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

      124k742117




      124k742117










      asked Jan 13 at 18:36









      mXdXmXdX

      1168




      1168






















          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












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          2 Answers
          2






          active

          oldest

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          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

          17k11633




          17k11633












          • $begingroup$
            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

          38.1k21634




          38.1k21634












          • $begingroup$
            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


















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