Mixing
Tricky problem of infinite harmonic sum of polynomials
2
$begingroup$
Let A be the set of all polynomials f with positive integral coefficients such that $f(n)|(2^n-1)$ for all $Ainmathbb{N}$. Then $$sum_{fin A} {1over f(2019)}=?$$
I tried to define a polynomial, but where is the degree? Hence I considered all polynomials that leave remainder $1$ mod $f.$ Hence $2^n-1$ is divisible. But actually I cannot guess how to proceed! Please help!
polynomials
edited Jan 20 at 16:01
Lee David Chung Lin
4,37031241
$endgroup$
|
show 3 more comments
2
$begingroup$
Let A be the set of all polynomials f with positive integral coefficients such that $f(n)|(2^n-1)$ for all $Ainmathbb{N}$. Then $$sum_{fin A} {1over f(2019)}=?$$
I tried to define a polynomial, but where is the degree? Hence I considered all polynomials that leave remainder $1$ mod $f.$ Hence $2^n-1$ is divisible. But actually I cannot guess how to proceed! Please help!
polynomials
edited Jan 20 at 16:01
Lee David Chung Lin
4,37031241
$endgroup$
$begingroup$
What is the source of all the questions you are asking?
$endgroup$
– lulu
Jan 20 at 15:36
$begingroup$
My unsolved problems in todays exam
$endgroup$
– user636268
Jan 20 at 15:36
$begingroup$
Do you mean ... Let $A$ be the set of all polynomials $f$ with positive integral coefficients such that $f(n)|(2^n-1)$ for all $n in mathbb{N}$. Then $$sum_{fin A} frac{1}{f(2019)}=text{?}$$ (Note that, along with formatting, I replaced "for all $Ainmathbb{N}$" with "for all $ninmathbb{N}$".)
$endgroup$
– Blue
Jan 20 at 15:45
$begingroup$
Yes! But i could not type it so replaced it
$endgroup$
– user636268
Jan 20 at 15:51
1
$begingroup$
Let us start by trying to understand the set $A$ better. According to the conditions on $f$ we must have $f(1) = 1$ (take $n=1$ and use that $f$ has positive coefficents). But since $f$ has positive, integral coefficients, this seems to indicate that $f$ must have the form $f(x) = x^k$. But in order for an $f$ on this form to meet the condition we would have to have that $k$ should divide $2^k -1$ for all $k$, which is clearly not true. So I suspect $A$ is empty, hence the sum is 0.
$endgroup$
– Testcase
Jan 20 at 17:29
|
show 3 more comments
2
2
2
$begingroup$
Let A be the set of all polynomials f with positive integral coefficients such that $f(n)|(2^n-1)$ for all $Ainmathbb{N}$. Then $$sum_{fin A} {1over f(2019)}=?$$
I tried to define a polynomial, but where is the degree? Hence I considered all polynomials that leave remainder $1$ mod $f.$ Hence $2^n-1$ is divisible. But actually I cannot guess how to proceed! Please help!
polynomials
edited Jan 20 at 16:01
Lee David Chung Lin
4,37031241
$endgroup$
Let A be the set of all polynomials f with positive integral coefficients such that $f(n)|(2^n-1)$ for all $Ainmathbb{N}$. Then $$sum_{fin A} {1over f(2019)}=?$$
I tried to define a polynomial, but where is the degree? Hence I considered all polynomials that leave remainder $1$ mod $f.$ Hence $2^n-1$ is divisible. But actually I cannot guess how to proceed! Please help!
polynomials
polynomials
edited Jan 20 at 16:01
Lee David Chung Lin
4,37031241
edited Jan 20 at 16:01
Lee David Chung Lin
4,37031241
edited Jan 20 at 16:01
Lee David Chung Lin
4,37031241
edited Jan 20 at 16:01
Lee David Chung Lin
4,37031241
edited Jan 20 at 16:01
Lee David Chung Lin
4,37031241
4,37031241
asked Jan 20 at 15:34
user636268
$begingroup$
What is the source of all the questions you are asking?
$endgroup$
– lulu
Jan 20 at 15:36
$begingroup$
My unsolved problems in todays exam
$endgroup$
– user636268
Jan 20 at 15:36
$begingroup$
Do you mean ... Let $A$ be the set of all polynomials $f$ with positive integral coefficients such that $f(n)|(2^n-1)$ for all $n in mathbb{N}$. Then $$sum_{fin A} frac{1}{f(2019)}=text{?}$$ (Note that, along with formatting, I replaced "for all $Ainmathbb{N}$" with "for all $ninmathbb{N}$".)
$endgroup$
– Blue
Jan 20 at 15:45
$begingroup$
Yes! But i could not type it so replaced it
$endgroup$
– user636268
Jan 20 at 15:51
1
$begingroup$
Let us start by trying to understand the set $A$ better. According to the conditions on $f$ we must have $f(1) = 1$ (take $n=1$ and use that $f$ has positive coefficents). But since $f$ has positive, integral coefficients, this seems to indicate that $f$ must have the form $f(x) = x^k$. But in order for an $f$ on this form to meet the condition we would have to have that $k$ should divide $2^k -1$ for all $k$, which is clearly not true. So I suspect $A$ is empty, hence the sum is 0.
$endgroup$
– Testcase
Jan 20 at 17:29
|
show 3 more comments
$begingroup$
What is the source of all the questions you are asking?
