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Advanced Engineering Mathematics 1 01006 Homework 1 30/9 - 2018 Atli Olsen - s130332 Exercise 1 > restart:

Problem A > >

>

We have found the roots to be z = 3+i and z = 3-i As we can see by checking the roots with Maple we get the same result. > P1:=(z)->z^2-6*z+10: > CheckP1:=RootOf(P1(z)): > allvalues({'z'=CheckP1}); (1.1)

Problem B We are asked to show that 2 is a root in the P2. We plug 2 into z's place to see if the equation equals zero. > > > We have confirmed that 2 is a root in P2. Now we will factorize the polynomial to be able to find the remainding roots more easily. We will use that 2 is a root to our advantage for the factorization. > > > > We solved the second term of the factorized polynomial so we have already all the solutions which

are as follows. > We can make a quick check to see if all the solutions are correct with Maple > P2:=(z)->(1/2)*z^3-4*z^2+11*z-10: > CheckP2:=RootOf(P2(z)): > allvalues({'z'=CheckP2}); (1.2)

Problem C We will now find all roots of P(z) and state their algebraic multiplicity and writhe P(z) in complete factorized form. > > >

We can read directly from the factorized P(z) all the solutions as we have solved the therms before. We have two terms and which means that the Algebraic multiplicity of the solutions is 2. We have the following solutions: > > > We can check with maple all the solutions for P(z) > P3:=(z)->P1(z)*P2(z): > 'P(z)'=P3(z); (1.3) > checkP:=RootOf(P3(z)): allvalues({'z'=checkP}); (1.4) We can also see the Algebraic Multiplicities of the solutions

Exercise 2 > restart:

Problem A We will first find the rectangular form of the exponential form to do this we will use Euler's formula

(Theorem 1.46) > > > The main arguments for the corresponding rectangular forms are. Absolute value and main argument of: > > > > So #Arg(z) We can check with maple. > '6*exp(-I*3*Pi/2)'=6*exp(-I*3*Pi/2); (2.1) > 'abs(z)'=abs(6*exp(-I*3*Pi/2)); (2.2) > 'argument(z)'=argument(6*exp(-I*3*Pi/2)) (2.3) Absolute value and main argument of: > > > >

We can check with maple. > '13*exp(I*13*Pi)'=13*exp(I*13*Pi); (2.4) > 'abs(z)'=abs(13*exp(I*13*Pi)); (2 5)

(2.5) > 'argument(z)'=argument(13*exp(I*13*Pi)) (2.6) Absolute value and main argument of: > >

>

> So Arg(z) = We can check with maple. > '1*exp(I*7*Pi/4)'=1*exp(I*7*Pi/4); (2.7) > 'abs(z)'=abs(1*exp(I*7*Pi/4)); (2.8) > 'argument(z)'=argument(1*exp(I*7*Pi/4)) (2.9)

Problem B Now we will write A,B,C in exponential form. > A:=2-2*I*sqrt(3); (2.10) > B:=sqrt(3)+I (2.11) > C:=B^12/A^6; (2.12) Starting with A We will first find the absolute value and main argument of A >

> > > So Arg(z) = Now we can use theorem 1.47 to write A on exponential form > (2.13) We can make a quick check with maple if we have done correctly > '4*exp(-I*Pi/3)'=4*exp(-I*Pi/3); (2.14) We can confirm that we have rewritten the rectangular form correctly in exponential form. Next B We will first find the absolute value and main argument > > > > So Arg(z) = Now we can use theorem 1.47 again. > We can make a quick check and as we can see below that we have made the conversion correctly from rectangular form to exponential form. > '2*exp(I*Pi/6)'=2*exp(I*Pi/6); (2.15) Next C > (sqrt(3)+I)^12/(2-(2*I)*sqrt(3))^6

12 6

We will write C in polar form and use the results from A and B to solve.

>

>

>

> We can divide the absolute value and we can subtract the arguments. >

Exercise 3 > restart:

Problem A We will solve the trigonometric equation > Then we look up in the unit circle and use equation (1-16) form the book chapter 1

> Now we can just state 3 positive solutions to the equation where p = 1,2,3. > arg(z)=-(1/3)*Pi+2*1*Pi; arg(z)=-(1/3)*Pi+2*2*Pi; arg(z)=-(1/3)*Pi+2*3*Pi;

(2.16)

(3.1)

Problem B We follow the procedure in Example 1.51 and theorem 1.52 to determine a solution. > > > > >

