Polynomial with 2 variables and complex coefficients, if $P(k,l)=0,forall k,lin mathbb{N}_0$ then $P=0$












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This is an excercise with 4 parts, I have difficulties with the third part. I will show you the results of the first two parts. I Need help with an inductionproof.



$(i)$Let $P(x,y)$ be a polynomial with the variables $x$ and $y$ and complex coefficients



For $ain mathbb{C}$ there are polynomials $Q_a(x,y)$ and $R_a(x)$ such that



$P(x,y) = (y-a)cdot Q_a(x,y)+R_a(x).$



Hint: Substitute $y = (y-a)+a$



My solution



$P(x,y) = sum_{i=0}^n c_i x^{alpha_i} y^{beta_i} =$



$sum_{i=0}^n c_i x^{alpha_i} ((y-a)+a)^{beta_i} = $



$sum_{i=0}^n c_i x^{alpha_i} (sum_{k=0}^{beta_i}binom{beta_i}{k}(y-a)^ka^{beta_i-k})=$



$sum_{i=0}^n c_i x^{alpha_i} (a^{beta_i}+(y-a)sum_{k=1}^{beta_i}binom{beta_i}{k}(y-a)^{k-1}a^{beta_i-k})=$



$sum_{i=0}^n a^{beta_i}c_i x^{alpha_i}+sum_{i=0}^n (y-a)sum_{k=1}^{beta_i}binom{beta_i}{k}c_i x^{alpha_i}(y-a)^{k-1}a^{beta_i-k})$



$R_a(x)+(y-a)Q_a(x,y)$



$(ii)$ If $P(k,a) = 0 forall kinmathbb{N}_0$ then $R_a$ is the zeropolynom, $R_a = 0$



Explaination: A polynomial is $= 0$, if all coefficients $= 0$



Hint: Use the first result and the fact that a polynomial with one variable and infinite zero-Points mus be the Zero-polynomial.



I did not understand the hint



but if the conditions of the Statement are satisfied then



$0=P(k,a)$,by assumption and then $P(k,a) = R_a(k)+(a-a)Q_a(k,a)= R_a(k)=0$



The last part is my Question,



There is also a hint:



Assume $Pneq 0$ and conclude with $(ii)$ inductively that there are $l_0 in mathbb{N}_0$ and $cinmathbb{C}$ such that $P(x,y) = ccdot Y cdot (y-1)cdot … cdot (y-l_0)$. Lead this to a contradiction.



I don't know how the induction would look like but if I could show the hypothesis then $P(n,l_0+1),ninmathbb{N}$ would not be equal to $0$ if $cneq 0$ which would contradict the Statement.



I think the Problem also might lay in the fact that I have not really solved $(ii)$, any help regarding (ii) and (iii) would be highly appreciated.






algebra-precalculus polynomials







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    $begingroup$


    This is an excercise with 4 parts, I have difficulties with the third part. I will show you the results of the first two parts. I Need help with an inductionproof.



    $(i)$Let $P(x,y)$ be a polynomial with the variables $x$ and $y$ and complex coefficients



    For $ain mathbb{C}$ there are polynomials $Q_a(x,y)$ and $R_a(x)$ such that



    $P(x,y) = (y-a)cdot Q_a(x,y)+R_a(x).$



    Hint: Substitute $y = (y-a)+a$



    My solution



    $P(x,y) = sum_{i=0}^n c_i x^{alpha_i} y^{beta_i} =$



    $sum_{i=0}^n c_i x^{alpha_i} ((y-a)+a)^{beta_i} = $



    $sum_{i=0}^n c_i x^{alpha_i} (sum_{k=0}^{beta_i}binom{beta_i}{k}(y-a)^ka^{beta_i-k})=$



    $sum_{i=0}^n c_i x^{alpha_i} (a^{beta_i}+(y-a)sum_{k=1}^{beta_i}binom{beta_i}{k}(y-a)^{k-1}a^{beta_i-k})=$



    $sum_{i=0}^n a^{beta_i}c_i x^{alpha_i}+sum_{i=0}^n (y-a)sum_{k=1}^{beta_i}binom{beta_i}{k}c_i x^{alpha_i}(y-a)^{k-1}a^{beta_i-k})$



    $R_a(x)+(y-a)Q_a(x,y)$



    $(ii)$ If $P(k,a) = 0 forall kinmathbb{N}_0$ then $R_a$ is the zeropolynom, $R_a = 0$



    Explaination: A polynomial is $= 0$, if all coefficients $= 0$



    Hint: Use the first result and the fact that a polynomial with one variable and infinite zero-Points mus be the Zero-polynomial.



    I did not understand the hint



    but if the conditions of the Statement are satisfied then



    $0=P(k,a)$,by assumption and then $P(k,a) = R_a(k)+(a-a)Q_a(k,a)= R_a(k)=0$



    The last part is my Question,



    There is also a hint:



    Assume $Pneq 0$ and conclude with $(ii)$ inductively that there are $l_0 in mathbb{N}_0$ and $cinmathbb{C}$ such that $P(x,y) = ccdot Y cdot (y-1)cdot … cdot (y-l_0)$. Lead this to a contradiction.



    I don't know how the induction would look like but if I could show the hypothesis then $P(n,l_0+1),ninmathbb{N}$ would not be equal to $0$ if $cneq 0$ which would contradict the Statement.



    I think the Problem also might lay in the fact that I have not really solved $(ii)$, any help regarding (ii) and (iii) would be highly appreciated.






    algebra-precalculus polynomials







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      This is an excercise with 4 parts, I have difficulties with the third part. I will show you the results of the first two parts. I Need help with an inductionproof.



