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Impara Challenge: Solving Nonlinear Equations | Optimization and Root Finding
Introduction to SciPy

bookChallenge: Solving Nonlinear Equations

In many scientific and engineering applications, you often encounter nonlinear equations that cannot be solved analytically and require numerical methods. The scipy.optimize module provides powerful algorithms to find the roots of such equations, enabling you to model and analyze real-world systems. In this challenge, you will apply your understanding of root-finding by solving a nonlinear equation that represents a physical process using scipy.optimize.root.

Compito

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Solve the nonlinear equation x^3 - 2x^2 + x - 1 = 0 to model a physical process. Use the provided physical_process_equation function for the equation.

  • Use scipy.optimize.root to numerically find a root of the equation, starting from an initial guess of 2.0.
  • Return the root value as a float from the solve_nonlinear_equation function.

Remember to extract the root from the result object using .x[0] and convert it to a float before returning. Make sure your function returns a float, not an array.

Soluzione

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Sezione 3. Capitolo 5
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bookChallenge: Solving Nonlinear Equations

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In many scientific and engineering applications, you often encounter nonlinear equations that cannot be solved analytically and require numerical methods. The scipy.optimize module provides powerful algorithms to find the roots of such equations, enabling you to model and analyze real-world systems. In this challenge, you will apply your understanding of root-finding by solving a nonlinear equation that represents a physical process using scipy.optimize.root.

Compito

Swipe to start coding

Solve the nonlinear equation x^3 - 2x^2 + x - 1 = 0 to model a physical process. Use the provided physical_process_equation function for the equation.

  • Use scipy.optimize.root to numerically find a root of the equation, starting from an initial guess of 2.0.
  • Return the root value as a float from the solve_nonlinear_equation function.

Remember to extract the root from the result object using .x[0] and convert it to a float before returning. Make sure your function returns a float, not an array.

Soluzione

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Tutto è chiaro?

Come possiamo migliorarlo?

Grazie per i tuoi commenti!

Sezione 3. Capitolo 5
single

single

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