Why Examine One Step in Isolation?
Forward Euler advances an approximation by using the slope at the beginning of a step. To understand the error introduced by that choice, first isolate a single step: begin at an exact solution value, take one Euler step, and compare the result with the exact value at the end of the step.
This comparison is called a local error because it measures the discrepancy created by one step under an exact starting value. It does not measure the difference between a numerical solution and the exact solution after many steps. That distinction matters: the one-step calculation identifies the error introduced locally, while accumulated errors are a separate question.
Definition and Sign Convention
Consider the initial-value problem \(y'=f(t,y)\), and let \(y(t)\) be an exact solution on an interval containing \(t\) and \(t+h\), where \(h>0\). Forward Euler, if started from the exact value \(y(t)\), produces the one-step value \(y(t)+h f(t,y(t))\). We define the signed, unscaled one-step local defect to be the exact endpoint value minus that Euler value.
A positive \(\delta(t,h)\) means the exact endpoint is greater than the Euler value; a negative one means it is smaller. There is a terminology convention worth making explicit: some texts call \(\delta(t,h)\) the local truncation error, while others call the error per unit step, \(\delta(t,h)/h\), the local truncation error. Here the unscaled defect is \(\delta(t,h)\), and we will name the normalized quantity separately.
The Taylor Expansion for One Euler Step
Assume that the exact solution \(y\) has three continuous derivatives on the interval between \(t\) and \(t+h\). Taylor’s theorem with remainder gives a point \(\xi\) between \(t\) and \(t+h\) such that
Because the exact solution satisfies \(y'(t)=f(t,y(t))\), the first two terms on the right are precisely the starting value and slope contribution used by Euler’s method. Subtracting the Euler value therefore leaves
This is the basic local-error formula. If \(y'''\) is bounded on the step, the remainder has magnitude at most a constant times \(h^3\). Thus, when \(y''(t)\) is nonzero, the leading term in the unscaled defect is \(\tfrac12 y''(t)h^2\). Dividing by \(h\) gives the corresponding normalized formula.
In particular, the unscaled defect is generally of size \(h^2\), while the normalized local truncation error is generally of size \(h\). These statements assume the indicated derivatives exist and are bounded on the step. If \(y''(t)=0\), the displayed leading term vanishes; the error can then be smaller, and its size is determined by later terms or by more detailed information about the solution.
Writing the Leading Term from the Differential Equation
For a scalar equation \(y'=f(t,y)\), suppose \(f\) has continuous first partial derivatives and \(y\) is a solution. The chain rule differentiates the composite function \(f(t,y(t))\), giving
Here \(f_t\) and \(f_y\) are the partial derivatives of \(f\) with respect to its first and second variables. Substitution into the Taylor formula expresses the leading defect using only the equation and its exact starting point.
The term \(f_t\) accounts for change in the slope as time changes, even if the solution value were held fixed. The term \(f_y f\) accounts for change in the slope caused by the solution itself changing. For an autonomous equation, \(f\) does not depend explicitly on \(t\), so \(f_t=0\) and \(y''=f_y f\).
Derivation: Why the Difference Starts at \(h^2\)
The first-order Taylor approximation uses the exact value and the exact slope at the start of the step. Forward Euler uses those same two pieces of information. Consequently, their difference cannot come from the constant or linear Taylor terms: those terms cancel exactly. The first remaining term is the quadratic term. The calculation below makes this cancellation explicit.
By Taylor’s theorem, \(y(t+h)=y(t)+h y'(t)+\tfrac12h^2y''(t)+\tfrac16h^3y'''(\xi)\).
The forward Euler value from the exact starting value is \(y(t)+h f(t,y(t))=y(t)+h y'(t)\).
Subtracting the Euler value removes \(y(t)\) and \(h y'(t)\), leaving \(\delta(t,h)=\tfrac12h^2y''(t)+\tfrac16h^3y'''(\xi)\).
The leading unscaled defect is proportional to \(h^2\); after division by \(h\), the leading normalized error is proportional to \(h\).
This argument requires enough smoothness for the Taylor formula with its stated remainder. If \(y\) has only two continuous derivatives, the Taylor expansion still yields \(\delta(t,h)=\tfrac12h^2y''(t)+o(h^2)\) as \(h\) tends to zero, but the explicit third-derivative remainder need not be available.
Worked Examples
Worked Example: A Time-Dependent Slope
Consider \(y'=t^2\) with \(y(1)=0\), and take one Euler step of size \(h=0.2\) from \(t=1\). The slope at the starting point is \(1^2=1\), so the Euler endpoint value is
An exact solution is \(y(t)=(t^3-1)/3\), since its derivative is \(t^2\) and its value at \(t=1\) is zero. At the endpoint \(t=1.2\),
The signed unscaled defect is therefore
To check the Taylor prediction, \(y''(t)=2t\), so \(y''(1)=2\), and \(y'''(t)=2\). The quadratic contribution is \(\tfrac12(0.2)^2(2)=0.04\), and the cubic remainder is \(\tfrac16(0.2)^3(2)=0.002666\ldots\). Their sum is \(0.042666\ldots\), exactly the defect found above. The normalized error is \(\delta/h=0.213333\ldots\).
