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!
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
+129899 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
+9481 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
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
+9481 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
\(\log_{b}a\)
See example: https://www.wolframalpha.com/input/?i=log(sqrt(2),32)
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
We have received your feedback regarding Wolfram|Alpha.
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.
+9481 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.
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.
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.
+9481 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.
