东华大学matlab数学实验-第四次答案

1

>> p=[1 1 1];x=roots(p)

x =

-0.5000 + 0.8660i

-0.5000 - 0.8660i

>> polyval(p,x)

ans =

1.0e-15 *

0.3331

0.3331

>> clear;

>> p=[3 0 -4 2 -1];x=roots(p)

x =

-1.4058 + 0.0000i

1.0000 + 0.0000i

0.2029 + 0.4427i

0.2029 - 0.4427i

>> polyval(p,x)

ans =

1.0e-13 *

-0.2498 + 0.0000i

-0.0266 + 0.0000i

0.0000 + 0.0072i

0.0000 - 0.0072i

>> clear;

>> p=[5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -6 8 0 0 0 -5 0 0];x=roots(p)

x =

0.0000 + 0.0000i

0.0000 + 0.0000i

0.9768 + 0.0000i

0.9388 + 0.2682i

0.9388 - 0.2682i

0.8554 + 0.5363i

0.8554 - 0.5363i

0.6615 + 0.8064i

0.6615 - 0.8064i

0.3516 + 0.9878i

0.3516 - 0.9878i

-0.0345 + 1.0150i

-0.0345 - 1.0150i

-0.4609 + 0.9458i

-0.4609 - 0.9458i

-0.1150 + 0.8340i

-0.1150 - 0.8340i

-0.7821 + 0.7376i

-0.7821 - 0.7376i

-0.9859 + 0.4106i

-0.9859 - 0.4106i

-1.0416 + 0.0000i

-0.7927 + 0.0000i

>> polyval(p,x)

ans =

1.0e-12 *

0.0000 + 0.0000i

0.0000 + 0.0000i

0.1305 + 0.0000i

0.0954 + 0.0245i

0.0954 - 0.0245i

0.0796 + 0.0419i

0.0796 - 0.0419i

-0.0643 + 0.0680i

-0.0643 - 0.0680i

0.1620 + 0.0550i

0.1620 - 0.0550i

-0.0120 + 0.0557i

-0.0120 - 0.0557i

-0.0978 + 0.3836i

-0.0978 - 0.3836i

0.0012 + 0.0121i

0.0012 - 0.0121i

0.0560 - 0.1047i

0.0560 + 0.1047i

0.2392 + 0.0861i

0.2392 - 0.0861i

-0.0636 + 0.0000i

0.0011 + 0.0000i

>> p1=[2 3];

>> p2=conv(p1,p1);

>> p=conv(p2,p1)

p =

8 36 54 27

>> p3=[8 36 54 23];x=roots(p3)

x =

-1.8969 + 0.6874i

-1.8969 - 0.6874i

-0.7063 + 0.0000i

>> polyval(p3,x)

ans =

1.0e-13 *

0.1066 + 0.0799i

0.1066 - 0.0799i

0.0355 + 0.0000i

>>

2

>> clear;

>> fun=@ (x)(x*log(sqrt(x^2-1)+x)-sqrt(x^2-1)-x/2);

>>fplot(fun,[0,10])

>> fzero(fun,2)

ans =

2.1155

3

>> fun=@ (x)[7/10*sin(x(1))+2/10*cos(x(2))-x(1),7/10*cos(x(1))-2/10*sin(x(2))-x(2)]; >> [x,f,h]=fsolve(fun,[0,0])

Equation solved.

fsolve completed because the vector of function values is near zero

as measured by the default value of the function tolerance, and

the problem appears regular as measured by the gradient.

<stopping criteria details>

x =

0.5265 0.5079

f =

1.0e-09 *

-0.2859 -0.4618

h =

1

>>

9(1)

clear;

>> xa=-2:1;ya=-7:1;

>> [x,y]=meshgrid(xa,ya);

>> z=y.^3/9+3*x.^2.*y+9*x.^2+y.^2+x.*y+9;

>> mesh(x,y,z)

(2)

>> clear;

>> fun=@ (x)x(2).^3/9+3*x(1).^2.*x(2)+9*x(1).^2+x(2).^2+x(1).*x(2)+9; >> [x,g]=fminsearch(fun,[0,0])

x =

0 0

g =

9

>> fun=@ (x)-(x(2).^3/9+3*x(1).^2.*x(2)+9*x(1).^2+x(2).^2+x(1).*x(2)+9); >> [x,g]=fminsearch(fun,[0,-6])

x =

-0.3333 -6.0000

g =

-22.0000

12

>> clear;

>> all=180*7500

all =

1350000

>> first=all*3/10

first =

405000

>> x=all*7/10

x =

945000

>> r=504/12/10000

r =

0.0042

>> n=20*12

n =

240

>> a=((1+r)^n*r*x)/((1+r)^n-1)

a =

6.2575e+03

>> x=x-100000

x =

845000

>> a=((1+r)^n*r*x)/((1+r)^n-1)

a =

5.5953e+03

>> r=405/10000/12

r =

0.0034

>> x=100000

x =

100000

>> a=((1+r)^n*r*x)/((1+r)^n-1)

a =

608.6182

>> 5.5953e+03+a

ans =

6.2039e+03

>>

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