Discover Euler's Method Examples

What are the problems that may arise by the application of Euler method?
2 problems are described. One of them:
1.It is written that from from the convergence order it can be concluded that halving the error means halving the step-size and thus doubling the number of grid points. This leads to the doubling of computational costs which might cause an enormous effort.
Do we halve the step-size in Euler method? For my imagination for Euler method we just define a step size as we wish. As smaller we take as more precise will be our approximation. But where does "halving" come from? Where is my confusion?

I am given the function $x(t)=4\arctan <!--\; \; -->t$ and told that routine computations will show that $x(0)=0$ and $x(1)=\mathrm{\pi <!--\; \pi \; -->}$.
I must determine a differential equation for $x(t)$ of the form $x^{\prime }(t)=f(t,x)$.
I must then use $f$ to approximate $\mathrm{\pi <!--\; \pi \; -->}$.
I will use Euler's Method to find minimum number of iterations ($n$) needed to yield the approximation $\mathrm{\pi <!--\; \pi \; -->}\approx 3.1400$, such that $n-1$ iterations is strictly less than 3.1400.
How do I find $n$? It seems to me that it can't be found given the instruction set that I am provided.

While studying Fourier analysis last semester, I saw an interesting identity:
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{n^2-<!--\; -\; -->{\mathrm{\alpha <!--\; \alpha \; -->}}^2}=\frac{1}{2{\mathrm{\alpha <!--\; \alpha \; -->}}^2}-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2\mathrm{\alpha <!--\; \alpha \; -->}\tan <!--\; \; -->\mathrm{\pi <!--\; \pi \; -->}\mathrm{\alpha <!--\; \alpha \; -->}}$
whenever $\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->\mathbb{C}\setminus <!--\; \setminus \; -->\mathbb{Z}$, which I learned two proofs using Fourier series and residue calculus.
More explicitly, we can deduce the theorem using Fourier series of $f(\mathrm{\theta <!--\; \theta \; -->})=e^{i(\mathrm{\pi <!--\; \pi \; -->}-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->})\mathrm{\alpha <!--\; \alpha \; -->}}$ on $[0,2\mathrm{\pi <!--\; \pi \; -->}]$ or contour integral of the function $g(z)=\frac{\mathrm{\pi <!--\; \pi \; -->}}{(z^2-<!--\; -\; -->{\mathrm{\alpha <!--\; \alpha \; -->}}^2)\tan <!--\; \; -->\mathrm{\pi <!--\; \pi \; -->}z}$ along large circles.
But these techniques, as long as I know, wasn't fully developed at Euler's time.
So what was Euler's method to prove this identity? Is there any proof at elementary level?

The problem I have is the initial value problem
$y^{‴}=x+y$
with
$y(1)=3,y^{\prime }(1)=2,y^{″}(1)=1$
that should be solved with Eulers method using the step length, $h=\frac{1}{2}$.

The iteration step for Eulers method is $y_{n+1}=y_n+hf(x_n,y_n)$. So I should need a system of equations of my initial values I have. I started with substituting:
$U_1=y,U_2=y^{\prime },U_3=y^{″}$
and then
$U_1^{\prime }=U_2,U_2^{\prime }=U_3,U_3^{\prime }=x+U_1$
And it's here I'm stuck, I don't understand how I should start iterate with the step-length from here, I have only encounteblack first-order problem with Eulers method so I would love if someone could point me in the right direction.

Why the staggeblack Euler (Euler-Backward) method is not runge-kutta method?

The method is given by
$x_{n+1}=x_n+hg(p_{n+1})$
$p_{n+1}=p_n+hg(x_n)$
I am not very familiar with the conditions of the Runge-kutta method, can someone help me with this? thank you.

Suppose we have a system of ODE's: a$a^{\prime }=-<!--\; -\; -->a-<!--\; -\; -->2b$ and $b^{\prime }=2a-<!--\; -\; -->b$ with initial conditions $a(0)=1$ and $b(0)=-<!--\; -\; -->1$.

