Looking for a proof for this counting problem












3














Consider all the coefficients in the expansion of
$$(x^a+x^b)^n$$
where $a,b,n$ are non negative integers.



Claim: The $color{Red}{textit{exponent}}$ with the $color{blue}{textit {maximum coefficient value}}$ in the expansion is $color{red}{lfloorfrac{n(a+b)}2rfloor}$

Example :
$(x^1+x^3)^4 = sumlimits_{k=0}^4binom{4}{k}x^{n-k}x^{3k} = x^{4} +4x^{6}+color{blue}{6}x^{color{red}{8}}+4x^{10}+x^{12}$



$color{red}{dfrac{4(1+3)}{2} = 8}$.



I feel this is a special case of central limit theorem in probability, but that theorem seems way more complicated to understand than my special case. So I'm trying to make sense of this result w/o using CLT. Any help ?






combinatorics







share|cite|improve this question




















  • 1




    Well you need to add something here, because this is false with $a=1$, $b=2$, and $n=1$.
    – BlarglFlarg
    Dec 2 '18 at 17:46










  • Oh sorry when n(a+b)/2 is not an integer, both the ceil and floor will have the same coefficients.. I'll add this in the question. Ty @BlarglFlarg
    – rsadhvika
    Dec 2 '18 at 17:49






  • 1




    It is just related to the binomial coefficients characteristics, nothing to do with a and b. Look at how they are calculated with Pascal' triangle
    – Damien
    Dec 2 '18 at 17:49






  • 1




    I know that the beauty of mathematics partially relies on the fact that it can connect seemingly unrelated topics but... how do you see a connection to CLT in this problem?
    – Taladris
    Dec 3 '18 at 0:31






  • 1




    @Taladris From what I understand it is (a bit) connected because $left(frac{x^a+x^b}{2}right)^n$ is the generating function of some sort of binomial distribution. (?)
    – Idéophage
    Dec 3 '18 at 4:47


















3














Consider all the coefficients in the expansion of
$$(x^a+x^b)^n$$
where $a,b,n$ are non negative integers.



Claim: The $color{Red}{textit{exponent}}$ with the $color{blue}{textit {maximum coefficient value}}$ in the expansion is $color{red}{lfloorfrac{n(a+b)}2rfloor}$

Example :
$(x^1+x^3)^4 = sumlimits_{k=0}^4binom{4}{k}x^{n-k}x^{3k} = x^{4} +4x^{6}+color{blue}{6}x^{color{red}{8}}+4x^{10}+x^{12}$



$color{red}{dfrac{4(1+3)}{2} = 8}$.



I feel this is a special case of central limit theorem in probability, but that theorem seems way more complicated to understand than my special case. So I'm trying to make sense of this result w/o using CLT. Any help ?






combinatorics







share|cite|improve this question




















  • 1




    Well you need to add something here, because this is false with $a=1$, $b=2$, and $n=1$.
    – BlarglFlarg
    Dec 2 '18 at 17:46










  • Oh sorry when n(a+b)/2 is not an integer, both the ceil and floor will have the same coefficients.. I'll add this in the question. Ty @BlarglFlarg
    – rsadhvika
    Dec 2 '18 at 17:49






  • 1




    It is just related to the binomial coefficients characteristics, nothing to do with a and b. Look at how they are calculated with Pascal' triangle
    – Damien
    Dec 2 '18 at 17:49






  • 1




    I know that the beauty of mathematics partially relies on the fact that it can connect seemingly unrelated topics but... how do you see a connection to CLT in this problem?
    – Taladris
    Dec 3 '18 at 0:31






  • 1




    @Taladris From what I understand it is (a bit) connected because $left(frac{x^a+x^b}{2}right)^n$ is the generating function of some sort of binomial distribution. (?)
    – Idéophage
    Dec 3 '18 at 4:47
















3












3








3







Consider all the coefficients in the expansion of
$$(x^a+x^b)^n$$
where $a,b,n$ are non negative integers.



Claim: The $color{Red}{textit{exponent}}$ with the $color{blue}{textit {maximum coefficient value}}$ in the expansion is $color{red}{lfloorfrac{n(a+b)}2rfloor}$

Example :
$(x^1+x^3)^4 = sumlimits_{k=0}^4binom{4}{k}x^{n-k}x^{3k} = x^{4} +4x^{6}+color{blue}{6}x^{color{red}{8}}+4x^{10}+x^{12}$



$color{red}{dfrac{4(1+3)}{2} = 8}$.



