= Euler method
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{title2=$y_{n+1}=y_n+h f(t_n,y_n)$}
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= Explicit Euler method
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= Forward Euler method
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The <Euler method> approximates an <initial value problem> $y'=f(t,y)$, $y(t_0)=y_0$, by evaluating the <derivative> at the current point and moving along that <tangent line> for a time $h$. With $t_n=t_0+nh$, its displayed update needs one evaluation of $f$ per step. This simple rule turns local slope information into a whole approximate trajectory. MIT's course notes derive it by dropping the second-order <Taylor remainder>.https://web.mit.edu/10.001/Web/Course_Notes/Differential_Equations_Notes/node3.html[MIT 10.001, Forward and Backward Euler Methods, equations 6–7]{ref}
Take $y'=-2y$, $y(0)=1$, whose exact solution is the <exponential function> $y(t)=e^{-2t}$. The <Euler method> gives
$$
y_{n+1}=(1-2h)y_n,\qquad y_n=(1-2h)^n.
$$
At $h=0.2$, the first values are $1,0.6,0.36,0.216$. At $t=1$, five steps give $0.6^5=0.07776$, compared with $e^{-2}\approx0.135335$. The result decays but is too small: <absolute stability> alone does not ensure accuracy. At $h=1.2$, the first values are $1,-1.4,1.96,-2.744$, and at $t=6$ five steps give $-5.37824$, while the exact solution is about $0.00000614421$.
\Image[/euler-stability-example.png]
{title=Euler approximations to exponential decay on the same finite time interval}
For the <Dahlquist test equation> $y'=\lambda y$, the <Euler method> has <stability function> $R(z)=1+z$ with $z=h\lambda$. Its <absolute stability> region is $|1+z|\leq1$, the closed disk centred at $-1$ with radius one. For the present real decay problem and $h>0$, bounded numerical values require $0<h\leq1$; strict decay requires $0<h<1$. At $h=1$ the values alternate $1,-1,1,\ldots$ without decay. These conditions follow directly from the <amplification factor> $1-2h$, matching MIT's criterion for $y'=-ay$.https://web.mit.edu/10.001/Web/Course_Notes/Differential_Equations_Notes/node3.html[MIT 10.001, Forward and Backward Euler Methods, linear decay example]{ref}
The <Euler method> is first order: with bounded second <derivative> and suitable <Lipschitz continuity>, its <global error> is $O(h)$ on a fixed interval. In this example the <logarithm> gives $\log(1-2h)=-2h-2h^2+O(h^3)$ for small positive $h$. At $t=nh$, this means $y_n=e^{-2t}(1-2th+O(h^2))$. It explains both the downward bias and why halving the step roughly halves the error once $h$ is sufficiently small.
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