How do I find the solutions of $|x-2|^10x^2-1=|x-2|^3x$?Sum of all real numbers $x$ such that $(textA quadratic)^textAnother quadratic=1$.How to find the roots of $x^4 +1$Find all real roots of $x^5+10x^3+20x - 4=0$Find real solutions for a mod equation with powerfinding integer solutions for a and bDetermine the form of set containing the real solutionsNature of roots of the equation $x^2-4qx+2q^2-r=0$How to tell if a function has double real roots or complex roots?How to manually calculate/approximate a specific value of the Lambert W functionSolving $x^4 -10x^3 + 26x^2 -10x +1 = 0$.Finding the minimum value.
How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?
Test if tikzmark exists on same page
What's the output of a record cartridge playing an out-of-speed record
Pattern match does not work in bash script
Why don't electron-positron collisions release infinite energy?
How old can references or sources in a thesis be?
TGV timetables / schedules?
can i play a electric guitar through a bass amp?
Why do I get two different answers for this counting problem?
Approximately how much travel time was saved by the opening of the Suez Canal in 1869?
How to format long polynomial?
Is it tax fraud for an individual to declare non-taxable revenue as taxable income? (US tax laws)
US citizen flying to France today and my passport expires in less than 2 months
Replacing matching entries in one column of a file by another column from a different file
How does strength of boric acid solution increase in presence of salicylic acid?
Do I have a twin with permutated remainders?
Is it legal for company to use my work email to pretend I still work there?
Are the number of citations and number of published articles the most important criteria for a tenure promotion?
Why are electrically insulating heatsinks so rare? Is it just cost?
Email Account under attack (really) - anything I can do?
Test whether all array elements are factors of a number
Modeling an IP Address
What defenses are there against being summoned by the Gate spell?
"to be prejudice towards/against someone" vs "to be prejudiced against/towards someone"
How do I find the solutions of $|x-2|^10x^2-1=|x-2|^3x$?
Sum of all real numbers $x$ such that $(textA quadratic)^textAnother quadratic=1$.How to find the roots of $x^4 +1$Find all real roots of $x^5+10x^3+20x - 4=0$Find real solutions for a mod equation with powerfinding integer solutions for a and bDetermine the form of set containing the real solutionsNature of roots of the equation $x^2-4qx+2q^2-r=0$How to tell if a function has double real roots or complex roots?How to manually calculate/approximate a specific value of the Lambert W functionSolving $x^4 -10x^3 + 26x^2 -10x +1 = 0$.Finding the minimum value.
$begingroup$
How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x ?$$
I found that this equation has 5 solutions, 4 positive and 1 negative by looking the graph:
Question: How do I compute the values of these roots manually?
algebra-precalculus logarithms
edited 6 hours ago
Jack
27.7k1782204
asked Mar 27 at 18:36
Namami ShankerNamami Shanker
171
$endgroup$
add a comment |
$begingroup$
How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x ?$$
I found that this equation has 5 solutions, 4 positive and 1 negative by looking the graph:
Question: How do I compute the values of these roots manually?
algebra-precalculus logarithms
edited 6 hours ago
Jack
27.7k1782204
asked Mar 27 at 18:36
Namami ShankerNamami Shanker
171
$endgroup$
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 27 at 19:27
add a comment |
$begingroup$
How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x ?$$
I found that this equation has 5 solutions, 4 positive and 1 negative by looking the graph:
Question: How do I compute the values of these roots manually?
algebra-precalculus logarithms
edited 6 hours ago
Jack
27.7k1782204
asked Mar 27 at 18:36
Namami ShankerNamami Shanker
171
$endgroup$
How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x ?$$
I found that this equation has 5 solutions, 4 positive and 1 negative by looking the graph:
Question: How do I compute the values of these roots manually?
