Fast way to manually compute exponents (interest, probability)?












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When doing financial, actuarial or probability calculations, often a person is faced with computing exponentiations.



For example, let's say the chance of a particular ship sinking is 1% per year. Then to compute the chance of the ship sinking over the course of a 30 year lifetime is 1 - 99% × 99% ... × 99% (30 times) or $1-0.99^{30}$. So, we have the problem of computing $frac1{99}$ raised to the 30th power. Is there a fast way to do this kind of calculation manually in the head or using a pencil and paper?






computational-mathematics







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    2












    $begingroup$


    When doing financial, actuarial or probability calculations, often a person is faced with computing exponentiations.



    For example, let's say the chance of a particular ship sinking is 1% per year. Then to compute the chance of the ship sinking over the course of a 30 year lifetime is 1 - 99% × 99% ... × 99% (30 times) or $1-0.99^{30}$. So, we have the problem of computing $frac1{99}$ raised to the 30th power. Is there a fast way to do this kind of calculation manually in the head or using a pencil and paper?






    computational-mathematics







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      When doing financial, actuarial or probability calculations, often a person is faced with computing exponentiations.



      For example, let's say the chance of a particular ship sinking is 1% per year. Then to compute the chance of the ship sinking over the course of a 30 year lifetime is 1 - 99% × 99% ... × 99% (30 times) or $1-0.99^{30}$. So, we have the problem of computing $frac1{99}$ raised to the 30th power. Is there a fast way to do this kind of calculation manually in the head or using a pencil and paper?






      computational-mathematics







      share|cite|improve this question











      $endgroup$




      When doing financial, actuarial or probability calculations, often a person is faced with computing exponentiations.



      For example, let's say the chance of a particular ship sinking is 1% per year. Then to compute the chance of the ship sinking over the course of a 30 year lifetime is 1 - 99% × 99% ... × 99% (30 times) or $1-0.99^{30}$. So, we have the problem of computing $frac1{99}$ raised to the 30th power. Is there a fast way to do this kind of calculation manually in the head or using a pencil and paper?






      computational-mathematics




      computational-mathematics






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 15 '18 at 15:58









      MJD

      47.5k29215396




      47.5k29215396










      asked Dec 15 '18 at 15:08









      Tyler DurdenTyler Durden

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          $begingroup$

          I am not sure about anything precise, but I can give the following approximate rule which turns out to be rather useful: if $n geq 10$, $(1-frac{1}{n})^n approx frac{1}{e} approx 0.37$.



          So in your example, you want to approximate now $0.37^{3/10} approx 0.37^{1/3}$.
          Now, $0.7^3 =0.343$, $0.8^3=0.512$ so, since the cube function grows faster as the argument increases (eg: a linear function has constant growth, so if the cube was linear, you would approximate $0.37 approx 0.715^3$ but it is convex, albeit not too strongly), you are likely to have some sort of $0.73^3 approx 0.37$.



          So your quantity can be guessed at $0.73$. The actual value is $sim 0.739$...



          Of course, the hundredth-precision is a stroke of luck, but the tenth-precision is not.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You've left a couple of dollar signs out near the beginning.
            $endgroup$
            – timtfj
            Dec 15 '18 at 16:11










          • $begingroup$
            Thanks, corrected!
            $endgroup$
            – Mindlack
            Dec 15 '18 at 16:12










          • $begingroup$
            Maybe it's also worth mentioning that the $0.37$ is an approximation to $frac1e$.
            $endgroup$
            – timtfj
            Dec 15 '18 at 17:09










          • $begingroup$
            You are right, I added it.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 17:11










          • $begingroup$
            I tried something a little different. Arriving at $log xapprox -frac3{10}$, which is almost the same as where you got, we now need $e^{-3/10}$. This is approximately $$1-frac3{10}+frac9{200}-frac{27}{6000}.$$ Doing just the first three terms is not too hard and gets us $xapprox frac{149}{200}=7.45$, which is not too far off. We are lucky with the next term, which is equal to $frac9{2000}$, and has an easy denominator. So it is not too hard to include that and get $7.415$, which is quite close.
            $endgroup$
            – MJD
            Dec 16 '18 at 15:28













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          1 Answer
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          $begingroup$

          I am not sure about anything precise, but I can give the following approximate rule which turns out to be rather useful: if $n geq 10$, $(1-frac{1}{n})^n approx frac{1}{e} approx 0.37$.



          So in your example, you want to approximate now $0.37^{3/10} approx 0.37^{1/3}$.
          Now, $0.7^3 =0.343$, $0.8^3=0.512$ so, since the cube function grows faster as the argument increases (eg: a linear function has constant growth, so if the cube was linear, you would approximate $0.37 approx 0.715^3$ but it is convex, albeit not too strongly), you are likely to have some sort of $0.73^3 approx 0.37$.



