What does the function $f(x,y)$ reduce to?
up vote
0
down vote
favorite
What does the function $f(x,y)$ reduce to?
$$f(x,y) = lim_{n to infty} sum_{i = 0}^{n} (((x bmod 2^{i-1})-(x bmod 2^i)) cdot ((y bmod 2^{i-1})-(y bmod 2^i)) cdot frac {1}{2^i})$$
limits summation
edited Nov 23 at 3:22
metamorphy
2,7471316
asked Nov 23 at 3:17
The Great Duck
10632047
add a comment |
up vote
0
down vote
favorite
What does the function $f(x,y)$ reduce to?
$$f(x,y) = lim_{n to infty} sum_{i = 0}^{n} (((x bmod 2^{i-1})-(x bmod 2^i)) cdot ((y bmod 2^{i-1})-(y bmod 2^i)) cdot frac {1}{2^i})$$
limits summation
edited Nov 23 at 3:22
metamorphy
2,7471316
asked Nov 23 at 3:17
The Great Duck
10632047
When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
– Semiclassical
Nov 24 at 3:52
@Semiclassical no it is correct as written.
– The Great Duck
Nov 24 at 5:36
@TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
– metamorphy
Nov 25 at 13:02
@metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
– The Great Duck
Nov 25 at 20:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What does the function $f(x,y)$ reduce to?
$$f(x,y) = lim_{n to infty} sum_{i = 0}^{n} (((x bmod 2^{i-1})-(x bmod 2^i)) cdot ((y bmod 2^{i-1})-(y bmod 2^i)) cdot frac {1}{2^i})$$
limits summation
edited Nov 23 at 3:22
metamorphy
2,7471316
asked Nov 23 at 3:17
The Great Duck
10632047
What does the function $f(x,y)$ reduce to?
$$f(x,y) = lim_{n to infty} sum_{i = 0}^{n} (((x bmod 2^{i-1})-(x bmod 2^i)) cdot ((y bmod 2^{i-1})-(y bmod 2^i)) cdot frac {1}{2^i})$$
limits summation
limits summation
edited Nov 23 at 3:22
metamorphy
2,7471316
asked Nov 23 at 3:17
The Great Duck
10632047
edited Nov 23 at 3:22
metamorphy
2,7471316
asked Nov 23 at 3:17
The Great Duck
10632047
edited Nov 23 at 3:22
metamorphy
2,7471316
edited Nov 23 at 3:22
metamorphy
2,7471316
edited Nov 23 at 3:22
metamorphy
2,7471316
2,7471316
asked Nov 23 at 3:17
The Great Duck
10632047
asked Nov 23 at 3:17
The Great Duck
10632047
asked Nov 23 at 3:17
The Great Duck
10632047
10632047
When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
– Semiclassical
Nov 24 at 3:52
@Semiclassical no it is correct as written.
– The Great Duck
Nov 24 at 5:36
@TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
– metamorphy
Nov 25 at 13:02
@metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
– The Great Duck
Nov 25 at 20:38
add a comment |
When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
– Semiclassical
Nov 24 at 3:52
@Semiclassical no it is correct as written.
– The Great Duck
Nov 24 at 5:36
@TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
– metamorphy
Nov 25 at 13:02
@metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
– The Great Duck
Nov 25 at 20:38
When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
– Semiclassical
Nov 24 at 3:52
When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
– Semiclassical
Nov 24 at 3:52
@Semiclassical no it is correct as written.
– The Great Duck
Nov 24 at 5:36
@Semiclassical no it is correct as written.
– The Great Duck
Nov 24 at 5:36
@TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
– metamorphy
Nov 25 at 13:02
@TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
– metamorphy
Nov 25 at 13:02
@metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
– The Great Duck
Nov 25 at 20:38
@metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
– The Great Duck
Nov 25 at 20:38
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009933%2fwhat-does-the-function-fx-y-reduce-to%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
– Semiclassical
Nov 24 at 3:52
@Semiclassical no it is correct as written.
– The Great Duck
Nov 24 at 5:36
@TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
– metamorphy
Nov 25 at 13:02
@metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
– The Great Duck
Nov 25 at 20:38