anonymous one year ago given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a

1. anonymous

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

2. Vocaloid

are you on mobile?

3. anonymous

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

4. dan815

what do those brackets mean

5. anonymous

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

6. Loser66

ceiling function?

7. anonymous

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

8. anonymous

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

9. Loser66

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

10. freckles

$x [x[x]]=15 \\ [x[x]]=\frac{15}{x} \\ \text{ so } \frac{15}{x} \text{ is an integer } \text{ and we have that } \frac{15}{x}-1<x[x]\le \frac{15}{x} \\ \\ ... \text{ still thinking...}$

11. freckles

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

12. anonymous

Where is the smallest integer bracket at?

13. anonymous

I'll just let parentheses be it.

14. freckles

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

15. freckles

$\lceil x \rceil$ oh yeah you can use \lceil x \rceil

16. anonymous

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

17. 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

18. anonymous

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

19. freckles

no

20. freckles

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

21. freckles

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

22. anonymous

oh, ok.. i see now

23. 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 :(

24. anonymous

25. anonymous

7?

26. anonymous

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

27. anonymous

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

28. anonymous

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

29. freckles

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

30. anonymous

so, just using trials any x ?

31. freckles

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

32. 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

33. anonymous

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

34. 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 $x [x[x]]=x[x(x)]=x[x^2]=x(x^2)=x^3 \\ x^3=15 \text{ so I think solving for the real x value here } \\ \text{ will give us a number closed to our solution }$

35. freckles

close*

36. 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

37. anonymous

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

38. freckles

yes i already said x cannot be an integer

39. freckles

x has to be a rational number

40. freckles

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

41. anonymous

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

42. anonymous

You wana show me a screenshot of this question.

43. freckles

so this is what we have so far $2<x<3 \\ \text{ and } \frac{15}{x} -1<x [x]\le \frac{15}{x} \\ \text{ where } \frac{15}{x} \text{ is an integer } \\ \text{ we also have } x \text{ is rational } \\ x=\frac{m}{n}, \text{ where } m \text{ and } n \text{ are integers not equal to 0}$ $\frac{m}{n}[\frac{m}{n}[\frac{m}{n}]]=15 \\ \text{ let's see } 2 <\frac{m}{n}<3 \\ \text{ so we have } [\frac{m}{n}]=3 \\ \frac{m}{n}[\frac{m}{n}(3)]=15 \\ \frac{m}{n}[ \frac{3 m}{n}]=15$ $6<\frac{3m}{n}<9 \text{ since } 2<\frac{m}{n}<3 \text{ so this means } [\frac{3m}{n}] \text{ could be } 7,8, \text{ or } 9 \\ \text{ let's go through the cases } \\$ $\text{ assume } [\frac{3m}{n}]=7 \text{ then } \frac{m}{n}(7)=15 \text{ and so } \frac{m}{n}=\frac{15}{7} \\ \text{ assume } [\frac{3m}{n}]=8 \text{ then } \frac{m}{n}(8)=15 \text{ so } \frac{m}{n}=\frac{15}{8} \\ \text{ assume } [\frac{3m}{n}]=9 \text{ then } \frac{m}{n}(9)=15 \text{ so } \frac{m}{n}=\frac{15}{9}$ I just figured out there was no point in writing x as m/n but I don't feel like changing them all back anyways you check your 3 cases to figure out if the solution

44. anonymous

I think the answer is somewhere along $2.99<x<3$

45. freckles

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

46. 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

47. freckles

oh I thought we were doing ceiling function

48. anonymous

Oh yea I reversed it.

49. anonymous

That one guy wrote 2.14 at the beginning was correct.

50. anonymous

it should be between 2.14 and something like an interval.

51. freckles

you do know 15/7 is approximately 2.14286

52. anonymous

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

53. freckles

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

54. anonymous

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

55. 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

56. anonymous

14.99998 would be the same 15.00002

57. anonymous

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

58. anonymous

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

59. anonymous

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

60. 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

61. anonymous

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

62. 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.

63. 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

64. anonymous

Actually it does. Lemme think of more complex ones.

65. anonymous

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

66. freckles

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

67. freckles

or x(x(x))=16.36?

68. 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

69. anonymous

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

70. 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.

71. anonymous

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

72. anonymous

Like a very fast way.

73. freckles

$x[x[x]]=16.36 \\ \text{ we know } x \text{ is not an integer } \\ \text{ but so we can get in the neighborhood } \\ \text{ we will solve } x^3=16.36 \\ \text{ which will actually give us a real number and not an integer }$ $x \approx 2.5 \\ \text{ so this means we know } \\ 2<x<3 \\ \text{ which means } [x]=3 \\ \\ \text{ so back \to the equation } \\ x[x[x]]=16.36 \\ x[x(3)]]=16.36 \\ x[3x]=16.36 \\ \text{ now recall } 2<x<3 \\ \text{ so we have } 6<3x<9 \\ \text{ so this means } \\ [3x]=7, 8, \text{ or } 9 \\ x(7)=16.36 \implies x=\frac{16.36}{7} \\ x(8)=16.36 \implies x=\frac{16.36}{8} \\ x(9)=16.36 \implies x=\frac{16.36}{9}$ we have 3 cases to check: checking: $\frac{16.36}{7}[\frac{16.36}{7}[\frac{16.36}{7}]]=\frac{16.36}{7}[\frac{16.36}{7}(3)]=\frac{16.36}{7}(8) \neq 16.36 \\$ $\frac{16.36}{8}[\frac{16.36}{8}[\frac{16.36}{8}]]=\frac{16.36}{8}[\frac{16.36}{8}(3)]=\frac{16.36}{8}(7) \neq 16.36 \\$ $\frac{16.36}{9}[\frac{16.36}{9}[\frac{16.36}{9}]]=\frac{16.36}{9}[\frac{16.36}{9}(2)]=\frac{16.36}{9}(4) \neq 16.36$ So I think there is no real x such that the equation is true since none of the cases worked

74. 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.

75. anonymous

76. 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

77. Loser66

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

78. 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.

79. Loser66

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

80. Loser66

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

81. Loser66

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82. Loser66

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83. Loser66

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84. Loser66

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