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May 14 |
awarded | Caucus |
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Mar 27 |
awarded | Informed |
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Mar 18 |
awarded | Scholar |
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Mar 18 |
comment |
Why is $f:\mathbb{R} \to S^1, f(t)=(2\pi \cos(t), 2\pi \sin(t))$ not closed? Couple of minutes after I have posted the question a friend of mine gave me a solution, which made me think exactly this solution. thanks :) |
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Mar 18 |
accepted | Why is $f:\mathbb{R} \to S^1, f(t)=(2\pi \cos(t), 2\pi \sin(t))$ not closed? |
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Mar 18 |
asked | Why is $f:\mathbb{R} \to S^1, f(t)=(2\pi \cos(t), 2\pi \sin(t))$ not closed? |
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Jan 11 |
awarded | Announcer |
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May 16 |
awarded | Yearling |
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Mar 10 |
awarded | Critic |
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Mar 9 |
comment |
Are there any calculus/complex numbers/etc proofs of the pythagorean theorem? I learned linear algebra this semester, and yes, this proof is really cool! |
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Sep 18 |
awarded | Nice Question |
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Sep 18 |
awarded | Commentator |
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Sep 18 |
comment |
Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$ I see, well it gives us information actually. From the symmetry of $1_\mathbb{Q}$, it doesn't matter which value does the function get around $x=0$ the change in the function is zero, just like the limit says. |
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Sep 18 |
comment |
Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$ I don't understand what you wrote, what does $1_\mathbb{Q}$ mean? |
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Sep 18 |
awarded | Editor |
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Sep 18 |
comment |
Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$ I have modified my question, I know that when the derivative exists this limit is equal to it, but can it be used when the derivative is not defined? |
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Sep 18 |
comment |
Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$ I know, I have modified my question to clarify this issue. |
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Sep 18 |
revised |
Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$ added more information |
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Sep 18 |
asked | Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$ |
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Sep 11 |
awarded | Autobiographer |
