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seen Nov 28 at 14:12

Currently enjoying learning the language of mathematics!


Dec
18
awarded  Popular Question
Nov
28
comment show that $(1+x^2)(1+x^4)(1+x^8)\cdots (1+x^{2^n}) = \frac{1-x^{2^{(n+1)}}}{1-x^2}$
Thank you for you clarification of Ilyas answer Arain. Much appreciated!
Nov
28
comment show that $(1+x^2)(1+x^4)(1+x^8)\cdots (1+x^{2^n}) = \frac{1-x^{2^{(n+1)}}}{1-x^2}$
Thank you for your answer! This was actually exactly what I was looking for!
Nov
28
accepted show that $(1+x^2)(1+x^4)(1+x^8)\cdots (1+x^{2^n}) = \frac{1-x^{2^{(n+1)}}}{1-x^2}$
Nov
28
comment show that $(1+x^2)(1+x^4)(1+x^8)\cdots (1+x^{2^n}) = \frac{1-x^{2^{(n+1)}}}{1-x^2}$
Hi Geoff! As far as I can understand this is the same approach that Ilya was taking.
Nov
28
comment show that $(1+x^2)(1+x^4)(1+x^8)\cdots (1+x^{2^n}) = \frac{1-x^{2^{(n+1)}}}{1-x^2}$
Thank you for your answer! Very much appreciated. Still not sure if I understand what you are going for here. If you have the time I would love a bit more explanation of your thoughts.
Nov
28
asked show that $(1+x^2)(1+x^4)(1+x^8)\cdots (1+x^{2^n}) = \frac{1-x^{2^{(n+1)}}}{1-x^2}$
Nov
20
awarded  Popular Question
Nov
18
comment Determine $h$ so that the linear system $Ax=b$ has infinitely many solutions.
Thank you! This was probably how the question was intended to be solved.
Nov
18
accepted Determine $h$ so that the linear system $Ax=b$ has infinitely many solutions.
Nov
18
comment Determine $h$ so that the linear system $Ax=b$ has infinitely many solutions.
Thank you for your comment! I will definitely look into this, although I think it should be possible to solve this question more intuitively.
Nov
18
comment Determine $h$ so that the linear system $Ax=b$ has infinitely many solutions.
Thank you for your answer @AlexR if you have time I would love to see how you would solve the calculations that you described.
Nov
18
asked Determine $h$ so that the linear system $Ax=b$ has infinitely many solutions.
Nov
7
comment Solve the equation: $z^2 - (7 +6i)z + 4 +22i = 0 $ in $\mathbb{C}$
Thank you very much! This was exactly what I was looking for.
Nov
7
accepted Solve the equation: $z^2 - (7 +6i)z + 4 +22i = 0 $ in $\mathbb{C}$
Nov
7
reviewed Approve Solve the equation: $z^2 - (7 +6i)z + 4 +22i = 0 $ in $\mathbb{C}$
Nov
7
comment Solve the equation: $z^2 - (7 +6i)z + 4 +22i = 0 $ in $\mathbb{C}$
Thank you for you comments, If anyone of you would have time to do the calculation, that would be much appreciated! I have a little trouble to get it to work correctly in this case...
Nov
7
asked Solve the equation: $z^2 - (7 +6i)z + 4 +22i = 0 $ in $\mathbb{C}$
Nov
3
accepted Describe how the number of solutions to an equation system depend on the variable $a$.
Nov
3
comment Describe how the number of solutions to an equation system depend on the variable $a$.
Thank you very much for your answer! Much appreciated!