Mixing
Find a function given a limit
$begingroup$
I'm coding some algorithms, but I need help to find a function.
Let' suppose these 3 functions:
$$c(x)=1-g(x)$$
$$r_1(x)=frac{sqrt{2}}{2^{left(x-1right)}}$$
$$r_2(x) = frac{sqrt{2}}{2^{left(x-1right)}}cdot g(x)$$
$$d(x)=frac{r_1(x)+r_2(x)}{2}cdotfrac{1}{c(x)}$$
My need is to find the function g that allows
$$lim_{xtoinfty}d(x)=1$$
(or at least, as close as possible to 1).
My constraints:
Being quickly close to the limit
$for xin[1; 128], g(x) in [0; 1[$
I put all these functions in a graph for clarity:
https://www.desmos.com/calculator/dgkmwtgis8
My level in mathematics in not very good, so for now I just try to adapt the following functions:
$$g_1=sqrt{frac{2^x-1}{2^x}}$$
$$g_2 =frac{2^x-1}{2^x}$$
g1 works for my code, but d(x) is too far from 1.
Thank you for any help you could provide !
limits functions
asked Feb 1 at 5:00
EalrannEalrann
32
$endgroup$
add a comment |
$begingroup$
I'm coding some algorithms, but I need help to find a function.
Let' suppose these 3 functions:
$$c(x)=1-g(x)$$
$$r_1(x)=frac{sqrt{2}}{2^{left(x-1right)}}$$
$$r_2(x) = frac{sqrt{2}}{2^{left(x-1right)}}cdot g(x)$$
$$d(x)=frac{r_1(x)+r_2(x)}{2}cdotfrac{1}{c(x)}$$
My need is to find the function g that allows
$$lim_{xtoinfty}d(x)=1$$
(or at least, as close as possible to 1).
My constraints:
Being quickly close to the limit
$for xin[1; 128], g(x) in [0; 1[$
I put all these functions in a graph for clarity:
https://www.desmos.com/calculator/dgkmwtgis8
My level in mathematics in not very good, so for now I just try to adapt the following functions:
$$g_1=sqrt{frac{2^x-1}{2^x}}$$
$$g_2 =frac{2^x-1}{2^x}$$
g1 works for my code, but d(x) is too far from 1.
Thank you for any help you could provide !
limits functions
asked Feb 1 at 5:00
EalrannEalrann
32
$endgroup$
$begingroup$
Do you mean that $d(x)=frac{r_1(x)+r_2(x)}{2c(x)}$, and do you want $lim_{xrightarrow infty} d(x) = 1$?
$endgroup$
– Doyun Nam
Feb 1 at 6:27
$begingroup$
Yes, exactly. I wasn't sure how to write that. I will edit my post accordingly.
$endgroup$
– Ealrann
Feb 1 at 6:30
$begingroup$
I find some function $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$. when applying this $g$ at desmos calculator you made above, the $d(x)$ is identical to $1$. Please check it. And because the process to get this $g$ is a bit tedious calculation, I skipped it to write down. However If you need it, please ask me for it.
$endgroup$
– Doyun Nam
Feb 1 at 6:41
1
$begingroup$
This solution seems perfect to me ! Thank you very, very much!
$endgroup$
– Ealrann
Feb 1 at 6:53
$begingroup$
You should copy this equation into an answer. At least I should give you a vote :D.
$endgroup$
– Ealrann
Feb 1 at 7:31
add a comment |
$begingroup$
I'm coding some algorithms, but I need help to find a function.
Let' suppose these 3 functions:
$$c(x)=1-g(x)$$
$$r_1(x)=frac{sqrt{2}}{2^{left(x-1right)}}$$
$$r_2(x) = frac{sqrt{2}}{2^{left(x-1right)}}cdot g(x)$$
$$d(x)=frac{r_1(x)+r_2(x)}{2}cdotfrac{1}{c(x)}$$
My need is to find the function g that allows
$$lim_{xtoinfty}d(x)=1$$
(or at least, as close as possible to 1).
