Texas Instruments TI-92+, Список советов

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cSolve((-1)^n=1,n)
returns n = 2*@n1, and
cZeros((-1)^n-1,n)
returns {2*@n1}; both of these results are correct.
As another example, consider
cSolve(ln(e^z_)=ln(z_^2))
As shown, with no guess, the returned result is
e^z_-z_^2=0
which doesn't help much. However, with an initial complex solution guess:
cSolve(ln(e^z_)=ln(z_^2),{z_=i})
cSolve() returns one complex result of 1.588... + 1.540...i. With a real guess:
cSolve(ln(e^z_)=ln(z_^2),{z_=0})
cSolve() returns one real result of -0.7034... Note that cSolve() returns these approximate floating-point
results even in Exact mode.
(Credit to Bhuvanesh Bhatt)
[11.4] Using solve() with multiple equations and solutions in programs
This rather involved tip relates to using the solve() function in programs, when multiple equations to be
solved are generated within the program. The basic idea is to build a string of the equations to be used
in the solve() function, evaluate the solve function string with expr(), then change the results into a
more useful list form. This tip only applies if you are interested in numeric solutions, not symbolic
solutions.
As an example, I'll use this system of equations to be solved:
ax
1
2
+ bx
1
+ c = y
1
ax
2
2
+ bx
2
+ c = y
2
ax
3
2
+ bx
3
+ c = y
3
where the unknown variables are a, b and c. x
1
, x
2
, x
3
, y
1
, y
2
and y
3
are given. Actually, since this is a
linear system, you wouldn't really use solve(); instead you would use the matrix algebra functions built
into the 89/92+. But it makes a good example.
Suppose that we know that
x
1
= 1 y
1
= -1
x
2
= 2 y
2
= -11
x
3
= 3 y
3
= -29
11 - 2