Problems involving the second Taylor Polynomial of $e^xcos x$












0












$begingroup$


I'm working on what seem to be very easy problems, but my answers aren't matching my textbook's.



1) Find the second Taylor polynomial of $f(x) = e^xcos x$ about $x_0 = 0$.



$P_2(x) = 1+x$. (correct)



2) Use $P_2(.5)$ to approximate $f(.5)$. Find an upper bound for the error $|f(.5)-P_2(.5)|$ and compare it to the actual answer.



The error can be no greater than $R_2(.5) = frac{e^.5(sin(.5)+cos(.5))}{24}=.0932$, the actual error is $.0531$. (correct)



Here is where my trouble starts:



3) Find a bound for the error $|f(x)-P_2(x)|$ in using $P_2(x)$ to approximate $f(x)$ on the interval $[0,1]$.



I've tried integrating the error function, which for this problem is $R_2(x) = dfrac{-2e^xi(sin(xi)+cos(xi)x^3}{6}$, from $0$ to $1$. I used the fact that the maximum value of $sin(xi)+cos(xi)=sqrt{2}$, with $xi = pi/4$, but this gave me an incorrect answer.



Any help appreciated.






numerical-methods







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I'm working on what seem to be very easy problems, but my answers aren't matching my textbook's.



    1) Find the second Taylor polynomial of $f(x) = e^xcos x$ about $x_0 = 0$.



    $P_2(x) = 1+x$. (correct)



    2) Use $P_2(.5)$ to approximate $f(.5)$. Find an upper bound for the error $|f(.5)-P_2(.5)|$ and compare it to the actual answer.



    The error can be no greater than $R_2(.5) = frac{e^.5(sin(.5)+cos(.5))}{24}=.0932$, the actual error is $.0531$. (correct)



    Here is where my trouble starts:



    3) Find a bound for the error $|f(x)-P_2(x)|$ in using $P_2(x)$ to approximate $f(x)$ on the interval $[0,1]$.



    I've tried integrating the error function, which for this problem is $R_2(x) = dfrac{-2e^xi(sin(xi)+cos(xi)x^3}{6}$, from $0$ to $1$. I used the fact that the maximum value of $sin(xi)+cos(xi)=sqrt{2}$, with $xi = pi/4$, but this gave me an incorrect answer.



    Any help appreciated.






    numerical-methods







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I'm working on what seem to be very easy problems, but my answers aren't matching my textbook's.



      1) Find the second Taylor polynomial of $f(x) = e^xcos x$ about $x_0 = 0$.



      $P_2(x) = 1+x$. (correct)



      2) Use $P_2(.5)$ to approximate $f(.5)$. Find an upper bound for the error $|f(.5)-P_2(.5)|$ and compare it to the actual answer.



      The error can be no greater than $R_2(.5) = frac{e^.5(sin(.5)+cos(.5))}{24}=.0932$, the actual error is $.0531$. (correct)



      Here is where my trouble starts:



      3) Find a bound for the error $|f(x)-P_2(x)|$ in using $P_2(x)$ to approximate $f(x)$ on the interval $[0,1]$.



      I've tried integrating the error function, which for this problem is $R_2(x) = dfrac{-2e^xi(sin(xi)+cos(xi)x^3}{6}$, from $0$ to $1$. I used the fact that the maximum value of $sin(xi)+cos(xi)=sqrt{2}$, with $xi = pi/4$, but this gave me an incorrect answer.



      Any help appreciated.






      numerical-methods







      share|cite|improve this question











      $endgroup$




      I'm working on what seem to be very easy problems, but my answers aren't matching my textbook's.



      1) Find the second Taylor polynomial of $f(x) = e^xcos x$ about $x_0 = 0$.



      $P_2(x) = 1+x$. (correct)



      2) Use $P_2(.5)$ to approximate $f(.5)$. Find an upper bound for the error $|f(.5)-P_2(.5)|$ and compare it to the actual answer.



      The error can be no greater than $R_2(.5) = frac{e^.5(sin(.5)+cos(.5))}{24}=.0932$, the actual error is $.0531$. (correct)



      Here is where my trouble starts:



      3) Find a bound for the error $|f(x)-P_2(x)|$ in using $P_2(x)$ to approximate $f(x)$ on the interval $[0,1]$.



