If a is small, then show that $[1/(1+a)]^3$ is nearly equal to $1-3a$.












1












$begingroup$


If a is small, then show that $dfrac{1}{(1+a)^3}$ is nearly equal to
$(1-3a)$. Also show that if $0<a<0.01$, then the error$<0.0007$.



Can we solve this without using calculus?



If so can anyone help me with the solution?






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What do you mean, without using calculus?
    $endgroup$
    – Martigan
    Jan 16 at 8:10










  • $begingroup$
    Just write the expression for the error and plug in the value of $a$. ...
    $endgroup$
    – Matti P.
    Jan 16 at 8:12










  • $begingroup$
    Is $a$ assumed to be positive?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 16 at 8:13






  • 2




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 8:15










  • $begingroup$
    Wolfram alpha answer
    $endgroup$
    – farruhota
    Jan 16 at 8:20
















1












$begingroup$


If a is small, then show that $dfrac{1}{(1+a)^3}$ is nearly equal to
$(1-3a)$. Also show that if $0<a<0.01$, then the error$<0.0007$.



Can we solve this without using calculus?



If so can anyone help me with the solution?






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What do you mean, without using calculus?
    $endgroup$
    – Martigan
    Jan 16 at 8:10










  • $begingroup$
    Just write the expression for the error and plug in the value of $a$. ...
    $endgroup$
    – Matti P.
    Jan 16 at 8:12










  • $begingroup$
    Is $a$ assumed to be positive?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 16 at 8:13






  • 2




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 8:15










  • $begingroup$
    Wolfram alpha answer
    $endgroup$
    – farruhota
    Jan 16 at 8:20














1












1








1


1



$begingroup$


If a is small, then show that $dfrac{1}{(1+a)^3}$ is nearly equal to
$(1-3a)$. Also show that if $0<a<0.01$, then the error$<0.0007$.



Can we solve this without using calculus?



If so can anyone help me with the solution?






calculus







share|cite|improve this question











$endgroup$




If a is small, then show that $dfrac{1}{(1+a)^3}$ is nearly equal to
$(1-3a)$. Also show that if $0<a<0.01$, then the error$<0.0007$.



Can we solve this without using calculus?



If so can anyone help me with the solution?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 8:42









Jyrki Lahtonen

109k13169372




109k13169372










asked Jan 16 at 8:06









SaeeSaee

387




387








  • 1




    $begingroup$
    What do you mean, without using calculus?
    $endgroup$
    – Martigan
    Jan 16 at 8:10










  • $begingroup$
    Just write the expression for the error and plug in the value of $a$. ...
    $endgroup$
    – Matti P.
    Jan 16 at 8:12










  • $begingroup$
    Is $a$ assumed to be positive?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 16 at 8:13






  • 2




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 8:15










  • $begingroup$
    Wolfram alpha answer
    $endgroup$
    – farruhota
    Jan 16 at 8:20














  • 1




    $begingroup$
    What do you mean, without using calculus?
    $endgroup$
    – Martigan
    Jan 16 at 8:10










  • $begingroup$
    Just write the expression for the error and plug in the value of $a$. ...
    $endgroup$
    – Matti P.
    Jan 16 at 8:12










  • $begingroup$
    Is $a$ assumed to be positive?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 16 at 8:13






  • 2




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 8:15










  • $begingroup$
    Wolfram alpha answer
    $endgroup$
    – farruhota
    Jan 16 at 8:20








1




1




$begingroup$
What do you mean, without using calculus?
$endgroup$
– Martigan
Jan 16 at 8:10




$begingroup$
What do you mean, without using calculus?
$endgroup$
– Martigan
Jan 16 at 8:10












$begingroup$
Just write the expression for the error and plug in the value of $a$. ...
$endgroup$
– Matti P.
Jan 16 at 8:12




$begingroup$
Just write the expression for the error and plug in the value of $a$. ...
$endgroup$
– Matti P.
Jan 16 at 8:12












$begingroup$
Is $a$ assumed to be positive?
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 8:13




$begingroup$
Is $a$ assumed to be positive?
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 8:13




2




2




$begingroup$
en.wikipedia.org/wiki/Binomial_series
$endgroup$
– lab bhattacharjee
Jan 16 at 8:15




$begingroup$
en.wikipedia.org/wiki/Binomial_series
$endgroup$
– lab bhattacharjee
Jan 16 at 8:15












$begingroup$
Wolfram alpha answer
$endgroup$
– farruhota
Jan 16 at 8:20




$begingroup$
Wolfram alpha answer
$endgroup$
– farruhota
Jan 16 at 8:20










3 Answers
3






active

oldest

votes


















1












$begingroup$

We know that $$(1+a)^3(1-a)^3=(1-a^2)^3$$
Therefore $$frac{1}{(1+a)^3}=(1-a)^3/(1-a^2)^3approx (1-3a+3a^2-a^3)approx1-3a$$
Now for small a we can neglect $a^2$ and$a^3$ that's because higher powers of a get very small when compared to 1 or a when a is small
Example $$a=0.001$$ then$$ a^2=0.00001$$ and$$
a^3=0.0000001$$ which are very small . $Q.E.D.$

Just plug in the value you need to calculate error.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Suppose $a$ is small. Then $(1+a)^x approx 1+xa$ holds and is known as the Binomial approximation. Thus $frac{1}{(1+a)^3} = (1+a)^{-3} approx 1+(-3)a = (1-3a)$.