I feel this is a special case of central limit theorem in probability, but that theorem seems way more complicated to understand than my special case. So I'm trying to make sense of this result w/o using CLT. Any help ?






combinatorics







share|cite|improve this question















Consider all the coefficients in the expansion of
$$(x^a+x^b)^n$$
where $a,b,n$ are non negative integers.



Claim: The $color{Red}{textit{exponent}}$ with the $color{blue}{textit {maximum coefficient value}}$ in the expansion is $color{red}{lfloorfrac{n(a+b)}2rfloor}$

Example :
$(x^1+x^3)^4 = sumlimits_{k=0}^4binom{4}{k}x^{n-k}x^{3k} = x^{4} +4x^{6}+color{blue}{6}x^{color{red}{8}}+4x^{10}+x^{12}$



$color{red}{dfrac{4(1+3)}{2} = 8}$.



I feel this is a special case of central limit theorem in probability, but that theorem seems way more complicated to understand than my special case. So I'm trying to make sense of this result w/o using CLT. Any help ?






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 21:46









LutzL

56k42054




56k42054










asked Dec 2 '18 at 17:41









rsadhvika

1,6631228




1,6631228








  • 1




    Well you need to add something here, because this is false with $a=1$, $b=2$, and $n=1$.
    – BlarglFlarg
    Dec 2 '18 at 17:46










  • Oh sorry when n(a+b)/2 is not an integer, both the ceil and floor will have the same coefficients.. I'll add this in the question. Ty @BlarglFlarg
    – rsadhvika
    Dec 2 '18 at 17:49






  • 1




    It is just related to the binomial coefficients characteristics, nothing to do with a and b. Look at how they are calculated with Pascal' triangle
    – Damien
    Dec 2 '18 at 17:49






  • 1




    I know that the beauty of mathematics partially relies on the fact that it can connect seemingly unrelated topics but... how do you see a connection to CLT in this problem?
    – Taladris
    Dec 3 '18 at 0:31






  • 1




    @Taladris From what I understand it is (a bit) connected because $left(frac{x^a+x^b}{2}right)^n$ is the generating function of some sort of binomial distribution. (?)
    – Idéophage
    Dec 3 '18 at 4:47
















  • 1




    Well you need to add something here, because this is false with $a=1$, $b=2$, and $n=1$.
    – BlarglFlarg
    Dec 2 '18 at 17:46










  • Oh sorry when n(a+b)/2 is not an integer, both the ceil and floor will have the same coefficients.. I'll add this in the question. Ty @BlarglFlarg
    – rsadhvika
    Dec 2 '18 at 17:49






  • 1




    It is just related to the binomial coefficients characteristics, nothing to do with a and b. Look at how they are calculated with Pascal' triangle
    – Damien
    Dec 2 '18 at 17:49






  • 1




    I know that the beauty of mathematics partially relies on the fact that it can connect seemingly unrelated topics but... how do you see a connection to CLT in this problem?
    – Taladris
    Dec 3 '18 at 0:31






  • 1




    @Taladris From what I understand it is (a bit) connected because $left(frac{x^a+x^b}{2}right)^n$ is the generating function of some sort of binomial distribution. (?)
    – Idéophage
    Dec 3 '18 at 4:47










1




1




Well you need to add something here, because this is false with $a=1$, $b=2$, and $n=1$.
– BlarglFlarg
Dec 2 '18 at 17:46




Well you need to add something here, because this is false with $a=1$, $b=2$, and $n=1$.
– BlarglFlarg
Dec 2 '18 at 17:46












Oh sorry when n(a+b)/2 is not an integer, both the ceil and floor will have the same coefficients.. I'll add this in the question. Ty @BlarglFlarg
– rsadhvika
Dec 2 '18 at 17:49




Oh sorry when n(a+b)/2 is not an integer, both the ceil and floor will have the same coefficients.. I'll add this in the question. Ty @BlarglFlarg
– rsadhvika
Dec 2 '18 at 17:49




1




1




It is just related to the binomial coefficients characteristics, nothing to do with a and b. Look at how they are calculated with Pascal' triangle
– Damien
Dec 2 '18 at 17:49




It is just related to the binomial coefficients characteristics, nothing to do with a and b. Look at how they are calculated with Pascal' triangle
– Damien
Dec 2 '18 at 17:49




1




1




I know that the beauty of mathematics partially relies on the fact that it can connect seemingly unrelated topics but... how do you see a connection to CLT in this problem?
– Taladris
Dec 3 '18 at 0:31




