+0

# two questions

0
58
15

1. Which positive real number x has the property that x, $$\lfloor x \rfloor,$$ and $$x - \lfloor x\rfloor$$ form a geometric progression (in that order)? (Recall that $$\lfloor x\rfloor$$ means the greatest integer less than or equal to x.)

2. Let $$N = \sum_{k = 1}^{1000}k(\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor).$$ Find N.

Thank you!

Jun 25, 2019

#1
0

2 -

sumfor(k, 1, 1000,  k* (ceil(logn(sqrt2,k)) -  floor(logn(sqrt2, k))) = 500,499
This has this "closed form":
a(n) =1 / 2 [n -1]*[n + 2] =500,499 when n=1000

Jun 25, 2019
#3
+25
0

This is incorrect

HelloWorld  Jun 25, 2019
#2
+101808
+2

1.  Let x  = p + a       let  floor (x)  =  p

So....we have the following series

(p + a) ,  p  ,  a

The common ratio, r,   =     a/p

So

(p + a) ( a/p)  =  p     ⇒   a + a^2/p  = p   ⇒   a^2 + pa - p^2 = 0

a =  -p +√[p^2 + 5p^2 ]          =        - p  + √5 p      =       p ( √5 - 1)    =    p * phi

_______________                 __________            _________

2                                             2                              2

Then  a/p  =  phi

Then

(p + a) (a/p)  =  p

(p  + p*phi) (phi)  =  p

p*phi + p*phi^2  =  p        if we let  p =1, then we have an identity for these powers of phi, namely :

phi + phi^2  =  1

And  a  = p*phi  =  1*phi  =  phi ≈  .618

So x  =   (a + p)  =  ( phi + 1 )  =  Phi  ≈  1.618

So...this is the common series  :

x , floor(x)  , x - floor(x)  =

Phi, 1 , phi

Jun 25, 2019
edited by CPhill  Jun 25, 2019
edited by CPhill  Jun 25, 2019
edited by CPhill  Jun 25, 2019
#15
0

Just to clarify, is phi the answer?

Guest Jun 26, 2019
#4
+8403
+2

When  k  is a power of  2 ,   $$\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor\ =\ 0$$

When  k  is not a power of 2 ,   $$\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor\ =\ 1$$

So  N  is the sum of all numbers which are not powers of  2  in the interval  [1, 1000]

$$N \quad=\quad \sum\limits_{k = 1}^{1000}k\quad-\quad \sum\limits_{n = 0}^{9}2^n\\~\\~\\ N \quad=\quad \sum\limits_{k = 1}^{1000}k\quad-\quad (1+2+4+8+16+32+64+128+256+512)\\~\\~\\ N \quad=\quad \sum\limits_{k = 1}^{1000}k\quad-\quad 1023\\~\\~\\ N \quad=\quad \Big(\frac{1000(1000+1)}{2}\Big)\quad-\quad1023\\~\\~\\ N \quad=\quad 500500\quad-\quad1023\\~\\~\\ N \quad=\quad 499477$$

It first says approximately 500497, then it comes back and says 499477

Jun 25, 2019
#5
0

Hectictar: The reason you got a different result is because of sqrt(2). The sqrt(2) is the BASE of log. In other words, it is meant to be : Log(base sqrt(2), k)). But you entered it in WA as:

k(ceil(log(sqrt(2), k))-floor(log(sqrt(2),k))). If you enter it the way he/she has it in LaTex, it should be entered like this:  k(ceil(log_(1.41421), k)) - floor(log_(1.41421,k)))= 500,499. See here:

https://www.wolframalpha.com/input/?i=%E2%88%91%5Bk*(ceil(log_(1.4142135,+k))+-+floor(log_(1.4142135,k))),+k,+1,+1000%5D

Jun 25, 2019
#6
+8403
+1

I tried to state the base using the underscore but it didn't work, so I used this notation

log(b, a)

which WA will interpret as

$$\log_{b}a$$

hectictar  Jun 25, 2019
#7
0

Thanks for that. I find that WA often has mistakes in it. Since I have a "Mathematica 11 Home Edition" , I communicate with them about errors I find in WA. Here is a recent reply I received from them about a mistake I had found:

The Wolfram|Alpha Team

Mon, May 6, 2:58 PM

to me

The issue you reported has been fixed on the site. See https://www.wolframalpha.com/input/?i=combinations+of+%7B1,+1,+1,+2,+2,+2,+3,+3,+4,+5%7D,+size+4

Thank you for helping us improve Wolfram|Alpha.

