\!\,}
{
ϕ
j
}
j
=
1
m
{\displaystyle \{\phi _{j}\}_{j=1}^{m}\!\,}
V
h
{\displaystyle V_{h}\!\,}
u
h
=
∑
j
=
1
n
u
h
j
ϕ
j
{\displaystyle u_{h}=\sum _{j=1}^{n}u_{hj}\phi _{j}\!\,}
u
h
=
(
u
h
1
,
…
,
u
h
m
)
t
{\displaystyle u_{h}=(u_{h1},\ldots ,u_{hm})^{t}\!\,}
a
(
u
h
,
ϕ
i
)
=
∑
j
=
1
m
u
h
j
∫
x
i
−
1
x
i
+
1
ϕ
j
′
ϕ
i
′
d
x
+
∫
x
i
−
1
x
i
+
1
e
∑
j
=
1
m
u
h
j
ϕ
j
ϕ
i
d
x
=
0
for
i
=
1
,
…
,
m
{\displaystyle a(u_{h},\phi _{i})=\sum _{j=1}^{m}u_{hj}\int _{x_{i-1}}^{x_{i+1}}\phi _{j}'\phi _{i}'dx+\int _{x_{i-1}}^{x_{i+1}}e^{\sum _{j=1}^{m}u_{hj}\phi _{j}}\phi _{i}dx=0{\mbox{ for }}i=1,\dots ,m\!\,}
Rewrite problem in matrix form
edit
The equivalent discrete problem for
i
=
1
,
2
,
…
,
m
{\displaystyle i=1,2,\ldots ,m\!\,}
∑
j
=
1
m
u
h
j
∫
x
i
−
1
x
i
+
1
ϕ
j
′
ϕ
i
′
d
x
+
∫
x
i
−
1
x
i
+
1
e
∑
j
=
1
m
u
h
j
ϕ
j
ϕ
i
d
x
=
0
{\displaystyle \sum _{j=1}^{m}u_{hj}\int _{x_{i-1}}^{x_{i+1}}\phi _{j}'\phi _{i}'dx+\int _{x_{i-1}}^{x_{i+1}}e^{\sum _{j=1}^{m}u_{hj}\phi _{j}}\phi _{i}dx=0\!\,}
[
2
h
−
1
h
…
⋱
⋱
⋱
−
1
h
2
h
−
1
h
−
1
h
2
h
]
⏟
A
[
u
h
1
⋮
⋮
u
h
m
]
⏟
U
h
+
h
[
e
u
h
1
e
u
h
2
⋮
e
u
h
m
]
⏟
F
h
(
U
h
)
{\displaystyle \underbrace {\begin{bmatrix}{\frac {2}{h}}&-{\frac {1}{h}}&&\ldots \\\ddots &\ddots &\ddots &\\&-{\frac {1}{h}}&{\frac {2}{h}}&-{\frac {1}{h}}\\&&-{\frac {1}{h}}&{\frac {2}{h}}\end{bmatrix}} _{A}\underbrace {\begin{bmatrix}u_{h1}\\\vdots \\\vdots \\u_{hm}\end{bmatrix}} _{U_{h}}+\underbrace {h{\begin{bmatrix}e^{u_{h1}}\\e^{u_{h2}}\\\vdots \\e^{u_{hm}}\end{bmatrix}}} _{F_{h}(U_{h})}\!\,}
∫
0
1
ϕ
j
′
ϕ
i
′
=
{
2
h
for
i
=
j
−
1
h
for
|
i
−
j
|
=
1
0
otherwise
{\displaystyle {\begin{aligned}\int _{0}^{1}\phi _{j}'\phi _{i}'&={\begin{cases}{\frac {2}{h}}&{\mbox{for }}i=j\\-{\frac {1}{h}}&{\mbox{for }}|i-j|=1\\0&{\mbox{otherwise}}\end{cases}}\end{aligned}}\!\,}
∫
x
i
−
1
x
i
+
1
e
u
h
ϕ
i
(
x
)
=
∫
x
i
−
1
x
i
e
u
h
ϕ
i
(
x
)
+
∫
x
i
x
i
+
1
e
u
h
ϕ
i
(
x
)
≈
(
e
u
h
(
x
i
−
1
)
ϕ
i
(
x
i
−
1
)
⏞
0
+
e
u
h
(
x
i
)
ϕ
i
(
x
i
)
2
)
h
+
(
e
u
h
(
x
i
)
ϕ
i
(
x
i
)
+
