Fixed point Iteration?

How to perform the Fixed point iteration method of (x-1)(x+10)???

3 Answers

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  • 9 years ago
    Favorite Answer

    there are many ways to set it up, and many of them do not converge (or converge exponentially slowly)

    here is one approach: G(x) = (x-1)(x+10) = 0

    x^2 + 9x - 10 = 0

    x_new = (1/9)(x^2 - 10) = f(x)

    pick a value of x, evalute f(x), use x_new as the next value of x to feed into f(x);

    repeat until |x_new - x| < epsilon

    [Newton-Raphson is another type of fixed point iteration you could use.]

  • 9 years ago

    there are many ways to set it up, and many of them do not converge (or converge exponentially slowly)

    here is one approach: G(x) = (x-1)(x+10) = 0

    x^2 + 9x - 10 = 0

    x_new = (1/9)(x^2 - 10) = f(x)

    pick a value of x, evalute f(x), use x_new as the next value of x to feed into f(x);

    repeat until |x_new - x| < epsilon

    [Newton-Raphson is another type of fixed point iteration you could use.]

  • riddle
    Lv 4
    4 years ago

    (i) fastened ingredient technology is for a answer of an equation f(x)=0 written interior the variety x=F(x) and iterating with x(n+a million,2)=F(xn)a million,2 beginning from x_0. Convergence calls for |F'(x)|<a million,2 interior the neighbourhood of the answer required. right here F(x)=a million,2+e^-x and F'(x)=-e^-x |F'(x)|=e^-x <5cf281c5be3da4eecf3bc6e8aac1bin the era [5cf281c5be3da4eecf3bc6e8aac1b5cf281c5be3da4eecf3bc6e8aac1b5cf281c5be3da4eecf3bc6e8aac1b]a million,2 so the approach could converge. x_0=a million,2.5cf281c5be3da4eecf3bc6e8aac1b5cf281c5be3da4eecf3bc6e8aac1b x_5cf281c5be3da4eecf3bc6e8aac1b=a million,2.333875cf281c5be3da4eecf3bc6e8aac1bx_5cf281c5be3da4eecf3bc6e8aac1b=a million,2.5cf281c5be3da4eecf3bc6e8aac1b6375cf281c5be3da4eecf3bc6e8aac1b5cf281c5be3da4eecf3bc6e8aac1bx_3=a million,2.5cf281c5be3da4eecf3bc6e8aac1b85cf281c5be3da4eecf3bc6e8aac1b605cf281c5be3da4eecf3bc6e8aac1b....... (ii)f(x)=tanx-5cf281c5be3da4eecf3bc6e8aac1b5cf281c5be3da4eecf3bc6e8aac1b f'(x)=sec^a million,2(x)a million,2 x_n+a million,2=x_n-f(x_n)/f'(x_n) x_5cf281c5be3da4eecf3bc6e8aac1b=a million,2.5cf281c5be3da4eecf3bc6e8aac1b5cf281c5be3da4eecf3bc6e8aac1b x_5cf281c5be3da4eecf3bc6e8aac1b=a million,2.a million,2-(tan(a million,2.a million,2)-a million,2)/sec^a million,2(a million,2.a million,2)=a million,2.5cf281c5be3da4eecf3bc6e8aac1b075cf281c5be3da4eecf3bc6e8aac1b55cf281c5be3da4eecf3bc6e8aac1b x_5cf281c5be3da4eecf3bc6e8aac1b=a million,2.5cf281c5be3da4eecf3bc6e8aac1b075cf281c5be3da4eecf3bc6e8aac1b487.. x_3=a million,2.5cf281c5be3da4eecf3bc6e8aac1b075cf281c5be3da4eecf3bc6e8aac1b487 so x=a million,2.5cf281c5be3da4eecf3bc6e8aac1b075cf281c5be3da4eecf3bc6e8aac1b5 to 5 DP (iii)

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