I know that the beauty of mathematics partially relies on the fact that it can connect seemingly unrelated topics but... how do you see a connection to CLT in this problem?
– Taladris
Dec 3 '18 at 0:31




1




1




@Taladris From what I understand it is (a bit) connected because $left(frac{x^a+x^b}{2}right)^n$ is the generating function of some sort of binomial distribution. (?)
– Idéophage
Dec 3 '18 at 4:47






@Taladris From what I understand it is (a bit) connected because $left(frac{x^a+x^b}{2}right)^n$ is the generating function of some sort of binomial distribution. (?)
– Idéophage
Dec 3 '18 at 4:47












3 Answers
3






active

oldest

votes


















6














If you know the binomial expansion (which you seem to), then this follows almost immediately. The key point to remember is that the maximum value of the coefficient is obtained at $nchoose k$, where $k$ is the largest integer less than or equal to $frac{n}{2}$. To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is greater than $1$. Plug in your value of $k$ and you get your result.



PS: Note that your exponents must actually appear in the expansion for this to happen.






share|cite|improve this answer























  • Ahh I kindof see now how this can be related to the max coefficient in binomial expansion. Ty :) Pretty sure you meant this : $$$$ To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is $color{red}{greater ~than~ 1}$.
    – rsadhvika
    Dec 2 '18 at 18:01












  • Ah yes you’re right, I meant that the binomial coefficient itself is increasing, not the ratio. I’ll correct that.
    – Boshu
    Dec 2 '18 at 18:05



















5














$(x^a+x^b)^n=x^{na}(1+x^c)^n=sum_{k=0}^n{n choose k}x^{na+k(b-a)}$. Now, ${n choose k}$ is maximum for $k=[n/2].$ So, answer is $na+[n/2](b-a)$ which matches with your answer if $n$ is even.






share|cite|improve this answer





















  • $c = b-a$; I like your approach more. It is neat and to the point, but I started working on the other answer first so had to mark it best as it helped me first. Thank you so much John_Wick you're awesome :)
    – rsadhvika
    Dec 2 '18 at 21:46





















3














The exponents of the development run from $na$ to $nb$, in increments $b-a$. By symmetry, the requested exponent can only be



$$frac{n(a+b)}{2}$$ when the numerator is even, and



$$frac{n(a+b)pm(b-a)}{2}$$



otherwise (there will be two consecutive terms with equal coefficient).






share|cite|improve this answer























    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022949%2flooking-for-a-proof-for-this-counting-problem%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    If you know the binomial expansion (which you seem to), then this follows almost immediately. The key point to remember is that the maximum value of the coefficient is obtained at $nchoose k$, where $k$ is the largest integer less than or equal to $frac{n}{2}$. To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is greater than $1$. Plug in your value of $k$ and you get your result.



    PS: Note that your exponents must actually appear in the expansion for this to happen.






    share|cite|improve this answer























    • Ahh I kindof see now how this can be related to the max coefficient in binomial expansion. Ty :) Pretty sure you meant this : $$$$ To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is $color{red}{greater ~than~ 1}$.
      – rsadhvika
      Dec 2 '18 at 18:01












    • Ah yes you’re right, I meant that the binomial coefficient itself is increasing, not the ratio. I’ll correct that.
      – Boshu
      Dec 2 '18 at 18:05
















    6














    If you know the binomial expansion (which you seem to), then this follows almost immediately. The key point to remember is that the maximum value of the coefficient is obtained at $nchoose k$, where $k$ is the largest integer less than or equal to $frac{n}{2}$. To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is greater than $1$. Plug in your value of $k$ and you get your result.



    PS: Note that your exponents must actually appear in the expansion for this to happen.






    share|cite|improve this answer























    • Ahh I kindof see now how this can be related to the max coefficient in binomial expansion. Ty :) Pretty sure you meant this : $$$$ To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is $color{red}{greater ~than~ 1}$.
      – rsadhvika
      Dec 2 '18 at 18:01












    • Ah yes you’re right, I meant that the binomial coefficient itself is increasing, not the ratio. I’ll correct that.
      – Boshu
      Dec 2 '18 at 18:05














    6












    6








    6






    If you know the binomial expansion (which you seem to), then this follows almost immediately. The key point to remember is that the maximum value of the coefficient is obtained at $nchoose k$, where $k$ is the largest integer less than or equal to $frac{n}{2}$. To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is greater than $1$. Plug in your value of $k$ and you get your result.



