Is there a generalized way to find an inverse of a function defined as a sum (if the inverse exists)?












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Suppose I had a summation for a monotonic increasing function, like $f(n)=sum_{k=1}^{n}k$. We already know this has a closed form solution in the form of $frac{n(n+1)}{2}$. If you want to find $n$, you could use the quadratic formula. But, what if you didn't know the closed formula? What if all you knew was some $sum_{k=1}^{n}g(k)$? Is there some way to invert this process to find $n$ like with $g^{-1}(n)$?






summation inverse







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    -1












    $begingroup$


    Suppose I had a summation for a monotonic increasing function, like $f(n)=sum_{k=1}^{n}k$. We already know this has a closed form solution in the form of $frac{n(n+1)}{2}$. If you want to find $n$, you could use the quadratic formula. But, what if you didn't know the closed formula? What if all you knew was some $sum_{k=1}^{n}g(k)$? Is there some way to invert this process to find $n$ like with $g^{-1}(n)$?






    summation inverse







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      Suppose I had a summation for a monotonic increasing function, like $f(n)=sum_{k=1}^{n}k$. We already know this has a closed form solution in the form of $frac{n(n+1)}{2}$. If you want to find $n$, you could use the quadratic formula. But, what if you didn't know the closed formula? What if all you knew was some $sum_{k=1}^{n}g(k)$? Is there some way to invert this process to find $n$ like with $g^{-1}(n)$?






      summation inverse







      share|cite|improve this question











      $endgroup$




      Suppose I had a summation for a monotonic increasing function, like $f(n)=sum_{k=1}^{n}k$. We already know this has a closed form solution in the form of $frac{n(n+1)}{2}$. If you want to find $n$, you could use the quadratic formula. But, what if you didn't know the closed formula? What if all you knew was some $sum_{k=1}^{n}g(k)$? Is there some way to invert this process to find $n$ like with $g^{-1}(n)$?






      summation inverse




      summation inverse






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 2 at 20:58







      user14554

















      asked Jan 2 at 8:13









      user14554user14554

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          $begingroup$

          It's possible to find many inverses of a sum. for example, if a+b=10, then $1+9$, $2+8$, $3+7$, $4+6$, $5+5$ are all solutions to the equation and there's also many negative solutions like $(-1)+11$.



          Good way to think about this, is that given 10 length sequence of successors, you split the sequence to two parts and the cut point is arbitrarily chosen, thus 0+10, 1+9, 2+8 etc are all good working solutions.



          So steps to solve this kind of problems($a+b=c$):




          • take $c$

          • split it to successors $1+1+1+..+1$

          • split the successors to many groups: $(1+1+1)+(1+1+1+1+1)$

          • mark the groups as variables $a+b$


          Now, to get more complexity, we should do $f(x)+f(y)+f(z)=d$, would become
          $$ d mapsto (a+b+c) mapsto (a,b,c) mapsto (f^{-1}(a),f^{-1}(b),f^{-1}(c)) mapsto (x=f^{-1}(a),y=f^{-1}(b),z=f^{-1}(c))$$
          i.e. after finding $a,b,c$ for solutions to $a+b+c=d$, just apply inverse of the function.



          Now if you have something like $f(1)+f(2)+f(3)=d$, then we need to use the previous logic, and add just ${x=1,y=2,z=3}$ to get $(f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            "It's possible to find many inverses of a sum" is a contradiction, isn't it? If I have defined a function, then an inverse must be unique if the function has an inverse. I suppose I could specify that: suppose the function is monotonic and increasing with consecutive indices.
            $endgroup$
            – user14554
            Jan 2 at 20:57








          • 1




            $begingroup$
            It's not a contradiction, but it's just an informal and straightforward way to say that it does not have a uniquely defined inverse function, only a preimage (fiber).
            $endgroup$
            – Niki Di Giano
            Jan 2 at 21:18














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          1 Answer
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          1 Answer
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          oldest

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          active

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          active

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          1












          $begingroup$

          It's possible to find many inverses of a sum. for example, if a+b=10, then $1+9$, $2+8$, $3+7$, $4+6$, $5+5$ are all solutions to the equation and there's also many negative solutions like $(-1)+11$.