> > > > > Now we will plug p into the exponential form above to get all the solutions. > z__0:='2*exp(I*(Pi/6+0*Pi/3))'; z__1:='2*exp(I*(Pi/6+1*Pi/3))'; z__2:='2*exp(I*(Pi/6+2*Pi/3))'; z__3:='2*exp(I*(Pi/6+3*Pi/3))'; z__4:='2*exp(I*(Pi/6+4*Pi/3))'; z__5:='2*exp(I*(Pi/6+5*Pi/3))';

(3.2) To see what solutions have a positive real part we rewrite from exponential form to rectangular form > z__0:='2*exp(I*(Pi/6+0*Pi/3))'=2*exp(I*(Pi/6+0*Pi/3)); z__1:='2*exp(I*(Pi/6+1*Pi/3))'=2*exp(I*(Pi/6+1*Pi/3));; z__2:='2*exp(I*(Pi/6+2*Pi/3))'=2*exp(I*(Pi/6+2*Pi/3));; z__3:='2*exp(I*(Pi/6+3*Pi/3))'=2*exp(I*(Pi/6+3*Pi/3));; z__4:='2*exp(I*(Pi/6+4*Pi/3))'=2*exp(I*(Pi/6+4*Pi/3));; z__5:='2*exp(I*(Pi/6+5*Pi/3))'=2*exp(I*(Pi/6+5*Pi/3));;

(3.3) We can see from my looking at the solutions we have that

and

have positive real parts. (3.4)

> Worth is also to mention that all six solutions lie on the unit circle with radius 2.

Problem C > > > > >

> > > z:=p->ln(4)+i*(-Pi/(6)+p*2*Pi); (3.5) > z(1); (3.6) > z(2); (3.7) > z(3); (3.8)

Exercise 4 > restart:with(plots):

Problem A We will show that the trigonomentric functions f(x) = cos(x) and g(x) = sin(x) fullfill the system of differential equations. > > Proof of differention > > > Thus showing that > And > > > And show that. >

With the given initial conditions > > Now we will investigate the hyperbolic functions f(x) = cosh(x) and g(x) = sinh(x). So we can show that. > > > >

And for g(x) > > > >

And with the given initial condition we get. > >

Problem B The radius of the circle equation is 1 and the length of cos(v) + I*sin(v) is also 1, because of the idiot rule. Therefore, for every value of v, the radius will always lay on the circle as illustrated below. > plot1C:=implicitplot(x^2 + y^2 =1,x=-1..1,y=-1..1,color=red, thickness=10): plot2C:=complexplot(cos(v)+I*sin(v),v=-Pi..Pi,style=point, symbol=solidcircle,symbolsize=10, linestyle=DOT, color=black ): display(plot1C,plot2C);

Problem C > restart:with(plots): > plot1:=complexplot(cosh(v)-I*sinh(v),v=-2.5..2.5,style=point, symbol=solidcircle,symbolsize=5, linestyle=DOT, color=black ): > plot2:=implicitplot(x^2 - y^2 =1,x=0..5,y=-5..5,color=red, thickness=10): If can be visually seen that for every real number lies on the unit hyperbola with the equation x^2 y^2 = 1. > display(plot2,plot1,view=[1..5,-5..5]);

Problem D Given the function. We will determine the approximating polynomials second degree and third degree for with the development point definition 4.6 > f:=x->cosh(x^2 - x);

and of for this we use

(4.1) We will first determine the approximation polynomial for We start by finding the first and second derivative of f(x) > diff(f(x),x); (4.2) > diff(f(x),x,x); (4.3) > >

> > > > > Now we have the approximation of

with development point

> (4.4) We can do a quick check with maple if the have determined the correct result, and as we can see below we have. > P__2:=unapply(mtaylor(f(x),x=0,3),x): 'P__2(x)'=P__2(x); (4.5) We will continue to determine the third approximation, but first we have to derive f for a third time. > diff(f(x),x,x,x);

>

> > > We can do a quick check with maple if the have determined the correct result, and as we can see below we have. > P__3:=unapply(mtaylor(f(x),x=0,4),x): 'P__3(x)'=P__3(x); (4.6) Now we will illustrate the approximations. > plot_f:=plot(f(x),x=-2..2,color=blue,legend=[f=f(x)]):

> plot_P2:=plot(P__2(x),x=-2..2,color=red,legend=[P__2=P__2(x)]): > plot_P3:=plot(P__3(x),x=-2..2,color=green,legend=[P__3=P__3(x)] ): > display(plot_f,plot_P2,plot_P3,view=-5..10,legendstyle=[font= ["HELVETICA",10],location=right]);

So we can see that the approximating polynomial of third degree follows f(x) more closely. Than of a second degree which also makes sense....