$endgroup$
– lulu
Jan 20 at 15:36
$begingroup$
My unsolved problems in todays exam
$endgroup$
– user636268
Jan 20 at 15:36
$begingroup$
Do you mean ... Let $A$ be the set of all polynomials $f$ with positive integral coefficients such that $f(n)|(2^n-1)$ for all $n in mathbb{N}$. Then $$sum_{fin A} frac{1}{f(2019)}=text{?}$$ (Note that, along with formatting, I replaced "for all $Ainmathbb{N}$" with "for all $ninmathbb{N}$".)
$endgroup$
– Blue
Jan 20 at 15:45
$begingroup$
Yes! But i could not type it so replaced it
$endgroup$
– user636268
Jan 20 at 15:51
1
$begingroup$
Let us start by trying to understand the set $A$ better. According to the conditions on $f$ we must have $f(1) = 1$ (take $n=1$ and use that $f$ has positive coefficents). But since $f$ has positive, integral coefficients, this seems to indicate that $f$ must have the form $f(x) = x^k$. But in order for an $f$ on this form to meet the condition we would have to have that $k$ should divide $2^k -1$ for all $k$, which is clearly not true. So I suspect $A$ is empty, hence the sum is 0.
$endgroup$
– Testcase
Jan 20 at 17:29
$begingroup$
What is the source of all the questions you are asking?
$endgroup$
– lulu
Jan 20 at 15:36
$begingroup$
What is the source of all the questions you are asking?
$endgroup$
– lulu
Jan 20 at 15:36
$begingroup$
My unsolved problems in todays exam
$endgroup$
– user636268
Jan 20 at 15:36
$begingroup$
My unsolved problems in todays exam
$endgroup$
– user636268
Jan 20 at 15:36
$begingroup$
Do you mean ... Let $A$ be the set of all polynomials $f$ with positive integral coefficients such that $f(n)|(2^n-1)$ for all $n in mathbb{N}$. Then $$sum_{fin A} frac{1}{f(2019)}=text{?}$$ (Note that, along with formatting, I replaced "for all $Ainmathbb{N}$" with "for all $ninmathbb{N}$".)
$endgroup$
– Blue
Jan 20 at 15:45
$begingroup$
Do you mean ... Let $A$ be the set of all polynomials $f$ with positive integral coefficients such that $f(n)|(2^n-1)$ for all $n in mathbb{N}$. Then $$sum_{fin A} frac{1}{f(2019)}=text{?}$$ (Note that, along with formatting, I replaced "for all $Ainmathbb{N}$" with "for all $ninmathbb{N}$".)
$endgroup$
– Blue
Jan 20 at 15:45
$begingroup$
Yes! But i could not type it so replaced it
$endgroup$
– user636268
Jan 20 at 15:51
$begingroup$
Yes! But i could not type it so replaced it
$endgroup$
– user636268
Jan 20 at 15:51
1
1
$begingroup$
Let us start by trying to understand the set $A$ better. According to the conditions on $f$ we must have $f(1) = 1$ (take $n=1$ and use that $f$ has positive coefficents). But since $f$ has positive, integral coefficients, this seems to indicate that $f$ must have the form $f(x) = x^k$. But in order for an $f$ on this form to meet the condition we would have to have that $k$ should divide $2^k -1$ for all $k$, which is clearly not true. So I suspect $A$ is empty, hence the sum is 0.
$endgroup$
– Testcase
Jan 20 at 17:29
$begingroup$
Let us start by trying to understand the set $A$ better. According to the conditions on $f$ we must have $f(1) = 1$ (take $n=1$ and use that $f$ has positive coefficents). But since $f$ has positive, integral coefficients, this seems to indicate that $f$ must have the form $f(x) = x^k$. But in order for an $f$ on this form to meet the condition we would have to have that $k$ should divide $2^k -1$ for all $k$, which is clearly not true. So I suspect $A$ is empty, hence the sum is 0.
$endgroup$
– Testcase
Jan 20 at 17:29
|
show 3 more comments
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$begingroup$
What is the source of all the questions you are asking?
$endgroup$
– lulu
Jan 20 at 15:36
$begingroup$
My unsolved problems in todays exam
$endgroup$
– user636268
Jan 20 at 15:36
$begingroup$
Do you mean ... Let $A$ be the set of all polynomials $f$ with positive integral coefficients such that $f(n)|(2^n-1)$ for all $n in mathbb{N}$. Then $$sum_{fin A} frac{1}{f(2019)}=text{?}$$ (Note that, along with formatting, I replaced "for all $Ainmathbb{N}$" with "for all $ninmathbb{N}$".)
$endgroup$
– Blue
Jan 20 at 15:45
$begingroup$
Yes! But i could not type it so replaced it
$endgroup$
– user636268
Jan 20 at 15:51
1
$begingroup$
Let us start by trying to understand the set $A$ better. According to the conditions on $f$ we must have $f(1) = 1$ (take $n=1$ and use that $f$ has positive coefficents). But since $f$ has positive, integral coefficients, this seems to indicate that $f$ must have the form $f(x) = x^k$. But in order for an $f$ on this form to meet the condition we would have to have that $k$ should divide $2^k -1$ for all $k$, which is clearly not true. So I suspect $A$ is empty, hence the sum is 0.
$endgroup$
– Testcase
Jan 20 at 17:29