given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a

- anonymous

given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a

- schrodinger

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- anonymous

what.... where is the tools drawing ? i cant draw the equations ._.

- Vocaloid

are you on mobile?

- anonymous

no. i just using my laptop. the problem like this
given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a

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## More answers

- dan815

what do those brackets mean

- anonymous

actually ⌈x⌉ mean the smallest integer which not less than x, right ?
help meh @dan :)

- Loser66

ceiling function?

- anonymous

yeah.... what's the start to do it, @Loser66 ?

- anonymous

the complete question is find all real a which satisfy f(a) = 15

- Loser66

I really don't know how to solve it :( , wolfram gives us the answer is 2.14, but how ??
@freckles

- freckles

\[x [x[x]]=15 \\ [x[x]]=\frac{15}{x} \\ \text{ so } \frac{15}{x} \text{ is an integer } \text{ and we have that } \frac{15}{x}-1

- freckles

\[\frac{15-x}{x}0 \\ \text{ we have } \frac{15-x}{x^2} <[x] \le \frac{15}{x^2}\]
I don't know what this means yet if anything can be concluded from it

- anonymous

Where is the smallest integer bracket at?

- anonymous

I'll just let parentheses be it.

- freckles

I just used brackets.
I think there is some code you can type in.

- freckles

\[\lceil x \rceil\]
oh yeah you can use \lceil x \rceil

- anonymous

Oh, I dont know much about symbols.
Thanks for pointing that out.

- anonymous

I'll just use parentheses.
I am not sure with this.
But I did this \[\sqrt[3]{15}=2.46\]
15/x=(x(x))
Isnt (2.46)=2.46?
So (2.46 x 2.46)=6.071
6.071=15/2.46

- anonymous

⌈x ⌈x⌉⌉ = 15/x
the right side is integer. so x must be an integer also right ?

- freckles

no

- freckles

15/x
I think x is between 0 and 15 though
for example x could be 15/7 see: 15/(15/7)=7

- freckles

so what I'm saying is x could be an integer or it could be rational

- anonymous

oh, ok.. i see now

- freckles

I'm just having problems showing x=15/7
we can certainly show 15/7 is the solution by pluggin it in
just having issues coming up with that answer :(

- anonymous

What is the answer?

- anonymous

7?

- anonymous

not sure. it is an essay question, no choices :v

- anonymous

a can be more than one answer.
Since you wrote "the complete question is find all real a which satisfy f(a) = 15"

- anonymous

yeah.. maybe a can be more than one answer, but how to get them ?

- freckles

\[x \cdot [x [x]]=15 \\ \text{ if } x=\frac{15}{7} \text{ then } \\ \frac{15}{7}[ \frac{15}{7}[\frac{15}{7}]]=\frac{15}{7}[ \frac{15}{7}(3)]=\frac{15}{7}[\frac{45}{7}]=\frac{15}{7}(7)=15 \\ \text{ I didn't say } x=7 \\ 7[7[7]]=7[7(7)]=7[49]=7(49) \neq 15\]

- anonymous

so, just using trials any x ?

- freckles

\[2.46 \cdot [2.46 [2.46]]=2.46[2.46(3)]=2.46(8) \neq 15\]

- anonymous

For 2.46
\[2\le x <3\]
x can be 2.46
then 2.46(2.46)
which is \[2.46 \times (2\le x <3)\] x can be 2.46=6.071
6.071=15/2.46
So I got x=2.46

- anonymous

Just kidding.
I dont know enough about ceiling and floor functions.

- freckles

But @Shalante I think your method helped us to find what our solution should be near
if we assume x is an an integer
then \[x [x[x]]=x[x(x)]=x[x^2]=x(x^2)=x^3 \\ x^3=15 \text{ so I think solving for the real x value here } \\ \text{ will give us a number closed to our solution }\]

- freckles

close*

- freckles

but solving for x here and plugging x in we will see that 15/x is not an integer
so x is not an integer
and we can conclude x is a rational near the real cube root value of 15

- anonymous

It doesnt need to be an integer.
The x is outside the ceiling function.

- freckles

yes i already said x cannot be an integer

- freckles

x has to be a rational number

- freckles

I was showing above by contradiction why x cannot be an integer

- anonymous

I take that back.
I'll try a new method (might take forever)

- anonymous

You wana show me a screenshot of this question.

- freckles

so this is what we have so far
\[2

- anonymous

I think the answer is somewhere along \[2.99

- freckles

if you check the three cases above you should see 15/7 is the solution @Shalante

- anonymous

Smallest integer of 2.99 is 2
2*2.999=5.998
Smallest integer of 5.998 is 5
5*2.999=14.995 which is close enough.
15/7 is 2.142
2.142 smallest integer is 2
2*2.142=4.284
Smallest integer of 4.284 is 4
2.142*4 is not 15

- freckles

oh I thought we were doing ceiling function

- anonymous

Oh yea I reversed it.

- anonymous

That one guy wrote 2.14 at the beginning was correct.

- anonymous

it should be between 2.14 and something like an interval.

- freckles

you do know 15/7 is approximately 2.14286

- anonymous

I kept thinking ceiling is a restraint meaning it has to be less.

