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visits member for 3 years, 8 months
seen Mar 28 at 23:11

Mar
29
comment Why does $a_n = (1+\frac{2}{n})^{n}$ converge to $e^2$?
Thank you very much for this answer. Just one question: why is the first limit going to zero rather than infinity?
Mar
29
asked Why does $a_n = (1+\frac{2}{n})^{n}$ converge to $e^2$?
Mar
13
accepted Proving elementary inverse image consequence
Mar
13
asked Proving elementary inverse image consequence
Mar
11
comment Is a linear tranformation onto or one-to-one?
oh ok, so it wouldn't have been a function in the first place... Thanks for clearing that up for me! And thanks for another great answer of course!
Mar
11
comment Is a linear tranformation onto or one-to-one?
Sorry if this is a stupid question, but is it always the case that whether a function is one-to-one depends only on the domain? For instance with the example $f(x) = x^2$ if the codomain were $0$, wouldn't the function still be one-to-one?
Mar
7
comment Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
Thank you for the answer, although I must say that it isn't the same without the bold HINT formatting ;)
Mar
7
comment Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
Definitely helps a lot, thank you!
Mar
7
accepted Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
Mar
7
comment Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
thank you for the reply, I can understand much more now. Just to be absolutely clear: in order to determine which terms are added/deleted one could replace each $n$ in the original LHS with $n+1$, correct?
Mar
7
comment Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
Unfortunately I still don't get it... with the induction, is it correct to sort of 'plug in' $n+1$ for each $n$? And somewhere in between $(2n+5)$ and $(4n-1)$ each term begins to be calculated differently? I can see why $2n+1$ is some odd number, but what is $4n-1$?
Mar
7
asked Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
Feb
28
awarded  Civic Duty
Feb
27
awarded  Suffrage
Feb
13
comment Why is the 'change-of-basis matrix' called such?
@Qiaochu, is 'in the basis given by' the same as 'relative to the basis'?
Feb
13
comment Why is the 'change-of-basis matrix' called such?
@Qiaochu: yes no problem, I hope I edited in the relevant definitions, sorry it took forever!
Feb
13
revised Why is the 'change-of-basis matrix' called such?
added 2211 characters in body; deleted 10 characters in body; deleted 4 characters in body
Feb
13
comment Why is the 'change-of-basis matrix' called such?
@Qiaochu: Sorry, I thought I was totally understanding it, but I am still confused by the part about some basis $e_1,...,e_n$ and a new basis $Pe_1,...,Pe_n$. Does this only work when $e$ is the usual basis? For instance, with $S=\{u_1,u_2\}=\{(1,-2),(3,-4)\}$ and $S' = \{v_1,v_2\}= \{(1,3), (3,8)\}$ the change of basis matrix $P = \left( \begin{array}{cc} -\frac{13}{2} & -18 \\ \frac{5}{2} & 7 \end{array} \right)$ although $Pu_1, Pu_2 \neq v_1, v_2$.
Feb
13
accepted Why is the 'change-of-basis matrix' called such?
Feb
12
asked Why is the 'change-of-basis matrix' called such?