Worked Example: Exponential Growth
For \(y'=y\) with \(y(0)=1\), take one step of size \(h=0.1\). The Euler value is \(1+0.1(1)=1.1\). The exact solution is \(y(t)=e^t\), so the exact endpoint value is \(e^{0.1}\). Thus
At the starting point \(y''(0)=1\), and \(y'''(t)=e^t\). Taylor’s theorem predicts a quadratic contribution of \(\tfrac12(0.1)^2(1)=0.005\), plus the remainder \(\tfrac16(0.1)^3e^\xi\) for some \(0<\xi<0.1\). The exact defect is greater than \(0.005\), consistent with that positive remainder. As a direct numerical check, the normalized error is \(0.005170918\ldots/0.1=0.05170918\ldots\), which is close to the leading prediction \(\tfrac12(0.1)y''(0)=0.05\).
Worked Example: A Nonlinear Dependence on the Solution
Consider \(y'=t+y\), \(y(0)=1\), with \(h=0.1\). At the starting point the slope is \(0+1=1\), so forward Euler gives \(y_{\mathrm E}=1+0.1(1)=1.1\). An exact solution is \(y(t)=2e^t-t-1\): it has \(y(0)=1\), and its derivative \(2e^t-1\) equals \(t+y(t)\). Therefore
For \(f(t,y)=t+y\), the partial derivatives are \(f_t=1\) and \(f_y=1\). At \((0,1)\), \(f=1\), so \(y''(0)=f_t+f_yf=1+1(1)=2\). The leading defect is \(\tfrac12(0.1)^2(2)=0.01\), in agreement with the computed defect. For a further check, \(y''(t)=2e^t\), so \(y'''(t)=2e^t\). Taylor’s formula gives the defect as \(0.01+\tfrac16(0.1)^3(2e^\xi)\) for some \(0<\xi<0.1\), a positive number slightly greater than \(0.01\), as computed.
A Common Pitfall: Losing a Factor of the Step Size
A frequent mistake is to report the normalized error when the question asks for the unscaled one-step defect, or vice versa. For the exponential example, the correct unscaled defect is \(e^{0.1}-1.1\), approximately \(0.005170918\). The normalized error is that number divided by \(0.1\), approximately \(0.05170918\). They are different quantities.
For instance, the incorrect calculation “the local defect is \(\tfrac12h y''(0)=0.05\)” has silently divided the unscaled Taylor term by \(h\). It is an approximation to the normalized error, not to the unscaled defect. The correction is to keep the definitions beside the calculation: \(\delta\) begins with \(\tfrac12h^2y''(t)\), while \(\tau=\delta/h\) begins with \(\tfrac12h y''(t)\).
Another important qualification is that a nonzero second derivative is needed for the quadratic term to be the leading term at the chosen starting point. If \(y''(t)=0\), it is not valid to conclude from the leading-term formula alone that the defect has a nonzero coefficient times \(h^2\). The full Taylor formula, including its remainder, tells what can be concluded.
What the Local Formula Tells You
The Taylor calculation explains why a single forward Euler step usually misses the exact endpoint by a quantity proportional to \(h^2\). If the step size is reduced, the leading unscaled defect changes approximately with the square of that step size, provided the solution’s curvature does not change substantially over the comparison. The normalized local truncation error has a leading term proportional to \(h\).
The coefficient also gives a useful interpretation: the local defect depends on the curvature \(y''(t)\) of the exact solution, not only on the slope \(y'(t)\). Where the solution is locally close to a straight line, \(y''(t)\) is small and the leading defect is small. Where the curvature is larger, the leading defect is larger. Its sign is governed by \(y''(t)\): a positive curvature gives a positive leading signed defect, and a negative curvature gives a negative one, for sufficiently small positive \(h\) when that curvature is nonzero.
This is a one-step result with an exact starting value. It does not, by itself, describe the error after repeated Euler steps, nor does it establish a global error bound. Those questions require additional reasoning about how errors behave from step to step. Here the central conclusion is narrower: Taylor expansion identifies the one-step defect and shows exactly where its dependence on the step size comes from.
Check Your Understanding
Answer these questions using the signed-defect convention defined above.
- Write the formula for the unscaled one-step defect when Euler starts from the exact value \(y(t)\).
- What are the leading powers of \(h\) in the unscaled defect and the normalized local truncation error?
- For \(y'=f(t,y)\), express \(y''\) in terms of the partial derivatives of \(f\) and the value of \(f\).
- Why does the constant and linear part of the Taylor expansion cancel when the Euler value is subtracted?
- If \(y''(t)=0\), what does the leading-term formula imply, and what does it not imply?
