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| visits | member for | 2 years, 6 months |
| seen | Mar 10 at 19:08 | |
| stats | profile views | 132 |
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Apr 17 |
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How to show certain things related to scalar products @Theo: yeah that was the 'workaround' I was trying out, but now I can go ahead and replace the alphas thanks to what you just taught me. |
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Apr 17 |
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How to show certain things related to scalar products added 2 characters in body |
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Apr 17 |
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How to show certain things related to scalar products @Theo: thank you so much for that! I was trying to figure out why it was such a mess... definitely valuable remarks! |
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Apr 17 |
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How to show certain things related to scalar products added 6 characters in body; added 114 characters in body; added 1 characters in body |
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Apr 17 |
asked | How to show certain things related to scalar products |
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Apr 17 |
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Subtraction and division with integers modulo 3 @Gerry: thanks for the explanation. I had copied that notation from Paul R. Halmos -Linear Algebra Problem Book... But, yeah I will definitely try to avoid using those sorts of fractions in the future. |
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Apr 17 |
accepted | How do you show this property of a differentiable function given information about the derivative? |
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Apr 17 |
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Subtraction and division with integers modulo 3 Thank you for this answer |
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Apr 17 |
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Subtraction and division with integers modulo 3 @Fabian: thanks for the helpful tips |
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Apr 17 |
accepted | Subtraction and division with integers modulo 3 |
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Apr 17 |
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Subtraction and division with integers modulo 3 @quanta: I'm sorry if I'm just completely missing something here, but are you saying there is a problem with the way I am trying to write down/express an idea, or with the idea itself that I am trying to carry out division on the integers modulo 5? |
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Apr 17 |
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Subtraction and division with integers modulo 3 @quanta: ok, yeah I meant by that 3 divided by 4 in integers modulo 5 (and I realize the way I used $x$ in my above comment was nonsense). But isn't that how you would carry out division of 3 by 4 in integers modulo 5? Since $4$ is the inverse of $4$, $3*4^{-1}=2$ in integers modulo 5, no? |
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Apr 17 |
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Subtraction and division with integers modulo 3 @quanta: thank you for the answer. I'm still digesting it, but I wanted to check if I am understanding part of the idea. If I wanted to divide on say integers modulo 5 (where there are a couple more examples) then for say $\frac{3}{4}=x$ I would first always calculate $4^{-1}$ and then multiply? And because of the gcd stuff you showed, I know that $1\equiv 4x \mod 5 \Rightarrow x = 4 \Rightarrow \frac{3}{4} = 2$? So in these cases it is about finding the inverses first? |
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Apr 17 |
asked | Subtraction and division with integers modulo 3 |
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Apr 16 |
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How do you show this property of a differentiable function given information about the derivative? @Qiaochu: thanks for that comment, I guess the derivative of a polynomial satisfies $f'(-x)=-f'(x) \Rightarrow f'$ is odd with degree $n \Rightarrow \int f'(x)dx$ is even with degree $n+1$..? Unfortunately I can't say much about a power series or a differentiable function in general... |
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Apr 16 |
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How do you show this property of a differentiable function given information about the derivative? $g(0) = f(0)-f(-0)=0$ ? |
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Apr 16 |
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How do you show this property of a differentiable function given information about the derivative? thank you for this answer. Unfortunately I can't manage to finish the problem with it yet. $g(x) = f(x)-f(-x) \Rightarrow g'(x)=f'(x)+f'(-x) \Rightarrow g'(x) =0 \Rightarrow g(x)=C \Rightarrow ?$ is the rest pretty much as Chris shows, or did you have something else in mind? (The only thing is that I don't know how I would have defined all these other functions and so on...) |
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Apr 16 |
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How do you show this property of a differentiable function given information about the derivative? @Fabian: yeah, careless mistake on my part, thank you for clearing that up |
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Apr 16 |
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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!) |
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Apr 16 |
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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... |