      I've tried integrating the error function, which for this problem is $R_2(x) = dfrac{-2e^xi(sin(xi)+cos(xi)x^3}{6}$, from $0$ to $1$. I used the fact that the maximum value of $sin(xi)+cos(xi)=sqrt{2}$, with $xi = pi/4$, but this gave me an incorrect answer.



      Any help appreciated.






      numerical-methods




      numerical-methods






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 13 at 19:34









      Bernard

      119k740113




      119k740113










      asked Jan 13 at 18:36









      mXdXmXdX

      556




      556






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$


          I've tried integrating the error function




          This isn't helpful. What we're looking for is an upper bound for the error - a largest possible value, not an integral. I would use that maximum for $sin+cos$; multiply it by the maximum for the exponential, and we have an upper bound for the relavant derivative. But then, we're not going to integrate anything.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            When I do this, I get ~1.281, but my textbook gives an answer of ~1.252. I'm not sure if I did something wrong or if it's a typo.
            $endgroup$
            – mXdX
            Jan 13 at 19:12






          • 1




            $begingroup$
            The problem says to find a bound. It does not say to find the best possible bound. You don't need to exactly match the answer to be right.
            $endgroup$
            – jmerry
            Jan 13 at 19:15






          • 1




            $begingroup$
            In more detail on that last comment: the $sin+cos$ has a maximum inside the interval, while the exponential keeps increasing to the end. Because they don't line up in the same place, the maximum of the product appears somewhere between, with a value less than the product of the maximums. We could find it exactly, but it's a lot of effort for not much gain.
            $endgroup$
            – jmerry
            Jan 13 at 19:33










          • $begingroup$
            I see. So we're taking the maximum of the exponential and the maximum of sinx+cosx at the the same time, even though their maximums don't occur at the same time. However, it still gives us some bound, even though there are smaller upper bounds. Right?
            $endgroup$
            – mXdX
            Jan 13 at 19:52








          • 1




            $begingroup$
            Yes, that's it.
            $endgroup$
            – jmerry
            Jan 13 at 19:57



















          1












          $begingroup$

          hint



          $$f(x)=P_2(x)+frac{x^3}{6}f'''(xi)$$



          with



          $$f'''(xi)=-2(cos(xi)+sin(xi))e^{xi}$$



          $$=-2sqrt{2}cos(xi-frac{pi}{4})e^xi$$
          thus



          $$|f(x)-P_2(x)|le frac{sqrt{2}}{3}e$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I have this already, I'm not sure where to go from here.
            $endgroup$
            – mXdX
            Jan 13 at 18:55










          • $begingroup$
            @mXdX Is it good now.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:00










          • $begingroup$
            Is this for (3)?
            $endgroup$
            – mXdX
            Jan 13 at 19:01










          • $begingroup$
            @mXdX Now, it is for 3.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:05










          • $begingroup$
            This answer is ~1.281, my textbook says the answer is 1.252. Do you think it's a typo?
            $endgroup$
            – mXdX
            Jan 13 at 19:08











          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%2f3072371%2fproblems-involving-the-second-taylor-polynomial-of-ex-cos-x%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$


          I've tried integrating the error function




          This isn't helpful. What we're looking for is an upper bound for the error - a largest possible value, not an integral. I would use that maximum for $sin+cos$; multiply it by the maximum for the exponential, and we have an upper bound for the relavant derivative. But then, we're not going to integrate anything.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            When I do this, I get ~1.281, but my textbook gives an answer of ~1.252. I'm not sure if I did something wrong or if it's a typo.
            $endgroup$
            – mXdX
            Jan 13 at 19:12






          • 1




            $begingroup$
            The problem says to find a bound. It does not say to find the best possible bound. You don't need to exactly match the answer to be right.
            $endgroup$
            – jmerry
            Jan 13 at 19:15






          • 1




            $begingroup$
            In more detail on that last comment: the $sin+cos$ has a maximum inside the interval, while the exponential keeps increasing to the end. Because they don't line up in the same place, the maximum of the product appears somewhere between, with a value less than the product of the maximums. We could find it exactly, but it's a lot of effort for not much gain.
            $endgroup$
            – jmerry
            Jan 13 at 19:33










          • $begingroup$
            I see. So we're taking the maximum of the exponential and the maximum of sinx+cosx at the the same time, even though their maximums don't occur at the same time. However, it still gives us some bound, even though there are smaller upper bounds. Right?
            $endgroup$
            – mXdX
            Jan 13 at 19:52








