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Euler’s Method. Leonhard Euler Pronounced like Leon-ard Oiler Swiss mathematician and physicist Introduced the idea of a function f(x) Wrote the work.

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Presentation on theme: "Euler’s Method. Leonhard Euler Pronounced like Leon-ard Oiler Swiss mathematician and physicist Introduced the idea of a function f(x) Wrote the work."— Presentation transcript:

1 Euler’s Method

2 Leonhard Euler Pronounced like Leon-ard Oiler Swiss mathematician and physicist Introduced the idea of a function f(x) Wrote the work Optics which became the preeminent theory on light until quantum light theory Nine point circle

3 What is Euler’s Method? Equation using slopes and values at various points of an equation to approximate the next value of said equation. Used when the given slope or can not be integrated

4 How It Works y new = y old + f’(x)*Δx A new y value can be approximated by taking the old value and adding the slope multiplied by the step size, which is the difference between two x values.

5 First Example! Given a point (1,2) and = x+y, approximate f(2) with two equal step sizes of.5 1.Write down equation: y new = y old + f’(x)*Δx f(1.5) = f(1) + f’(1)(0.5)

6 First Example, Cont’d. 2. Plug in relative values: f(1.5) = 2 + (1+2)(.5) 3. Solve: f(1.5) = 3.5 or 7/2 4. Plug in for new values: f(2) = 3.5 + f’(1.5)(.5) f(2) = 3.5 + 5(0.5) = 6

7 Second Example Using two equal step sizes of.1, approximate f(.8) given = and f(1)=2

8 How to Do it 1.Write down the equation y new =y old +(m)Δx 2.Plug in values for f(.9) 1.y new =2-(.1)(.5)(3) 1.f(.9)= 1.85 3.Plug in values for f(.8) 1.y new =1.85-(.1)(.925)(3.85) 1.f(.8)=1.493875

9 Be Careful! Using Euler’s method gives an approximation, not an actual solution. Depending on whether or not the equation is concave up or down, the Euler’s solution could be higher or lower than the actual one. Oh no! So what about that last example?

10 Concavity In the last example, f’(x) = x + y. f”(x), therefore, is equal to 1. The concavity is always going to be positive, which means that the approximate value given by Euler’s method will be lower than the real value.

11 Now it’s Your Turn! f’(x) = Find f(3) given point (1,3) Two equal step sizes

12 Did You Get it Right? f(2) = f(1) + f’(1) * (1) – f’(1) = 1 + 2 = 0.5 2(1)(3) f(2) = 3 + 0.5 = 3.5 f(3) = f(2) + f’(2) * (1) – f’(2) = 3.5 + 2 = 11 2(2)(3.5) 28 f(3) = 3.5 + 11/28 = 109/28

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