Simplification of $ sqrt{(1-x^2)}$ to $(1-frac{x^2}{2})$












1












$begingroup$


While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:



$$
sqrt{1-x^2} approx left(1-frac{x^2}{2}right)
$$



I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).






calculus approximation







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  • 3




    $begingroup$
    It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
    $endgroup$
    – Galc127
    Dec 16 '18 at 13:52


















1












$begingroup$


While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:



$$
sqrt{1-x^2} approx left(1-frac{x^2}{2}right)
$$



I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).






calculus approximation







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
    $endgroup$
    – Galc127
    Dec 16 '18 at 13:52
















1












1








1





$begingroup$


While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:



$$
sqrt{1-x^2} approx left(1-frac{x^2}{2}right)
$$



I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).






calculus approximation







share|cite|improve this question











$endgroup$




While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:



$$
sqrt{1-x^2} approx left(1-frac{x^2}{2}right)
$$



I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).






calculus approximation




calculus approximation






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share|cite|improve this question













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edited Dec 16 '18 at 14:00









Viktor Glombik

9381527




9381527










asked Dec 16 '18 at 13:48









eosuseosus

82




82








  • 3




    $begingroup$
    It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
    $endgroup$
    – Galc127
    Dec 16 '18 at 13:52
















  • 3




    $begingroup$
    It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
    $endgroup$
    – Galc127
    Dec 16 '18 at 13:52










3




3




$begingroup$
It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
$endgroup$
– Galc127
Dec 16 '18 at 13:52






$begingroup$
It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
$endgroup$
– Galc127
Dec 16 '18 at 13:52












3 Answers
3






active

oldest

votes


















9












$begingroup$

Term to look for: linear approximation



In general, the best linear approximation for a differentiable function near a point $c$ is
$$
f(x) approx f(c) + f'(c);(x-c)
$$
This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.



Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
$$
sqrt{1-x} approx 1 - frac{x}{2}
$$
To get your case, substitute $x^2$ for $x$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
    $endgroup$
    – Botond
    Dec 16 '18 at 14:04



















3












$begingroup$

Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).



$$sqrt{1-t}approx 1-frac t2.$$



Hence for small $x$,



$$sqrt{1-x^2}approx 1-frac{x^2}2.$$



enter image description here



The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)



$$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
and
$$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$



enter image description here






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      9












      $begingroup$

      Term to look for: linear approximation



      In general, the best linear approximation for a differentiable function near a point $c$ is
      $$
      f(x) approx f(c) + f'(c);(x-c)
      $$
      This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.



      Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
      $$
      sqrt{1-x} approx 1 - frac{x}{2}
      $$
      To get your case, substitute $x^2$ for $x$.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
        $endgroup$
        – Botond
        Dec 16 '18 at 14:04
















      9












      $begingroup$

      Term to look for: linear approximation



      In general, the best linear approximation for a differentiable function near a point $c$ is
      $$
      f(x) approx f(c) + f'(c);(x-c)
      $$
      This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.



      Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
      $$
      sqrt{1-x} approx 1 - frac{x}{2}
      $$
      To get your case, substitute $x^2$ for $x$.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
        $endgroup$
        – Botond
        Dec 16 '18 at 14:04














      9












      9








      9





      $begingroup$

      Term to look for: linear approximation



      In general, the best linear approximation for a differentiable function near a point $c$ is
      $$
      f(x) approx f(c) + f'(c);(x-c)
      $$
      This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.



      Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
      $$
      sqrt{1-x} approx 1 - frac{x}{2}
      $$
      To get your case, substitute $x^2$ for $x$.






      share|cite|improve this answer











      $endgroup$



      Term to look for: linear approximation



      In general, the best linear approximation for a differentiable function near a point $c$ is
      $$
      f(x) approx f(c) + f'(c);(x-c)
      $$
      This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.



      Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
      $$
      sqrt{1-x} approx 1 - frac{x}{2}
      $$
      To get your case, substitute $x^2$ for $x$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 16 '18 at 13:58

























      answered Dec 16 '18 at 13:53









      GEdgarGEdgar

      62.5k267171




      62.5k267171








      • 1




        $begingroup$
        I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
        $endgroup$
        – Botond
        Dec 16 '18 at 14:04














      • 1




        $begingroup$
        I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
        $endgroup$
        – Botond
        Dec 16 '18 at 14:04








      1




      1




      $begingroup$
      I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
      $endgroup$
      – Botond
      Dec 16 '18 at 14:04




      $begingroup$
      I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
      $endgroup$
      – Botond
      Dec 16 '18 at 14:04











      3












      $begingroup$

      Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).



      $$sqrt{1-t}approx 1-frac t2.$$



      Hence for small $x$,



      $$sqrt{1-x^2}approx 1-frac{x^2}2.$$



      enter image description here



      The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)



      $$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
      and
      $$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$



      enter image description here






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).



        $$sqrt{1-t}approx 1-frac t2.$$



        Hence for small $x$,



        $$sqrt{1-x^2}approx 1-frac{x^2}2.$$



        enter image description here



        The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)



        $$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
        and
        $$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$



        enter image description here






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).



          $$sqrt{1-t}approx 1-frac t2.$$



          Hence for small $x$,



          $$sqrt{1-x^2}approx 1-frac{x^2}2.$$



          enter image description here



          The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)



          $$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
          and
          $$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$



          enter image description here






          share|cite|improve this answer











          $endgroup$



          Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).



          $$sqrt{1-t}approx 1-frac t2.$$



          Hence for small $x$,



          $$sqrt{1-x^2}approx 1-frac{x^2}2.$$



          enter image description here



          The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)



          $$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
          and
          $$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 16 '18 at 14:10

























          answered Dec 16 '18 at 14:00









          Yves DaoustYves Daoust

          128k675227




          128k675227























              2












              $begingroup$

              Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.






                  share|cite|improve this answer











                  $endgroup$



                  Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 16 '18 at 14:18

























                  answered Dec 16 '18 at 13:59









                  Chris CusterChris Custer

                  13.7k3827




                  13.7k3827






























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