My constraints:
Being quickly close to the limit
$for xin[1; 128], g(x) in [0; 1[$
I put all these functions in a graph for clarity:
https://www.desmos.com/calculator/dgkmwtgis8
My level in mathematics in not very good, so for now I just try to adapt the following functions:
$$g_1=sqrt{frac{2^x-1}{2^x}}$$
$$g_2 =frac{2^x-1}{2^x}$$
g1 works for my code, but d(x) is too far from 1.
Thank you for any help you could provide !
limits functions
asked Feb 1 at 5:00
EalrannEalrann
32
$endgroup$
I'm coding some algorithms, but I need help to find a function.
Let' suppose these 3 functions:
$$c(x)=1-g(x)$$
$$r_1(x)=frac{sqrt{2}}{2^{left(x-1right)}}$$
$$r_2(x) = frac{sqrt{2}}{2^{left(x-1right)}}cdot g(x)$$
$$d(x)=frac{r_1(x)+r_2(x)}{2}cdotfrac{1}{c(x)}$$
My need is to find the function g that allows
$$lim_{xtoinfty}d(x)=1$$
(or at least, as close as possible to 1).
My constraints:
Being quickly close to the limit
$for xin[1; 128], g(x) in [0; 1[$
I put all these functions in a graph for clarity:
https://www.desmos.com/calculator/dgkmwtgis8
My level in mathematics in not very good, so for now I just try to adapt the following functions:
$$g_1=sqrt{frac{2^x-1}{2^x}}$$
$$g_2 =frac{2^x-1}{2^x}$$
g1 works for my code, but d(x) is too far from 1.
Thank you for any help you could provide !
limits functions
limits functions
asked Feb 1 at 5:00
EalrannEalrann
32
asked Feb 1 at 5:00
EalrannEalrann
32
edited Feb 1 at 6:35
Ealrann
asked Feb 1 at 5:00
EalrannEalrann
32
asked Feb 1 at 5:00
EalrannEalrann
32
asked Feb 1 at 5:00
EalrannEalrann
32
32
$begingroup$
Do you mean that $d(x)=frac{r_1(x)+r_2(x)}{2c(x)}$, and do you want $lim_{xrightarrow infty} d(x) = 1$?
$endgroup$
– Doyun Nam
Feb 1 at 6:27
$begingroup$
Yes, exactly. I wasn't sure how to write that. I will edit my post accordingly.
$endgroup$
– Ealrann
Feb 1 at 6:30
$begingroup$
I find some function $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$. when applying this $g$ at desmos calculator you made above, the $d(x)$ is identical to $1$. Please check it. And because the process to get this $g$ is a bit tedious calculation, I skipped it to write down. However If you need it, please ask me for it.
$endgroup$
– Doyun Nam
Feb 1 at 6:41
1
$begingroup$
This solution seems perfect to me ! Thank you very, very much!
$endgroup$
– Ealrann
Feb 1 at 6:53
$begingroup$
You should copy this equation into an answer. At least I should give you a vote :D.
$endgroup$
– Ealrann
Feb 1 at 7:31
add a comment |
$begingroup$
Do you mean that $d(x)=frac{r_1(x)+r_2(x)}{2c(x)}$, and do you want $lim_{xrightarrow infty} d(x) = 1$?
$endgroup$
– Doyun Nam
Feb 1 at 6:27
$begingroup$
Yes, exactly. I wasn't sure how to write that. I will edit my post accordingly.
$endgroup$
– Ealrann
Feb 1 at 6:30
$begingroup$
I find some function $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$. when applying this $g$ at desmos calculator you made above, the $d(x)$ is identical to $1$. Please check it. And because the process to get this $g$ is a bit tedious calculation, I skipped it to write down. However If you need it, please ask me for it.
$endgroup$
– Doyun Nam
Feb 1 at 6:41
1
$begingroup$
This solution seems perfect to me ! Thank you very, very much!
$endgroup$
– Ealrann
Feb 1 at 6:53
$begingroup$
You should copy this equation into an answer. At least I should give you a vote :D.
$endgroup$
– Ealrann
Feb 1 at 7:31
$begingroup$
Do you mean that $d(x)=frac{r_1(x)+r_2(x)}{2c(x)}$, and do you want $lim_{xrightarrow infty} d(x) = 1$?