e
u
h
(
x
i
+
1
)
ϕ
i
(
x
i
+
1
)
⏞
0
2
)
h
=
h
e
u
h
(
x
i
)
ϕ
i
(
x
i
)
⏞
1
=
h
e
u
h
(
x
i
)
=
h
e
∑
j
=
1
m
u
h
j
ϕ
j
(
x
i
)
=
h
e
u
h
i
{\displaystyle {\begin{aligned}\int _{x_{i-1}}^{x_{i+1}}e^{u_{h}}\phi _{i}(x)&=\int _{x_{i-1}}^{x_{i}}e^{u_{h}}\phi _{i}(x)+\int _{x_{i}}^{x_{i+1}}e^{u_{h}}\phi _{i}(x)\\&\approx \left({\frac {e^{u_{h}(x_{i-1})}\overbrace {\phi _{i}(x_{i-1})} ^{0}+e^{u_{h}(x_{i})}\phi _{i}(x_{i})}{2}}\right)h+\left({\frac {e^{u_{h}(x_{i})}\phi _{i}(x_{i})+e^{u_{h}(x_{i+1})}\overbrace {\phi _{i}(x_{i+1})} ^{0}}{2}}\right)h\\&=he^{u_{h}(x_{i})}\overbrace {\phi _{i}(x_{i})} ^{1}\\&=he^{u_{h}(x_{i})}\\&=he^{\sum _{j=1}^{m}u_{hj}\phi _{j}(x_{i})}\\&=he^{u_{hi}}\end{aligned}}\!\,}
u
h
0
=
u
h
(
m
+
1
)
=
0
{\displaystyle u_{h0}=u_{h(m+1)}=0\!\,}
Problem 6
edit
Determine the local order of accuracy and the stability properties of the two-step scheme
x
i
+
2
−
3
x
i
+
1
+
2
x
i
=
Δ
t
⋅
[
13
12
f
(
t
i
+
2
,
x
i
+
2
)
−
5
3
f
(
t
i
+
1
,
x
i
+
1
)
−
5
12
f
(
t
i
,
x
i
)
]
{\displaystyle x_{i+2}-3x_{i+1}+2x_{i}=\Delta t\cdot \left[{\frac {13}{12}}f(t_{i+2},x_{i+2})-{\frac {5}{3}}f(t_{i+1},x_{i+1})-{\frac {5}{12}}f(t_{i},x_{i})\right]\!\,}
x
′
(
t
)
=
f
(
t
,
x
)
{\displaystyle x'(t)=f(t,x)\!\,}
Solution 6
edit
Local Order of Accuracy
edit
Note that
x
i
≈
x
(
t
i
)
{\displaystyle x_{i}\approx x(t_{i})\!\,}
Δ
t
{\displaystyle \Delta t\!\,}
t
i
+
1
=
t
i
+
h
{\displaystyle t_{i+1}=t_{i}+h\!\,}
t
i
+
2
=
t
i
+
2
h
{\displaystyle t_{i+2}=t_{i}+2h\!\,}
x
(
t
i
+
2
h
)
−
3
x
(
t
i
+
h
)
+
2
x
(
t
i
)
≈
h
[
13
12
x
′
(
t
i
+
2
h
)
−
5
3
x
′
(
t
i
+
h
)
−
5
12
x
′
(
t
i
)
]
{\displaystyle x(t_{i}+2h)-3x(t_{i}+h)+2x(t_{i})\approx h\left[{\frac {13}{12}}x'(t_{i}+2h)-{\frac {5}{3}}x'(t_{i}+h)-{\frac {5}{12}}x'(t_{i})\right]\!\,}
Expand Left Hand Side Using Taylor Expansion
edit
Expanding about
t
i
{\displaystyle t_{i}\!\,}
Order
x
(
t
i
+
2
h
)
−
3
x
(
t
i
+
h
)
2
x
(
t
i
)
Σ
0
x
(
t
i
)
−
3
x
(
t
i
)
2
x
(
t
i
)
0
1
2
x
′
(
t
i
)
h
−
3
x
′
(
t
i
)
h
0
−
x
′
(
t
i
)
h
2
4
2
!
