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Published byGeorgina Wells Modified over 8 years ago
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Euler’s Method If we have a formula for the derivative of a function and we know the value of the function at one point, Euler’s method lets us build an approximation to the function f. tt Euler’s method is numerical antidifferentiation.
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Point of View tt tt Area = f’(point)* t y = f’(point)* t
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Total Change Furthermore, adding the y’s to the original y 0 in Euler’s method, yields the final y- value. (Why?) The sum of the y’s is a left Riemann sum approximation to the (signed) area under the graph of f ’. That is, to say, the sum of the y’s in Euler’s method is an approximation of the total change in the function f over the entire interval.
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The sum of the y’s is a left Riemann sum approximation to the (signed) area under the graph of f ’. The sum of the y’s in Euler’s method is and approximation of the total change in the function f over the entire interval. The integral of f’ over the interval [a,b] represents both the (signed) area under the graph of f’ and the total change in the function f over [a,b].
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Suppose the formula for the derivative of y=f(t) is given in terms of t only. (E.g. y’ = sin(t 2 ).) At each stage of Euler’s method, we compute the change in y by multiplying the slope of function at the (left) point by t. This same quantity represents the area of the left Riemann rectangle at the corresponding point on the graph of f ’! Euler’s method computes the total change in f over the interval. The left Riemann Sums of f’ compute the same thing. Euler’s Method and Riemann Sums
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