Floor and Ceiling Functions in an Exponential Equation

This video is about a floor and a ceiling equation
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EXPLORE:

[goodplacetostop]

8:21 I know some guy who loves the floor function... I think you should submit this equation

[tamarpeer]

opt 1: floor(x)=x=ceil(x)
2n^n=175/8, no sol.
opt 2: floor(x)+1=ceil(x)

if x is negsative, |x^floor(x)|<=1 and |x^ceil(x)|<=1 so their sum is smaller than 175/8

so x is positive
we can look at the equation y^floor(y)=175/8
for y>=3, y^floor(y)>=3^3>175/8
so x<y<3
since x is smaller than 3 we can now look at the equation u^floor(u)(1+3)=175/8
for u<=2, we have u^floor(u)<2^2<175/32
so 2<x<3
we now floor(x) and ceil(x)
x^2+x^3=175/8 is a cubic equation
according to wolfram alpha, there is a single solution x=5/2

[DrQuatsch]

By process of elimination I found that indeed only the interval 2 < x < 3 has a possible solution. So x can be written as x = 2 + a with 0 < a < 1.
we will get (2 + a)² + (2 + a)³ = 175/8. Expanding this gives us: a³ + 7a² + 16a + 12 = 175/8. Moving everything to one side gives us: a³ + 7a² + 16a - 79/8 = 0. Multiply everything by 8 to get rid of the fraction: 8a³ + 56a² + 128a - 79 = 0. Now by taking b = 2a, we get: b³ + 14b² + 64b - 79 = 0. The sum of the factors in this polynomial is 0, so b = 1 is a solution. You can factor it out and (by long division or any other method) you'll get: (b - 1)(b² + 15b + 79) = 0.
So b - 1 = 0 or b² + 15b + 79 = 0. For the quadratic we'll have a determinant of 15² - 4 x 79, which is less than 0. So the quadratic doesn't give a real solution. Therefore b = 1 is the only solution. Now 2a = b, so a = b/2 = 1/2. We had x = 2 + a, so that will be 2 + 1/2 = 5/2. Checking the solution by filling it in in the original equation checks out, so x = 5/2 is the solution to this equation.

[mcwulf]

And if your guess of u=5 doesn't work, you have a cubic to solve.

[manjunathbhat]

Here's another approach:
x^floor(x) + x^ceil(x) = x^floor(x) * (1 + x) when x is not an integer, and we know that x is not an integer for sure. We know that x is a rational number because RHS is rational. So, x = p/q, where p, q are coprime integers. Hence,
(p/q)^floor(x)*(1 + p/q) = 175/2^3
p^floor(x)*(q + p)/q^(floor(x) + 1) = 175/2^3, comparing denominators, we have q = 2, floor(x) + 1 = 3 => floor(x) = 2. Now comparing numerators, we have (2 + p)*p^2 = 175, and we also know that floor(p/2) = 2, hence p = 5. Thus x = 5/2

[timetraveller]

Nice problem can i get a Calculus one? especially a limit one.

[leecherlarry]

compi had to try various functions but this one worked:

[Ssilki_V_Profile]

For this function x can be negative. But for negative values function may be positive.

[italixgaming]

My solution (of course way faster):
First of all, we eliminate the case where floor(x)=ceiling(x). This case would mean that x is an integer. The equation woud become 2.x^x=175/8, which is impossible since the LHS of the equation is an integer.
We also eliminate the negative values since if x<0 then
Our equation can be rewritten: x^(floor(x))(1+x)=175/8
Let's discuss on floor(x): if floor(x)=0 then our LHS is 1+x<2<175/8: there is no solution.
If floor(x)=1 then our LHS is x(1+x)<2.3=6<175/8: there is no solution.
If floor(x)=2 then the LHS is x²(1+x) : the minimal value of this expression on [2;3[ is 12 and the maximum value is the limit on 3-, which is 9.4=36. We notice that 12<175/8<36 and since the function is continuous we know that we have a solution somewhere.
Before finding it, let's just eliminate the other values of floor(x). If floor(x)>2 (x>3), then so there is no solution.
So our equation becomes: x²(1+x)=175/8 where 2<x<3.
Since the RHS is from he form a/8, we want to test a solution from the form b/2 but there is only one candidate to test, which is 5/2.
(5/2)².7/2=25/4.7/2=175/8.
Now we conclude, saying that the function product of two functions that are stricly rising for positive values, is strictly rising so the solution is unique.

[manojsurya]

Helllo syber math, I want to know whether u r a math professor in a institution or not?

[srijanbhowmick]

Here's my solution :----










First I wanted to bound x
Now, if x >= 3, floor(x) >= 3
Thus,
x^floor(x) >= 27 but 27 < 175/8
So x < 3
Also x is obviously non-negative
So 0 <= x < 3
Then I did case-work when floor(x) = 0, 1, 2
Only the last case yielded a solution which is x = 5 / 2

[neuralwarp]

So basically try all the possible answers. I don't find this satisfactory as a generic methodology. You might as well use Newton-Raphson.