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- 首页 >> 其他 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
1therefore 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.
3
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
1therefore 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.
3