    PS: Note that your exponents must actually appear in the expansion for this to happen.






    share|cite|improve this answer














    If you know the binomial expansion (which you seem to), then this follows almost immediately. The key point to remember is that the maximum value of the coefficient is obtained at $nchoose k$, where $k$ is the largest integer less than or equal to $frac{n}{2}$. To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is greater than $1$. Plug in your value of $k$ and you get your result.



    PS: Note that your exponents must actually appear in the expansion for this to happen.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 2 '18 at 18:06

























    answered Dec 2 '18 at 17:47









    Boshu

    705315




    705315












    • Ahh I kindof see now how this can be related to the max coefficient in binomial expansion. Ty :) Pretty sure you meant this : $$$$ To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is $color{red}{greater ~than~ 1}$.
      – rsadhvika
      Dec 2 '18 at 18:01












    • Ah yes you’re right, I meant that the binomial coefficient itself is increasing, not the ratio. I’ll correct that.
      – Boshu
      Dec 2 '18 at 18:05


















    • Ahh I kindof see now how this can be related to the max coefficient in binomial expansion. Ty :) Pretty sure you meant this : $$$$ To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is $color{red}{greater ~than~ 1}$.
      – rsadhvika
      Dec 2 '18 at 18:01












    • Ah yes you’re right, I meant that the binomial coefficient itself is increasing, not the ratio. I’ll correct that.
      – Boshu
      Dec 2 '18 at 18:05
















    Ahh I kindof see now how this can be related to the max coefficient in binomial expansion. Ty :) Pretty sure you meant this : $$$$ To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is $color{red}{greater ~than~ 1}$.
    – rsadhvika
    Dec 2 '18 at 18:01






    Ahh I kindof see now how this can be related to the max coefficient in binomial expansion. Ty :) Pretty sure you meant this : $$$$ To prove this, consider the ratios $frac{n choose k+1}{nchoose k}$ and see for what range this is $color{red}{greater ~than~ 1}$.
    – rsadhvika
    Dec 2 '18 at 18:01














    Ah yes you’re right, I meant that the binomial coefficient itself is increasing, not the ratio. I’ll correct that.
    – Boshu
    Dec 2 '18 at 18:05




    Ah yes you’re right, I meant that the binomial coefficient itself is increasing, not the ratio. I’ll correct that.
    – Boshu
    Dec 2 '18 at 18:05











    5














    $(x^a+x^b)^n=x^{na}(1+x^c)^n=sum_{k=0}^n{n choose k}x^{na+k(b-a)}$. Now, ${n choose k}$ is maximum for $k=[n/2].$ So, answer is $na+[n/2](b-a)$ which matches with your answer if $n$ is even.






    share|cite|improve this answer





















    • $c = b-a$; I like your approach more. It is neat and to the point, but I started working on the other answer first so had to mark it best as it helped me first. Thank you so much John_Wick you're awesome :)
      – rsadhvika
      Dec 2 '18 at 21:46


















    5














    $(x^a+x^b)^n=x^{na}(1+x^c)^n=sum_{k=0}^n{n choose k}x^{na+k(b-a)}$. Now, ${n choose k}$ is maximum for $k=[n/2].$ So, answer is $na+[n/2](b-a)$ which matches with your answer if $n$ is even.






    share|cite|improve this answer





















    • $c = b-a$; I like your approach more. It is neat and to the point, but I started working on the other answer first so had to mark it best as it helped me first. Thank you so much John_Wick you're awesome :)
      – rsadhvika
      Dec 2 '18 at 21:46
















    5












    5








    5






    $(x^a+x^b)^n=x^{na}(1+x^c)^n=sum_{k=0}^n{n choose k}x^{na+k(b-a)}$. Now, ${n choose k}$ is maximum for $k=[n/2].$ So, answer is $na+[n/2](b-a)$ which matches with your answer if $n$ is even.






    share|cite|improve this answer












    $(x^a+x^b)^n=x^{na}(1+x^c)^n=sum_{k=0}^n{n choose k}x^{na+k(b-a)}$. Now, ${n choose k}$ is maximum for $k=[n/2].$ So, answer is $na+[n/2](b-a)$ which matches with your answer if $n$ is even.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 17:51









    John_Wick

    1,366111




    1,366111












    • $c = b-a$; I like your approach more. It is neat and to the point, but I started working on the other answer first so had to mark it best as it helped me first. Thank you so much John_Wick you're awesome :)
      – rsadhvika
      Dec 2 '18 at 21:46




















