What exactly do you mean by "it"?

Sorry about leaving that out:

newBCs := {F(0) = 0, T(10) = 0, D(F)(0) = 0, D(F)(10) = 0, D(T)(0) = -lambda};

Note that for lambda = 1 this gives the original boundary condition D(T)(0) = -1.  With D(T)(0) = lambda, as omabbasi tried, it would be attempting to solve the equation with D(T)(0) = 1.  I'm guessing that that BVP may not have a solution.

Sorry about leaving that out:

newBCs := {F(0) = 0, T(10) = 0, D(F)(0) = 0, D(F)(10) = 0, D(T)(0) = -lambda};

Note that for lambda = 1 this gives the original boundary condition D(T)(0) = -1.  With D(T)(0) = lambda, as omabbasi tried, it would be attempting to solve the equation with D(T)(0) = 1.  I'm guessing that that BVP may not have a solution.

I assume desp is talking about the ring Z[sqrt(7)], not a field.  PolynomialIdeals is for polynomial ideals over a field, so this won't help.
 

What will work is

> numtheory[factorEQ](30, 7);

``(-8+3*7^(1/2))*``(3+7^(1/2))^2*``(2+7^(1/2))*``(2-7^(1/2))*``(5)

I assume desp is talking about the ring Z[sqrt(7)], not a field.  PolynomialIdeals is for polynomial ideals over a field, so this won't help.
 

What will work is

> numtheory[factorEQ](30, 7);

``(-8+3*7^(1/2))*``(3+7^(1/2))^2*``(2+7^(1/2))*``(2-7^(1/2))*``(5)

I agree that it's better to use maximize or Optimization:-Maximize if those work.  The trouble is that they might not always work.  The plot, while far from perfect, is a rough-and-ready resource.   For example:

> g:= x*sum(i^(-5/2)*exp(-i*x),i=1..infinity);

> maximize(g,x=0..2,location);

maximize(x*sum(1/i^(5/2)*exp(-i*x),i = 1 .. infinity),x = 0 .. 2,location),{}

> Optimization:-Maximize(g,x=0..2);

Error, (in Optimization:-NLPSolve) summation variable previously assigned, second argument evaluates to 1. = 1 .. infinity
 

(Yes, a bug, which I'll report)

> Optimization:-Maximize(g,x=0..2,method=branchandbound);

Error, (in Optimization:-NLPSolve) method=branchandbound option is available for univariate unconstrained bounded problems only

(Possibly a different bug)

> plot(g, x = 0 .. 2);

> A:= op([1,1],%):
   maxvalue:= max(map2(op,2,A));

maxvalue := .3971541099

> select(has, A, maxvalue);

[[.9172052292, .3971541099]]

The y value is pretty close to the true maximum, I think; the x value is very rough, but further refinement is possible.

I agree that it's better to use maximize or Optimization:-Maximize if those work.  The trouble is that they might not always work.  The plot, while far from perfect, is a rough-and-ready resource.   For example:

> g:= x*sum(i^(-5/2)*exp(-i*x),i=1..infinity);

> maximize(g,x=0..2,location);

maximize(x*sum(1/i^(5/2)*exp(-i*x),i = 1 .. infinity),x = 0 .. 2,location),{}

> Optimization:-Maximize(g,x=0..2);

Error, (in Optimization:-NLPSolve) summation variable previously assigned, second argument evaluates to 1. = 1 .. infinity
 

(Yes, a bug, which I'll report)

> Optimization:-Maximize(g,x=0..2,method=branchandbound);

Error, (in Optimization:-NLPSolve) method=branchandbound option is available for univariate unconstrained bounded problems only

(Possibly a different bug)

> plot(g, x = 0 .. 2);

> A:= op([1,1],%):
   maxvalue:= max(map2(op,2,A));

maxvalue := .3971541099

> select(has, A, maxvalue);

[[.9172052292, .3971541099]]

The y value is pretty close to the true maximum, I think; the x value is very rough, but further refinement is possible.

OddEven := L -> [seq([L[],1$-(-1)^nops(L),L[2..-1][]][2*i-1],i=1..nops(L))];

OddEven := L -> [seq([L[],1$-(-1)^nops(L),L[2..-1][]][2*i-1],i=1..nops(L))];

A bit less elegant?:

OddEven := L -> [seq([L[],op(select(type,{nops(L)},odd)),L[2..-1][]][2*i-1],i=1..nops(L))];

A bit less elegant?:

OddEven := L -> [seq([L[],op(select(type,{nops(L)},odd)),L[2..-1][]][2*i-1],i=1..nops(L))];

The first method doesn't work correctly unless the length of the integer is divisible by 3.

> f(1234);

     [123, 4]

I would do it this way:

> h := n -> ListTools[Reverse](convert(n,base,1000));

> h(1234567890);

  [1, 234, 567, 890]

The first method doesn't work correctly unless the length of the integer is divisible by 3.

> f(1234);

     [123, 4]

I would do it this way:

> h := n -> ListTools[Reverse](convert(n,base,1000));

> h(1234567890);

  [1, 234, 567, 890]

The add and mul functions were introduced in Maple V Release 4: see the help page ?updatesR4,language.  I'm sure convert(...,`+`) goes way back, long before I became involved with Maple.  It's mentioned in the Maple V Library Reference Manual (that's for Maple V Release 1, copyright 1991)

The add and mul functions were introduced in Maple V Release 4: see the help page ?updatesR4,language.  I'm sure convert(...,`+`) goes way back, long before I became involved with Maple.  It's mentioned in the Maple V Library Reference Manual (that's for Maple V Release 1, copyright 1991)