What is the main difference between pointwise and uniform convergence as defined here?












3














I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.



Let $E$ be a non-empty subset of $Bbb{R}$. A sequence of functions ${f_n}_{nin Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



On the other hand ${f_n}_{nin Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



QUESTION:



Why is $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?






real-analysis analysis definition uniform-convergence pointwise-convergence







share|cite|improve this question




















  • 2




    Please refer to the original definition, not the altered version. In your post, these are identical.
    – xbh
    1 hour ago
















3














I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.



Let $E$ be a non-empty subset of $Bbb{R}$. A sequence of functions ${f_n}_{nin Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



On the other hand ${f_n}_{nin Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



QUESTION:



Why is $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?






real-analysis analysis definition uniform-convergence pointwise-convergence







share|cite|improve this question




















  • 2




    Please refer to the original definition, not the altered version. In your post, these are identical.
    – xbh
    1 hour ago














3












3








3


1





I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.



Let $E$ be a non-empty subset of $Bbb{R}$. A sequence of functions ${f_n}_{nin Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



On the other hand ${f_n}_{nin Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



QUESTION:



Why is $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?






real-analysis analysis definition uniform-convergence pointwise-convergence







share|cite|improve this question















I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.



Let $E$ be a non-empty subset of $Bbb{R}$. A sequence of functions ${f_n}_{nin Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



On the other hand ${f_n}_{nin Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if begin{align}f_n(x)to f(x),;forall,xin E.end{align}



QUESTION:



Why is $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?






real-analysis analysis definition uniform-convergence pointwise-convergence




real-analysis analysis definition uniform-convergence pointwise-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago

























asked 2 hours ago









Mike

1,498321




1,498321








  • 2




    Please refer to the original definition, not the altered version. In your post, these are identical.
    – xbh
    1 hour ago














  • 2




    Please refer to the original definition, not the altered version. In your post, these are identical.
    – xbh
    1 hour ago








2




2




Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
1 hour ago




Please refer to the original definition, not the altered version. In your post, these are identical.
– xbh
1 hour ago










3 Answers
3






active

oldest

votes


















4














$f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



$f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.






share|cite|improve this answer





















  • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
    – Mike
    2 hours ago










  • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
    – Tsemo Aristide
    2 hours ago










  • That's so true.
    – Mike
    2 hours ago










  • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
    – Mike
    2 hours ago












  • $f_n(n)=1, f_n(x)=0$ if $xneq n$
    – Tsemo Aristide
    2 hours ago





















4














Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
$$
f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
$$
This is strictly stronger than pointwise convergence.



Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.






share|cite|improve this answer

















  • 1




    Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
    – Matt A Pelto
    43 mins ago



















1














Sorry, but yes, you probably are missing something important, because the second statement in your post




On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
$$f_n(x)to f(x),;forall,xin E.$$




is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



Your best bet is to check the original source to find out what exactly it says there.






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%2f3063473%2fwhat-is-the-main-difference-between-pointwise-and-uniform-convergence-as-defined%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









    4














    $f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



    $f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



    In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.






    share|cite|improve this answer





















    • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
      – Mike
      2 hours ago










    • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
      – Tsemo Aristide
      2 hours ago










    • That's so true.
      – Mike
      2 hours ago










    • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
      – Mike
      2 hours ago












    • $f_n(n)=1, f_n(x)=0$ if $xneq n$
      – Tsemo Aristide
      2 hours ago


















    4














    $f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



    $f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



    In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.






    share|cite|improve this answer





















    • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
      – Mike
      2 hours ago










    • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
      – Tsemo Aristide
      2 hours ago










    • That's so true.
      – Mike
      2 hours ago










    • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
      – Mike
      2 hours ago












    • $f_n(n)=1, f_n(x)=0$ if $xneq n$
      – Tsemo Aristide
      2 hours ago
















    4












    4








    4






    $f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



    $f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



    In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.






    share|cite|improve this answer












    $f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$



    $f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.



    In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    Tsemo Aristide

    56.2k11444




    56.2k11444












    • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
      – Mike
      2 hours ago










    • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
      – Tsemo Aristide
      2 hours ago










    • That's so true.
      – Mike
      2 hours ago










    • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
      – Mike
      2 hours ago












    • $f_n(n)=1, f_n(x)=0$ if $xneq n$
      – Tsemo Aristide
      2 hours ago




















    • (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
      – Mike
      2 hours ago










    • If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
      – Tsemo Aristide
      2 hours ago










    • That's so true.
      – Mike
      2 hours ago










    • Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
      – Mike
      2 hours ago












    • $f_n(n)=1, f_n(x)=0$ if $xneq n$
      – Tsemo Aristide
      2 hours ago


















    (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
    – Mike
    2 hours ago




    (+1) I already know this but want to know why $f_n(x)to f(x),;forall,xin E,$ is used for both uniform and pointwise convergence.
    – Mike
    2 hours ago












    If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
    – Tsemo Aristide
    2 hours ago




    If $f_n$ converges uniformly, it convergence pointwise, so we can write $f_n(x)rightarrow f(x)$ for the both case. The difference is that for uniform convergence, if you draw the picture, you will see that the distance between the graph of $f_n$ and $f$ tends to $0$, this is not necessarily true for simply convergence.
    – Tsemo Aristide
    2 hours ago












    That's so true.
    – Mike
    2 hours ago




    That's so true.
    – Mike
    2 hours ago












    Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
    – Mike
    2 hours ago






    Kindly check your post. Do you mean $f_n(n)=0$ or $f_n(x)=0$?
    – Mike
    2 hours ago














    $f_n(n)=1, f_n(x)=0$ if $xneq n$
    – Tsemo Aristide
    2 hours ago






    $f_n(n)=1, f_n(x)=0$ if $xneq n$
    – Tsemo Aristide
    2 hours ago













    4














    Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
    $$
    f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
    $$
    This is strictly stronger than pointwise convergence.



    Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.






    share|cite|improve this answer

















    • 1




      Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
      – Matt A Pelto
      43 mins ago
















    4














    Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
    $$
    f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
    $$
    This is strictly stronger than pointwise convergence.



    Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.






    share|cite|improve this answer

















    • 1




      Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
      – Matt A Pelto
      43 mins ago














    4












    4








    4






    Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
    $$
    f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
    $$
    This is strictly stronger than pointwise convergence.



    Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.






    share|cite|improve this answer












    Uniform convergence is actually $mathcal L^infty$ convergence, i.e.
    $$
    f_n rightrightarrows f [x in E]!! iff !! sup_{x in E} vert f_n - fvert(x) to 0[n to infty].
    $$
    This is strictly stronger than pointwise convergence.



    Alternatively, uniform convergence implies pointwise convergence, so $f_n to f$ in both cases.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    xbh

    5,7451522




    5,7451522








    • 1




      Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
      – Matt A Pelto
      43 mins ago














    • 1




      Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
      – Matt A Pelto
      43 mins ago








    1




    1




    Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
    – Matt A Pelto
    43 mins ago




    Essential supremum and supremum coincide in the context of continuous functions. Pointwise convergence does imply uniform convergence in some particular context such as that of a sequence of $l$-Lipschitz functions where $l in mathbb R$.
    – Matt A Pelto
    43 mins ago











    1














    Sorry, but yes, you probably are missing something important, because the second statement in your post




    On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
    $$f_n(x)to f(x),;forall,xin E.$$




    is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



    Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



    Your best bet is to check the original source to find out what exactly it says there.






    share|cite|improve this answer


























      1














      Sorry, but yes, you probably are missing something important, because the second statement in your post




      On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
      $$f_n(x)to f(x),;forall,xin E.$$




      is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



      Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



      Your best bet is to check the original source to find out what exactly it says there.






      share|cite|improve this answer
























        1












        1








        1






        Sorry, but yes, you probably are missing something important, because the second statement in your post




        On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
        $$f_n(x)to f(x),;forall,xin E.$$




        is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



        Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



        Your best bet is to check the original source to find out what exactly it says there.






        share|cite|improve this answer












        Sorry, but yes, you probably are missing something important, because the second statement in your post




        On the other hand ${f_n}_{ninmathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
        $$f_n(x)to f(x),;forall,xin E.$$




        is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.



        Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $varepsilon/delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.



        Your best bet is to check the original source to find out what exactly it says there.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 29 mins ago









        zipirovich

        11.1k11631




        11.1k11631






























            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%2f3063473%2fwhat-is-the-main-difference-between-pointwise-and-uniform-convergence-as-defined%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

            What visual should I use to simply compare current year value vs last year in Power BI desktop

            Image
            0 I have column SubToBind that shows current year ratio and column PY SubToBind which shows previous year ratio. How can I simply create a card that would be displaying them both in a nice way? Something like that: I cannot achieve that with regular card, I also tried Card with States by OKViz but also unable to see both values. In KPI visual - I dont like the word "Goal", it will confuse end users. Are there any cards or other visuals that would do the job? Thanks powerbi dax powerbi-embedded share | improve this question edited Jun 11 '18 at 20:11 Oleg ...
            Read more

            How to ignore python UserWarning in pytest?

            Image
            up vote 1 down vote favorite I am using openpyxl for parsing .xlsm files, and pytest for testing. When I open a file, I get the: OpenPyxl -> UserWarning: Data Validation extension is not supported and will be removed That is not really a problem bcs. program works and I can't change the .xlsm file to fix that. But... When I run pytest with something like: def test_wrong_file_format(): assert check_excel(open_excel('file.xlsm')) == True I will get the warning i mentioned altought check_excel(open_excel('file.xlsm')) returns True and the test should suceed... Is there a nice way to tell the pytest that "It's not a bug it's a feature" and tests should pass even when they get this warning? Is there other way than using something like: with pytest.warns(UserWarning): ...
            Read more

            Alexandru Averescu

            Image
            Alexandre Averescu Alexandru Averescu Naissance 3 avril 1859 Babele, Ismail  (ro) près de Izmaïl (principauté de Moldavie) Décès 2 octobre 1938 (à 79 ans) [ 1 ] Bucarest Origine   Roumanie Arme cavalerie Grade Maréchal Années de service 1901-1917 Commandement chef d'état-major général 2 e  armée 3 e  armée Groupe d'armée sud armée du Danube Conflits Guerre russo-turque de 1877-1878 Deuxième guerre balkanique Première Guerre mondiale Faits d'armes Bataille de Turtucaia/Tutrakan Bataille de Mărăști Distinctions Autres fonctions 3 fois président du Conseil des ministres ministre des Affaires étrangères ministre sans portefeuille conseiller de la Couronne modifier   Alexandre Averescu (en roumain : Alexandru Averescu .mw-parser-output .prononciation>a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/8/8a/Loudspeaker.svg/11px-Loudspeaker.svg.png")center ...
            Read more