How can we find the maximum value of the step size such that the norm a solution of the system goes to zero (if we apply the forward Euler formula)?

Edit: the main part is to calculate the eigenvalues of the following matrix, based on the Euler method, this becomes
$\left(\begin{array}{cc}-<!--\; -\; -->1-<!--\; -\; -->h& -<!--\; -\; -->2-<!--\; -\; -->2h\\ 2+2h& -<!--\; -\; -->1-<!--\; -\; -->h\end{array}\right)$
The eigenvalues are $(-<!--\; -\; -->1+2i)(1+h)$ and $(-<!--\; -\; -->1-<!--\; -\; -->2i)(1+h)$

I was messing around with a program I wrote to trace through slope fields using Euler's method, and I noticed this equation looked particularly funny:
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{y-<!--\; -\; -->x}{y\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(y)}}$
Is there any solution, implicit or explicit?

$y^{\prime }(t)=f(t,y),a\le <!--\; \le \; -->t\le <!--\; \le \; -->b,y(a)=\mathrm{\alpha <!--\; \alpha \; -->}$
apply for the extrapolation technique
we let $h_0=\frac{h}{2}$ and use Euler method with $w_0=\mathrm{\alpha <!--\; \alpha \; -->}$
$w_1=w_0+h_{0}f(a,w_0)$
and then midpoint method
$t_{i-<!--\; -\; -->1}=a,t_i=a+h_0=a+\frac{h}{2}$ to produce
approximation to $y(a+2h_0)$
$w_2=w_0+2h_{0}f(a+h_0,w_1)$
$y_{1,1}=\frac{1}{2}[w_2+w_1+h_{0}f(a+2h_0,w_2)]$
then why this form results $O(h_0^2)$ approximation to $y(t_1)$

One method consists in computing the numerical solution for an arbitrary $h$ and then $2h$. The Richardson extrapolation gives an estimate of $e=\underset{t}{max}|y(t,2h)-<!--\; -\; -->y(t,h)|$ of the error. When the error is smaller than the tolerance, we keep the result and start from $2y(t,h)-<!--\; -\; -->y(t,h)$. If the error is larger we restart with $h/2$ until we reach the tolerance.
( $y(t,2h)$ means approximation with $2h$)
I don't understand why the Richardson extrapolation is mentioned. For what do I have to use it? Can I not just calculate $y(t,2h)$ and $y(t,h)$ and see the error?

A ball is dropped from the top of a tall building. If air resistance is taken into account, recall that the downward velocity v of the ball is modeled by the differential equation:
$dv/dt=g-<!--\; -\; -->pv$
where g = 9.8 $m/se{c}^2$ is the acceleration due to gravity, and p = 0.1 is the drag coefficient. Assuming the initial velocity of the ball is zero, use Euler’s method with a step size of h = .25 sec to estimate the velocity of the ball after one second. Keep track of three decimal places during the calculation.

I just started learning the Euler's method. Can someone help me with this problem. I know that:
$dv/dt=f(v,t)=g-<!--\; -\; -->pv$
I plug the numbers given into the equation: $dv/dt=9.8-<!--\; -\; -->.25v$
change of x = step size (h) = .25
I think $(v(0),t(0))=(0,0)$??

In particular, if Euler's method is implemented on a computer, what's the minimum step size that can be used before rounding errors cause the Euler approximations to become completely unreliable?
I presume it's when step size reaches the machine epsilon? E.g. if machine epsilon is e-16, then once step size is roughly e-16, the Euler approximations are unreliable.

