Sum Challenge 001
Let the divisors of n be A={d1,d2,d3,….,dn} where
d1=1 and dn =n. Let Pd =∑i σ(di), d Є A, d/di.
Remember that σ(x)=sum of the divisors of x.
Prove that, I
∑d/n μ(d)Pd=1.
Sum Challenge
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- Phlembac Adib Hasan
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Re: Sum Challenge
The problem statement is confusing. Please clearly define $P_d$.Is it $P_{d_i} $? Otherwise it seems that $ P_d $ has only one value for every $n$.If you can't use LaTeX, then please explain it.SANZEED wrote:Sum Challenge 001
Let the divisors of n be A={d1,d2,d3,….,dn} where
d1=1 and dn =n. Let Pd =∑i σ(di), d Є A, d/di.
Remember that σ(x)=sum of the divisors of x.
Prove that, I
∑d/n μ(d)Pd=1.
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- nafistiham
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Re: Sum Challenge
i was requested by SANZEED to latex his problem.unfortunately, now i myself don't understand what he wants to ask
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: Sum Challenge
YES,$p_{d}$ means $p_{d_{i}}$.let $d$ be adivisor of n.Then take the value of the divisors of n for $d_{i}$ which are divisible by $d$.
Moderation Note: $L^AT_EX$ed correctly.
Moderation Note: $L^AT_EX$ed correctly.
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- Phlembac Adib Hasan
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Re: Sum Challenge
বড় সংখ্যার জন্য কাজ করতেসে না। ক্যালকুলেশনেই ভুল করলাম কিনা
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Re: Sum Challenge
ADB! you should try vinogradov's theoremPhlembac Adib Hasan wrote:বড় সংখ্যার জন্য কাজ করতেসে না। ক্যালকুলেশনেই ভুল করলাম কিনা
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
Re: Sum Challenge
Have you found errors?But I checked my use of Vinogradov's......now Iam also confused....
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