network profile
| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year |
| seen | Jul 23 '12 at 12:21 | |
| stats | profile views | 3 |
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Jul 7 |
comment |
Quick question about comparing functions. I can sure have $h(x)=f(x)-g(x)$ but the problems is, I wouldn't know in what interval would it be nonnegative or nonpositive. It's more interesting and convenient if I'd know if it is nonnegative or nonpositive when I $ \underline{already} $ have an interval. |
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Jul 7 |
awarded | Supporter |
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Jul 7 |
awarded | Scholar |
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Jul 7 |
awarded | Commentator |
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Jul 7 |
accepted | Quick question about comparing functions. |
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Jul 7 |
comment |
Quick question about comparing functions. Oh yeah, I've forgotten that stipulation. Many thanks, you just saved me from all that strenuous graphing work. |
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Jul 7 |
comment |
Quick question about comparing functions. Oh, yeah I'm getting what you're trying to say. This is similar to the first derivative test. From what I understood: If the derivative of f is greater than g for all values of x in the given interval, then we could say that the values of f is greater than g in the given interval. Is that it? |
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Jul 7 |
asked | Quick question about comparing functions. |
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May 26 |
comment |
Conventions for antiderivatives notation. Oh thanks, I guess it's just the notation. For some time, I've wondered how could mathematicians come up with such brilliant notations. I just came around asking; maybe I got it all wrong in my head. |
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May 26 |
asked | Conventions for antiderivatives notation. |
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May 13 |
comment |
Proving a theorem on limits that approach infinity. Yeah, I was too focused on getting a value for delta. I'm not very comfortable with assuming a value; but it works anyways. |
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May 13 |
revised |
Proving a theorem on limits that approach infinity. There shouldn't be a zero in -x<(delta) |
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May 13 |
asked | Proving a theorem on limits that approach infinity. |
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May 13 |
comment |
About limit theorems I mean $|(f(x)-K)|+|(g(x)-L)|<\epsilon $ Latex is making me dizzy. |
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May 13 |
comment |
About limit theorems Does that mean $|f(x)+g(x)-(K+L)|<|(f(x)-K)|+|g(x)-K)|$ Thus $|f(x)+g(x)-(K+L)|<\epsilon $ since $|(f(x)-K)|+|g(x)-K)|=\epsilon $ |
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May 13 |
comment |
About limit theorems It's ambiguous. |
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May 12 |
comment |
About limit theorems Right. Last one though, for those who actually looked at the link, how did we get $<\frac{1}{2}\epsilon + \frac{1}{2}\epsilon $ then $=\epsilon$? |
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May 12 |
revised |
About limit theorems added 73 characters in body |
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May 12 |
comment |
About limit theorems Wow, that's pretty cool. I still can't think of it in general terms though, since the example was quite specific but that certainly did help. Thanks. |
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May 12 |
awarded | Student |