    To find the error, note that as $a rightarrow 0$, $frac{1}{(1+a)^3} rightarrow 1$ and $(1-3a) rightarrow 1$. When $a=0.01$, the error between the two expressions is $|frac{1}{(1+a)^3} - (1-3a)| = |frac{1}{[1+(0.01)]^3} - [1-3(0.01)]| approx (0.9706) - (0.9700) = 0.0006$, which is less than $0.0007$. Noting that the difference in the expressions is monotonically increasing in the interval $0<a<0.01$, we can say that the error is thus less than $0.0007$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      $$
      begin{align}
      frac1{(1+a)^3}-(1-3a)
      &=frac{6+8a+3a^2}{(1+a)^3}cdot a^2\[6pt]
      &leleft{begin{array}{}
      22a^2&text{if }age-frac12\
      6a^2&text{if }age0
      end{array}right.
      end{align}
      $$
      since $frac{6+8a+3a^2}{(1+a)^3}=frac3{1+a}+frac2{(1+a)^2}+frac1{(1+a)^3}$ is decreasing for $agt-1$.



      If $0le ale0.01$, then the error is at most $0.0006$.






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        We know that $$(1+a)^3(1-a)^3=(1-a^2)^3$$
        Therefore $$frac{1}{(1+a)^3}=(1-a)^3/(1-a^2)^3approx (1-3a+3a^2-a^3)approx1-3a$$
        Now for small a we can neglect $a^2$ and$a^3$ that's because higher powers of a get very small when compared to 1 or a when a is small
        Example $$a=0.001$$ then$$ a^2=0.00001$$ and$$
        a^3=0.0000001$$         which are very small . $Q.E.D.$

        Just plug in the value you need to calculate error.






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          We know that $$(1+a)^3(1-a)^3=(1-a^2)^3$$
          Therefore $$frac{1}{(1+a)^3}=(1-a)^3/(1-a^2)^3approx (1-3a+3a^2-a^3)approx1-3a$$
          Now for small a we can neglect $a^2$ and$a^3$ that's because higher powers of a get very small when compared to 1 or a when a is small
          Example $$a=0.001$$ then$$ a^2=0.00001$$ and$$
          a^3=0.0000001$$           which are very small . $Q.E.D.$

          Just plug in the value you need to calculate error.






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            We know that $$(1+a)^3(1-a)^3=(1-a^2)^3$$
            Therefore $$frac{1}{(1+a)^3}=(1-a)^3/(1-a^2)^3approx (1-3a+3a^2-a^3)approx1-3a$$
            Now for small a we can neglect $a^2$ and$a^3$ that's because higher powers of a get very small when compared to 1 or a when a is small
            Example $$a=0.001$$ then$$ a^2=0.00001$$ and$$
            a^3=0.0000001$$             which are very small . $Q.E.D.$

            Just plug in the value you need to calculate error.






            share|cite|improve this answer











            $endgroup$



            We know that $$(1+a)^3(1-a)^3=(1-a^2)^3$$
            Therefore $$frac{1}{(1+a)^3}=(1-a)^3/(1-a^2)^3approx (1-3a+3a^2-a^3)approx1-3a$$
            Now for small a we can neglect $a^2$ and$a^3$ that's because higher powers of a get very small when compared to 1 or a when a is small
            Example $$a=0.001$$ then$$ a^2=0.00001$$ and$$
            a^3=0.0000001$$             which are very small . $Q.E.D.$

            Just plug in the value you need to calculate error.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 18 at 9:26

























            answered Jan 18 at 8:57









            KarthikKarthik

            13012




            13012























                1












                $begingroup$

                Suppose $a$ is small. Then $(1+a)^x approx 1+xa$ holds and is known as the Binomial approximation. Thus $frac{1}{(1+a)^3} = (1+a)^{-3} approx 1+(-3)a = (1-3a)$.



                To find the error, note that as $a rightarrow 0$, $frac{1}{(1+a)^3} rightarrow 1$ and $(1-3a) rightarrow 1$. When $a=0.01$, the error between the two expressions is $|frac{1}{(1+a)^3} - (1-3a)| = |frac{1}{[1+(0.01)]^3} - [1-3(0.01)]| approx (0.9706) - (0.9700) = 0.0006$, which is less than $0.0007$. Noting that the difference in the expressions is monotonically increasing in the interval $0<a<0.01$, we can say that the error is thus less than $0.0007$.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Suppose $a$ is small. Then $(1+a)^x approx 1+xa$ holds and is known as the Binomial approximation. Thus $frac{1}{(1+a)^3} = (1+a)^{-3} approx 1+(-3)a = (1-3a)$.



