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# 数学代写|数值分析代写Numerical analysis代考|MATH/CS514 One-step methods

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## 数学代写数值分析代写Numerical analysis代考|One-step methods

We are prepared to discuss some numerical methods to approximate the solution to all these ODEs. To simplify the notation, we present our methods in the context of the scalar equation
$$x^{\prime}(t)=f(t, x(t))$$
with the initial condition $x\left(t_0\right)=x_0$. All the algorithms generalize trivially to systems: simply replace scalars with vectors.

When computing approximate solutions to the initial value problem, we will not obtain the solution for every value of $t>t_0$, but only on a discrete grid. In particular, we will generate approximate solutions at some regular grid of time steps
$$t_k=t_0+k h$$
for some constant step-size $h$. (The methods we consider in this subsection allow $h$ to change with each step size, so one actually has $t_k=t_{k-1}+h_k$. For simplicity of notation, we will assume for now that the step-size is fixed.)

The approximation to $x$ at the time $t_k$ is denoted by $x_k$, so hopefully
$$x_k \approx x\left(t_k\right)$$
Of course, the initial data is exact:
$$x_0=x\left(t_0\right) .$$

## 数学代写|数值分析代写Numerical analysis代考|Euler’s method

We need some approximation that will advance from the exact point on the solution curve, $\left(t_0, x_0\right)$ to time $t_1$. From basic calculus we know that
$$x^{\prime}(t)=\lim _{h \rightarrow 0} \frac{x(t+h)-x(t)}{h} .$$
This definition of the derivative inspires our first method. Apply it at time $t_0$ with a small but finite time step $h>0$ to obtain
$$x^{\prime}\left(t_0\right) \approx \frac{x\left(t_0+h\right)-x\left(t_0\right)}{h} .$$
Since $x^{\prime}\left(t_0\right)=f\left(t_0, x\left(t_0\right)\right)=f\left(t_0, x_0\right)$, we know the quantity on the left hand side of this approximation. The only quantity we don’t know is $x\left(t_0+h\right)=x\left(t_1\right)$. Rearrange the above to put $x\left(t_1\right)$ on the left hand side:
$$x\left(t_1\right) \approx x\left(t_0\right)+h x^{\prime}\left(t_0\right)=x_0+h f\left(t_0, x_0\right) .$$
This approximation is precisely the one suggested by the direction field discussion in Section 5.1.1. There, to progress from the starting point $\left(t_0, x_0\right)$, we followed the line of slope $f\left(t_0, x_0\right)$ some distance, which in the present context is our step size, $h$. To progress from the new point, $\left(t_1, x_1\right)$, we follow a new slope, $f\left(t_1, x_1\right)$, giving the iteration
$$x_2=x_1+h f\left(t_1, x_1\right) \text {. }$$

## 数学代写数值分析代写Numerical analysis代考|One-step methods

$$x^{\prime}(t)=f(t, x(t))$$

$$t_k=t_0+k h$$

$$x_k \approx x\left(t_k\right)$$

$$x_0=x\left(t_0\right)$$

## 数学代写|数值分析代写Numerical analysis代考|Euler’s method

$$x^{\prime}(t)=\lim _{h \rightarrow 0} \frac{x(t+h)-x(t)}{h} .$$

$$x^{\prime}\left(t_0\right) \approx \frac{x\left(t_0+h\right)-x\left(t_0\right)}{h} .$$

$$x\left(t_1\right) \approx x\left(t_0\right)+h x^{\prime}\left(t_0\right)=x_0+h f\left(t_0, x_0\right) .$$

$$x_2=x_1+h f\left(t_1, x_1\right) .$$

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