          • 1




            $begingroup$
            Yes, that's it.
            $endgroup$
            – jmerry
            Jan 13 at 19:57
















          0












          $begingroup$


          I've tried integrating the error function




          This isn't helpful. What we're looking for is an upper bound for the error - a largest possible value, not an integral. I would use that maximum for $sin+cos$; multiply it by the maximum for the exponential, and we have an upper bound for the relavant derivative. But then, we're not going to integrate anything.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            When I do this, I get ~1.281, but my textbook gives an answer of ~1.252. I'm not sure if I did something wrong or if it's a typo.
            $endgroup$
            – mXdX
            Jan 13 at 19:12






          • 1




            $begingroup$
            The problem says to find a bound. It does not say to find the best possible bound. You don't need to exactly match the answer to be right.
            $endgroup$
            – jmerry
            Jan 13 at 19:15






          • 1




            $begingroup$
            In more detail on that last comment: the $sin+cos$ has a maximum inside the interval, while the exponential keeps increasing to the end. Because they don't line up in the same place, the maximum of the product appears somewhere between, with a value less than the product of the maximums. We could find it exactly, but it's a lot of effort for not much gain.
            $endgroup$
            – jmerry
            Jan 13 at 19:33










          • $begingroup$
            I see. So we're taking the maximum of the exponential and the maximum of sinx+cosx at the the same time, even though their maximums don't occur at the same time. However, it still gives us some bound, even though there are smaller upper bounds. Right?
            $endgroup$
            – mXdX
            Jan 13 at 19:52








          • 1




            $begingroup$
            Yes, that's it.
            $endgroup$
            – jmerry
            Jan 13 at 19:57














          0












          0








          0





          $begingroup$


          I've tried integrating the error function




          This isn't helpful. What we're looking for is an upper bound for the error - a largest possible value, not an integral. I would use that maximum for $sin+cos$; multiply it by the maximum for the exponential, and we have an upper bound for the relavant derivative. But then, we're not going to integrate anything.






          share|cite|improve this answer









          $endgroup$




          I've tried integrating the error function




          This isn't helpful. What we're looking for is an upper bound for the error - a largest possible value, not an integral. I would use that maximum for $sin+cos$; multiply it by the maximum for the exponential, and we have an upper bound for the relavant derivative. But then, we're not going to integrate anything.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 19:02









          jmerryjmerry

          6,067718




          6,067718












          • $begingroup$
            When I do this, I get ~1.281, but my textbook gives an answer of ~1.252. I'm not sure if I did something wrong or if it's a typo.
            $endgroup$
            – mXdX
            Jan 13 at 19:12






          • 1




            $begingroup$
            The problem says to find a bound. It does not say to find the best possible bound. You don't need to exactly match the answer to be right.
            $endgroup$
            – jmerry
            Jan 13 at 19:15






          • 1




            $begingroup$
            In more detail on that last comment: the $sin+cos$ has a maximum inside the interval, while the exponential keeps increasing to the end. Because they don't line up in the same place, the maximum of the product appears somewhere between, with a value less than the product of the maximums. We could find it exactly, but it's a lot of effort for not much gain.
            $endgroup$
            – jmerry
            Jan 13 at 19:33










          • $begingroup$
            I see. So we're taking the maximum of the exponential and the maximum of sinx+cosx at the the same time, even though their maximums don't occur at the same time. However, it still gives us some bound, even though there are smaller upper bounds. Right?
            $endgroup$
            – mXdX
            Jan 13 at 19:52








          • 1




            $begingroup$
            Yes, that's it.
            $endgroup$
            – jmerry
            Jan 13 at 19:57


















          • $begingroup$
            When I do this, I get ~1.281, but my textbook gives an answer of ~1.252. I'm not sure if I did something wrong or if it's a typo.
            $endgroup$
            – mXdX
            Jan 13 at 19:12






          • 1




            $begingroup$
            The problem says to find a bound. It does not say to find the best possible bound. You don't need to exactly match the answer to be right.
            $endgroup$
            – jmerry
            Jan 13 at 19:15






          • 1




            $begingroup$
            In more detail on that last comment: the $sin+cos$ has a maximum inside the interval, while the exponential keeps increasing to the end. Because they don't line up in the same place, the maximum of the product appears somewhere between, with a value less than the product of the maximums. We could find it exactly, but it's a lot of effort for not much gain.
            $endgroup$
            – jmerry
            Jan 13 at 19:33










