Find all the real solutions to cubic equation, Mathematics

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Find all the real solutions to cubic equation x^3 + 4x^2 - 10 =0. Use the cubic equation x^3 + 4x^2 - 10 =0 and perform the following call to the bisection method [0, 1, 30]

Use the fixed point iteration to find the fixed point(s) for the function g(x) = 1 + x - x^2/3

Find all the real solutions to cubic equation x^3 +4x^2-10=0. Use the cubic equation x^3 + 4x^2 - 10 =0 and perform the following call to the regulaFalsi [0, 1, 30]

Use newton's method to find the three roots of a cubic polynomial f(x) = 4x^3 - 15x^2 + 17x-6. Determine the Newton-raphson iteration formula g[x] = x - (f(x)/f'(x)) that is used. Show details of the computation for the starting value p0 = 3.

Use the secant method to find the three roots of cubic polynomial f[x]=4x^3 - 16x^2 + 17x - 4. Determine the secant iterative formula g[x] = x - (f[x]/f'[x]) that is used. Show details of the computation for the starting value p0=3 and p1=2.8

Use appropiate Lagrange interpolating polynomials of degrees one, two, and three to approximate each of the following:

A) f(8.4) if f(8.1)= 16.94410, f(8.3)=17.56492, f(8.6)=18.50515, and f(8.7)=18.82091

B)f(1/3) if f(-0.75)= -0.07181250, f(-0.5) = -0.02475000, f(-0.025) = 0.33493750, and f(0)=18.82091

Use the newton forward divided-difference formula is used to approximate f(0.3) given the following data

X        0.0     0.2     0.4     0.6

F(x)  15.0   21.0   30.0   51.0

Suppose it is discovered that f(0.4) was understand by 10 and f(0.6) was overstated by 5. By what amount should the approximation to f(0.3) be changed?

Using the error formulas

|f(x)-P1(x)| ≤ 1/8 max (f(x))h2, linear interpolation

|f(x)-P2(x)| ≤ 1/9√3 max (f(x))h3, quadratic interpolation

A)  what is an appropriate size for the interpolation table for the function tan x on the interval [0,1] in order that linear interpolation produce an error no larger than 0.5 x 10^6

B)   Answer A)

A) Using taylor series expansions derive the O(h^2) central difference approximation

F'(x)= (f(x+h)-f(x-h))/2h

B)  using richardson extrapolation and taylor series expansions derive the O (h4) derivative approximation

F'(x)= (-f(x+2h)+8f(x+h)-8f(x-h)+f(x-2h))/12h

Consider the richardson table for derivatives in the form step size table

Step size                       table

H                                  D(0,0)

H/2                               D(1,0)        D(1,1)

H/2^2                           D(2,0)        D(2,1)        D(2,2)

H/2^3                           D(3,0)        D(3,1)        D(2,3)        D(3,3)

.

.

Where the central difference formula

(h)   = (f(x+h)-f(x-h)) /2h

Is used to construct the first column using

D(n,0)= (h/2^n)

And the following formula

D(n,m)= (4^mD(n,m-1)-D(n-1,m-1))/4^m-1 (use for hand calculations)

D(n,m-1)+((D(n,m-1)-D(n-1,m-1)/(4^m-1)) (use for programming)

Is used, for n≥m, to obtain entries in other columns in terms of the entry to their left and the entry above this entry. For example, D(2,1) is obtained in terms of D(2,0) and D(1,0) and D(3,2) is obtained in terms of D(3,1) and D(2,1)

A) construct the table for the derivative of tan x at x=0.5. Choose an initial step size of h=1 and calculate 4 rows by hand using a calculator

B) use maple procedure richardson in file richardson.txt to calculate 6 rows of the richardson extrapolation table.

----------------------------------------------------------------------------------------------

# lip.txt:

#Symbolic calculation of LIP

#(Lagrange interpolating polynomial)

#

#Arguments

#

#xp   list [x0,x1,....,xn] of nodes

#yp   list[y0,y1,.....,yn] of function values at nodes

#x     symbolic variable for the polynomial

#

#lists xp amd yp have n+1 elements and begin at subscript 1

#so the interpolating polynomial is of degree n

----------------------------------------------------------------------------------------------

lip := proc(xp,yp,x)

         local n,s,p,k,j;

         N := nops(xp) -1; #nops(xp) gives number of elements in xp

         S := 0;

         For k from 0 to n do

                   P := yp[k+1];

                   For j from 0 to n do

                            If j<>k then

                                     P := p*(x-xp[j+1])/(xp[k+1]-xp[j+1]);

                            Fi;

                   Od;

                   S := s=p;

         Od;

         Return s;

End proc:


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