          Good way to think about this, is that given 10 length sequence of successors, you split the sequence to two parts and the cut point is arbitrarily chosen, thus 0+10, 1+9, 2+8 etc are all good working solutions.



          So steps to solve this kind of problems($a+b=c$):




          • take $c$

          • split it to successors $1+1+1+..+1$

          • split the successors to many groups: $(1+1+1)+(1+1+1+1+1)$

          • mark the groups as variables $a+b$


          Now, to get more complexity, we should do $f(x)+f(y)+f(z)=d$, would become
          $$ d mapsto (a+b+c) mapsto (a,b,c) mapsto (f^{-1}(a),f^{-1}(b),f^{-1}(c)) mapsto (x=f^{-1}(a),y=f^{-1}(b),z=f^{-1}(c))$$
          i.e. after finding $a,b,c$ for solutions to $a+b+c=d$, just apply inverse of the function.



          Now if you have something like $f(1)+f(2)+f(3)=d$, then we need to use the previous logic, and add just ${x=1,y=2,z=3}$ to get $(f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            "It's possible to find many inverses of a sum" is a contradiction, isn't it? If I have defined a function, then an inverse must be unique if the function has an inverse. I suppose I could specify that: suppose the function is monotonic and increasing with consecutive indices.
            $endgroup$
            – user14554
            Jan 2 at 20:57








          • 1




            $begingroup$
            It's not a contradiction, but it's just an informal and straightforward way to say that it does not have a uniquely defined inverse function, only a preimage (fiber).
            $endgroup$
            – Niki Di Giano
            Jan 2 at 21:18


















          1












          $begingroup$

          It's possible to find many inverses of a sum. for example, if a+b=10, then $1+9$, $2+8$, $3+7$, $4+6$, $5+5$ are all solutions to the equation and there's also many negative solutions like $(-1)+11$.



          Good way to think about this, is that given 10 length sequence of successors, you split the sequence to two parts and the cut point is arbitrarily chosen, thus 0+10, 1+9, 2+8 etc are all good working solutions.



          So steps to solve this kind of problems($a+b=c$):




          • take $c$

          • split it to successors $1+1+1+..+1$

          • split the successors to many groups: $(1+1+1)+(1+1+1+1+1)$

          • mark the groups as variables $a+b$


          Now, to get more complexity, we should do $f(x)+f(y)+f(z)=d$, would become
          $$ d mapsto (a+b+c) mapsto (a,b,c) mapsto (f^{-1}(a),f^{-1}(b),f^{-1}(c)) mapsto (x=f^{-1}(a),y=f^{-1}(b),z=f^{-1}(c))$$
          i.e. after finding $a,b,c$ for solutions to $a+b+c=d$, just apply inverse of the function.



          Now if you have something like $f(1)+f(2)+f(3)=d$, then we need to use the previous logic, and add just ${x=1,y=2,z=3}$ to get $(f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            "It's possible to find many inverses of a sum" is a contradiction, isn't it? If I have defined a function, then an inverse must be unique if the function has an inverse. I suppose I could specify that: suppose the function is monotonic and increasing with consecutive indices.
            $endgroup$
            – user14554
            Jan 2 at 20:57








          • 1




            $begingroup$
            It's not a contradiction, but it's just an informal and straightforward way to say that it does not have a uniquely defined inverse function, only a preimage (fiber).
            $endgroup$
            – Niki Di Giano
            Jan 2 at 21:18
















          1












          1








          1





          $begingroup$

          It's possible to find many inverses of a sum. for example, if a+b=10, then $1+9$, $2+8$, $3+7$, $4+6$, $5+5$ are all solutions to the equation and there's also many negative solutions like $(-1)+11$.



          Good way to think about this, is that given 10 length sequence of successors, you split the sequence to two parts and the cut point is arbitrarily chosen, thus 0+10, 1+9, 2+8 etc are all good working solutions.