Best wishes,

Lindsay
The Wolfram|Alpha Team
www.wolframalpha.com

________________________________________________________________________

Since I can code in 2 languages{C++ and Python}, I generally write my own code for many math problem I encounter. Here is a short computer code written in C++ taking sqrt(2) as the BASE of log:
k=1;p=0;  a=k*(ceil(log k / log(sqrt(2))) - floor(log k / log(sqrt(2))));p=p+a; k++; if(k<1001, goto2, discard=0;printp
Total = 500,499.

Jun 25, 2019
edited by Guest  Jun 25, 2019
#8
+8403
+2

Okay I think I understand what is happening.

$$\sum\limits_{k=1}^{1000}\ k(\lceil\log_{1.41421}k\rceil-\lfloor\log_{1.41421}k\rfloor)$$     IS equal to  500,499

When  k = 1 ,   $$\lceil\log_{1.41421}k\rceil-\lfloor\log_{1.41421}k\rfloor\ =\ \lceil\log_{1.41421}1\rceil-\lfloor\log_{1.41421}1\rfloor\ =\ 0$$

But for every other value of  k , the difference is not zero.

For example:

$$\lceil\log_{1.41421}32\rceil-\lfloor\log_{1.41421}32\rfloor\ =\ 1$$

BUT

$$\lceil\log_{\sqrt2}32\rceil-\lfloor\log_{\sqrt2}32\rfloor\ =\ 0$$

So it makes perfect sense that

$$\sum\limits_{k=1}^{1000}\ k(\lceil\log_{1.41421}k\rceil-\lfloor\log_{1.41421}k\rfloor)\ =\ 500500-1\ =\ 500499$$

while

$$\sum\limits_{k=1}^{1000}\ k(\lceil\log_{\sqrt2}k\rceil-\lfloor\log_{\sqrt2}k\rfloor)\ =\ 500500-1023\ =\ 499477$$

Basically, if you estimate the base first you will not get the same answer.

hectictar  Jun 25, 2019
edited by hectictar  Jun 25, 2019
#14
0

Thank you so much, Guest!

Guest Jun 26, 2019
#9
0

Thanks hectictar: All this was because of that "troublemaker" who said "This is incorrect" !!. The way to understand it is that for every number other than 1, the difference between the "ceiling" and the "floor" is always 1 (in this case). So, basically the problem boils down to:[1000 x 1001] / 2 - 1 = 500,499.

Jun 25, 2019
#10
+8403
+1

"for every number other than 1, the difference between the "ceiling" and the "floor" is always 1 (in this case)"

That is not true in the case where the base is  $$\sqrt2$$

See example:

$$\lceil\log_{\sqrt2}32\rceil-\lfloor\log_{\sqrt2}32\rfloor\ =\ 0$$

hectictar  Jun 25, 2019
#11
0

Sorry, you are right. I neglected to say also when the result of "floor" and the "ceiling" is the same whole integer as the case is in the example you gave.

Jun 25, 2019
#12
+8403
+2

Just one final thing in case it is unclear...

It is true that

$$\sum\limits_{k=1}^{1000} k(\lceil\log_{1.41421}k\rceil-\lfloor\log_{1.41421}k\rfloor)\ =\ 500499$$

It is also true that

$$\sum\limits_{k=1}^{1000}k(\lceil\log_{\sqrt2}k\rceil-\lfloor\log_{\sqrt2}k\rfloor)\ =\ 499477$$

The question asks for the second sum, not the first one.

Jun 25, 2019
edited by hectictar  Jun 25, 2019
#13
+1

Thank you very much! I really appreciate all your help :)

Guest Jun 26, 2019