Kitonum

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17 years, 235 days

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These are replies submitted by Kitonum

@Maple_lover1  I corrected my answer.

@BrettKnoss  See my solution above.

@BrettKnoss  I didn't understand the meaning of your question. If a series is given, then the general term of the series  u(n)  must be known for any  .

@BrettKnoss  c(n)  is just the coefficient in front of  n_th  term of the series:

restart;
S := Sum((x-2)^n/(n^2-1)^2, n = 2 .. infinity);
op(1,S);
coeff(%, (x-2)^n);

                       

 

@matviiv10 

restart;
h:=0.01: r:=0.02: N:=10: f:=y->y^5: K:=(x,y)->sin(x)*y^5:
for n from 1 to N do 
x[n]:=sin(h*n);
od:
for m from 1 to N do
y[m]:=cos(r*m);
od:
n:='n': m:='m':
Sys:={seq(sum(K(x[n],y[m])*U[n],n=1..N)=f(y[m]), m=1..N)};
solve(Sys);
assign(%);
plot([seq([x[n],U[n]], n=1..N)]);

 

@ahmeng  Above, I showed only one solution (in my opinion, the most direct and reliable). Here are 2 more ways:


 

restart

``

"f(x):=18*log10(x);"

proc (x) options operator, arrow, function_assign; 18*log10(x) end proc

(1)

"g(x):=1/(2) x^(3)-8*x^(2)+(69/(2))^()*x-27;"

proc (x) options operator, arrow, function_assign; (1/2)*x^3-8*x^2+(69/2)*x-27 end proc

(2)

plot([f(x), g(x)], x = -1 .. 11)

 

``

`assuming`([solve(f(x)-g(x), x)], [x::real])

exp(RootOf(-(exp(_Z))^3*ln(10)+16*(exp(_Z))^2*ln(10)-69*exp(_Z)*ln(10)+54*ln(10)+36*_Z))

(3)

allvalues(exp(RootOf(-(exp(_Z))^3*ln(10)+16*(exp(_Z))^2*ln(10)-69*exp(_Z)*ln(10)+54*ln(10)+36*_Z)))

exp(RootOf(-(exp(_Z))^3*ln(10)+16*(exp(_Z))^2*ln(10)-69*exp(_Z)*ln(10)+54*ln(10)+36*_Z, 1.505446443)), 1, exp(RootOf(-(exp(_Z))^3*ln(10)+16*(exp(_Z))^2*ln(10)-69*exp(_Z)*ln(10)+54*ln(10)+36*_Z, -3.291052648)), exp(RootOf(-(exp(_Z))^3*ln(10)+16*(exp(_Z))^2*ln(10)-69*exp(_Z)*ln(10)+54*ln(10)+36*_Z, 2.302585093)), exp(RootOf(-(exp(_Z))^3*ln(10)+16*(exp(_Z))^2*ln(10)-69*exp(_Z)*ln(10)+54*ln(10)+36*_Z, 1.788246136-5.696578776*I)), exp(RootOf(-(exp(_Z))^3*ln(10)+16*(exp(_Z))^2*ln(10)-69*exp(_Z)*ln(10)+54*ln(10)+36*_Z, .6642129866+7.816965006*I)), exp(RootOf(-(exp(_Z))^3*ln(10)+16*(exp(_Z))^2*ln(10)-69*exp(_Z)*ln(10)+54*ln(10)+36*_Z, 1.788246136+5.696578776*I)), exp(RootOf(-(exp(_Z))^3*ln(10)+16*(exp(_Z))^2*ln(10)-69*exp(_Z)*ln(10)+54*ln(10)+36*_Z, .6642129866-7.816965006*I))

(4)

evalf(%[1 .. 4])

4.506164927, 1., 0.3721465484e-1, 10.00000000

(5)

Student:-Calculus1:-Roots(f(x)-g(x), x)

[0.3721465485e-1, 1, 4.506164928, 10.00000000]

(6)

``


 

Download more_ways.mw

@ibrahimcbx 

1. You can do this with either separate commands or with a single command.

2. I didn't understand the meaning of your second question.

@Carl Love  It seems that your solution will be correct only in the segment  p in [0, 3-2*sqrt(2)]

 

@salma951  I didn't notice that 3 appears twice and 6 is missing. So write  randperm([1,2,3,3,4,5]) 

@jalal  Replace  range = -3 .. 10  by  (for example)  range = 1 .. 10  .

 

@acer  I think in all intermediate calculations it is better to use exact mathematics. Of course, we can present the final results in any desired form.

For example:

[solve({eqy3, x>=-5, x<=5})];

           

 

nm 4588  We can easily generalize your method to other exponential functions (in addition to exp). We can also use the  applyop  command (many shorter if the position of the corresponding terms is known):

restart;
expr := -2*exp(x)*x - exp(x) + 2*exp(x)^2*x + exp(x)^2 +(3^x)^2+ sin(x)^3 + cos(x)^2;
subsindets(expr,{exp(anything)^anything,(anything^anything)^anything},combine); # The first way
applyop(combine,{$1..5},expr); # The second way

          

 

@jalal  It looks even better if the multiplication sign is implied, but not displayed:

4%/5*``(2*x-5);

                              

 

@nm  assuming positive greatly limits the scope of this identity, and if this condition is violated, the identity may be incorrect. For example,  -cos(A) - sin(A) <> sqrt(2)*sin(A+Pi/4)

@jalal 

restart;
randomize():
do
V:=LinearAlgebra[RandomVector](6,generator=10..20);
if nops(convert(V,set))=6 then break fi;
od:
V;

or

restart;
with(combinat):
randomize():
convert(randperm(randcomb([$ 10..20], 6)), Vector);

 

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