Product of roots of a polynomial decreased from one

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I got a question recently, and have been unable to solve it.




If $1,alpha_1,alpha_2,dots,alpha_n-1$ are the roots of $X^n-1=0$, then find the value of $prod(1-alpha_i)$.




My solution:



The product will expand to $$1-sum_cycalpha_i+sum_cycalpha_ialpha_i+1-dots$$



This means I can use the relations of roots of a polynomial to its coefficients, thus getting this equal to



$1+0+0+dots+0+0pm1$, depending on whether $n$ is even or odd.



But the book gives the answer as $n$.



How does this happen? Please help.




algebra-precalculus




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    up vote
    4
    down vote

    favorite












    I got a question recently, and have been unable to solve it.




    If $1,alpha_1,alpha_2,dots,alpha_n-1$ are the roots of $X^n-1=0$, then find the value of $prod(1-alpha_i)$.




    My solution:



    The product will expand to $$1-sum_cycalpha_i+sum_cycalpha_ialpha_i+1-dots$$



    This means I can use the relations of roots of a polynomial to its coefficients, thus getting this equal to



    $1+0+0+dots+0+0pm1$, depending on whether $n$ is even or odd.



    But the book gives the answer as $n$.



    How does this happen? Please help.




    algebra-precalculus




    share|cite|improve this question





















      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      I got a question recently, and have been unable to solve it.




      If $1,alpha_1,alpha_2,dots,alpha_n-1$ are the roots of $X^n-1=0$, then find the value of $prod(1-alpha_i)$.




      My solution:



      The product will expand to $$1-sum_cycalpha_i+sum_cycalpha_ialpha_i+1-dots$$



      This means I can use the relations of roots of a polynomial to its coefficients, thus getting this equal to



      $1+0+0+dots+0+0pm1$, depending on whether $n$ is even or odd.



      But the book gives the answer as $n$.



      How does this happen? Please help.




      algebra-precalculus




      share|cite|improve this question











      I got a question recently, and have been unable to solve it.




      If $1,alpha_1,alpha_2,dots,alpha_n-1$ are the roots of $X^n-1=0$, then find the value of $prod(1-alpha_i)$.




      My solution:



      The product will expand to $$1-sum_cycalpha_i+sum_cycalpha_ialpha_i+1-dots$$



      This means I can use the relations of roots of a polynomial to its coefficients, thus getting this equal to



      $1+0+0+dots+0+0pm1$, depending on whether $n$ is even or odd.



      But the book gives the answer as $n$.



      How does this happen? Please help.





      algebra-precalculus





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 28 at 8:04









      MalayTheDynamo

      2,052833




      2,052833




















          2 Answers
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          up vote
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          Notice that



          $$prod_i=1^n-1 (X - alpha_i) = fracX^n-1X-1 = X^n-1 + X^n-2 + cdots + X + 1$$



          so for $X = 1$ we get



          $$prod_i=1^n-1 (1 - alpha_i)= 1^n-1 + 1^n-2 + cdots + 1 + 1 = n$$



          Your approach can also be used but notice that $alpha_1, ldots, alpha_n-1$ are roots of $X^n-1 + X^n-2 + cdots + X + 1$, and not $X^n-1$ because the root $1$ is missing.



          So



          $$prod_i=1^n-1(1-alpha_i) = 1-sum_cycalpha_i+sum_cycalpha_ialpha_i+1-cdots = 1 - (-1) + 1 - cdots = n$$






          share|cite|improve this answer























          • What do you mean by $X-alpha_n$? Is it $alpha_i$? And is this true for all $X,alpha$?
            – MalayTheDynamo
            Jul 28 at 8:19











          • @MalayTheDynamo Yes, $alpha_i$ of course.
            – mechanodroid
            Jul 28 at 8:19

















          up vote
          3
          down vote













          The main flaw in your argument is that $sum_i=1^n-1 alpha_i$ is not the sum of the roots of the polynomial: you are missing the root 1. Similarly $sum_i=1^n-1sum_j=1^n-1alpha_ialpha_j neq 0.$



          Here's a hint: given that the $alpha_i$ are roots of $x^n-1$, can you write down a polynomial whose roots are $1-alpha_i$?






