iteration留学生代写、代做C/C++语言、代写real axis、代做C/C++代写

日期：2019-04-06 09:04

1. Let g (x) = 1 + 7

10 cos (x). Prove, that the equation g (x) = x has exactly one solution

in (1;1)

and compute it by xed point iteration with an error of at most 0:01.

Solution: As the application is for the entire real axis, we only need to check

Therefore the function g (x) has exactly one ?xed point on the real axis. We can begin

with x0 = 0 and consider the sequence given by

xn+1 = 1 + 7=10 cos (xn):

If we denote the limit by c, we have

x1 = 1:7; x2 = 0:909808854; x3 = 1:429727654; x5 = :::

After frequent iterations we get

xn = 1:232389373:

cos (1:232389373) = xn

to the accuracy allowed by our calculation, so :

therefore we may use c = 1: 232 389 373 as a su￠ ciently good approximation.

2. Find the quadratic interpolation polynomial p(x) for p1 + x

with the interpolation

points x0 = 0; x1 = 1; x2 = 2 :

Solution: The function g (x) = p

1 has the property that g (0) = 0; g (1) =21; g (2) = p51. If we want to use the Lagrange form of the interpolation

polynomial, we therefore do not need L0. Now

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therefore the polynomial is 0:210 393 136 0x + 0:203 820 426 4x2:

3. Consider a function f on the interval [1;

1]. Assume it has two continuos derivatives

and jf00 (x)j 1

10 in this interval. Let p (x) be the linear interpolation polynomial

with nodes x0 = 1;

x1 = 1: What is the most precise information about the maximal

absolute value of the error in the interval you can give?

Solution: We know that

is non-negative in [1;1] and positive in (1;1). Therefore it

has a positive maximum in the open interval, as it is zero on the boundary, and there its derivative 2x

equals zero. Thus the maximum is at x = 0, and we have as the

best possible estimate that

5x + 3: Both are cubic

polynomials. We have

s1 (1) = 13 + 12 + 1 + 1 = 4; s2 (1) = 1

Thus s is continuous at x = 1. Also

Thus s is continuously di§erentiable in [0; 2]. Finally

Therefore s is a cubic spline.

5. Write the interpolation polynomial of degree at most two for the function f(x) = x2

with the nodes 1;

0; 1 in the Lagrange form. What is the maximum error ?

Solution: For these points

therefore the interpolation polynomial is

(1)2

L0 (x) + 0L1 (x) + 12L2 (x) = 1

Therefore the error is zero. We could have known this without any calculations, as

we are using a quadratic polynomial to interpolate a quadratic polynomial, so the

interpolation polynomial is the same as the original polynomial.

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