          So your quantity can be guessed at $0.73$. The actual value is $sim 0.739$...



          Of course, the hundredth-precision is a stroke of luck, but the tenth-precision is not.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You've left a couple of dollar signs out near the beginning.
            $endgroup$
            – timtfj
            Dec 15 '18 at 16:11










          • $begingroup$
            Thanks, corrected!
            $endgroup$
            – Mindlack
            Dec 15 '18 at 16:12










          • $begingroup$
            Maybe it's also worth mentioning that the $0.37$ is an approximation to $frac1e$.
            $endgroup$
            – timtfj
            Dec 15 '18 at 17:09










          • $begingroup$
            You are right, I added it.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 17:11










          • $begingroup$
            I tried something a little different. Arriving at $log xapprox -frac3{10}$, which is almost the same as where you got, we now need $e^{-3/10}$. This is approximately $$1-frac3{10}+frac9{200}-frac{27}{6000}.$$ Doing just the first three terms is not too hard and gets us $xapprox frac{149}{200}=7.45$, which is not too far off. We are lucky with the next term, which is equal to $frac9{2000}$, and has an easy denominator. So it is not too hard to include that and get $7.415$, which is quite close.
            $endgroup$
            – MJD
            Dec 16 '18 at 15:28


















          2












          $begingroup$

          I am not sure about anything precise, but I can give the following approximate rule which turns out to be rather useful: if $n geq 10$, $(1-frac{1}{n})^n approx frac{1}{e} approx 0.37$.



          So in your example, you want to approximate now $0.37^{3/10} approx 0.37^{1/3}$.
          Now, $0.7^3 =0.343$, $0.8^3=0.512$ so, since the cube function grows faster as the argument increases (eg: a linear function has constant growth, so if the cube was linear, you would approximate $0.37 approx 0.715^3$ but it is convex, albeit not too strongly), you are likely to have some sort of $0.73^3 approx 0.37$.



          So your quantity can be guessed at $0.73$. The actual value is $sim 0.739$...



          Of course, the hundredth-precision is a stroke of luck, but the tenth-precision is not.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You've left a couple of dollar signs out near the beginning.
            $endgroup$
            – timtfj
            Dec 15 '18 at 16:11










          • $begingroup$
            Thanks, corrected!
            $endgroup$
            – Mindlack
            Dec 15 '18 at 16:12










          • $begingroup$
            Maybe it's also worth mentioning that the $0.37$ is an approximation to $frac1e$.
            $endgroup$
            – timtfj
            Dec 15 '18 at 17:09










          • $begingroup$
            You are right, I added it.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 17:11










          • $begingroup$
            I tried something a little different. Arriving at $log xapprox -frac3{10}$, which is almost the same as where you got, we now need $e^{-3/10}$. This is approximately $$1-frac3{10}+frac9{200}-frac{27}{6000}.$$ Doing just the first three terms is not too hard and gets us $xapprox frac{149}{200}=7.45$, which is not too far off. We are lucky with the next term, which is equal to $frac9{2000}$, and has an easy denominator. So it is not too hard to include that and get $7.415$, which is quite close.
            $endgroup$
            – MJD
            Dec 16 '18 at 15:28
















          2












          2








          2





          $begingroup$

          I am not sure about anything precise, but I can give the following approximate rule which turns out to be rather useful: if $n geq 10$, $(1-frac{1}{n})^n approx frac{1}{e} approx 0.37$.



          So in your example, you want to approximate now $0.37^{3/10} approx 0.37^{1/3}$.
          Now, $0.7^3 =0.343$, $0.8^3=0.512$ so, since the cube function grows faster as the argument increases (eg: a linear function has constant growth, so if the cube was linear, you would approximate $0.37 approx 0.715^3$ but it is convex, albeit not too strongly), you are likely to have some sort of $0.73^3 approx 0.37$.



          So your quantity can be guessed at $0.73$. The actual value is $sim 0.739$...



          Of course, the hundredth-precision is a stroke of luck, but the tenth-precision is not.






          share|cite|improve this answer











          $endgroup$



          I am not sure about anything precise, but I can give the following approximate rule which turns out to be rather useful: if $n geq 10$, $(1-frac{1}{n})^n approx frac{1}{e} approx 0.37$.



          So in your example, you want to approximate now $0.37^{3/10} approx 0.37^{1/3}$.
          Now, $0.7^3 =0.343$, $0.8^3=0.512$ so, since the cube function grows faster as the argument increases (eg: a linear function has constant growth, so if the cube was linear, you would approximate $0.37 approx 0.715^3$ but it is convex, albeit not too strongly), you are likely to have some sort of $0.73^3 approx 0.37$.



          So your quantity can be guessed at $0.73$. The actual value is $sim 0.739$...