x
″
(
t
i
)
h
2
−
3
2
!
x
″
(
t
i
)
h
2
0
1
2
x
″
(
t
i
)
h
2
3
8
3
!
x
‴
(
t
i
)
h
3
−
3
3
!
x
‴
(
t
i
)
h
3
0
5
6
x
‴
(
t
i
)
h
3
4
O
(
h
4
)
O
(
h
4
)
0
O
(
h
4
)
{\displaystyle {\begin{array}{|c|c|c|c|c|}{\mbox{Order}}&x(t_{i}+2h)&-3x(t_{i}+h)&2x(t_{i})&\Sigma \\\hline &&&&\\0&x(t_{i})&-3x(t_{i})&2x(t_{i})&0\\&&&&\\1&2x'(t_{i})h&-3x'(t_{i})h&0&-x'(t_{i})h\\&&&&\\2&{\frac {4}{2!}}x''(t_{i})h^{2}&-{\frac {3}{2!}}x''(t_{i})h^{2}&0&{\frac {1}{2}}x''(t_{i})h^{2}\\&&&&\\3&{\frac {8}{3!}}x'''(t_{i})h^{3}&-{\frac {3}{3!}}x'''(t_{i})h^{3}&0&{\frac {5}{6}}x'''(t_{i})h^{3}\\&&&&\\4&{\mathcal {O}}(h^{4})&{\mathcal {O}}(h^{4})&0&{\mathcal {O}}(h^{4})\\&&&&\\\hline \end{array}}}
Expand Right Hand Side Using Taylor Expansion
edit
Also expanding about
t
i
{\displaystyle t_{i}\!\,}
Order
13
12
h
x
′
(
t
i
+
2
h
)
−
5
3
h
x
′
(
t
i
+
h
)
−
5
12
h
x
′
(
t
i
)
Σ
0
0
0
0
0
1
13
12
h
x
′
(
t
i
)
−
5
3
h
x
′
(
t
i
)
−
5
12
h
x
′
(
t
i
)
−
1
⋅
h
x
′
(
t
i
)
2
13
12
(
2
h
)
h
x
″
(
t
i
)
−
5
3
h
h
x
″
(
t
i
)
0
1
2
h
2
x
″
(
t
i
)
3
13
12
(
2
h
)
2
h
x
‴
(
t
i
)
2
−
5
3
h
2
h
x
‴
(
t
i
)
2
0
4
3
h
3
x
‴
(
t
i
)
4
O
(
h
4
)
O
(
h
4
)
0
O
(
h
4
)
{\displaystyle {\begin{array}{|c|c|c|c|c|}{\mbox{Order}}&{\frac {13}{12}}hx'(t_{i}+2h)&-{\frac {5}{3}}hx'(t_{i}+h)&-{\frac {5}{12}}hx'(t_{i})&\Sigma \\\hline &&&&\\0&0&0&0&0\\&&&&\\1&{\frac {13}{12}}hx'(t_{i})&-{\frac {5}{3}}hx'(t_{i})&-{\frac {5}{12}}hx'(t_{i})&-1\cdot hx'(t_{i})\\&&&&\\2&{\frac {13}{12}}(2h)hx''(t_{i})&-{\frac {5}{3}}hhx''(t_{i})&0&{\frac {1}{2}}h^{2}x''(t_{i})\\&&&&\\3&{\frac {13}{12}}(2h)^{2}{\frac {hx'''(t_{i})}{2}}&-{\frac {5}{3}}h^{2}{\frac {hx'''(t_{i})}{2}}&0&{\frac {4}{3}}h^{3}x'''(t_{i})\\&&&&\\4&{\mathcal {O}}(h^{4})&{\mathcal {O}}(h^{4})&0&{\mathcal {O}}(h^{4})\\\hline \end{array}}}
Compare Terms to Determine Order
edit
Taking the difference of the left and right hand side Taylor expansions show that the two step equation is of order two since the terms match up to order 2.
