Proof that $sin {x}$ is infinitely continuously differentiable over $[m,n]$












4












$begingroup$


I am trying to prove that $sin {x}$ is infinitely continuously differentiable over $[m,n]$ where $m$ and $n$ are real numbers. Here is my attempt at doing so. Is my proof complete? If not, what can I do to improve it? Thank you in advance.



Since,



$frac{d}{dx}sin{x} = cos{x}$,



$frac{d^2}{dx^2}sin{x} = -sin{x}$,



$frac{d^3}{dx^3}sin{x} = -cos{x}$,



and



$frac{d^4}{dx^4}sin{x} = sin{x}$,



the derivatives of $sin{x}$, are periodic. Since the first four derivatives of $sin{x}$ are continuous over $[m,n]$ where $m$ and $n$ are real numbers, $sin{x}$ must be differentiable an infinite amount of times over $[m,n]$.






calculus trigonometry proof-verification proof-writing







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
    $endgroup$
    – user29418
    Jan 7 at 21:06
















4












$begingroup$


I am trying to prove that $sin {x}$ is infinitely continuously differentiable over $[m,n]$ where $m$ and $n$ are real numbers. Here is my attempt at doing so. Is my proof complete? If not, what can I do to improve it? Thank you in advance.



Since,



$frac{d}{dx}sin{x} = cos{x}$,



$frac{d^2}{dx^2}sin{x} = -sin{x}$,



$frac{d^3}{dx^3}sin{x} = -cos{x}$,



and



$frac{d^4}{dx^4}sin{x} = sin{x}$,



the derivatives of $sin{x}$, are periodic. Since the first four derivatives of $sin{x}$ are continuous over $[m,n]$ where $m$ and $n$ are real numbers, $sin{x}$ must be differentiable an infinite amount of times over $[m,n]$.






calculus trigonometry proof-verification proof-writing







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
    $endgroup$
    – user29418
    Jan 7 at 21:06














4












4








4


1



$begingroup$


I am trying to prove that $sin {x}$ is infinitely continuously differentiable over $[m,n]$ where $m$ and $n$ are real numbers. Here is my attempt at doing so. Is my proof complete? If not, what can I do to improve it? Thank you in advance.



Since,



$frac{d}{dx}sin{x} = cos{x}$,



$frac{d^2}{dx^2}sin{x} = -sin{x}$,



$frac{d^3}{dx^3}sin{x} = -cos{x}$,



and



$frac{d^4}{dx^4}sin{x} = sin{x}$,



the derivatives of $sin{x}$, are periodic. Since the first four derivatives of $sin{x}$ are continuous over $[m,n]$ where $m$ and $n$ are real numbers, $sin{x}$ must be differentiable an infinite amount of times over $[m,n]$.






calculus trigonometry proof-verification proof-writing







share|cite|improve this question









$endgroup$




I am trying to prove that $sin {x}$ is infinitely continuously differentiable over $[m,n]$ where $m$ and $n$ are real numbers. Here is my attempt at doing so. Is my proof complete? If not, what can I do to improve it? Thank you in advance.



Since,



$frac{d}{dx}sin{x} = cos{x}$,



$frac{d^2}{dx^2}sin{x} = -sin{x}$,



$frac{d^3}{dx^3}sin{x} = -cos{x}$,



and



$frac{d^4}{dx^4}sin{x} = sin{x}$,



the derivatives of $sin{x}$, are periodic. Since the first four derivatives of $sin{x}$ are continuous over $[m,n]$ where $m$ and $n$ are real numbers, $sin{x}$ must be differentiable an infinite amount of times over $[m,n]$.






calculus trigonometry proof-verification proof-writing




calculus trigonometry proof-verification proof-writing






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 21:02









GnumbertesterGnumbertester

37518




37518








  • 3




    $begingroup$
    Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
    $endgroup$
    – user29418
    Jan 7 at 21:06














  • 3




    $begingroup$
    Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
    $endgroup$
    – user29418
    Jan 7 at 21:06








3




3




$begingroup$
Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
$endgroup$
– user29418
Jan 7 at 21:06




$begingroup$
Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
$endgroup$
– user29418
Jan 7 at 21:06










1 Answer
1






active

oldest

votes


















2












$begingroup$

A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.






share|cite|improve this answer









$endgroup$













    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%2f3065484%2fproof-that-sin-x-is-infinitely-continuously-differentiable-over-m-n%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









    2












    $begingroup$

    A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.






        share|cite|improve this answer









        $endgroup$



        A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 21:19









        D.B.D.B.

        1,2128




        1,2128






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065484%2fproof-that-sin-x-is-infinitely-continuously-differentiable-over-m-n%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

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            Display a custom attribute below product name in the front-end Magento 1.9.3.8

            Image
            0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
            Read more