MATLAB: Solve: floor(x/5) – floor(x/7) = 1 in Matlab

eq_floorsolve

Sir/Mam, i want to find all the solutions of the equation: floor(x/5) – floor(x/7) = 1 ;where x belongs to Natural Number, so how can i find all its solutions in Matlab? with Thanks

Best Answer

Okay, if this is not a homework, than you find enough information here to solve the problem already. I summarize:

Matt J showed, that the results must be 0 <= x <= 35.
You can easily check all values:

x      = 0:35;
match  = (floor(x / 5) - floor(x / 7) == 1);
result = x(match)

And here you get your 17 values.

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MATLAB: How to implement a Matlab function, findtbracket.m, that, when a random function is inputed, the file will find the initial interval of the function

Mónica

prueba lo siguiente:

x0=4.333  
a=x0
b=x0
k=1
f=@(x) sin(x)
k3=1
while ~(floor(x0-2^(-k3-1))==floor(x0))
k3=k3+1
end
k4=1
while ~(floor(x0-2^-k4)==floor(x0))
k4=k4+1
end
k=max(k3,k4)
d=2^-k
while (sign(f(a))==sign(f(b)))
  a=a-d
  b=b+d
  d=2*d
end

a =   2.583333000000000
b =   6.083333000000000

me parece que el enunciado que has adjuntado, implementado tal y como se lee, se cuelga para ciertos valores de x0, por ejemplo, tanto

k2=1
while ~(floor(x0-2^(-k2-1))==x0 && floor(x0-2^-k2)<x0) 
  k2=k2+1
end

como

k2=1
while ~(floor(x0-2^(-k2-1))==floor(x0) && floor(x0-2^-k2)<floor(x0))  
  k2=k2+1
end

se cuelgan para x0=4.9

por lo que quizás sea posible modificar el enunciado así:

Si esta respuesta te resulta de utilidad para solucionar tu pregunta, te agradeceria me votes haciendo click en el link que tiene el pulgar arriba al lado de esta respuesta.

Atentamente

John

jgb2012@sky.com

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MATLAB: How to implement a Matlab function, findtbracket.m, that, when a random function is inputed, the file will find the initial interval of the function

MATLAB: How to “accept” an answer to your question

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