Ricardo Berger

2022-04-30

I am approximating a solution to a first order LODE using Euler's method. I made two tables, one using a step size of .01 and another using .05 ( I had to start at x=0 and end at x=1). I am not understanding the directions for the second part of my assignment:

It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $C{h}^{n}$, where $C$ is a generally unknown constant and $n$ is the order. For Euler's method, plot the points:

$(0.1,\varphi (1)-{y}_{10}),$

$(.05,\varphi (1)-{y}_{20}),$

$(.025,\varphi (1)-{y}_{40}),$

$(.0125,\varphi (1)-{y}_{80}),$

$(.00625,\varphi (1)-{y}_{160})$

And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?

Any clarification is appreciated.

Thanks

Edit:

The LODE I am given is ${y}^{\prime}=x+2y,y(0)=1$ and the exact solution I found was $\varphi (x)=\frac{1}{4}(-2x+5{e}^{2x}-1)$.

It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $C{h}^{n}$, where $C$ is a generally unknown constant and $n$ is the order. For Euler's method, plot the points:

$(0.1,\varphi (1)-{y}_{10}),$

$(.05,\varphi (1)-{y}_{20}),$

$(.025,\varphi (1)-{y}_{40}),$

$(.0125,\varphi (1)-{y}_{80}),$

$(.00625,\varphi (1)-{y}_{160})$

And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?

Any clarification is appreciated.

Thanks

Edit:

The LODE I am given is ${y}^{\prime}=x+2y,y(0)=1$ and the exact solution I found was $\varphi (x)=\frac{1}{4}(-2x+5{e}^{2x}-1)$.

Gianna Travis

Beginner2022-05-01Added 11 answers

From the way I see it, if $\varphi $ is the exact solution, the teacher probably wants you to compare the errors at $x=1$ for different step sizes.

Take for example the one before the last

$(0.0125,\varphi (1)-{y}_{80})$

the step size to use here is h such that $n=80$ and $0+nh=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}h=1/80=0.0125$ (I didn't see that coming).

The first number in the bracket is the step size to use for each point. For each point evaluate the approximated value of $\varphi (1)\approx {y}_{n}$ Then ploy the graph of points $({h}_{n},{\u03f5}_{n})$ where ${\u03f5}_{n}=\varphi (1)-{y}_{n}$ and ${h}_{n}$ is the equivalent step size.

Take for example the one before the last

$(0.0125,\varphi (1)-{y}_{80})$

the step size to use here is h such that $n=80$ and $0+nh=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}h=1/80=0.0125$ (I didn't see that coming).

The first number in the bracket is the step size to use for each point. For each point evaluate the approximated value of $\varphi (1)\approx {y}_{n}$ Then ploy the graph of points $({h}_{n},{\u03f5}_{n})$ where ${\u03f5}_{n}=\varphi (1)-{y}_{n}$ and ${h}_{n}$ is the equivalent step size.

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