x
i
+
2
−
3
x
i
+
1
+
2
x
i
−
Δ
t
⋅
[
13
12
f
(
t
i
+
2
,
x
i
+
2
)
−
5
3
f
(
t
i
+
1
,
x
i
+
1
)
−
5
12
f
(
t
i
,
x
i
)
]
=
O
(
Δ
t
3
)
{\displaystyle x_{i+2}-3x_{i+1}+2x_{i}-\Delta t\cdot \left[{\frac {13}{12}}f(t_{i+2},x_{i+2})-{\frac {5}{3}}f(t_{i+1},x_{i+1})-{\frac {5}{12}}f(t_{i},x_{i})\right]={\mathcal {O}}(\Delta t^{3})\!\,}
5
6
≠
4
3
{\displaystyle {\frac {5}{6}}\neq {\frac {4}{3}}\!\,}
Stability Condition
edit
Our two-step equation is stable if the roots of the equation
λ
2
−
3
λ
+
2
=
0
{\displaystyle \lambda ^{2}-3\lambda +2=0\!\,}
|
λ
j
|
≤
1
{\displaystyle |\lambda _{j}|\leq 1\!\,}
|
λ
j
|
=
1
{\displaystyle |\lambda _{j}|=1\!\,}
λ
2
−
3
λ
+
2
=
(
λ
−
1
)
(
λ
−
2
)
=
0
{\displaystyle \lambda ^{2}-3\lambda +2=(\lambda -1)(\lambda -2)=0\!\,}
Convergence
edit
A multi-step method is convergent if and only if it is stable and consistent. Our two-step equation is not stable, hence not guaranteed to converge.
![{\displaystyle [0,1]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b26fc5fee30dada6f15cc080115cae582606d7f3)
![{\displaystyle V_{h}=\{v\in V:\left.v\right|_{[x_{i-1},x_{i}]}\in {\mathcal {P}}_{1}([x_{i-1},x_{i}]),i=1,\dots ,m+1,v(0)=v(1)=0\}\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8b81ea4266098e485c27de3c2878513ee2ecac2)
![{\displaystyle \phi _{j}(x)={\begin{cases}{\frac {x-x_{j-1}}{h}}&{\mbox{for }}x\in [x_{j-1},x_{j}]\\{\frac {x_{j+1}-x}{h}}&{\mbox{for }}x\in [x_{j},x_{j+1}]\\0&{\mbox{otherwise}}\end{cases}}\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e560654509e1fde1e76a33b04b42c4d8dc1c90d)
![{\displaystyle \phi _{j}'(x)={\begin{cases}{\frac {1}{h}}&{\mbox{for }}x\in [x_{j-1},x_{j}]\\-{\frac {1}{h}}&{\mbox{for }}x\in [x_{j},x_{j+1}]\\0&{\mbox{otherwise}}\end{cases}}\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce07be9836a28d50878c62470d828ee9128d3e7c)
![{\displaystyle x_{i+2}-3x_{i+1}+2x_{i}=\Delta t\cdot \left[{\frac {13}{12}}f(t_{i+2},x_{i+2})-{\frac {5}{3}}f(t_{i+1},x_{i+1})-{\frac {5}{12}}f(t_{i},x_{i})\right]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d38745bdaf8a8d151ca4d7f9423c5ef07c414c93)
![{\displaystyle x(t_{i}+2h)-3x(t_{i}+h)+2x(t_{i})\approx h\left[{\frac {13}{12}}x'(t_{i}+2h)-{\frac {5}{3}}x'(t_{i}+h)-{\frac {5}{12}}x'(t_{i})\right]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4cfef726050f99bd552051e47004cecba8fde56)
![{\displaystyle x_{i+2}-3x_{i+1}+2x_{i}-\Delta t\cdot \left[{\frac {13}{12}}f(t_{i+2},x_{i+2})-{\frac {5}{3}}f(t_{i+1},x_{i+1})-{\frac {5}{12}}f(t_{i},x_{i})\right]={\mathcal {O}}(\Delta t^{3})\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/689447336d26a301b7d764908c0cb4ee00c85ccf)