          share|cite|improve this answer





















          • Oh. I get it. I missed adding the $1$. Now, that's embarrassing. Thanks!
            – MalayTheDynamo
            Jul 28 at 8:17










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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

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          up vote
          4
          down vote













          Notice that



          $$prod_i=1^n-1 (X - alpha_i) = fracX^n-1X-1 = X^n-1 + X^n-2 + cdots + X + 1$$



          so for $X = 1$ we get



          $$prod_i=1^n-1 (1 - alpha_i)= 1^n-1 + 1^n-2 + cdots + 1 + 1 = n$$



          Your approach can also be used but notice that $alpha_1, ldots, alpha_n-1$ are roots of $X^n-1 + X^n-2 + cdots + X + 1$, and not $X^n-1$ because the root $1$ is missing.



          So



          $$prod_i=1^n-1(1-alpha_i) = 1-sum_cycalpha_i+sum_cycalpha_ialpha_i+1-cdots = 1 - (-1) + 1 - cdots = n$$






          share|cite|improve this answer























          • What do you mean by $X-alpha_n$? Is it $alpha_i$? And is this true for all $X,alpha$?
            – MalayTheDynamo
            Jul 28 at 8:19











          • @MalayTheDynamo Yes, $alpha_i$ of course.
            – mechanodroid
            Jul 28 at 8:19














          up vote
          4
          down vote













          Notice that



          $$prod_i=1^n-1 (X - alpha_i) = fracX^n-1X-1 = X^n-1 + X^n-2 + cdots + X + 1$$



          so for $X = 1$ we get



          $$prod_i=1^n-1 (1 - alpha_i)= 1^n-1 + 1^n-2 + cdots + 1 + 1 = n$$



          Your approach can also be used but notice that $alpha_1, ldots, alpha_n-1$ are roots of $X^n-1 + X^n-2 + cdots + X + 1$, and not $X^n-1$ because the root $1$ is missing.



          So



          $$prod_i=1^n-1(1-alpha_i) = 1-sum_cycalpha_i+sum_cycalpha_ialpha_i+1-cdots = 1 - (-1) + 1 - cdots = n$$






          share|cite|improve this answer























          • What do you mean by $X-alpha_n$? Is it $alpha_i$? And is this true for all $X,alpha$?
            – MalayTheDynamo
            Jul 28 at 8:19











          • @MalayTheDynamo Yes, $alpha_i$ of course.
            – mechanodroid
            Jul 28 at 8:19












          up vote
          4
          down vote










          up vote
          4
          down vote









          Notice that



          $$prod_i=1^n-1 (X - alpha_i) = fracX^n-1X-1 = X^n-1 + X^n-2 + cdots + X + 1$$



          so for $X = 1$ we get



          $$prod_i=1^n-1 (1 - alpha_i)= 1^n-1 + 1^n-2 + cdots + 1 + 1 = n$$



          Your approach can also be used but notice that $alpha_1, ldots, alpha_n-1$ are roots of $X^n-1 + X^n-2 + cdots + X + 1$, and not $X^n-1$ because the root $1$ is missing.



          So



          $$prod_i=1^n-1(1-alpha_i) = 1-sum_cycalpha_i+sum_cycalpha_ialpha_i+1-cdots = 1 - (-1) + 1 - cdots = n$$






          share|cite|improve this answer















          Notice that



          $$prod_i=1^n-1 (X - alpha_i) = fracX^n-1X-1 = X^n-1 + X^n-2 + cdots + X + 1$$



          so for $X = 1$ we get



          $$prod_i=1^n-1 (1 - alpha_i)= 1^n-1 + 1^n-2 + cdots + 1 + 1 = n$$



          Your approach can also be used but notice that $alpha_1, ldots, alpha_n-1$ are roots of $X^n-1 + X^n-2 + cdots + X + 1$, and not $X^n-1$ because the root $1$ is missing.