Consider the boundary-value problem
$\begin{array}{r}{y}^{″}=y^3+x,y(a)=\mathrm{\alpha <!--\; \alpha \; -->},y(b)=\mathrm{\beta <!--\; \beta \; -->},a\le <!--\; \le \; -->x\le <!--\; \le \; -->b\end{array}$
To use the shooting method to solve this problem, one needs a starting guess for the initial slope $y\prime (a)$. One way to obtain such a starting guess for the initial slope is, in effect, to do a ”preliminary shooting” in which we take a single step of Euler’s method with $h=b-a$.
(a) Using this approach, write out the resulting algebraic equation for the initial slope.
(b) What starting value for the initial slope results from this approach?
I'm really not sure how to begin here; my idea was to write the 2nd order ODE using a new variable, such as: $y^{\prime }=u$, $u^{\prime }=y^{″}=y^3+x$. Then, maybe I could use Euler's method to solve for $u=y^{\prime }$; however, I'm not sure how to go about doing this. Any help appreciated.

Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized:
$y^{″}-<!--\; -\; -->2y^{\prime }+y=\sin <!--\; \; -->(t)$
The way that I solved this doesn't involve Euler's formula, and I was wondering how I might use the formula here.

My approach:
The formula can be written as $y(t)=y_h(t)+y_p(t)$ where $y_h(t)$ is the "homogeneous version" of the ODE and $y_p(t)$ is the particular solution that we'll obtain via the basic rule of the method of undetermined coefficients.

$y_h(t)$:
Putting $r(t)=\sin <!--\; \; -->(t)=0$ in the original equation, the ODE we need to solve is:
$y^{″}-<!--\; -\; -->2y^{\prime }+y=0$
where we can set the general solution as $y=e^{\mathrm{\lambda <!--\; \lambda \; -->}t}$ and obtain the characteristic equation:
${\mathrm{\lambda <!--\; \lambda \; -->}}^2-<!--\; -\; -->2\mathrm{\lambda <!--\; \lambda \; -->}+1=0$
which has a real double root, hence giving us the solution:
$y_h(t)=(c_1+c_{2}t)e^t$

$y_p(t)$:
Judging by the fact that $r(t)$ is shape $k\sin <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t)$ and we know that $\mathrm{\omega <!--\; \omega \; -->}=1$ we can set the general solution to be of form:
$\begin{array}{rl}{y}_p(t)& =\phantom{-<!--\; -\; -->}K\cos <!--\; \; -->(t)+M\sin <!--\; \; -->(t)\\ y_p^{\prime }(t)& =-<!--\; -\; -->K\sin <!--\; \; -->(t)+M\cos <!--\; \; -->(t)\\ y_p^{″}(t)& =-<!--\; -\; -->K\cos <!--\; \; -->(t)-<!--\; -\; -->M\sin <!--\; \; -->(t)\end{array}$
substituting these equations into the original equation and then simplifying gives us:
$y_p(t)=\frac{1}{2}\cos <!--\; \; -->(t)$
And in conclusion, we can write that the solution to the given ODE is:
$\begin{array}{rl}y(t)& =y_h(t)+y_p(t)\\ & =(c_1+c_{2}t)e^t+\frac{1}{2}\cos <!--\; \; -->(t)\end{array}$
How would we be able to derive this conclusion via Euler's formula? Thanks in advance.

I have the following initial value problem on $t\in <!--\; \in \; -->[0,T]$:
$y^{\prime }=y^2(1-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}y)$
I wish to use the Trapezoidal method to solve this I.V.P., and I cannot use simply use the pblackictor-corrector method, i.e. improved Euler's method. The trapezoidal method is defined by:
$y_{n+1}=y_n+\frac{h}{2}(f_{n+1}+f_n)$
Plugging in the value from the O.D.E., we have after some simplification:
$y_{n+1}=y_n+\frac{h}{2}(y_{n+1}^2+y_n^2-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}{y}_{n+1}^3-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}{y}_n^3)$
This is an implicit method, so I must use a root finding method at each iteration to find the $y_{n+1}$ on the right side of the equation. I have read online that Newton's method is usually used for such purposes, but I cannot figure out how to implement it.
$y_{n+1}=y_n-<!--\; -\; -->\frac{f(y_n)}{f^{\prime }(y_n)}$
For example, what values would $f(y_n)$ and $f^{\prime }(y_n)$ take in this context? Thank you in advance!