                  To find the error, note that as $a rightarrow 0$, $frac{1}{(1+a)^3} rightarrow 1$ and $(1-3a) rightarrow 1$. When $a=0.01$, the error between the two expressions is $|frac{1}{(1+a)^3} - (1-3a)| = |frac{1}{[1+(0.01)]^3} - [1-3(0.01)]| approx (0.9706) - (0.9700) = 0.0006$, which is less than $0.0007$. Noting that the difference in the expressions is monotonically increasing in the interval $0<a<0.01$, we can say that the error is thus less than $0.0007$.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Suppose $a$ is small. Then $(1+a)^x approx 1+xa$ holds and is known as the Binomial approximation. Thus $frac{1}{(1+a)^3} = (1+a)^{-3} approx 1+(-3)a = (1-3a)$.



                    To find the error, note that as $a rightarrow 0$, $frac{1}{(1+a)^3} rightarrow 1$ and $(1-3a) rightarrow 1$. When $a=0.01$, the error between the two expressions is $|frac{1}{(1+a)^3} - (1-3a)| = |frac{1}{[1+(0.01)]^3} - [1-3(0.01)]| approx (0.9706) - (0.9700) = 0.0006$, which is less than $0.0007$. Noting that the difference in the expressions is monotonically increasing in the interval $0<a<0.01$, we can say that the error is thus less than $0.0007$.






                    share|cite|improve this answer









                    $endgroup$



                    Suppose $a$ is small. Then $(1+a)^x approx 1+xa$ holds and is known as the Binomial approximation. Thus $frac{1}{(1+a)^3} = (1+a)^{-3} approx 1+(-3)a = (1-3a)$.



                    To find the error, note that as $a rightarrow 0$, $frac{1}{(1+a)^3} rightarrow 1$ and $(1-3a) rightarrow 1$. When $a=0.01$, the error between the two expressions is $|frac{1}{(1+a)^3} - (1-3a)| = |frac{1}{[1+(0.01)]^3} - [1-3(0.01)]| approx (0.9706) - (0.9700) = 0.0006$, which is less than $0.0007$. Noting that the difference in the expressions is monotonically increasing in the interval $0<a<0.01$, we can say that the error is thus less than $0.0007$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 18 at 9:12









                    DohlemanDohleman

                    395112




                    395112























                        1












                        $begingroup$

                        $$
                        begin{align}
                        frac1{(1+a)^3}-(1-3a)
                        &=frac{6+8a+3a^2}{(1+a)^3}cdot a^2\[6pt]
                        &leleft{begin{array}{}
                        22a^2&text{if }age-frac12\
                        6a^2&text{if }age0
                        end{array}right.
                        end{align}
                        $$
                        since $frac{6+8a+3a^2}{(1+a)^3}=frac3{1+a}+frac2{(1+a)^2}+frac1{(1+a)^3}$ is decreasing for $agt-1$.



                        If $0le ale0.01$, then the error is at most $0.0006$.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          $$
                          begin{align}
                          frac1{(1+a)^3}-(1-3a)
                          &=frac{6+8a+3a^2}{(1+a)^3}cdot a^2\[6pt]
                          &leleft{begin{array}{}
                          22a^2&text{if }age-frac12\
                          6a^2&text{if }age0
                          end{array}right.
                          end{align}
                          $$
                          since $frac{6+8a+3a^2}{(1+a)^3}=frac3{1+a}+frac2{(1+a)^2}+frac1{(1+a)^3}$ is decreasing for $agt-1$.



                          If $0le ale0.01$, then the error is at most $0.0006$.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            $$
                            begin{align}
                            frac1{(1+a)^3}-(1-3a)
                            &=frac{6+8a+3a^2}{(1+a)^3}cdot a^2\[6pt]
                            &leleft{begin{array}{}
                            22a^2&text{if }age-frac12\
                            6a^2&text{if }age0
                            end{array}right.
                            end{align}
                            $$
                            since $frac{6+8a+3a^2}{(1+a)^3}=frac3{1+a}+frac2{(1+a)^2}+frac1{(1+a)^3}$ is decreasing for $agt-1$.



                            If $0le ale0.01$, then the error is at most $0.0006$.






                            share|cite|improve this answer











                            $endgroup$



                            $$
                            begin{align}
                            frac1{(1+a)^3}-(1-3a)
                            &=frac{6+8a+3a^2}{(1+a)^3}cdot a^2\[6pt]
                            &leleft{begin{array}{}
                            22a^2&text{if }age-frac12\
                            6a^2&text{if }age0
                            end{array}right.
                            end{align}
                            $$
                            since $frac{6+8a+3a^2}{(1+a)^3}=frac3{1+a}+frac2{(1+a)^2}+frac1{(1+a)^3}$ is decreasing for $agt-1$.



                            If $0le ale0.01$, then the error is at most $0.0006$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 18 at 11:48

























                            answered Jan 18 at 11:37









                            robjohn♦robjohn

                            267k27308631




                            267k27308631






























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