The error of a Taylor polynomial (relativistic kinetic energy)












1












$begingroup$


I have been trying to utilize the formula:
enter image description here



I simply cannot figure out how to determine the error when using



enter image description here



Instead of:



enter image description here



I have made a Taylor polynomial around 0 of grade 4, and I cannot find the deviation when using the Taylor polynomial instead of the original function.



The Taylor polynomial is:
enter image description here



And it was made of the Lorentz factor:



enter image description here






real-analysis taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you have a look at Taylor-Lagrange formula ?
    $endgroup$
    – Damien
    Dec 16 '18 at 8:23










  • $begingroup$
    Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
    $endgroup$
    – Nikolai
    Dec 16 '18 at 10:58










  • $begingroup$
    You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
    $endgroup$
    – Damien
    Dec 16 '18 at 12:28










  • $begingroup$
    Well i have made the taylor series, but don't understand how i get the bound
    $endgroup$
    – Nikolai
    Dec 16 '18 at 15:58
















1












$begingroup$


I have been trying to utilize the formula:
enter image description here



I simply cannot figure out how to determine the error when using



enter image description here



Instead of:



enter image description here



I have made a Taylor polynomial around 0 of grade 4, and I cannot find the deviation when using the Taylor polynomial instead of the original function.



The Taylor polynomial is:
enter image description here



And it was made of the Lorentz factor:



enter image description here






real-analysis taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you have a look at Taylor-Lagrange formula ?
    $endgroup$
    – Damien
    Dec 16 '18 at 8:23










  • $begingroup$
    Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
    $endgroup$
    – Nikolai
    Dec 16 '18 at 10:58










  • $begingroup$
    You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
    $endgroup$
    – Damien
    Dec 16 '18 at 12:28










  • $begingroup$
    Well i have made the taylor series, but don't understand how i get the bound
    $endgroup$
    – Nikolai
    Dec 16 '18 at 15:58














1












1








1





$begingroup$


I have been trying to utilize the formula:
enter image description here



I simply cannot figure out how to determine the error when using



enter image description here



Instead of:



enter image description here



I have made a Taylor polynomial around 0 of grade 4, and I cannot find the deviation when using the Taylor polynomial instead of the original function.



The Taylor polynomial is:
enter image description here



And it was made of the Lorentz factor:



enter image description here






real-analysis taylor-expansion







share|cite|improve this question











$endgroup$




I have been trying to utilize the formula:
enter image description here



I simply cannot figure out how to determine the error when using



enter image description here



Instead of:



enter image description here



I have made a Taylor polynomial around 0 of grade 4, and I cannot find the deviation when using the Taylor polynomial instead of the original function.



The Taylor polynomial is:
enter image description here



And it was made of the Lorentz factor:



enter image description here






real-analysis taylor-expansion




real-analysis taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 6:07









Moo

5,62131020




5,62131020










asked Dec 16 '18 at 1:22









Nikolai Nikolai

205




205












  • $begingroup$
    Did you have a look at Taylor-Lagrange formula ?
    $endgroup$
    – Damien
    Dec 16 '18 at 8:23










  • $begingroup$
    Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
    $endgroup$
    – Nikolai
    Dec 16 '18 at 10:58










  • $begingroup$
    You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
    $endgroup$
    – Damien
    Dec 16 '18 at 12:28










  • $begingroup$
    Well i have made the taylor series, but don't understand how i get the bound
    $endgroup$
    – Nikolai
    Dec 16 '18 at 15:58


















  • $begingroup$
    Did you have a look at Taylor-Lagrange formula ?
    $endgroup$
    – Damien
    Dec 16 '18 at 8:23










  • $begingroup$
    Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
    $endgroup$
    – Nikolai
    Dec 16 '18 at 10:58










  • $begingroup$
    You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
    $endgroup$
    – Damien
    Dec 16 '18 at 12:28










  • $begingroup$
    Well i have made the taylor series, but don't understand how i get the bound
    $endgroup$
    – Nikolai
    Dec 16 '18 at 15:58
















