Taylor Derivation of Euler's Method

From a Differential Equation to a Step Rule

An initial-value problem specifies a differential equation together with a value of the solution at a starting time. In its standard scalar form, write it as

y'(t)=f(t,y(t)), \qquad y(t_0)=y_0.

The equation tells us the slope of the solution at any point $(t,y(t))$. If the exact solution were already known, its slope could be used to estimate how much the solution changes over a short time interval. The forward Euler method turns that idea into a recurrence: at each step, it uses the slope at the current point to estimate the next solution value.

The recurrence is not an arbitrary rule. It follows by keeping the first two terms of the exact solution’s Taylor expansion in time. The derivation also clarifies which slope belongs in the update, what information the method uses, and why the result is an approximation rather than an exact identity.

The Taylor Expansion at the Current Time

Suppose the exact solution $y$ has two continuous derivatives on an interval containing $t$ and $t+h$. Taylor’s theorem, expanded about $t$, gives

y(t+h) = y(t)+h y'(t)+\frac{h^2}{2}y''(\xi)

for some $\xi$ between $t$ and $t+h$. For a forward step, $h>0$, so $t<\xi<t+h$. The formula separates the change over the step into the current value, the first-derivative contribution, and a remaining term involving the second derivative.

To obtain a first-order update, retain the terms through the first derivative and discard the second-derivative term. The differential equation gives $y'(t)=f(t,y(t))$. Therefore the retained expression is

y(t+h)\approx y(t)+h f(t,y(t)).

This expression still refers to the exact solution at $t$. A numerical method instead stores an approximation to that value and uses it to calculate the next approximation.

Definition: The forward Euler method for $y'=f(t,y)$, $y(t_0)=y_0$, uses a fixed step size $h>0$, grid points $t_n=t_0+nh$, and the recurrence $$ y_{n+1}=y_n+h f(t_n,y_n), \qquad y_0=y(t_0). $$ Here $y_n$ is the numerical approximation at time $t_n$; it is not generally the exact value $y(t_n)$.

The word forward describes the move from the current grid point $t_n$ to the later point $t_{n+1}=t_n+h$. The update uses only information available at the start of that step: $t_n$, $y_n$, and the slope $f(t_n,y_n)$.

Why the Taylor Argument Gives This Recurrence

The derivation has two stages. First, Taylor’s theorem describes what the exact solution does over a step. Second, the method replaces the exact value at the start of the step with the value currently stored by the computation. That replacement is what makes the rule usable without knowing the exact solution.

Expand the exact solution.
At a grid point $t_n$, Taylor’s theorem gives $y(t_n+h)=y(t_n)+h y'(t_n)+\frac{h^2}{2}y''(\xi_n)$, for some $\xi_n$ between $t_n$ and $t_n+h$.

Use the differential equation.
The initial-value problem states $y'(t_n)=f(t_n,y(t_n))$. Substituting this into the expansion gives $y(t_n+h)=y(t_n)+h f(t_n,y(t_n))+\frac{h^2}{2}y''(\xi_n)$.

Keep the first-order terms.
Truncating after the first derivative leaves $y(t_n+h)\approx y(t_n)+h f(t_n,y(t_n))$. This is the Taylor-based estimate for the next solution value.

Make the estimate recursive.
Use the computed value $y_n$ in place of the unavailable exact value $y(t_n)$, and set $t_{n+1}=t_n+h$. The result is $y_{n+1}=y_n+h f(t_n,y_n)$.

This derivation explains both the name and the structure of the method. It is a first-order Taylor truncation because it retains the value and first-derivative terms. It is an explicit update because the right-hand side uses the known pair $(t_n,y_n)$, not an unknown value at the end of the step.

For a smooth right-hand side $f$, the second derivative of the exact solution can also be expressed using the chain rule. Along a solution,

y''(t) = \frac{d}{dt}f(t,y(t)) = \frac{\partial f}{\partial t}(t,y(t)) + \frac{\partial f}{\partial y}(t,y(t))\,y'(t) = \frac{\partial f}{\partial t}(t,y(t)) + \frac{\partial f}{\partial y}(t,y(t))\,f(t,y(t)).

This identity illustrates what the discarded Taylor term depends on. The Euler update keeps the first-derivative contribution and does not include that second-derivative contribution. The derivation requires a solution smooth enough for the Taylor expansion; for the displayed chain-rule identity, the relevant partial derivatives of $f$ must also exist and be continuous.

Worked Examples

Worked Example: Two Forward Euler Steps for a Time-Dependent Equation

Apply forward Euler with $h=0.1$ to $y'=t+y$, $y(0)=1$. The initial grid point and value are $t_0=0$ and $y_0=1$.

At the first point, the slope is $f(t_0,y_0)=0+1=1$. Thus

y_1=y_0+h f(t_0,y_0) =1+0.1(1) =1.1, \qquad t_1=0+0.1=0.1.

At the next point, use the new time and the computed value. The slope is $f(t_1,y_1)=0.1+1.1=1.2$, so

y_2=y_1+h f(t_1,y_1) =1.1+0.1(1.2) =1.1+0.12 =1.22, \qquad t_2=0.1+0.1=0.2.