          So steps to solve this kind of problems($a+b=c$):




          • take $c$

          • split it to successors $1+1+1+..+1$

          • split the successors to many groups: $(1+1+1)+(1+1+1+1+1)$

          • mark the groups as variables $a+b$


          Now, to get more complexity, we should do $f(x)+f(y)+f(z)=d$, would become
          $$ d mapsto (a+b+c) mapsto (a,b,c) mapsto (f^{-1}(a),f^{-1}(b),f^{-1}(c)) mapsto (x=f^{-1}(a),y=f^{-1}(b),z=f^{-1}(c))$$
          i.e. after finding $a,b,c$ for solutions to $a+b+c=d$, just apply inverse of the function.



          Now if you have something like $f(1)+f(2)+f(3)=d$, then we need to use the previous logic, and add just ${x=1,y=2,z=3}$ to get $(f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3)$.






          share|cite|improve this answer











          $endgroup$



          It's possible to find many inverses of a sum. for example, if a+b=10, then $1+9$, $2+8$, $3+7$, $4+6$, $5+5$ are all solutions to the equation and there's also many negative solutions like $(-1)+11$.



          Good way to think about this, is that given 10 length sequence of successors, you split the sequence to two parts and the cut point is arbitrarily chosen, thus 0+10, 1+9, 2+8 etc are all good working solutions.



          So steps to solve this kind of problems($a+b=c$):




          • take $c$

          • split it to successors $1+1+1+..+1$

          • split the successors to many groups: $(1+1+1)+(1+1+1+1+1)$

          • mark the groups as variables $a+b$


          Now, to get more complexity, we should do $f(x)+f(y)+f(z)=d$, would become
          $$ d mapsto (a+b+c) mapsto (a,b,c) mapsto (f^{-1}(a),f^{-1}(b),f^{-1}(c)) mapsto (x=f^{-1}(a),y=f^{-1}(b),z=f^{-1}(c))$$
          i.e. after finding $a,b,c$ for solutions to $a+b+c=d$, just apply inverse of the function.



          Now if you have something like $f(1)+f(2)+f(3)=d$, then we need to use the previous logic, and add just ${x=1,y=2,z=3}$ to get $(f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 10:27

























          answered Jan 2 at 9:51









          tp1tp1

          34648




          34648












          • $begingroup$
            "It's possible to find many inverses of a sum" is a contradiction, isn't it? If I have defined a function, then an inverse must be unique if the function has an inverse. I suppose I could specify that: suppose the function is monotonic and increasing with consecutive indices.
            $endgroup$
            – user14554
            Jan 2 at 20:57








          • 1




            $begingroup$
            It's not a contradiction, but it's just an informal and straightforward way to say that it does not have a uniquely defined inverse function, only a preimage (fiber).
            $endgroup$
            – Niki Di Giano
            Jan 2 at 21:18




















          • $begingroup$
            "It's possible to find many inverses of a sum" is a contradiction, isn't it? If I have defined a function, then an inverse must be unique if the function has an inverse. I suppose I could specify that: suppose the function is monotonic and increasing with consecutive indices.
            $endgroup$
            – user14554
            Jan 2 at 20:57








          • 1




            $begingroup$
            It's not a contradiction, but it's just an informal and straightforward way to say that it does not have a uniquely defined inverse function, only a preimage (fiber).
            $endgroup$
            – Niki Di Giano
            Jan 2 at 21:18


















          $begingroup$
          "It's possible to find many inverses of a sum" is a contradiction, isn't it? If I have defined a function, then an inverse must be unique if the function has an inverse. I suppose I could specify that: suppose the function is monotonic and increasing with consecutive indices.
          $endgroup$
          – user14554
          Jan 2 at 20:57






          $begingroup$
          "It's possible to find many inverses of a sum" is a contradiction, isn't it? If I have defined a function, then an inverse must be unique if the function has an inverse. I suppose I could specify that: suppose the function is monotonic and increasing with consecutive indices.
          $endgroup$
          – user14554
          Jan 2 at 20:57






          1




          1




          $begingroup$
          It's not a contradiction, but it's just an informal and straightforward way to say that it does not have a uniquely defined inverse function, only a preimage (fiber).
          $endgroup$
          – Niki Di Giano
          Jan 2 at 21:18






          $begingroup$
          It's not a contradiction, but it's just an informal and straightforward way to say that it does not have a uniquely defined inverse function, only a preimage (fiber).
          $endgroup$
          – Niki Di Giano
          Jan 2 at 21:18




















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