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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 4 months |
| seen | yesterday | |
| stats | profile views | 33 |
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Apr 11 |
comment |
Taylor series $ \sqrt{\frac{t}{t+1}}$ nice. thanks you |
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Apr 11 |
comment |
Taylor series $ \sqrt{\frac{t}{t+1}}$ but I dont know if calculated $\sqrt {1-x}=1+\frac{x}{2}-\frac{1}{4}x^2+\ldots$ if it is good. Maybe $\sqrt {1-x+x^2}$ have factor at $x$ |
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Apr 11 |
accepted | Taylor series $ \sqrt{\frac{t}{t+1}}$ |
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Apr 11 |
asked | Taylor series $ \sqrt{\frac{t}{t+1}}$ |
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Apr 11 |
accepted | limit with $\arctan$ |
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Apr 11 |
comment |
limit with $\arctan$ the $\frac{1}{2}$ isnt for arctan |
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Apr 11 |
revised |
limit with $\arctan$ added 4 characters in body |
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Apr 11 |
comment |
limit with $\arctan$ this is for $(n+1)$ |
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Apr 11 |
asked | limit with $\arctan$ |
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Apr 10 |
comment |
Sum of the roots equation thanks for the hint, but more important for me is how to find the $x_1+x_2$ |
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Apr 8 |
comment |
Sum of the roots equation Thats true, I need to show that there are two roots on the interval and calculate only the sum |
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Apr 8 |
revised |
Sum of the roots equation edited body |
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Apr 8 |
revised |
Sum of the roots equation added 5 characters in body |
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Apr 8 |
asked | Sum of the roots equation |
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Apr 7 |
comment |
Uniform continuity of $\ln(x)$ ok sorry, my mistake. So the interval where $\ln(x)$ is uniformly continuous and where satisfy Lipschitz condition is this same? $A=[a,\infty)$ with $a>0$ But lipschitz and uniformly continuous isnt equivalent right? Is any exemple function uniformly continuous but not satisfy lipschitz? |
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Apr 7 |
comment |
Uniform continuity of $\ln(x)$ let us continue this discussion in chat |
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Apr 7 |
comment |
Uniform continuity of $\ln(x)$ So why the function isnt uniform comtinuity on $[a,+\infty)$ where satisfy lipschitz but is uniform continuity on interval $[b,+\infty) where b>1 |
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Apr 7 |
comment |
Uniform continuity of $\ln(x)$ The interval where f is uniform continuity and satisfy lipschitz condition is the same? |
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Apr 7 |
comment |
Uniform continuity of $\ln(x)$ For me $\frac{1}{x}$ is bounded on$ [a,+\infty) $ where $a>0$ by $f(a)$ but the $\ln$ isnt uniform continouity on the interval. And i thought that lipschitz implies uniform continuity. |
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Apr 6 |
asked | Uniform continuity of $\ln(x)$ |
