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I know how to approach this with a higher degree in the numerator with division but not the other way around. Can someone give me a hint? $$f(x) =...
Asked on 12/27/2020 by sabrinapat
1 answerI want to discuss a particular part of the proof of $cl(cl(A))=cl(A)$ in a topological space: One way to proof this includes the following reasoning: $xin cl(cl(A)) Rightarrow...
Asked on 12/27/2020 by Averroes2
1 answerLet's take two real numbers $a,b$. The distance between $a$ and $b$ is $|a-b|$. Let ${}$ denote fractional part. Then for any $a$ and...
Asked on 12/26/2020 by MathGeek
1 answerSay I have $f(z) = sum_n^infty a_n z^n$ and $g(z)= sum_n^infty b_n z^n$. Now suppose that$$f(z)=g(z)$$ for infinitely many $z$ that are not discrete. Why...
Asked on 12/26/2020 by Maths Wizzard
2 answerTo prove:$$text{If } A_1subseteq A_2 subseteq ... subseteq A_ntext{ ,then } bigcup_{i=1}^n A_i = A_n$$ using the axioms of ZFC Set Theory.Honestly, this statement is very obvious, but I...
Asked on 12/26/2020
2 answerCan this equation be solved with analytical methods, or is it only numeric methods since current mathematical tools don't go that far? Its complex roots are the same as the...
Asked on 12/26/2020 by user694069
1 answerShow that the Riemann Integral $I(y)=int_0^1 frac{x-y}{(x+y)^3}dx$ exists for $yin(0,1]$.$A=int_0^1(int_0^1 frac{x-y}{(x+y)^3}dx)dy$, $B=int_0^1(int_0^1 frac{x-y}{(x+y)^3}dy)dx$ Show that $A$ and $B$ exist and $Aneq B$....
Asked on 12/26/2020
1 answerI tried to use the Cosine theorem and get that |(PA)^2 + (PB)^2 - (AB)^2|<=|2PAPB|. Can someone explain to me what should I do? Thank you!...
Asked on 12/26/2020 by HestiaCranel
2 answerYou are in the back of a NYC taxi, and know exactly how fare your trip was in miles (m), and exactly how long it took in hours (t). The...
Asked on 12/25/2020 by Cameron Chandler
1 answerI've seen the calculation for a Schreier transversal and basis for $[F_2,F_2]lhd F_2=langle x,yrangle$, but these groups aren't so complex that the calculations were particularly illuminating. I was wondering...
Asked on 12/25/2020 by Makenzie
2 answerGet help from others!
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