## 15359 Reputation

12 years, 100 days

## Exact solution...

It is easy to see that the product of the members of which are equidistant from the ends is 4, for example

simplify((sqrt(3)+tan(Pi/180))*(sqrt(3)+tan(29*Pi/180)));
4

So, we obtain

mul(simplify((sqrt(3)+tan(Pi*k/180))*(sqrt(3)+tan((30-k)*Pi/180))), k=1..14)*simplify(sqrt(3)+tan(15*Pi/180));
536870912

## Possible variant...

z:=(x,y)->5/(1+x^2+y^2):

A:=plot3d(z(x,y), x=-4..7, y=-4..4, numpoints=3000):

x:=t->2+3*cos(t): y:=t->-1+2*sin(t):

N:=100:

B:=seq(plots[spacecurve]([x(t),y(t),z(x(t), y(t))], t=0..2*Pi*k/N, color=black, thickness=3), k=0..N):

bug:=seq(plottools[sphere]([x(2*Pi*k/N),y(2*Pi*k/N),z(x(2*Pi*k/N), y(2*Pi*k/N))], 0.15, style=surface, color=red),  k=0..N):

C:=seq(plots[display](A, B[k], bug[k]), k=1..N+1):

plots[display](C, insequence=true, axes=normal, view=[-4.7..7.7, -4.7..4.7, -1..5.7]);

## Drunken sailor...

Your task is similar to the problem of wandering drunk sailor. Do a keyword search drunken sailor.

## Re...

Vector(10,[seq(sin(i), i=-0.5..-0.1, 0.1), seq(sin(i), i=0.1..0.5, 0.1)]);

or

Vector(10, [seq(sin(i), i in {seq(-0.5..0.5, 0.1)} minus {0.})]);

## Re...

subs(a=`3`, sqrt(a)/a);

## Solution...

To find the center of rotation is sufficient to use two points:

restart; with(geometry):

point(X, x0, y0), point(A, 4, -2), point(B, 5, -4):   #  X - the center of rotation

L:=coordinates(rotation(A1, A, alpha, 'counterclockwise', X)):

M:=coordinates(rotation(B1, B, alpha, 'counterclockwise', X)):

solve({L[1]=4, L[2]=2, M[1]=6, M[2]=3});

## Possible variant...

You can use  seq  command.

Example:

X:=[seq(i, i=0..20)]:
Y:=[seq((-1)^i*x, i=0..19)]:
y:=piecewise(seq(op([X[i]<x and x<=X[i+1], Y[i]]), i=1..20)):
plot(y, x=0..20);

## Re...

The last line should be

dsolve({eq, bcs}, v(x));

## Re...

Matrix([seq([seq(a[k], k = x+m .. y+m)], m = 0 .. 2)]);

## Example...

a:=1:  c:=2:

fsolve(5*b^5+(60-5*a)*b^4+(125+50*c-80*a)*b^3+(594*c-445*a-775)*b^2+(2324*c-1005*a-3270)*b+3000*c-750*a-3000=0, b);

-5.950891678, -4.378215842, -3.371729567

PS. We find the real roots. If also complex roots are needed, then

fsolve(5*b^5+(60-5*a)*b^4+(125+50*c-80*a)*b^3+(594*c-445*a-775)*b^2+(2324*c-1005*a-3270)*b+3000*c-750*a-3000=0, b, complex);

-5.950891678, -4.378215842, -3.371729567,
1.350418543 - 1.816274549 I, 1.350418543 + 1.816274549 I

## Possible approach...

Consider the function  f:=x->sqrt(a* x + b) +  sqrt(c*x + d) . If  a*c>=0  then  f   is a monotonic function. Therefore, the equation   sqrt(a* x + b) +  sqrt(c*x + d) = m  can not have two solutions. So the necessary condition is   a*c<0 . For definiteness, let  a>0 . From these conditions and from the conditions a*x+b>=0,  c*x+d>=0  we get  -b/a<=x<=-d/c . Therefore, in the limited ranges of parameters  a, b, c, d  all the solutions can be found by the usual brute force.

The following code finds all equations with integers  a=1 .. 10,  b=-10 .. 10,  c=-10 .. -1,  d=-10 .. 10, each of which has exactly two integer solutions:

L:=[]:

for a to 10 do

for b from -10 to 10 do

for c from -10 to -1 do

for d from -10 to 10 do

if -b/a<=-d/c then s:=floor(-b/a): t:=ceil(-d/c):

M:=[]:

for x from s to t do

u:=sqrt(a*x+b): v:=sqrt(c*x+d):

if type(u, integer) and type(v, integer) then M:=[op(M), [x, u+v]]: fi:

od:

if nops(M)=2 and M[1,2]=M[2,2] then L:=[op(L), [a,b,c,d,M[2,2],[M[1,1],M[2,1]]]]: fi: fi:

od: od: od: od:

nops(L);

L[1..50];

ListTools[Search]([1, 5, -1, 8, 5, [-1,4]], L);

For example, the list  [1, -9, -1, 10, 1, [9, 10]]  corresponds to the equation  sqrt(x-9)+sqrt(-x+10)=1  with the roots 9 and 10. The list  L  contains all the solutions in the specified ranges. Received 319 solutions. Displayed first 50 solutions. The original equation is also in the list  L  at position 82.

## Syntax...

Check your syntax! I have no problem:

## No solutions...

It makes no sense to look for what does not exist, because no global minimum and global maximum. A point  (x,y,z)  lies on the sphere of radius  sqrt(5)  and the plane  x+y+z+1=0   intersects the sphere. Therefore, the denominator of the fraction  96/(x+y+z+1)  can be arbitrarily close to 0 and may be positive or negative.

This can be seen clearly in the plot:

phi:=Pi/4:
A:=subs(x=sqrt(5)*sin(theta)*cos(phi), y=sqrt(5)*sin(theta)*sin(phi), z=sqrt(5)*cos(theta), 1/2*x^2*y^2 + y^2*z^2 + z^2 *x^2 + 96/(x + y + z + 1)):
plot(A, theta=0..Pi, view=[0..Pi, -1000..1000], thickness=2, discont=true);

## Correction...

Should be

u:=(x,t)->exp(-Pi^2*t)*cos(Pi*x):

## Brute force...

Your problem can be solved by the usual brute force. The program returns a list of all suitable  [a, b​​, c]  and the corresponding sequences  t[k], k=-1..20

t[-1]:=0: t[0]:=1: L:=[]: T:=[]:

for a to 20 do

for b to 20 do

for c to 20 do

for n from 0 to 19 do

t[n+1]:=((a*n^2+a*n+b)*t[n]+c*(n^2)*t[n-1])/(n+1)^2:

od:

if convert([seq(type(t[k], integer), k=1..20)], `and`) then L:=[op(L), [a,b,c]]:

T:=[op(T), [seq(t[k], k=-1..20)]] fi:

od: od: od:

L; T;

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