dharr

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Name: Dr. David Harrington

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Member for: 20 years, 335 days

Company/Institution: University of Victoria

Title: Professor or university staff

Location: Victoria, British Columbia, Canada

Website: http://web.uvic.ca/~dharr

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I am a retired professor of chemistry at the University of Victoria, BC, Canada. My research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are answers submitted by dharr

missing multiplication...

Answered: dharr 8205

December 05 2016

0 2

Another 2-D math problem. There needs to be a multiplication between the c^2 and the parenthesis. Then it easily plots, and fsolve gives the two solutions, which are obviously x=+/-R.

y(x)2.mw

Not unique...

Answered: dharr 8205

November 30 2016

1 4

The companion matrix can be used - for monic polynomials. The determinant is not unique and may not be the one you want.

>	with(LinearAlgebra):

>	p:=w^2 + u*w + v; #monic polynomial

(1)

>	w*IdentityMatrix(degree(p,w))-CompanionMatrix(p,w);

Matrix([[w, v], [-1, w+u]])

(2)

Download Companion.mw

There is no exact solution, only a close...

Answered: dharr 8205

November 02 2016

2 2

There is no exact solution to the accuracy that Maple usually works at. You need to find a close "near enough" solution, which means you can't use fsolve. The easiest way is to minimize a sum of squares as in the worksheet below.

>	eqs := {(2^2/a^2+2^2/c^2-2(22)cos(x)/(ac))/sin(x)^2 = 1/2.2393^2, (2^2/a^2+4^2/c^2+4(22)cos(x)/(ac))/sin(x)^2 = 1/1.5968^2, (1^2/a^2+sin(x)^2/b^2+2^2/c^2+(22)cos(x)/(ac))/sin(x)^2 = 1/2.7896^2, 1/sin(x)^2(2^2*sin(x)^2/b^2) = 1/2.8650^2};

${(4/a^2+4/c^2-8*cos(x)/(a*c))/sin(x)^2 = .1994230893, (4/a^2+16/c^2+16*cos(x)/(a*c))/sin(x)^2 = .3921922000, (1/a^2+sin(x)^2/b^2+4/c^2+4*cos(x)/(a*c))/sin(x)^2 = .1285038476, 4/b^2 = .1218290191}$

(1)

Move the sin(x)^2 over to the right hand side

>	eqs2:=eqs*~sin(x)^2;

${4*sin(x)^2/b^2 = .1218290191*sin(x)^2, 4/a^2+4/c^2-8*cos(x)/(a*c) = .1994230893*sin(x)^2, 4/a^2+16/c^2+16*cos(x)/(a*c) = .3921922000*sin(x)^2, 1/a^2+sin(x)^2/b^2+4/c^2+4*cos(x)/(a*c) = .1285038476*sin(x)^2}$

(2)

Sunm of squares of differences to minimize

>	S:=add((lhs(eq)-rhs(eq))^2,eq in eqs2);

(4*sin(x)^2/b^2-.1218290191*sin(x)^2)^2+(4/a^2+4/c^2-8*cos(x)/(a*c)-.1994230893*sin(x)^2)^2+(4/a^2+16/c^2+16*cos(x)/(a*c)-.3921922000*sin(x)^2)^2+(1/a^2+sin(x)^2/b^2+4/c^2+4*cos(x)/(a*c)-.1285038476*sin(x)^2)^2

(3)

Minimise subject to some constraints

>	Optimization:-Minimize(S,{a>=0,b>=0,c>=0,x>=0,x<=Pi});

(4)

Gives a close solution, but not the one you wanted (though b is right). Probably there are several close solutions; it will be hard to decide - perhaps there are several combinations of a and c and x that work

Download Crystal.mw

fsolve only returns one answer...

Answered: dharr 8205

November 01 2016

1 2

In general fsolve only returns one answer. If you want others, you can specify a range to look for roots in, e..g, x=187.5..188.5 or give an initial guess, e.g, x=188. See the help for fsolve

Loop rather than fsolve...

Answered: dharr 8205

October 25 2016

2 2

Using fsolve to find the number of terms at which the error is 0.3 won't work because the error for an integer number of terms won't exactly be 0.3, so there is no solution. You know that the error must decrease so it is efficient to just add an extra term on and then recalculate the error in a loop.

The integration method _d01akc works well for these sorts of integrals.

restart;

>	f:=x->x; y:=(n,x)->exp(-x/2)sin(nPi*x); w:=(x)->exp(x);

(1)

Note the normalizing integral is always 1/2:

>	d:=int(w(x)*y(n,x)^2,x=0..1) assuming n::posint;

(2)

And we can get a general formula for the coefficients

>	int(w(x)f(x)y(n,x),x=0..1)/d assuming n::posint; c:=unapply(%,n);

-8*n*Pi*(4*Pi^2*exp(1/2)*(-1)^n*n^2+3*(-1)^(1+n)*exp(1/2)+4)/(16*Pi^4*n^4+8*Pi^2*n^2+1)

proc (n) options operator, arrow; -8*n*Pi*(4*Pi^2*exp(1/2)*(-1)^n*n^2+3*(-1)^(n+1)*exp(1/2)+4)/(16*Pi^4*n^4+8*Pi^2*n^2+1) end proc

(3)

>	Fourierf:=0:Lerror:=infinity: for n while Lerror>0.3 do Fourierf:=Fourierf+c(n)y(n,x); Lerror:=evalf(sqrt(Int((f(x)-Fourierf)^2w(x),x=0..1,method='_d01akc'))); #edit: test after adding term print(n,Lerror); od:

(4)

Download Eigenfn_error.mw

Works in Maple 2015...

Answered: dharr 8205

September 06 2016

0 0

In Maple 2015.2 evalf(Int(...)) quickly returns the answer you give.

expand within sin...

Answered: dharr 8205

July 28 2016

2 4

Bits package...

Answered: dharr 8205

April 02 2016

1 2

Bits:-GetBits(19,-1..0,bits=8);

0, 0, 0, 1, 0, 0, 1, 1

Bits:-GetBits(-(19+1),-1..0,bits=8);

1, 1, 1, 0, 1, 1, 0, 0 #ones complement

Bits:-GetBits(-19,-1..0,bits=8);

1, 1, 1, 0, 1, 1, 0, 1 #twos complement

Nullspace...

Answered: dharr 8205

March 25 2016

2 0

Since you know an eigenvalue of (square) M is 1, then

NullSpace(M-1*IdentityMatrix(RowDimension(M)));

returns the set of eigenvector(s) corresponding to eigenvalue 1.

Bits package...

Answered: dharr 8205

March 23 2016

1 1

map(x->add(i,i in Bits:-Split(x)),mylist);

or even simpler in Maple 2015 (wthout the hyperlink formatting):

map(add@Bits:-Split,mylist);

Seems to work...

Answered: dharr 8205

March 17 2016

1 3

In Maple 2015:

f:=x->x^2;g:=x->x^3;
A:=Array([f,g]);
A(1)(m);

gives m^2

which I think is what you want.

implied multiplication...

Answered: dharr 8205

March 16 2016

2 3

The code works for me in Maple 2015 in 1-D math. It seems you have a space between print and ( which is interpreted as multiplication - remove it and it should work.

Shift-enter...

Answered: dharr 8205

March 11 2016

1 1

Use shift-enter to make a new line in the same execution group. (In more recent versions, the error message gives you this advice).

remove x=0 solution...

Answered: dharr 8205

March 04 2016

0 0

Following the observation from @tomleslie, you can just divide your equation through by x to get x^3+x*z+y and then implicitplot gives you what you want.