Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Callisto
Group Title
In the domain [0, 100], how many solutions are there for the equation \[x^2 \lfloor x \rfloor x = 81.25\]
How to start?
 one year ago
 one year ago
Callisto Group Title
In the domain [0, 100], how many solutions are there for the equation \[x^2 \lfloor x \rfloor x = 81.25\] How to start?
 one year ago
 one year ago

This Question is Closed

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
Instead of thinking of it as one function defined on [1,100], maybe think of it as 100 different function defined on the intervals [n,n+1] as n goes from 0 to 99.
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
That way you can minimize the frustration the floor function brings to the table.
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
er, i guess the interval should really be [n,n+1)
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
Then on each interval, the question is really:\[x^2+nx81.25=0\]where x ranges from [n,n+1). Not sure if this leads to a solution, but it might be a better way of thinking of the problem.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Domain = [0, 100] :S
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
wait, the function\[x^2+\lfloor x \rfloor x\]is strictly increasing, isnt it?
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
then it can have at most one solution.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
That means it has 0 :'(
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
strictly increasing on the interval [0,100] anyways.
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
well, im not 100% sure on the strictly increasing part, im a little fuzzy on what happens when you jump from one unit interval to the other.
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
no, its strictly increasing for sure.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Sorry!!! Wait... a mistake in the question posted here!!
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
ah ok, that makes more sense.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
I'm sorry!!!! :(
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
if you look at the function:\[x^2\lfloor x \rfloor x\]restricted to an interval {n,n+1) (open on the right), note that the values start at 0 when x = n, and as x approaches n+1, the values also approach n+1:dw:1357103977725:dw
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
So the equation \[x^2\lfloor x \rfloor x=81.25\]only has a solution when \[x\le 81\]the interval {81,82) is going to be the first time the parabola piece gets as large as 81.25.
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
less than >.> i meant greater than lol.
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
\[x\ge 81.25\]
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
>.<\[x\ge 81\]
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
So there are 19 solutions, one in each of the intervals [81, 82), [82,83),....., [99,100)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Thanks! I've never thought about this way to look at the problem.. :(
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
The idea is that the floor function on large intervals in a pain, but on unit intervals, its just a the constant function. an interesting problem indeed :)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
I'll remember it! Thanks again!!
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
x[x] = {x} right.. ? {x} = fractional part of x so if we simplify a little, we see we have x*{x} = 81.25 does this help ?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
x= 81.25 /{x} x can go max to 100 so fractional part can be minimum 0.8125 and max 0.999999...
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
just substitute in new values of {x} to get new values of x so according to me,,infinite solutions..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
ohh wait,,did i make a mistake in logic,,i am getting a feeling,,!??
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
@joemath314159 @mukushla @calculusfunctions
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
if you look at the function:\[\frac{81.25}{\{x\}}\] since the fractional part of x can only go from 0 to 1, the only time that will have a solution is when x is bigger than 81.
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
if you look at domain, say, [79, 80), this equation:\[x=\frac{81.25}{\{x\}}\]wont have a solution. since the smallest\[\frac{81.25}{\{x\}}\]can be is 81.25
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
maybe this is where am flawed: since we are dealing with fractional part separately, x , cannot have a fractional part..!! like for example if we set {x} = 0.9, x = 90.2777.. hence contradiction > 2 different fractional parts..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
i see,,only solutions will be when x is an integer..so that the contradiction with fractional part doesnt come into play and since x has to be greater than 81.25 as @joemath314159 rightly pointed, x can take values 82,83...100 >19 solutions..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
i mean x can not take those values,, but there'll be a solution each in [82,83) and so on.. so yes,,19 solutions..
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
problem says how many solutions so its done...and this is just for fun :) we can find exact solutions:\[[x]+\{x\}=\frac{81.25}{\{x\}}\]\[[x]=\frac{81.25}{\{x\}}\{x\}\]now we know that \[[x]=\{81,82,...,99\}\]so for example for the solution that lies on the interval [81,82) we have\[81=\frac{81.25}{\{x\}}\{x\}\]this is a quadratic for \(\{x\}\)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
yep..i understand better now..thank you..
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.