The linear system $y^{\prime }=Ay,y(0)=y_0$, where $A$ is a symmetric matrix, is solved by the Euler method.
Let $e_n=y_n-<!--\; -\; -->y(nh)$
Where $y_n$ denotes the Euler approximation and $y(nh)$ the exact solution ($h$ is Euler step size).
Prove that $||e_n|{|}_2=||y_0|{|}_{2}max|(1+h\mathrm{\lambda <!--\; \lambda \; -->}{)}^n-<!--\; -\; -->e^{nh\mathrm{\lambda <!--\; \lambda \; -->}}|$
Where $\mathrm{\lambda <!--\; \lambda \; -->}\in <!--\; \in \; -->\mathrm{\sigma <!--\; \sigma \; -->}(A)$ where $\mathrm{\sigma <!--\; \sigma \; -->}(A)$ is the set of eigenvalues of $A$.
I have tried various approaches such as writing $e_n$ as the error bound of the Euler method and taking the norm, but I can't seem to get $||y_0|{|}_2$ in my answers.

Using the Euler's method (with $h={10}^{-<!--\; -\; -->n}$) to find $y(1)$
$y^{\prime }=\frac{\sin <!--\; \; -->(x)}{x}$
Since
$y^{\prime }(x)=\frac{\sin <!--\; \; -->(x)}{x}$
then
$f(x,y)=\frac{\sin <!--\; \; -->(x)}{x}$
I know
$y_{n+1}=y_n+h\cdot <!--\; \cdot \; -->f(x_n,y_n)$
and given $y(0)=0$, so
$x_0=y_0=0$
Therefore,
$x_1=x_0+h=0+{10}^{-<!--\; -\; -->0}=1\Rightarrow <!--\; \Rightarrow \; -->y_1=y(x_1)=y(1)$
Then,
$y_1=0+{10}^{-<!--\; -\; -->0}\cdot <!--\; \cdot \; -->f(x_0,y_0)$
$y_1=f(0,0)$
But
$f(0,0)=\frac{\sin <!--\; \; -->(0)}{0}=\frac{0}{0}$
How am I supposed to do it?

I'm asked to decide if I should solve the system
$\stackrel{\dot{}<!--\; \dot{}\; -->}{y}=\left(\begin{array}{cc}-<!--\; -\; -->600& 400\\ 400& -<!--\; -\; -->600\end{array}\right)y,t\in <!--\; \in \; -->[t_0,t_e],y(t_0)=y_0$
with either the explicit Euler method or the implicit Euler method.

Using the explicit Euler method I would get the updating scheme
$y_{n+1}=\left(\begin{array}{cc}1-<!--\; -\; -->600h& 400h\\ 400h& 1-<!--\; -\; -->600h\end{array}\right)y_n$
where the eigenvalues of the driving matrix is
${\mathrm{\lambda <!--\; \lambda \; -->}}_1=401-<!--\; -\; -->600h,$
${\mathrm{\lambda <!--\; \lambda \; -->}}_2=-<!--\; -\; -->399-<!--\; -\; -->600h.$
For the solution to be stable these need to be less than one which gives the conditions
$h\ge <!--\; \ge \; -->\frac{4}{6}=\frac{2}{3},$
$h\ge <!--\; \ge \; -->-<!--\; -\; -->\frac{4}{6}=-<!--\; -\; -->\frac{2}{3}.$
The last condition doesn't say anything but the first condition seems restrictive since I can't choose $h$ as small as possible.

If I instead were to use the implicit Euler method I would get the updating scheme
$y_{n+1}={\left(\begin{array}{cc}1-<!--\; -\; -->600h& 400h\\ 400h& 1-<!--\; -\; -->600h\end{array}\right)}^{-<!--\; -\; -->1}{y}_n.$
Now I can't solve for the eigenvalues of this system but I've heard the implicit Euler is unconditionally stable so it shouldn't matter.