Check each update by recovering its increment: $y_1-y_0=1.1-1=0.1=0.1(1)$, and $y_2-y_1=1.22-1.1=0.12=0.1(1.2)$. Each increment is the step size multiplied by the slope at the corresponding starting point.

Worked Example: Two Steps for an Autonomous Equation

Use $h=0.2$ to take two forward Euler steps for $y'=y(1-y)$, $y(0)=0.5$. This equation is autonomous: its right-hand side depends on $y$, but not explicitly on $t$.

At the initial point, the slope is

$$ f(0,0.5)=0.5(1-0.5)=0.5(0.5)=0.25. $$

The first update gives $t_1=0+0.2=0.2$ and

$$ y_1=0.5+0.2(0.25)=0.5+0.05=0.55. $$

At the second starting point, the slope is $f(t_1,y_1)=0.55(1-0.55)=0.55(0.45)=0.2475$. The second update is

t_2=0.2+0.2=0.4, \qquad y_2=0.55+0.2(0.2475) =0.55+0.0495 =0.5995.

For a direct check, $y_1-y_0=0.55-0.5=0.05=0.2(0.25)$, and $y_2-y_1=0.5995-0.55=0.0495=0.2(0.2475)$. These are precisely the increments prescribed by the recurrence.

Worked Example: Updating Both the Time and the Solution Value

Take two forward Euler steps with $h=0.25$ for $y'=t^2+2y$, starting from $t_0=1$, $y_0=1$. Because the equation depends explicitly on time, each slope must be evaluated using the current time as well as the current solution approximation.

The first slope is $f(1,1)=1^2+2(1)=1+2=3$. Therefore

t_1=1+0.25=1.25, \qquad y_1=1+0.25(3)=1.75.

At $(t_1,y_1)=(1.25,1.75)$, the slope is

$$ f(1.25,1.75) =(1.25)^2+2(1.75) =1.5625+3.5 =5.0625. $$

Hence the second step gives

t_2=1.25+0.25=1.5, \qquad y_2=1.75+0.25(5.0625) =1.75+1.265625 =3.015625.

Check the increments: $1.75-1=0.75=0.25(3)$, and $3.015625-1.75=1.265625=0.25(5.0625)$. In the second slope, the time contribution is $(1.25)^2=1.5625$, not $1^2$; the time coordinate advances along with the solution approximation.

A Common Mistake: Using the Wrong Slope

A tempting but incorrect variation is to evaluate the slope at the end of the step while still treating the rule as forward Euler. For the first example, $y'=t+y$, $y(0)=1$, and $h=0.1$, the starting-point slope is $f(0,1)=1$. The correct update is therefore $1+0.1(1)=1.1$.

If the endpoint time is substituted into the slope while keeping the old solution value, the calculation becomes

$$ 1+0.1 f(0.1,1) = 1+0.1(0.1+1) = 1.11. $$

This is not the update derived from Taylor expansion at $t=0$: the first-derivative term in that expansion is $h y'(0)=h f(0,y(0))$. For the forward Euler rule, evaluate $f$ at the start of the step and use the current numerical value $y_n$. A different choice of where to evaluate the slope would require a separate derivation.

Another simple error is to omit the factor $h$. The slope describes change per unit of time; the product $h f(t_n,y_n)$ estimates the change over a step of length $h$. For example, a slope of $3$ over a step of $0.25$ contributes $0.25(3)=0.75$, not $3$, to the update.

What the Derivation Is Used For

The Taylor derivation makes forward Euler a natural first method for turning a differential equation into a sequence of arithmetic steps. It also gives a reliable way to construct the recurrence for a new initial-value problem: identify the right-hand side $f$, choose the grid spacing $h$, evaluate the slope at the current pair, and add $h$ times that slope to the current approximation.

It is important to keep the roles of the exact and numerical quantities separate. Taylor’s theorem is applied to the exact solution $y(t)$; the method then uses the same first-order expression with computed values $y_n$. The recurrence defines a numerical sequence, and the Taylor argument explains why that sequence is a reasonable step-by-step approximation. The recurrence itself does not assert that $y_{n+1}$ equals the exact value $y(t_{n+1})$.

Key takeaway: Expanding the exact solution at $t_n$, substituting $y'(t_n)=f(t_n,y(t_n))$, and retaining terms through first order gives the forward Euler rule $y_{n+1}=y_n+h f(t_n,y_n)$.

Check Your Understanding

Use the Taylor derivation and the update rule to answer the following questions.

For $y'=f(t,y)$, what terms are retained when the exact solution is expanded about $t_n$ and truncated after its first derivative?
Why does the forward Euler recurrence evaluate $f$ at $(t_n,y_n)$ rather than at the endpoint?
With $h=0.1$, $y'=t+y$, and current values $t_n=0.2$, $y_n=1.3$, calculate the next time and solution approximation.
For $y'=y^2$, $y_0=0.5$, and $h=0.2$, calculate one forward Euler update and check its increment.
What smoothness condition on the exact solution allows the Taylor formula with a second-derivative remainder to be used?