          • $begingroup$
            I see. So we're taking the maximum of the exponential and the maximum of sinx+cosx at the the same time, even though their maximums don't occur at the same time. However, it still gives us some bound, even though there are smaller upper bounds. Right?
            $endgroup$
            – mXdX
            Jan 13 at 19:52








          • 1




            $begingroup$
            Yes, that's it.
            $endgroup$
            – jmerry
            Jan 13 at 19:57
















          $begingroup$
          When I do this, I get ~1.281, but my textbook gives an answer of ~1.252. I'm not sure if I did something wrong or if it's a typo.
          $endgroup$
          – mXdX
          Jan 13 at 19:12




          $begingroup$
          When I do this, I get ~1.281, but my textbook gives an answer of ~1.252. I'm not sure if I did something wrong or if it's a typo.
          $endgroup$
          – mXdX
          Jan 13 at 19:12




          1




          1




          $begingroup$
          The problem says to find a bound. It does not say to find the best possible bound. You don't need to exactly match the answer to be right.
          $endgroup$
          – jmerry
          Jan 13 at 19:15




          $begingroup$
          The problem says to find a bound. It does not say to find the best possible bound. You don't need to exactly match the answer to be right.
          $endgroup$
          – jmerry
          Jan 13 at 19:15




          1




          1




          $begingroup$
          In more detail on that last comment: the $sin+cos$ has a maximum inside the interval, while the exponential keeps increasing to the end. Because they don't line up in the same place, the maximum of the product appears somewhere between, with a value less than the product of the maximums. We could find it exactly, but it's a lot of effort for not much gain.
          $endgroup$
          – jmerry
          Jan 13 at 19:33




          $begingroup$
          In more detail on that last comment: the $sin+cos$ has a maximum inside the interval, while the exponential keeps increasing to the end. Because they don't line up in the same place, the maximum of the product appears somewhere between, with a value less than the product of the maximums. We could find it exactly, but it's a lot of effort for not much gain.
          $endgroup$
          – jmerry
          Jan 13 at 19:33












          $begingroup$
          I see. So we're taking the maximum of the exponential and the maximum of sinx+cosx at the the same time, even though their maximums don't occur at the same time. However, it still gives us some bound, even though there are smaller upper bounds. Right?
          $endgroup$
          – mXdX
          Jan 13 at 19:52






          $begingroup$
          I see. So we're taking the maximum of the exponential and the maximum of sinx+cosx at the the same time, even though their maximums don't occur at the same time. However, it still gives us some bound, even though there are smaller upper bounds. Right?
          $endgroup$
          – mXdX
          Jan 13 at 19:52






          1




          1




          $begingroup$
          Yes, that's it.
          $endgroup$
          – jmerry
          Jan 13 at 19:57




          $begingroup$
          Yes, that's it.
          $endgroup$
          – jmerry
          Jan 13 at 19:57











          1












          $begingroup$

          hint



          $$f(x)=P_2(x)+frac{x^3}{6}f'''(xi)$$



          with



          $$f'''(xi)=-2(cos(xi)+sin(xi))e^{xi}$$



          $$=-2sqrt{2}cos(xi-frac{pi}{4})e^xi$$
          thus



          $$|f(x)-P_2(x)|le frac{sqrt{2}}{3}e$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I have this already, I'm not sure where to go from here.
            $endgroup$
            – mXdX
            Jan 13 at 18:55










          • $begingroup$
            @mXdX Is it good now.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:00










          • $begingroup$
            Is this for (3)?
            $endgroup$
            – mXdX
            Jan 13 at 19:01










          • $begingroup$
            @mXdX Now, it is for 3.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:05










          • $begingroup$
            This answer is ~1.281, my textbook says the answer is 1.252. Do you think it's a typo?
            $endgroup$
            – mXdX
            Jan 13 at 19:08
















          1












          $begingroup$

          hint



          $$f(x)=P_2(x)+frac{x^3}{6}f'''(xi)$$



          with



          $$f'''(xi)=-2(cos(xi)+sin(xi))e^{xi}$$



          $$=-2sqrt{2}cos(xi-frac{pi}{4})e^xi$$
          thus



          $$|f(x)-P_2(x)|le frac{sqrt{2}}{3}e$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I have this already, I'm not sure where to go from here.
            $endgroup$
            – mXdX
            Jan 13 at 18:55










