Euler's method in Python (Q1680): Difference between revisions
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| Property / code source: def euler(f, x0, y0, h, n):\n """Advance y' = f(x, y), y(x0) = y0 by n Euler steps of size h."""\n xs, ys = [x0], [y0]\n x, y = x0, y0\n for _ in range(n):\n y = y + h * f(x, y)\n x = x + h\n xs.append(x)\n ys.append(y)\n return xs, ys\n\n# y' = y, y(0) = 1, ten steps of h = 0.1 on [0, 1]; exact value e\nxs, ys = euler(lambda x, y: y, 0.0, 1.0, 0.1, 10)\nprint(ys[-1]) # 2.5937424601... versus e = 2.718281828... / rank | |||
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| Property / programming language: Python / rank | |||
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| Property / instance of: code snippet / rank | |||
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| Property / code source | |||
def euler(f, x0, y0, h, n):\n """Advance y' = f(x, y), y(x0) = y0 by n Euler steps of size h."""\n xs, ys = [x0], [y0]\n x, y = x0, y0\n for _ in range(n):\n y = y + h * f(x, y)\n x = x + h\n xs.append(x)\n ys.append(y)\n return xs, ys\n\n# y' = y, y(0) = 1, ten steps of h = 0.1 on [0, 1]; exact value e\nxs, ys = euler(lambda x, y: y, 0.0, 1.0, 0.1, 10)\nprint(ys[-1]) # 2.5937424601... versus e = 2.718281828... | |||
| Property / code source: def euler(f, x0, y0, h, n):\n """Advance y' = f(x, y), y(x0) = y0 by n Euler steps of size h."""\n xs, ys = [x0], [y0]\n x, y = x0, y0\n for _ in range(n):\n y = y + h * f(x, y)\n x = x + h\n xs.append(x)\n ys.append(y)\n return xs, ys\n\n# y' = y, y(0) = 1, ten steps of h = 0.1 on [0, 1]; exact value e\nxs, ys = euler(lambda x, y: y, 0.0, 1.0, 0.1, 10)\nprint(ys[-1]) # 2.5937424601... versus e = 2.718281828... / rank | |||
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| Property / programming language: Python / rank | |||
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Latest revision as of 07:28, 4 September 2026
No description defined
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Euler's method in Python |
No description defined |
Statements
def euler(f, x0, y0, h, n):\n """Advance y' = f(x, y), y(x0) = y0 by n Euler steps of size h."""\n xs, ys = [x0], [y0]\n x, y = x0, y0\n for _ in range(n):\n y = y + h * f(x, y)\n x = x + h\n xs.append(x)\n ys.append(y)\n return xs, ys\n\n# y' = y, y(0) = 1, ten steps of h = 0.1 on [0, 1]; exact value e\nxs, ys = euler(lambda x, y: y, 0.0, 1.0, 0.1, 10)\nprint(ys[-1]) # 2.5937424601... versus e = 2.718281828...
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