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Apr
16
comment How do you show this property of a differentiable function given information about the derivative?
@Fabian: Thank you, I noticed that as well, but wasn't sure if it would turn out to be a 'trick.' As for deducing something: what more than $f(-x) = -f(x)$ or maybe even $f(-x)+f(x)=0$ should I see? Also, why is it ok to evaluate the definite integral here? (sorry if i'm missing very obvious stuff!)
Apr
16
comment How do you show this property of a differentiable function given information about the derivative?
@Thomas: It's not homework, but thanks for the hint anyway. If I integrate both sides of the above equation I get $f(-x) + C = -\int f(x)dx$ right? I am not sure what to do with that and in I general get thrown off by the constant whenever I try to use integration...
Apr
16
revised How do you show this property of a differentiable function given information about the derivative?
edited title
Apr
16
asked How do you show this property of a differentiable function given information about the derivative?
Apr
10
accepted What is the best way to show that no positive powers of this matrix will be the identity matrix?
Apr
10
asked What is the best way to show that no positive powers of this matrix will be the identity matrix?
Mar
30
awarded  Cleanup
Mar
30
revised Why does $a_n = (1+\frac{2}{n})^{n}$ converge to $e^2$?
rolled back to a previous revision
Mar
29
accepted Why does $a_n = (1+\frac{2}{n})^{n}$ converge to $e^2$?
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 $4n-1$?