Alternating polynomial of even degree is always greater than 0












1












$begingroup$


Is it true that
$$x^{2k} - x^{2k-1} + x^{2k-2} + .... + x^2 - x geq 0$$
for any $x$ real, and $kgeq 3$ a positive integer?
It seems to be true for $kgeq 3$, but it is not true for k=2. I thought about grouping the terms two by two but I didn't have much success.



Edit: Nevermind it obviously fails for x in $(0,1)$. Ignore the question.



Edit 2: Actually, it seems that
$$x^{2k} - x^{2k-1} + x^{2k-2} + .... + x^2 - x + 1 geq 0$$
for all positive integers $k$. Any ideas on this one?






algebra-precalculus inequality polynomials







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    $begingroup$
    You can delete the question if you wish
    $endgroup$
    – user574848
    Jan 6 at 22:51










  • $begingroup$
    $$xcdot frac{x^{2k}-1}{x+1} $$
    $endgroup$
    – Hagen von Eitzen
    Jan 6 at 22:51


















1












$begingroup$


Is it true that
$$x^{2k} - x^{2k-1} + x^{2k-2} + .... + x^2 - x geq 0$$
for any $x$ real, and $kgeq 3$ a positive integer?
It seems to be true for $kgeq 3$, but it is not true for k=2. I thought about grouping the terms two by two but I didn't have much success.



Edit: Nevermind it obviously fails for x in $(0,1)$. Ignore the question.



Edit 2: Actually, it seems that
$$x^{2k} - x^{2k-1} + x^{2k-2} + .... + x^2 - x + 1 geq 0$$
for all positive integers $k$. Any ideas on this one?






algebra-precalculus inequality polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can delete the question if you wish
    $endgroup$
    – user574848
    Jan 6 at 22:51










  • $begingroup$
    $$xcdot frac{x^{2k}-1}{x+1} $$
    $endgroup$
    – Hagen von Eitzen
    Jan 6 at 22:51
















1












1








1


1



$begingroup$


Is it true that
$$x^{2k} - x^{2k-1} + x^{2k-2} + .... + x^2 - x geq 0$$
for any $x$ real, and $kgeq 3$ a positive integer?
It seems to be true for $kgeq 3$, but it is not true for k=2. I thought about grouping the terms two by two but I didn't have much success.



Edit: Nevermind it obviously fails for x in $(0,1)$. Ignore the question.



Edit 2: Actually, it seems that
$$x^{2k} - x^{2k-1} + x^{2k-2} + .... + x^2 - x + 1 geq 0$$
for all positive integers $k$. Any ideas on this one?






algebra-precalculus inequality polynomials







share|cite|improve this question











$endgroup$




Is it true that
$$x^{2k} - x^{2k-1} + x^{2k-2} + .... + x^2 - x geq 0$$
for any $x$ real, and $kgeq 3$ a positive integer?
It seems to be true for $kgeq 3$, but it is not true for k=2. I thought about grouping the terms two by two but I didn't have much success.



Edit: Nevermind it obviously fails for x in $(0,1)$. Ignore the question.



Edit 2: Actually, it seems that
$$x^{2k} - x^{2k-1} + x^{2k-2} + .... + x^2 - x + 1 geq 0$$
for all positive integers $k$. Any ideas on this one?






algebra-precalculus inequality polynomials




algebra-precalculus inequality polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 22:51







Tanny Sieben

















asked Jan 6 at 22:47









Tanny SiebenTanny Sieben

34318




34318








  • 1




    $begingroup$
    You can delete the question if you wish
    $endgroup$
    – user574848
    Jan 6 at 22:51










  • $begingroup$
    $$xcdot frac{x^{2k}-1}{x+1} $$
    $endgroup$
    – Hagen von Eitzen
    Jan 6 at 22:51
















  • 1




    $begingroup$
    You can delete the question if you wish
    $endgroup$
    – user574848
    Jan 6 at 22:51










  • $begingroup$
    $$xcdot frac{x^{2k}-1}{x+1} $$
    $endgroup$
    – Hagen von Eitzen
    Jan 6 at 22:51










1




1




$begingroup$
You can delete the question if you wish
$endgroup$
– user574848
Jan 6 at 22:51




$begingroup$
You can delete the question if you wish
$endgroup$
– user574848
Jan 6 at 22:51












$begingroup$
$$xcdot frac{x^{2k}-1}{x+1} $$
$endgroup$
– Hagen von Eitzen
Jan 6 at 22:51






$begingroup$
$$xcdot frac{x^{2k}-1}{x+1} $$
$endgroup$
– Hagen von Eitzen
Jan 6 at 22:51












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$begingroup$

Hint: Unless $x=-1$, your expression is equal to$$frac{x^{2k+1}+1}{x+1}.$$






share|cite|improve this answer









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    $begingroup$

    Hint: Unless $x=-1$, your expression is equal to$$frac{x^{2k+1}+1}{x+1}.$$






    share|cite|improve this answer









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      $begingroup$

      Hint: Unless $x=-1$, your expression is equal to$$frac{x^{2k+1}+1}{x+1}.$$






      share|cite|improve this answer









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        $begingroup$

        Hint: Unless $x=-1$, your expression is equal to$$frac{x^{2k+1}+1}{x+1}.$$






        share|cite|improve this answer









        $endgroup$



        Hint: Unless $x=-1$, your expression is equal to$$frac{x^{2k+1}+1}{x+1}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 22:54









        José Carlos SantosJosé Carlos Santos

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        155k22124227






























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