algebra-precalculus logarithms
algebra-precalculus logarithms
edited 6 hours ago
Jack
27.7k1782204
asked Mar 27 at 18:36
Namami ShankerNamami Shanker
171
edited 6 hours ago
Jack
27.7k1782204
asked Mar 27 at 18:36
Namami ShankerNamami Shanker
171
edited 6 hours ago
Jack
27.7k1782204
edited 6 hours ago
Jack
27.7k1782204
edited 6 hours ago
Jack
27.7k1782204
27.7k1782204
asked Mar 27 at 18:36
Namami ShankerNamami Shanker
171
asked Mar 27 at 18:36
Namami ShankerNamami Shanker
171
asked Mar 27 at 18:36
Namami ShankerNamami Shanker
171
171
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 27 at 19:27
add a comment |
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 27 at 19:27
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 27 at 19:27
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 27 at 19:27
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited Mar 27 at 20:39
Moo
5,64031020
answered Mar 27 at 18:43
Maria MazurMaria Mazur
49.9k1361124
$endgroup$
add a comment |
$begingroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered Mar 27 at 18:45
Peter ForemanPeter Foreman
5,8791317
$endgroup$
add a comment |
$begingroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered Mar 27 at 19:07
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited Mar 28 at 2:08
Solomon Ucko
14719
answered Mar 27 at 18:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.6k42867
$endgroup$
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
Mar 27 at 18:45
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
Mar 27 at 18:47
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
Mar 27 at 18:54
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%2f3164927%2fhow-do-i-find-the-solutions-of-x-210x2-1-x-23x%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited Mar 27 at 20:39
Moo
5,64031020
answered Mar 27 at 18:43
Maria MazurMaria Mazur
49.9k1361124
$endgroup$
add a comment |
$begingroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited Mar 27 at 20:39
Moo
5,64031020
answered Mar 27 at 18:43
Maria MazurMaria Mazur
49.9k1361124
$endgroup$
add a comment |
$begingroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited Mar 27 at 20:39
Moo
5,64031020
answered Mar 27 at 18:43
Maria MazurMaria Mazur
49.9k1361124
$endgroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited Mar 27 at 20:39
Moo
5,64031020
answered Mar 27 at 18:43
Maria MazurMaria Mazur
49.9k1361124
edited Mar 27 at 20:39
Moo
5,64031020
edited Mar 27 at 20:39
Moo
5,64031020
edited Mar 27 at 20:39
Moo
5,64031020
5,64031020
answered Mar 27 at 18:43
Maria MazurMaria Mazur
49.9k1361124
answered Mar 27 at 18:43
Maria MazurMaria Mazur
49.9k1361124
answered Mar 27 at 18:43
Maria MazurMaria Mazur
49.9k1361124
49.9k1361124
add a comment |
add a comment |
$begingroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered Mar 27 at 18:45
Peter ForemanPeter Foreman
5,8791317
$endgroup$
add a comment |
$begingroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered Mar 27 at 18:45
Peter ForemanPeter Foreman
5,8791317
$endgroup$
add a comment |
$begingroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered Mar 27 at 18:45
Peter ForemanPeter Foreman
5,8791317
$endgroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered Mar 27 at 18:45
Peter ForemanPeter Foreman
5,8791317
answered Mar 27 at 18:45
Peter ForemanPeter Foreman
5,8791317
answered Mar 27 at 18:45
Peter ForemanPeter Foreman
5,8791317
answered Mar 27 at 18:45
Peter ForemanPeter Foreman
5,8791317
5,8791317
add a comment |
add a comment |
$begingroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered Mar 27 at 19:07
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered Mar 27 at 19:07
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered Mar 27 at 19:07
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered Mar 27 at 19:07
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 27 at 19:07
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 27 at 19:07
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 27 at 19:07
Mostafa AyazMostafa Ayaz
18.2k31040
18.2k31040
add a comment |
add a comment |
$begingroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited Mar 28 at 2:08
Solomon Ucko
14719
answered Mar 27 at 18:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.6k42867
$endgroup$
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
Mar 27 at 18:45
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
Mar 27 at 18:47
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
Mar 27 at 18:54
add a comment |
$begingroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited Mar 28 at 2:08
Solomon Ucko
14719
answered Mar 27 at 18:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.6k42867
$endgroup$
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
Mar 27 at 18:45
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
Mar 27 at 18:47
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
Mar 27 at 18:54
add a comment |
$begingroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited Mar 28 at 2:08
Solomon Ucko
14719
answered Mar 27 at 18:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.6k42867
$endgroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited Mar 28 at 2:08
Solomon Ucko
14719
answered Mar 27 at 18:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.6k42867
edited Mar 28 at 2:08
Solomon Ucko
14719
edited Mar 28 at 2:08
Solomon Ucko
14719
edited Mar 28 at 2:08
Solomon Ucko
14719
14719
answered Mar 27 at 18:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.6k42867
answered Mar 27 at 18:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.6k42867
answered Mar 27 at 18:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.6k42867
78.6k42867
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
Mar 27 at 18:45
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
Mar 27 at 18:47
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
Mar 27 at 18:54
add a comment |
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
Mar 27 at 18:45
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
Mar 27 at 18:47
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
Mar 27 at 18:54
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
Mar 27 at 18:45
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
Mar 27 at 18:45
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
Mar 27 at 18:47
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
Mar 27 at 18:47
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
Mar 27 at 18:54
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
Mar 27 at 18:54
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%2f3164927%2fhow-do-i-find-the-solutions-of-x-210x2-1-x-23x%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
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 27 at 19:27