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| bio | website | |
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| location | Dubuque, IA | |
| age | ||
| visits | member for | 2 years, 9 months |
| seen | 22 mins ago | |
| stats | profile views | 11,856 |
As of August 2011 I am an assistant professor of mathematics at Loras College in Dubuque, Iowa, USA.
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15h |
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Does $f:\mathbb{R}\to\mathbb{R}$ mean that $f$ maps to all reals? This is still incorrect. If $f:X\to Y$ is bijective, then for all $y\in Y$, there is a unique $x\in X$ such that $f(x)=y$. You have it the other way around. (As noted by Thomas Andrews, 1-1 is not the same as bijective, but you did not accurately describe either.) If you have $f:X\to Y$, assuming no other properties of $f$, then this already means that for each $x\in X$ there is a unique $y\in Y$ such that $f(x)=y$. |
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16h |
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number of zeros of function $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)-1$ deleted 5 characters in body |
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16h |
answered | number of zeros of function $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)-1$ |
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16h |
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number of zeros of function $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)-1$ Rush: Why not tell us that before? Anyway, thank you very much for clarifying. |
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16h |
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number of zeros of function $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)-1$ That product is famous. Do you know an elementary function representation of it? Where is the problem from? |
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22h |
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Why do injective holomorphic functions have nonzero derivative See math.stackexchange.com/questions/35304/… and math.stackexchange.com/questions/83577/… |
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1d |
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Solving the equation $\dfrac{(1+x)^{36} -1}{x} =20142.9/420$ for $x$. edited title |
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1d |
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Solving the equation $\dfrac{(1+x)^{36} -1}{x} =20142.9/420$ for $x$. added 2 characters in body |
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1d |
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Solving the equation $\dfrac{(1+x)^{36} -1}{x} =20142.9/420$ for $x$. Out of curiosity, where did the equation come from? |
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1d |
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How to prove to be an irrational number? Like $\sqrt{2}$ $\sqrt{3}$ or $\sum\limits_{k=1}^{\infty} \frac{1}{n^2}=\pi^2/6$ Internet searches easily lead to answers to many of these questions. In particular, you'd probably find some proofs of irrationality on Wikipedia. Regarding the last question, see the related questions math.stackexchange.com/q/159350 and math.stackexchange.com/q/28243. |
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2d |
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The negative square root of $-1$ as the value of $i$ @Glen0: What you just wrote only makes sense for square roots of nonnegative numbers. In particular, note that $|x|$ is not a square root of $x^2$ unless $x$ is real. |
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2d |
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The negative square root of $-1$ as the value of $i$ If we have a definition of $\sqrt{-1}$ in the first place (as a particular number whose square is $-1$), but have not yet designated our definition of $i$, then yes, it would make perfect sense to define $i=-\sqrt{-1}$ (the other number whose square is $-1$), and then one could carry out all of the theory of complex numbers with this choice. So there is a bit more to the question in that sense. |
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2d |
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The negative square root of $-1$ as the value of $i$ I agree with the part about having to define $i$ to be one of the square roots of $-1$ and sticking with the choice. However, I don't see how the part about being careful with operations is particularly relevant to the question. You say "counterexample," but it is not a counterexample to anything in the question as far as I can see, but rather to the property $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ which would hold for the usual square root function if $a\geq 0$ and $b>0$. |
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2d |
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The negative square root of $-1$ as the value of $i$ If $\sqrt{-1}$ and $i$ are used to mean the same thing, as they are in many contexts including the beginning of the question above, then the equation $\sqrt{-1}=-i$ is incorrect. You seem to be using the square root symbol to denote a 2-valued function, but this does not clear up the confusion in the question as far as I can see. |
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2d |
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The negative square root of $-1$ as the value of $i$ Could you please be more specific as to what operations you are referring to that we cannot do? |
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2d |
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The negative square root of $-1$ as the value of $i$ deleted 2 characters in body |
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2d |
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Bounded sequence in Hilbert space contains weak convergent subsequence @FrankMcGovern: Separability isn't needed, but if desired one could reduce to the separable case by considering the Hilbert subspace generated by $\{x_n\}$. |
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2d |
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Bounded sequence in Hilbert space contains weak convergent subsequence Shorter than what? Do you have a long proof? |
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May 19 |
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On the pH scale, each unit change in pH represents a tenfold increase in acidity or alkalinity. Could you answer this if instead of $3.2$ it were $6$? $5$? $4$? |
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May 19 |
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Why is 987654321/123456789 = 8.0000000729? @cobaltduck: It is a terrible approximation given how many digits you're putting in. It has more than $10$ times the error of $\sqrt{9.87}$ |