    • $c = b-a$; I like your approach more. It is neat and to the point, but I started working on the other answer first so had to mark it best as it helped me first. Thank you so much John_Wick you're awesome :)
      – rsadhvika
      Dec 2 '18 at 21:46


















    $c = b-a$; I like your approach more. It is neat and to the point, but I started working on the other answer first so had to mark it best as it helped me first. Thank you so much John_Wick you're awesome :)
    – rsadhvika
    Dec 2 '18 at 21:46






    $c = b-a$; I like your approach more. It is neat and to the point, but I started working on the other answer first so had to mark it best as it helped me first. Thank you so much John_Wick you're awesome :)
    – rsadhvika
    Dec 2 '18 at 21:46













    3














    The exponents of the development run from $na$ to $nb$, in increments $b-a$. By symmetry, the requested exponent can only be



    $$frac{n(a+b)}{2}$$ when the numerator is even, and



    $$frac{n(a+b)pm(b-a)}{2}$$



    otherwise (there will be two consecutive terms with equal coefficient).






    share|cite|improve this answer




























      3














      The exponents of the development run from $na$ to $nb$, in increments $b-a$. By symmetry, the requested exponent can only be



      $$frac{n(a+b)}{2}$$ when the numerator is even, and



      $$frac{n(a+b)pm(b-a)}{2}$$



      otherwise (there will be two consecutive terms with equal coefficient).






      share|cite|improve this answer


























        3












        3








        3






        The exponents of the development run from $na$ to $nb$, in increments $b-a$. By symmetry, the requested exponent can only be



        $$frac{n(a+b)}{2}$$ when the numerator is even, and



        $$frac{n(a+b)pm(b-a)}{2}$$



        otherwise (there will be two consecutive terms with equal coefficient).






        share|cite|improve this answer














        The exponents of the development run from $na$ to $nb$, in increments $b-a$. By symmetry, the requested exponent can only be



        $$frac{n(a+b)}{2}$$ when the numerator is even, and



        $$frac{n(a+b)pm(b-a)}{2}$$



        otherwise (there will be two consecutive terms with equal coefficient).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 '18 at 17:58

























        answered Dec 2 '18 at 17:51









        Yves Daoust

        124k671221




        124k671221






























            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














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022949%2flooking-for-a-proof-for-this-counting-problem%23new-answer', 'question_page');
            }
            );

            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







            -

            Popular posts from this blog

            AnyDesk - Fatal Program Failure

            Image
            up vote 0 down vote favorite Good day all. I have a window that pops up saying Fatal Program Failure (AnyDesk). This is what I have done so far: Uninstalled AnyDesk and reinstalled it. ( The window still appeared ) I disabled AnyDesk start up in Msconfig and the window seems to have disapeared. Lastly I made sure that AnyDesk worked by remoting into it, and it worked. Today the same window appeared. How do I get rid of this window? remote-desktop share | improve this question edited 10 hours ago Tomasz Jakub Rup 552 3 13 ...
            Read more

            How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

            Image
            up vote 0 down vote favorite I need help with an issue we're facing in a project where we are moving from a Windows-based POS environment to one running on CentOS7. The problem consists of a number of factors, please find the details below: The supplier of our POS application will only deliver an application that is hard-coded to run in a 1024:768 4:3 format. The hardware we are using is a POS unit from HP that has a built-in 16:9 touch-screen. The employees using this solution is not supposed to see nor have access to anything but the POS application. Our sales organization demands that touch is an available feature when going live with the new POS environment. Going further into this, allow me to explain what we have encountered on the way trying to achieve the goal of having a touch-screen: We've simply change...
            Read more

            QoS: MAC-Priority for clients behind a repeater

            Image
            1 1 I've setup a wireless network using a Linksys-Router (DD-WRT) as an AP and a TP-Link repeater for range extension. To manage bandwith for specific users, I've setup QoS-Rules on the AP (DD-WRT) using MAC-Priority: However the AP shows only the Repeater's MAC and the MACs of users inside the AP-Range. The MAC addresses of users behind the Repeater are not listed : The purple MAC is not listed in the AP-Clientlist. Its related to a device connected via the TP-Link repeater. Q: How can I setup MAC-Priority for repeater-clients? Do I simply add the purple MAC in AP-MAC priority list although it's not listed as a client? I want to setup specific priority instead of giving one priority for all clients behind the repeater by adding the repeater's MAC in the MAC priority list in the AP as ...
            Read more