          So



          $$prod_i=1^n-1(1-alpha_i) = 1-sum_cycalpha_i+sum_cycalpha_ialpha_i+1-cdots = 1 - (-1) + 1 - cdots = n$$







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 28 at 8:19


























          answered Jul 28 at 8:16









          mechanodroid

          22.2k52041




          22.2k52041











          • What do you mean by $X-alpha_n$? Is it $alpha_i$? And is this true for all $X,alpha$?
            – MalayTheDynamo
            Jul 28 at 8:19











          • @MalayTheDynamo Yes, $alpha_i$ of course.
            – mechanodroid
            Jul 28 at 8:19
















          • What do you mean by $X-alpha_n$? Is it $alpha_i$? And is this true for all $X,alpha$?
            – MalayTheDynamo
            Jul 28 at 8:19











          • @MalayTheDynamo Yes, $alpha_i$ of course.
            – mechanodroid
            Jul 28 at 8:19















          What do you mean by $X-alpha_n$? Is it $alpha_i$? And is this true for all $X,alpha$?
          – MalayTheDynamo
          Jul 28 at 8:19





          What do you mean by $X-alpha_n$? Is it $alpha_i$? And is this true for all $X,alpha$?
          – MalayTheDynamo
          Jul 28 at 8:19













          @MalayTheDynamo Yes, $alpha_i$ of course.
          – mechanodroid
          Jul 28 at 8:19




          @MalayTheDynamo Yes, $alpha_i$ of course.
          – mechanodroid
          Jul 28 at 8:19










          up vote
          3
          down vote













          The main flaw in your argument is that $sum_i=1^n-1 alpha_i$ is not the sum of the roots of the polynomial: you are missing the root 1. Similarly $sum_i=1^n-1sum_j=1^n-1alpha_ialpha_j neq 0.$



          Here's a hint: given that the $alpha_i$ are roots of $x^n-1$, can you write down a polynomial whose roots are $1-alpha_i$?






          share|cite|improve this answer





















          • Oh. I get it. I missed adding the $1$. Now, that's embarrassing. Thanks!
            – MalayTheDynamo
            Jul 28 at 8:17














          up vote
          3
          down vote













          The main flaw in your argument is that $sum_i=1^n-1 alpha_i$ is not the sum of the roots of the polynomial: you are missing the root 1. Similarly $sum_i=1^n-1sum_j=1^n-1alpha_ialpha_j neq 0.$



          Here's a hint: given that the $alpha_i$ are roots of $x^n-1$, can you write down a polynomial whose roots are $1-alpha_i$?






          share|cite|improve this answer





















          • Oh. I get it. I missed adding the $1$. Now, that's embarrassing. Thanks!
            – MalayTheDynamo
            Jul 28 at 8:17












          up vote
          3
          down vote










          up vote
          3
          down vote









          The main flaw in your argument is that $sum_i=1^n-1 alpha_i$ is not the sum of the roots of the polynomial: you are missing the root 1. Similarly $sum_i=1^n-1sum_j=1^n-1alpha_ialpha_j neq 0.$



          Here's a hint: given that the $alpha_i$ are roots of $x^n-1$, can you write down a polynomial whose roots are $1-alpha_i$?






          share|cite|improve this answer













          The main flaw in your argument is that $sum_i=1^n-1 alpha_i$ is not the sum of the roots of the polynomial: you are missing the root 1. Similarly $sum_i=1^n-1sum_j=1^n-1alpha_ialpha_j neq 0.$



          Here's a hint: given that the $alpha_i$ are roots of $x^n-1$, can you write down a polynomial whose roots are $1-alpha_i$?







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 28 at 8:15









          user7530

          33.3k558109




          33.3k558109











          • Oh. I get it. I missed adding the $1$. Now, that's embarrassing. Thanks!
            – MalayTheDynamo
            Jul 28 at 8:17
















          • Oh. I get it. I missed adding the $1$. Now, that's embarrassing. Thanks!
            – MalayTheDynamo
            Jul 28 at 8:17















          Oh. I get it. I missed adding the $1$. Now, that's embarrassing. Thanks!
          – MalayTheDynamo
          Jul 28 at 8:17




          Oh. I get it. I missed adding the $1$. Now, that's embarrassing. Thanks!
          – MalayTheDynamo
          Jul 28 at 8:17












           

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