          • $begingroup$
            @mXdX Is it good now.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:00










          • $begingroup$
            Is this for (3)?
            $endgroup$
            – mXdX
            Jan 13 at 19:01










          • $begingroup$
            @mXdX Now, it is for 3.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:05










          • $begingroup$
            This answer is ~1.281, my textbook says the answer is 1.252. Do you think it's a typo?
            $endgroup$
            – mXdX
            Jan 13 at 19:08














          1












          1








          1





          $begingroup$

          hint



          $$f(x)=P_2(x)+frac{x^3}{6}f'''(xi)$$



          with



          $$f'''(xi)=-2(cos(xi)+sin(xi))e^{xi}$$



          $$=-2sqrt{2}cos(xi-frac{pi}{4})e^xi$$
          thus



          $$|f(x)-P_2(x)|le frac{sqrt{2}}{3}e$$






          share|cite|improve this answer











          $endgroup$



          hint



          $$f(x)=P_2(x)+frac{x^3}{6}f'''(xi)$$



          with



          $$f'''(xi)=-2(cos(xi)+sin(xi))e^{xi}$$



          $$=-2sqrt{2}cos(xi-frac{pi}{4})e^xi$$
          thus



          $$|f(x)-P_2(x)|le frac{sqrt{2}}{3}e$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 13 at 19:04

























          answered Jan 13 at 18:42









          hamam_Abdallahhamam_Abdallah

          38k21634




          38k21634












          • $begingroup$
            I have this already, I'm not sure where to go from here.
            $endgroup$
            – mXdX
            Jan 13 at 18:55










          • $begingroup$
            @mXdX Is it good now.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:00










          • $begingroup$
            Is this for (3)?
            $endgroup$
            – mXdX
            Jan 13 at 19:01










          • $begingroup$
            @mXdX Now, it is for 3.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:05










          • $begingroup$
            This answer is ~1.281, my textbook says the answer is 1.252. Do you think it's a typo?
            $endgroup$
            – mXdX
            Jan 13 at 19:08


















          • $begingroup$
            I have this already, I'm not sure where to go from here.
            $endgroup$
            – mXdX
            Jan 13 at 18:55










          • $begingroup$
            @mXdX Is it good now.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:00










          • $begingroup$
            Is this for (3)?
            $endgroup$
            – mXdX
            Jan 13 at 19:01










          • $begingroup$
            @mXdX Now, it is for 3.
            $endgroup$
            – hamam_Abdallah
            Jan 13 at 19:05










          • $begingroup$
            This answer is ~1.281, my textbook says the answer is 1.252. Do you think it's a typo?
            $endgroup$
            – mXdX
            Jan 13 at 19:08
















          $begingroup$
          I have this already, I'm not sure where to go from here.
          $endgroup$
          – mXdX
          Jan 13 at 18:55




          $begingroup$
          I have this already, I'm not sure where to go from here.
          $endgroup$
          – mXdX
          Jan 13 at 18:55












          $begingroup$
          @mXdX Is it good now.
          $endgroup$
          – hamam_Abdallah
          Jan 13 at 19:00




          $begingroup$
          @mXdX Is it good now.
          $endgroup$
          – hamam_Abdallah
          Jan 13 at 19:00












          $begingroup$
          Is this for (3)?
          $endgroup$
          – mXdX
          Jan 13 at 19:01




          $begingroup$
          Is this for (3)?
          $endgroup$
          – mXdX
          Jan 13 at 19:01












          $begingroup$
          @mXdX Now, it is for 3.
          $endgroup$
          – hamam_Abdallah
          Jan 13 at 19:05




          $begingroup$
          @mXdX Now, it is for 3.
          $endgroup$
          – hamam_Abdallah
          Jan 13 at 19:05












          $begingroup$
          This answer is ~1.281, my textbook says the answer is 1.252. Do you think it's a typo?
          $endgroup$
          – mXdX
          Jan 13 at 19:08




          $begingroup$
          This answer is ~1.281, my textbook says the answer is 1.252. Do you think it's a typo?
          $endgroup$
          – mXdX
          Jan 13 at 19:08


















          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%2f3072371%2fproblems-involving-the-second-taylor-polynomial-of-ex-cos-x%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

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
          Read more

          The Binding of Isaac: Rebirth/Afterbirth

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
          Read more

          What does “Dominus providebit” mean?

          Image
          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
          Read more