$begingroup$
Did you have a look at Taylor-Lagrange formula ?
$endgroup$
– Damien
Dec 16 '18 at 8:23




$begingroup$
Did you have a look at Taylor-Lagrange formula ?
$endgroup$
– Damien
Dec 16 '18 at 8:23












$begingroup$
Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
$endgroup$
– Nikolai
Dec 16 '18 at 10:58




$begingroup$
Yes I have, but I cannot figure out how to implement it. Every example of using it that I can find, uses the the natural exponential function or a trigonometric function, and their bound is easy to figure out.
$endgroup$
– Nikolai
Dec 16 '18 at 10:58












$begingroup$
You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
$endgroup$
– Damien
Dec 16 '18 at 12:28




$begingroup$
You can get a bound on the error with Taylor-Lagrange as soon as you can calculate the Taylor series. What is blocking you ?
$endgroup$
– Damien
Dec 16 '18 at 12:28












$begingroup$
Well i have made the taylor series, but don't understand how i get the bound
$endgroup$
– Nikolai
Dec 16 '18 at 15:58




$begingroup$
Well i have made the taylor series, but don't understand how i get the bound
$endgroup$
– Nikolai
Dec 16 '18 at 15:58










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let us call $f(t)$ the function
$$f(t) = frac{1}{sqrt{1-t}} $$



The $k^{th}$ derivative function of $f()$ is equal to
$$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
This last relation can for example be demonstrated by recurrence.



According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



$$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



It follows



$$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



Now, one just have to replace $t$ :
$$t = x^2 = left(frac{v}{c}right)^2$$






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%2f3042135%2fthe-error-of-a-taylor-polynomial-relativistic-kinetic-energy%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









    1












    $begingroup$

    Let us call $f(t)$ the function
    $$f(t) = frac{1}{sqrt{1-t}} $$



    The $k^{th}$ derivative function of $f()$ is equal to
    $$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
    This last relation can for example be demonstrated by recurrence.



    According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



    $$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



    It follows



    $$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



    Now, one just have to replace $t$ :
    $$t = x^2 = left(frac{v}{c}right)^2$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let us call $f(t)$ the function
      $$f(t) = frac{1}{sqrt{1-t}} $$



      The $k^{th}$ derivative function of $f()$ is equal to
      $$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
      This last relation can for example be demonstrated by recurrence.



      According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



      $$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



      It follows



      $$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



      Now, one just have to replace $t$ :
      $$t = x^2 = left(frac{v}{c}right)^2$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let us call $f(t)$ the function
        $$f(t) = frac{1}{sqrt{1-t}} $$



        The $k^{th}$ derivative function of $f()$ is equal to
        $$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
        This last relation can for example be demonstrated by recurrence.



        According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



        $$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



        It follows



        $$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



        Now, one just have to replace $t$ :
        $$t = x^2 = left(frac{v}{c}right)^2$$






        share|cite|improve this answer











        $endgroup$



        Let us call $f(t)$ the function
        $$f(t) = frac{1}{sqrt{1-t}} $$



        The $k^{th}$ derivative function of $f()$ is equal to
        $$f^{k}(t) = frac{prod_{i=1}^k{(2i-1)}}{2^k} (1-t)^{-frac{2k+1}{2}} $$
        This last relation can for example be demonstrated by recurrence.



        According to Taylor-Lagrange formula, there exists $xi$ between $0$ and $t$ such that



        $$f(t) = 1 + frac{t}{2} + frac{3}{8}t^2 + frac{5}{16}(1-xi)^{-7/2},t^3 = 1 + frac{t}{2} + frac{3}{8}t^2 + R(t)$$



        It follows



        $$0 < R(t) < frac{5}{16}(1-t)^{-7/2},t^3 $$



        Now, one just have to replace $t$ :
        $$t = x^2 = left(frac{v}{c}right)^2$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 22:00

























        answered Dec 16 '18 at 21:52









        DamienDamien

        59714




        59714






























            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%2f3042135%2fthe-error-of-a-taylor-polynomial-relativistic-kinetic-energy%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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more