          Of course, the hundredth-precision is a stroke of luck, but the tenth-precision is not.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 '18 at 17:10

























          answered Dec 15 '18 at 15:48









          MindlackMindlack

          4,830210




          4,830210












          • $begingroup$
            You've left a couple of dollar signs out near the beginning.
            $endgroup$
            – timtfj
            Dec 15 '18 at 16:11










          • $begingroup$
            Thanks, corrected!
            $endgroup$
            – Mindlack
            Dec 15 '18 at 16:12










          • $begingroup$
            Maybe it's also worth mentioning that the $0.37$ is an approximation to $frac1e$.
            $endgroup$
            – timtfj
            Dec 15 '18 at 17:09










          • $begingroup$
            You are right, I added it.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 17:11










          • $begingroup$
            I tried something a little different. Arriving at $log xapprox -frac3{10}$, which is almost the same as where you got, we now need $e^{-3/10}$. This is approximately $$1-frac3{10}+frac9{200}-frac{27}{6000}.$$ Doing just the first three terms is not too hard and gets us $xapprox frac{149}{200}=7.45$, which is not too far off. We are lucky with the next term, which is equal to $frac9{2000}$, and has an easy denominator. So it is not too hard to include that and get $7.415$, which is quite close.
            $endgroup$
            – MJD
            Dec 16 '18 at 15:28




















          • $begingroup$
            You've left a couple of dollar signs out near the beginning.
            $endgroup$
            – timtfj
            Dec 15 '18 at 16:11










          • $begingroup$
            Thanks, corrected!
            $endgroup$
            – Mindlack
            Dec 15 '18 at 16:12










          • $begingroup$
            Maybe it's also worth mentioning that the $0.37$ is an approximation to $frac1e$.
            $endgroup$
            – timtfj
            Dec 15 '18 at 17:09










          • $begingroup$
            You are right, I added it.
            $endgroup$
            – Mindlack
            Dec 15 '18 at 17:11










          • $begingroup$
            I tried something a little different. Arriving at $log xapprox -frac3{10}$, which is almost the same as where you got, we now need $e^{-3/10}$. This is approximately $$1-frac3{10}+frac9{200}-frac{27}{6000}.$$ Doing just the first three terms is not too hard and gets us $xapprox frac{149}{200}=7.45$, which is not too far off. We are lucky with the next term, which is equal to $frac9{2000}$, and has an easy denominator. So it is not too hard to include that and get $7.415$, which is quite close.
            $endgroup$
            – MJD
            Dec 16 '18 at 15:28


















          $begingroup$
          You've left a couple of dollar signs out near the beginning.
          $endgroup$
          – timtfj
          Dec 15 '18 at 16:11




          $begingroup$
          You've left a couple of dollar signs out near the beginning.
          $endgroup$
          – timtfj
          Dec 15 '18 at 16:11












          $begingroup$
          Thanks, corrected!
          $endgroup$
          – Mindlack
          Dec 15 '18 at 16:12




          $begingroup$
          Thanks, corrected!
          $endgroup$
          – Mindlack
          Dec 15 '18 at 16:12












          $begingroup$
          Maybe it's also worth mentioning that the $0.37$ is an approximation to $frac1e$.
          $endgroup$
          – timtfj
          Dec 15 '18 at 17:09




          $begingroup$
          Maybe it's also worth mentioning that the $0.37$ is an approximation to $frac1e$.
          $endgroup$
          – timtfj
          Dec 15 '18 at 17:09












          $begingroup$
          You are right, I added it.
          $endgroup$
          – Mindlack
          Dec 15 '18 at 17:11




          $begingroup$
          You are right, I added it.
          $endgroup$
          – Mindlack
          Dec 15 '18 at 17:11












          $begingroup$
          I tried something a little different. Arriving at $log xapprox -frac3{10}$, which is almost the same as where you got, we now need $e^{-3/10}$. This is approximately $$1-frac3{10}+frac9{200}-frac{27}{6000}.$$ Doing just the first three terms is not too hard and gets us $xapprox frac{149}{200}=7.45$, which is not too far off. We are lucky with the next term, which is equal to $frac9{2000}$, and has an easy denominator. So it is not too hard to include that and get $7.415$, which is quite close.
          $endgroup$
          – MJD
          Dec 16 '18 at 15:28






          $begingroup$
          I tried something a little different. Arriving at $log xapprox -frac3{10}$, which is almost the same as where you got, we now need $e^{-3/10}$. This is approximately $$1-frac3{10}+frac9{200}-frac{27}{6000}.$$ Doing just the first three terms is not too hard and gets us $xapprox frac{149}{200}=7.45$, which is not too far off. We are lucky with the next term, which is equal to $frac9{2000}$, and has an easy denominator. So it is not too hard to include that and get $7.415$, which is quite close.
          $endgroup$
          – MJD
          Dec 16 '18 at 15:28




















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