      $(i)$Let $P(x,y)$ be a polynomial with the variables $x$ and $y$ and complex coefficients



      For $ain mathbb{C}$ there are polynomials $Q_a(x,y)$ and $R_a(x)$ such that



      $P(x,y) = (y-a)cdot Q_a(x,y)+R_a(x).$



      Hint: Substitute $y = (y-a)+a$



      My solution



      $P(x,y) = sum_{i=0}^n c_i x^{alpha_i} y^{beta_i} =$



      $sum_{i=0}^n c_i x^{alpha_i} ((y-a)+a)^{beta_i} = $



      $sum_{i=0}^n c_i x^{alpha_i} (sum_{k=0}^{beta_i}binom{beta_i}{k}(y-a)^ka^{beta_i-k})=$



      $sum_{i=0}^n c_i x^{alpha_i} (a^{beta_i}+(y-a)sum_{k=1}^{beta_i}binom{beta_i}{k}(y-a)^{k-1}a^{beta_i-k})=$



      $sum_{i=0}^n a^{beta_i}c_i x^{alpha_i}+sum_{i=0}^n (y-a)sum_{k=1}^{beta_i}binom{beta_i}{k}c_i x^{alpha_i}(y-a)^{k-1}a^{beta_i-k})$



      $R_a(x)+(y-a)Q_a(x,y)$



      $(ii)$ If $P(k,a) = 0 forall kinmathbb{N}_0$ then $R_a$ is the zeropolynom, $R_a = 0$



      Explaination: A polynomial is $= 0$, if all coefficients $= 0$



      Hint: Use the first result and the fact that a polynomial with one variable and infinite zero-Points mus be the Zero-polynomial.



      I did not understand the hint



      but if the conditions of the Statement are satisfied then



      $0=P(k,a)$,by assumption and then $P(k,a) = R_a(k)+(a-a)Q_a(k,a)= R_a(k)=0$



      The last part is my Question,



      There is also a hint:



      Assume $Pneq 0$ and conclude with $(ii)$ inductively that there are $l_0 in mathbb{N}_0$ and $cinmathbb{C}$ such that $P(x,y) = ccdot Y cdot (y-1)cdot … cdot (y-l_0)$. Lead this to a contradiction.



      I don't know how the induction would look like but if I could show the hypothesis then $P(n,l_0+1),ninmathbb{N}$ would not be equal to $0$ if $cneq 0$ which would contradict the Statement.



      I think the Problem also might lay in the fact that I have not really solved $(ii)$, any help regarding (ii) and (iii) would be highly appreciated.






      algebra-precalculus polynomials







      share|cite|improve this question











      $endgroup$




      This is an excercise with 4 parts, I have difficulties with the third part. I will show you the results of the first two parts. I Need help with an inductionproof.



      $(i)$Let $P(x,y)$ be a polynomial with the variables $x$ and $y$ and complex coefficients



      For $ain mathbb{C}$ there are polynomials $Q_a(x,y)$ and $R_a(x)$ such that



      $P(x,y) = (y-a)cdot Q_a(x,y)+R_a(x).$



      Hint: Substitute $y = (y-a)+a$



      My solution



      $P(x,y) = sum_{i=0}^n c_i x^{alpha_i} y^{beta_i} =$



      $sum_{i=0}^n c_i x^{alpha_i} ((y-a)+a)^{beta_i} = $



      $sum_{i=0}^n c_i x^{alpha_i} (sum_{k=0}^{beta_i}binom{beta_i}{k}(y-a)^ka^{beta_i-k})=$



      $sum_{i=0}^n c_i x^{alpha_i} (a^{beta_i}+(y-a)sum_{k=1}^{beta_i}binom{beta_i}{k}(y-a)^{k-1}a^{beta_i-k})=$



      $sum_{i=0}^n a^{beta_i}c_i x^{alpha_i}+sum_{i=0}^n (y-a)sum_{k=1}^{beta_i}binom{beta_i}{k}c_i x^{alpha_i}(y-a)^{k-1}a^{beta_i-k})$



      $R_a(x)+(y-a)Q_a(x,y)$



      $(ii)$ If $P(k,a) = 0 forall kinmathbb{N}_0$ then $R_a$ is the zeropolynom, $R_a = 0$



      Explaination: A polynomial is $= 0$, if all coefficients $= 0$



      Hint: Use the first result and the fact that a polynomial with one variable and infinite zero-Points mus be the Zero-polynomial.



      I did not understand the hint



      but if the conditions of the Statement are satisfied then



      $0=P(k,a)$,by assumption and then $P(k,a) = R_a(k)+(a-a)Q_a(k,a)= R_a(k)=0$



      The last part is my Question,



      There is also a hint:



      Assume $Pneq 0$ and conclude with $(ii)$ inductively that there are $l_0 in mathbb{N}_0$ and $cinmathbb{C}$ such that $P(x,y) = ccdot Y cdot (y-1)cdot … cdot (y-l_0)$. Lead this to a contradiction.



      I don't know how the induction would look like but if I could show the hypothesis then $P(n,l_0+1),ninmathbb{N}$ would not be equal to $0$ if $cneq 0$ which would contradict the Statement.



      I think the Problem also might lay in the fact that I have not really solved $(ii)$, any help regarding (ii) and (iii) would be highly appreciated.






      algebra-precalculus polynomials




      algebra-precalculus polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 12:13







      RM777

















      asked Dec 16 '18 at 12:02









      RM777RM777

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