Proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$












1














I'm looking for a simple proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$.



My first attempt was to express the exponential as a Taylor series:



$$frac{1+x^2}{n^2} geq frac{x^2}{n^2}-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... , .$$



Obviously



$$frac{1+x^2}{n^2} geq frac{x^2}{n^2},$$



so if I can show



$$-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... <0,$$



then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.






inequality







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    1














    I'm looking for a simple proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$.



    My first attempt was to express the exponential as a Taylor series:



    $$frac{1+x^2}{n^2} geq frac{x^2}{n^2}-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... , .$$



    Obviously



    $$frac{1+x^2}{n^2} geq frac{x^2}{n^2},$$



    so if I can show



    $$-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... <0,$$



    then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.






    inequality







    share|cite|improve this question

























      1












      1








      1







      I'm looking for a simple proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$.



      My first attempt was to express the exponential as a Taylor series:



      $$frac{1+x^2}{n^2} geq frac{x^2}{n^2}-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... , .$$



      Obviously



      $$frac{1+x^2}{n^2} geq frac{x^2}{n^2},$$



      so if I can show



      $$-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... <0,$$



      then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.






      inequality







      share|cite|improve this question













      I'm looking for a simple proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$.



      My first attempt was to express the exponential as a Taylor series:



      $$frac{1+x^2}{n^2} geq frac{x^2}{n^2}-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... , .$$



      Obviously



      $$frac{1+x^2}{n^2} geq frac{x^2}{n^2},$$



      so if I can show



      $$-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... <0,$$



      then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.






      inequality




      inequality






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      share|cite|improve this question











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      asked Nov 29 at 23:14









      WillG

      46038




      46038






















          3 Answers
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          1














          It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :



          $$e^{-x^2/n^2} geq 1 - x^2/n^2$$



          But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :



          $$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$



          $$Leftrightarrow$$



          $$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$






          share|cite|improve this answer































            3














            For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.






            share|cite|improve this answer





















            • +1 by me since we answered at the same time and the answer is as precise !
              – Rebellos
              Nov 29 at 23:37



















            0














            Set $y=x^2/n^2$. Then you want to show that
            $$
            frac{1}{n^2}+yge 1-e^{-y}
            $$
            Note that $yge0$. A standard process is to consider
            $$
            f(y)=frac{1}{n^2}+y-1+e^{-y}
            $$
            and note that $f(0)=1/n^2>0$. Also
            $$
            f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
            $$
            for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
            $$
            f(y)>0
            $$
            for $yge0$.






            share|cite|improve this answer





















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              3 Answers
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              3 Answers
              3






              active

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              active

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              active

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              1














              It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :



              $$e^{-x^2/n^2} geq 1 - x^2/n^2$$



              But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :



              $$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$



              $$Leftrightarrow$$



              $$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$






              share|cite|improve this answer




























                1














                It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :



                $$e^{-x^2/n^2} geq 1 - x^2/n^2$$



                But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :



                $$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$



                $$Leftrightarrow$$



                $$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$






                share|cite|improve this answer


























                  1












                  1








                  1






                  It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :



                  $$e^{-x^2/n^2} geq 1 - x^2/n^2$$



                  But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :



                  $$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$



                  $$Leftrightarrow$$



                  $$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$






                  share|cite|improve this answer














                  It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :



                  $$e^{-x^2/n^2} geq 1 - x^2/n^2$$



                  But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :



                  $$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$



                  $$Leftrightarrow$$



                  $$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 29 at 23:34

























                  answered Nov 29 at 23:29









                  Rebellos

                  14.3k31245




                  14.3k31245























                      3














                      For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.






                      share|cite|improve this answer





















                      • +1 by me since we answered at the same time and the answer is as precise !
                        – Rebellos
                        Nov 29 at 23:37
















                      3














                      For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.






                      share|cite|improve this answer





















                      • +1 by me since we answered at the same time and the answer is as precise !
                        – Rebellos
                        Nov 29 at 23:37














                      3












                      3








                      3






                      For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.






                      share|cite|improve this answer












                      For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 29 at 23:29









                      Kavi Rama Murthy

                      49.3k31854




                      49.3k31854












                      • +1 by me since we answered at the same time and the answer is as precise !
                        – Rebellos
                        Nov 29 at 23:37


















                      • +1 by me since we answered at the same time and the answer is as precise !
                        – Rebellos
                        Nov 29 at 23:37
















                      +1 by me since we answered at the same time and the answer is as precise !
                      – Rebellos
                      Nov 29 at 23:37




                      +1 by me since we answered at the same time and the answer is as precise !
                      – Rebellos
                      Nov 29 at 23:37











                      0














                      Set $y=x^2/n^2$. Then you want to show that
                      $$
                      frac{1}{n^2}+yge 1-e^{-y}
                      $$
                      Note that $yge0$. A standard process is to consider
                      $$
                      f(y)=frac{1}{n^2}+y-1+e^{-y}
                      $$
                      and note that $f(0)=1/n^2>0$. Also
                      $$
                      f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
                      $$
                      for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
                      $$
                      f(y)>0
                      $$
                      for $yge0$.






                      share|cite|improve this answer


























                        0














                        Set $y=x^2/n^2$. Then you want to show that
                        $$
                        frac{1}{n^2}+yge 1-e^{-y}
                        $$
                        Note that $yge0$. A standard process is to consider
                        $$
                        f(y)=frac{1}{n^2}+y-1+e^{-y}
                        $$
                        and note that $f(0)=1/n^2>0$. Also
                        $$
                        f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
                        $$
                        for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
                        $$
                        f(y)>0
                        $$
                        for $yge0$.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          Set $y=x^2/n^2$. Then you want to show that
                          $$
                          frac{1}{n^2}+yge 1-e^{-y}
                          $$
                          Note that $yge0$. A standard process is to consider
                          $$
                          f(y)=frac{1}{n^2}+y-1+e^{-y}
                          $$
                          and note that $f(0)=1/n^2>0$. Also
                          $$
                          f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
                          $$
                          for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
                          $$
                          f(y)>0
                          $$
                          for $yge0$.






                          share|cite|improve this answer












                          Set $y=x^2/n^2$. Then you want to show that
                          $$
                          frac{1}{n^2}+yge 1-e^{-y}
                          $$
                          Note that $yge0$. A standard process is to consider
                          $$
                          f(y)=frac{1}{n^2}+y-1+e^{-y}
                          $$
                          and note that $f(0)=1/n^2>0$. Also
                          $$
                          f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
                          $$
                          for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
                          $$
                          f(y)>0
                          $$
                          for $yge0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 29 at 23:41









                          egreg

                          177k1484200




                          177k1484200






























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