$endgroup$
– Doyun Nam
Feb 1 at 6:27
$begingroup$
Do you mean that $d(x)=frac{r_1(x)+r_2(x)}{2c(x)}$, and do you want $lim_{xrightarrow infty} d(x) = 1$?
$endgroup$
– Doyun Nam
Feb 1 at 6:27
$begingroup$
Yes, exactly. I wasn't sure how to write that. I will edit my post accordingly.
$endgroup$
– Ealrann
Feb 1 at 6:30
$begingroup$
Yes, exactly. I wasn't sure how to write that. I will edit my post accordingly.
$endgroup$
– Ealrann
Feb 1 at 6:30
$begingroup$
I find some function $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$. when applying this $g$ at desmos calculator you made above, the $d(x)$ is identical to $1$. Please check it. And because the process to get this $g$ is a bit tedious calculation, I skipped it to write down. However If you need it, please ask me for it.
$endgroup$
– Doyun Nam
Feb 1 at 6:41
$begingroup$
I find some function $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$. when applying this $g$ at desmos calculator you made above, the $d(x)$ is identical to $1$. Please check it. And because the process to get this $g$ is a bit tedious calculation, I skipped it to write down. However If you need it, please ask me for it.
$endgroup$
– Doyun Nam
Feb 1 at 6:41
1
1
$begingroup$
This solution seems perfect to me ! Thank you very, very much!
$endgroup$
– Ealrann
Feb 1 at 6:53
$begingroup$
This solution seems perfect to me ! Thank you very, very much!
$endgroup$
– Ealrann
Feb 1 at 6:53
$begingroup$
You should copy this equation into an answer. At least I should give you a vote :D.
$endgroup$
– Ealrann
Feb 1 at 7:31
$begingroup$
You should copy this equation into an answer. At least I should give you a vote :D.
$endgroup$
– Ealrann
Feb 1 at 7:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If we assume $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$, then $d(x)$ is identical to $1$.
answered Feb 1 at 7:39
Doyun NamDoyun Nam
68119
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3095846%2ffind-a-function-given-a-limit%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If we assume $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$, then $d(x)$ is identical to $1$.
answered Feb 1 at 7:39
Doyun NamDoyun Nam
68119
$endgroup$
add a comment |
$begingroup$
If we assume $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$, then $d(x)$ is identical to $1$.
answered Feb 1 at 7:39
Doyun NamDoyun Nam
68119
$endgroup$
add a comment |
$begingroup$
If we assume $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$, then $d(x)$ is identical to $1$.
answered Feb 1 at 7:39
Doyun NamDoyun Nam
68119
$endgroup$
If we assume $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$, then $d(x)$ is identical to $1$.
answered Feb 1 at 7:39
Doyun NamDoyun Nam
68119
answered Feb 1 at 7:39
Doyun NamDoyun Nam
68119
answered Feb 1 at 7:39
Doyun NamDoyun Nam
68119
answered Feb 1 at 7:39
Doyun NamDoyun Nam
68119
68119
add a comment |
add a comment |
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.
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%2f3095846%2ffind-a-function-given-a-limit%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
$begingroup$
Do you mean that $d(x)=frac{r_1(x)+r_2(x)}{2c(x)}$, and do you want $lim_{xrightarrow infty} d(x) = 1$?
$endgroup$
– Doyun Nam
Feb 1 at 6:27
$begingroup$
Yes, exactly. I wasn't sure how to write that. I will edit my post accordingly.
$endgroup$
– Ealrann
Feb 1 at 6:30
$begingroup$
I find some function $g(x)=frac{2^x -sqrt{2}}{2^x + sqrt{2}}$. when applying this $g$ at desmos calculator you made above, the $d(x)$ is identical to $1$. Please check it. And because the process to get this $g$ is a bit tedious calculation, I skipped it to write down. However If you need it, please ask me for it.
$endgroup$
– Doyun Nam
Feb 1 at 6:41
1
$begingroup$
This solution seems perfect to me ! Thank you very, very much!
$endgroup$
– Ealrann
Feb 1 at 6:53
$begingroup$
You should copy this equation into an answer. At least I should give you a vote :D.
$endgroup$
– Ealrann
Feb 1 at 7:31