- freckles

but 2.14 is only an approximation to actual answer which is 15/7

- anonymous

I got 15.00002
0.0
The question asked for all a values.
An interval would help I guess.

- freckles

well I narrowed it down to 3 cases for a
you just check the 3 cases to see which is the answer
I think there is only one solution

- anonymous

14.99998 would be the same 15.00002

- anonymous

I thinks theres more than one.
Its close enough.
You know chemistry?
Its like a 0.001 percent error?

- anonymous

Decimals are close enough especially in the thousandth.
But it says find all, so I would use interval.

- anonymous

Are you talking about 15/8 and 15/9?
I dont think it would fit at all.

- freckles

I come up with three cases
a=15/8
a=15/9
a=15/7
using that a is between 2 and 3
which means the ceiling(a)=3
and so ceiling(3a)=7,8, or 9 since 3a is between 6 and 9
so this gave us three possibilities
7a=15
8a=15
9a=15
but pluggin back in we see only one of these is successful
therefore the only solution is a=15/7 since that is the only success

- anonymous

I got 2.999 earlier doing it the floor way by just plugging and chugging.
Its actually not bad either.

- anonymous

I wasnt familiar with greatest integer function until like 25 minutes ago.
Thats when I start to understand and keep plugging.
Its like guessing a number between 2-3 in the hundredth decimal and they kept telling you if its lower or higher.

- anonymous

Not sure if your method worked since the problem is an easier one or thats the way it is.
Because I am sure, I would just plug and chug in the more complex one and it does take that long either.
like x(x(x))=15.4

- anonymous

Actually it does.
Lemme think of more complex ones.

- anonymous

Try 16.36.
I couldnt do it dividing it by 6 to 10.

- freckles

you are trying to solve x(x(x))=15.4?

- freckles

or x(x(x))=16.36?

- anonymous

I solved 15.4 already using your method of dividing it like this 15.4/7.
But I cant do it at 16.36

- anonymous

Because I want to know if there is a rule that works on all of them.
Or is it just some certain numbers?

- anonymous

In math besides complex numbers or some other I dont know, almost all rules works on all numbers.
Pretty interesting to know for this one.

- anonymous

16.36/7 16.36/8 16.36/9
Besides plug and chug.

- anonymous

Like a very fast way.

- freckles

\[x[x[x]]=16.36 \\ \text{ we know } x \text{ is not an integer } \\ \text{ but so we can get in the neighborhood } \\ \text{ we will solve } x^3=16.36 \\ \text{ which will actually give us a real number and not an integer }\]
\[x \approx 2.5 \\ \text{ so this means we know } \\ 2

- anonymous

But 15.4 is not an integer and worked.
15.4/7=2.2
2.2>>3
2.2*3=6.2
6.2>>7
7*2.2=15.4
So this method only works for specific numbers then
Maybe there is a pattern to it just like some only work on odd,even, prime, etc.
Interesting.

- anonymous

You already got your answer.
No need to bump this.

- Loser66

I have an idea: if x is an integer, then \(\sqrt[3] {15}=2.46\), and we know that \(2\leq x\leq 2.46\)
\(f(2) =2\lceil 2\lceil 2\rceil \rceil =2\lceil 2*2\rceil=2*4 =8\)
\(f(2.46) = 2.46\lceil 2.46\lceil 2.46\rceil \rceil =2.46\lceil 2.46*3\rceil=2.46*8 =19.86\)
therefore, x must go to the left of 2.46.
Now, by taking half way, \(\dfrac{2.46-2}{2}= 0.23\), we test 2.23
\(f(2.23)=2.23\lceil 2.23\lceil 2.23\rceil \rceil =2.23\lceil 2.23*3\rceil=2.23*7 =15.61\)
The result is still higher than what we want, hence , we need "half way" again.
\(\dfrac{2.23-2}{2}= 0.115\), we test 2.115
\(f(2.115)= 2.115\lceil 2.115\lceil 2.115\rceil \rceil =2.115\lceil 2.115*3\rceil=2.115*7 =14.805\)
It is a little bit smaller than what we need, but now, we take "half way" to the right from 2.115
\(\dfrac{2.23-2.115}{2}=0.0575, so, we need test 2.115 + 0.0575=2.1725
\(f(2.1725) = 2.1725\lceil 2.1725\lceil 2.1725\rceil \rceil =2.1725\lceil 2.1725*3\rceil=2.1725*7 =15.2075\)
it is still a little bit higher than what we need, but now, take half way to the left from 2.1725

- Loser66

typo at the red mark, it is
\(f(2.1725) = 2.1725\lceil 2.1725\lceil 2.1725\rceil \rceil =2.1725\lceil 2.1725*3\rceil=2.1725*7 =15.2075\)

- Loser66

It is still a little bit higher than what we need, now, take half way between 2. 115 and 2.1725, it is 2.14375
Test :
\(f(2.14375) =2.14375\lceil 2.14375\lceil 2.14375\rceil \rceil\\ =2.14375\lceil 2.14375*3\rceil=2.14375*7 =15.00625\)
I think it is good enough.

- Loser66

Hahahaha.... That is power of "sleeping". !! I meant after taking a snap, we can solve the problem we couldn't solve before.

- Loser66

In general, this is |dw:1443617046497:dw|

- Loser66

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- Loser66

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- Loser66

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- Loser66

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