So is the answer that I should choose implicit Euler because it is unconditionally stable or am I missing something? The order of consistency of both is 1 so that should not matter.

I have already posted question where I was asking about sketching Euler method.
The explicit Euler method for numerically solving the begining values of differential equation $x\prime =f(t,x),x(t_0)=x_0$ on the interval $I=[t_0,T]$ is given by
$x_{k+1}=x_k+hf(t_k,x_k),k=0,\dots ,N-1$ with $h=(T-<!--\; -\; -->t_0)/N,N\in N.$
$X_k$ is an approximation of the exact solution $x(t)$ of the begining values at time $t_k:=t_0+kh,k=0,...,N.$ By linear interpolation between the points $(t_k,x_k)$ and $(t_{k+1},x_{k+1}),k=0,...,N-<!--\; -\; -->1,$ we obtain a approximation solution $x_h(t)$.
I need to calculate an approximation of the solution of the begining values at the point $t=1$
$x^{\prime }=-<!--\; -\; -->t/x,x(0)=1$
I need to use h = 0.5. Specify $x_h$(1) and calculate the error, that is difference $x_h$(1)−$x$(1), where $x$(1) is the value of the exact solution.
I really don't know how to start. How can I calulate $x$ or $x_h$ at all?

Approximating $y^{\prime }(t^n)$ at the relation $y^{\prime }(t^n)=f(t^n,y(t^n))$ with the difference quotient $\left[\frac{y(t^{n+1})-<!--\; -\; -->y(t^n)}{h}\right]$ we get to the Euler method.
Approximating the same derivative with the quotient $\left[\frac{y(t^n)-<!--\; -\; -->y(t^{n-<!--\; -\; -->1})}{h}\right]$ we get to the backward Euler method
$y^{n+1}=y^n+hf(t^{n+1},y^{n+1}),n=0,\dots <!--\; \dots \; -->,N-<!--\; -\; -->1$
where $y^0:=y_0$.
In order to find the formula for the forward Euler method, we use the limit $\underset{h\to <!--\; \to \; -->0}{lim}\frac{y(x_0+h)-<!--\; -\; -->y(x_0)}{h}$ for $x_0=t^n,h=t^{n+1}-<!--\; -\; -->t^n$.
In order to find the formula for the backward Euler method, could we pick $h=t^{n-<!--\; -\; -->1}-<!--\; -\; -->t^n$ although it is negative?
Or how do we get otherwise to the approximation:
$y^{\prime }(t^n)\approx <!--\; \approx \; -->\frac{y(t^n)-<!--\; -\; -->y(t^{n-<!--\; -\; -->1})}{h}$

i have this differential equation and actually i am not sure if i have solved it right with Euler's improved method :
$z^{″}=f_z-<!--\; -\; -->C_z\ast <!--\; \ast \; -->|z^{\prime }|\ast <!--\; \ast \; -->z^{\prime }$
$z^{\prime }=u$
$u^{\prime }=z^{″}=f_z-<!--\; -\; -->C_z\ast <!--\; \ast \; -->|u|\ast <!--\; \ast \; -->u$
Improved Euler Method :
$K_1=fz-<!--\; -\; -->Cz\ast <!--\; \ast \; -->|u_n|\ast <!--\; \ast \; -->u_n$
$K_2=fz-<!--\; -\; -->(C_z\ast <!--\; \ast \; -->|u_n+h\ast <!--\; \ast \; -->K_1|)\ast <!--\; \ast \; -->(u_n+h\ast <!--\; \ast \; -->K_1)$
so we get :
$z_{n+1}=zn+h\ast <!--\; \ast \; -->u_n$
$u_{n+1}=u_n+\frac{h}{2}\ast <!--\; \ast \; -->(K_1+h\ast <!--\; \ast \; -->K_2)$
Is this right ?
P.S (fz , Cz are just variables with numbers not function inputs)