## 15202 Reputation

12 years, 70 days

## Global minimum...

Quote: "

With m = 1, we have f(x) = 3*x^2+2*x-7.

minimize(h(x),x>=1);

we get minimum is -2"

A strange argument!  The global minimum of the function of two variables f(x,m)=3*x^2+2*m*x+2*m^2-3*m-6  in the region 1<= x, m=-infinity..infinity,  equals  -25/8  and it is reached at  x=1, m=1/4 .

## Global minimum...

Quote: "

With m = 1, we have f(x) = 3*x^2+2*x-7.

minimize(h(x),x>=1);

we get minimum is -2"

A strange argument!  The global minimum of the function of two variables f(x,m)=3*x^2+2*m*x+2*m^2-3*m-6  in the region 1<= x, m=-infinity..infinity,  equals  -25/8  and it is reached at  x=1, m=1/4 .

## Re...

The program of Joe Riel solves another problem: it shows that 1st, 2nd, 3rd members of the sequence are identical with 25th, 26th and 27th members. Thus, this sequence is periodic with period 24.

## Re...

The program of Joe Riel solves another problem: it shows that 1st, 2nd, 3rd members of the sequence are identical with 25th, 26th and 27th members. Thus, this sequence is periodic with period 24.

convert(M,list);  geom3d[point]('M',op(%));

[1, -1, 3]

M

convert(M,list);  geom3d[point]('M',op(%));

[1, -1, 3]

M

## Right...

@toandhsp It seems all right!

## Comment...

Why do you think that the point H is the projection of the midpoint of AB on the plane P? If this is true, then your solution is also true!

Dear Mr. toandhsp! I've already written for you several procedures. Now learn to do it yourself!

Dear Mr. toandhsp! I've already written for you several procedures. Now learn to do it yourself!

## Error...

Dear Mr. Hirnik! Unfortunately, the calculations made ​​by you in the package DirectSearch are wrong! Look at the screenshot and all become clear. First, using the procedure P1 i found true minimum for the first 10 numbers from your list, then for next 15. From the inequality |a+b|<=|a|+|b| obtained a rough estimate of initial minimum. I think that actually it is even much less!

## Error...

Dear Mr. Hirnik! Unfortunately, the calculations made ​​by you in the package DirectSearch are wrong! Look at the screenshot and all become clear. First, using the procedure P1 i found true minimum for the first 10 numbers from your list, then for next 15. From the inequality |a+b|<=|a|+|b| obtained a rough estimate of initial minimum. I think that actually it is even much less!

## Correction...

Your code does not work because:

1) For the command DotProduct the option conjugate=false is required.

2) The system has not one but two solutions.

The corrected code:

restart:

with(LinearAlgebra):

A:=<-1,3,6>: B:=<2,2,-0>: N:=<x,y,z>: o:=<1,-1,7>:

Sys:={Norm(o-N,2)^2=9, DotProduct(N - o, N-A,conjugate=false) = 0, DotProduct(N - o, N-B,conjugate=false) = 0}:

Sol:=[solve(Sys)]:

N1:=[seq(rhs(Sol[1,i]),i=1..3)];

N2:=[seq(rhs(Sol[2,i]),i=1..3)];

## Correction...

Your code does not work because:

1) For the command DotProduct the option conjugate=false is required.

2) The system has not one but two solutions.

The corrected code:

restart:

with(LinearAlgebra):

A:=<-1,3,6>: B:=<2,2,-0>: N:=<x,y,z>: o:=<1,-1,7>:

Sys:={Norm(o-N,2)^2=9, DotProduct(N - o, N-A,conjugate=false) = 0, DotProduct(N - o, N-B,conjugate=false) = 0}:

Sol:=[solve(Sys)]:

N1:=[seq(rhs(Sol[1,i]),i=1..3)];

N2:=[seq(rhs(Sol[2,i]),i=1..3)];

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