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Nov
17
comment Functional Equation. $f(mn)=f(m)f(n)$ and …
From your third line. Condition 2) is an inequality, not an equality.
Nov
17
awarded  Editor
Nov
17
revised Functional Equation. $f(mn)=f(m)f(n)$ and …
edited title
Nov
17
awarded  Student
Nov
17
comment What is the integer part of $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} +\cdots + \frac{1}{\sqrt{(2n+1)^2}}$
I meant $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{(2n+1)^2 - 2}} + \frac{1}{\sqrt{(2n+1)^2}}$. Maybe we cannot compute this for any case??
Nov
17
asked Functional Equation. $f(mn)=f(m)f(n)$ and …
Nov
17
asked What is the integer part of $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} +\cdots + \frac{1}{\sqrt{(2n+1)^2}}$