sidra ali

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3 years, 340 days

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These are questions asked by sidra ali

how i can simplify

(f(x[n])/Df(x[n]));
in code

restart;
taylor(f(x), x = gamma, 8);
f(x[n]) := subs([x-gamma = e[n], f(gamma) = 0, seq(((D@@k)(f))(gamma) = factorial(k)*c[k]*(D(f))(gamma), k = 1 .. 1000)], %);

1 2
f(gamma) + D(f)(gamma) (x - gamma) + - @@(D, 2)(f)(gamma) (x - gamma)
2

1 3 1 4
+ - @@(D, 3)(f)(gamma) (x - gamma) + -- @@(D, 4)(f)(gamma) (x - gamma)
6 24

1 5 1 6
+ --- @@(D, 5)(f)(gamma) (x - gamma) + --- @@(D, 6)(f)(gamma) (x - gamma)
120 720

1 7 / 8\
+ ---- @@(D, 7)(f)(gamma) (x - gamma) + O\(x - gamma) /
5040
2 3
c[1] D(f)(gamma) e[n] + c[2] D(f)(gamma) e[n] + c[3] D(f)(gamma) e[n]

4 5 6
+ c[4] D(f)(gamma) e[n] + c[5] D(f)(gamma) e[n] + c[6] D(f)(gamma) e[n]

7 / 8\
+ c[7] D(f)(gamma) e[n] + O\e[n] /

taylor(D(f)(x), x = gamma, 8);
Df(x[n]) := subs([x-gamma = e[n], f(gamma) = 0, seq(((D@@k)(f))(gamma) = factorial(k)*c[k]*(D(f))(gamma), k = 2 .. 1000)], %);

D(f)(gamma) + @@(D, 2)(f)(gamma) (x - gamma)

1 2 1 3
+ - @@(D, 3)(f)(gamma) (x - gamma) + - @@(D, 4)(f)(gamma) (x - gamma)
2 6

1 4 1 5
+ -- @@(D, 5)(f)(gamma) (x - gamma) + --- @@(D, 6)(f)(gamma) (x - gamma)
24 120

1 6
+ --- @@(D, 7)(f)(gamma) (x - gamma)
720

1 7 / 8\
+ ---- @@(D, 8)(f)(gamma) (x - gamma) + O\(x - gamma) /
5040
2
D(f)(gamma) + 2 c[2] D(f)(gamma) e[n] + 3 c[3] D(f)(gamma) e[n]

3 4
+ 4 c[4] D(f)(gamma) e[n] + 5 c[5] D(f)(gamma) e[n]

5 6
+ 6 c[6] D(f)(gamma) e[n] + 7 c[7] D(f)(gamma) e[n]

7 / 8\
+ 8 c[8] D(f)(gamma) e[n] + O\e[n] /

(f(x[n])/Df(x[n]));
this last term did not use f(x[n]) value from above to solve it. plxx help if any one can solve it...

how i can find order of convergence of newton method by expanding taylor series?? plz send me code???

 

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