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MIT18_03S10_c02

# MIT18_03S10_c02 - 18.03 Class 2 Feb 3 2010 Numerical...

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18.03 Class 2 , Feb 3, 2010 Numerical Methods [1] How do you know the value of e? [2] Euler's method [3] Sources of error [4] Higher order methods [1] The study of differential equations has three parts: . Analytic, exact, symbolic methods . Quantitative methods (direction fields, isoclines .... ) . Numerical methods Even if we can solve symbolically, the question of computing values remains. The number e is the value y(1) of the solution to y' = y with y(0) = 1. But how do you find that in fact e = 2.718282828459045 .... ? The answer is: numerical methods. (Euler already computed e to at least 18 decimal places. It is now known to some 200 billion places, but I won't write them all out here.) As an example, take the first order ODE y' = y^2 - x = F(x,y) with initial condition y(0) = -1 -- an Initial Value Problem, IVP. For example: what is y(1) ? I invoked the Euler's Method Mathlet, and selected F(x,y) = y^2 - x . The slope field appears. I selected the initial condition (0,-1) , and then invoked "actual." This solution is one of those trapped in the funnel, so for large x , the graph of y(x) is close to the graph of - \sqrt(x) : y(100) is very close to -10 . But what about y(1) ? The listing says y(1) = -.83 . How did the computer know this? [2] The tangent line approximation gives one approach. Recall: We have a function y(x), and a point (a,y0) on its graph (so y(a) = y0 ). The tangent line to the graph at this point has slope y'(a), which I will denote m_0 . The tangent line approximation is y(a+h) ~ y0 + m0 h The right hand side is the y-coordinate of a point on the tangent line directly above or below (a+h , y(a+h)).

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In our case, we do know the derivative of y at x = 0 : y'(0) = F(0,-1) = (-1)^2 + 0 = 1 So the tangent line at (0,-1) has slope 1 , and we find the tangent line approximation y(1) ~ -1 + 1 = 0. I showed this on the applet.
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MIT18_03S10_c02 - 18.03 Class 2 Feb 3 2010 Numerical...

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