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Eric Towers

 Discussion on question by TareTar: Co…

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Eric Towers
Sun 11:48
If you don't say $x$ is a function of $t$ and $y$ is a function of $t$, I don't know how to interpret $\partial x/\partial t$ and $\partial y/\partial t$. Partial derivatives are with respect to "slots" in the function declaration. So, if you write $\partial g(a,b,c)/\partial c$ I know that we hold the first two slots fixed and vary the third slot. If you write $\partial g(a,b,c)/\partial t$, there's no interpretation because $g$ doesn't have a slot labelled "$t$". If you write $g(a(t),b,c)$, then $\partial g/\partial t$ can only mean $\partial g/\partial a \ \partial a/\partial t$.
Eric Towers
Sun 11:48
"How can $f$ change at all if $x$ and $y$ are being held constant"... In the partial derivative $\partial f/\partial x$, only $y$ is being held constant and in the partial derivative $\partial f/\partial y$ only $x$ is being held constant. There is no partial derivative in which both are simultaneously held constant.
 

 Discussion on question by Eric Needha…

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Eric Towers
Jun 17 15:05
I tell my students that "approximation" is meaningless without precision and accuracy guarantees.
 

 Discussion on question by Datsosweet:…

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Eric Towers
Feb 26 13:45
Maybe math.stackexchange.com/questions/2351731/inverse-outer-produ‌​ct is relevant.
Eric Towers
Feb 26 13:45
Not clear what you intend as inputs and what you intend as outputs. Not clear how $y_k = x_i x_j$ relates to $\nu_k = \lambda_i \lambda_j$ (although it may be that you have not clarified which of the $\nu$s or the $\lambda$s are inputs and which are outputs).
 

 Discussion on question by Steven Thom…

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Eric Towers
Jan 1 09:52
Also, there are non-Euclidean affine spaces (which necessarily cannot be embedded in a (real) vector space). See math.stackexchange.com/questions/2159756/… .
Eric Towers
Jan 1 09:52
Well, no. The first definition doesn't require an ambient space, so can be applied in settings where there is no ambient vector space to host the affine space. Also, relative to your definition, the first definition can put the space in "impossible" places, for example, $i$ units away origin in the host space.
 

 Discussion on question by Conor OMall…

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Eric Towers
Dec 15, 2024 20:12
Are the $a$s, $b$s, and $c$s real, complex, something else?
Eric Towers
Dec 15, 2024 20:12
There is a risk that $-b_i x + c_i \approx 0$, causing out of range problems. How do you want to handle that?
Eric Towers
Dec 15, 2024 20:12
Should we think of the sequences $(a_i)_i$, $(b_i)_i$, and $(c_i)_i$ as fixed or rarely changing and $x$ as changing frequently? (I.e., is it acceptable to do some sort of precomputation on these sequences of coefficients, which precomputation can be used repeatedly for some set of $x$s?)
 

 Discussion on question by Mike_bb: Ho…

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Eric Towers
Dec 12, 2024 15:43
This might help: Field extension. Note that one does not try to prove that an extension element exists -- one just extends by it, like extending from the reals to polynomials with real coefficients (extending the reals by a formal symbol, $x$).
 

 Discussion on question by ojas dessai…

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Eric Towers
Dec 6, 2024 18:45
No one should conflate phi, "$\phi$", with the empty set, "$\emptyset$" or "$\varnothing$".
 

 Discussion on question by Anu: The no…

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Eric Towers
Dec 4, 2024 10:44
Seems unlikely if $A$ is a large multiple of the identity, $B = C = 0$, and $D$ is a (subunit) multiple of the identity.
 

 Discussion on question by Joshua Till…

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Eric Towers
Nov 14, 2024 19:13
If $\mathrm{char}(k) \neq 2$, commutative and anticommutative together imply nilpotent of index 2.
Eric Towers
Nov 14, 2024 19:13
Nilpotent algebra: en.wikipedia.org/wiki/Nilpotent_algebra
Eric Towers
Nov 14, 2024 19:13
The first two relations suggest "boolean algebra" and the last, commutator, relation suggests non-commutative, so maybe a skew Boolean algebra. Reference: Antonio Bucciarelli, Antonino Salibra.
 

 Discussion on question by Bao Yu Bo: …

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Eric Towers
Nov 5, 2024 11:19
Also, for degrees, the limit is $\displaystyle \frac{\pi}{180^\circ}$. But this is a form of $1$: $\pi$ radians is $180^\circ$.
 

 Discussion on question by Seth: Quick…

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Eric Towers
Jun 30, 2024 13:26
Can you expilcitly write down a generic element of $\langle 2 \rangle \cap \langle x \rangle$? I.e. $$\langle 2 \rangle \cap \langle x \rangle = \left\{\text{(some form)} \mid \text{(some condition(s))} \right\} \text{.} $$
Eric Towers
Jun 30, 2024 13:26
Just because you are working through abstract algebra does not mean all prior knowledge of linear algebra, especially the various mental tools for thinking about a problem, have vanished from your mind. You still have to translate those means back into the tools you have in abstract algebra -- but forgetting your tools for exploring problems is a terrible idea.
Eric Towers
Jun 30, 2024 13:26
For all of these, it may be more helpful to think of $\Bbb{Z}[x]$ as having the "basis" $\{1, x, x^2, x^3, \dots\}$ and ask what the quotient does to a scalar multiple of each basis element. For $z_i \in \Bbb{Z}$, $z_0 \cdot 1$ modulo $2x\Bbb{Z}$ is? $z_1 \cdot x$ modulo $2x\Bbb{Z}$ is? $z_2 \cdot x^2$ modulo $2x\Bbb{Z}$ is?
 

 Discussion on question by Mobius: Def…

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Eric Towers
Jan 20, 2022 15:44
Related: it appears that OP has finally correctly quoted Gallian.... and I'm not continuing in Chat because I have to blind-type all of this -- I don't get to see a text input box.
Eric Towers
Jan 20, 2022 15:44
If you tell me that $\Bbb{Q}$ is a particular set of ordered pairs of integers, then any ordered pair of integers that is not in that set is not a rational number. If you tell me that $\Bbb{Q}$ is a set of equivalence classes of ordered pairs of integers then any element of any equivalence class is a rational number. This is literally the distinction that has been made repeatedly.
Eric Towers
Jan 20, 2022 14:54
@BillDubuque : I don't claim it isn't a member of Hungerford's $\Bbb{Z}/n\Bbb{Z}$. I claim it is a number in Hungerford's $\Bbb{Z}/n\Bbb{Z}$. However, the integer $n$ is none of $\{0, \dots, n-1\} \subset \Bbb[Z}$, so $n$ is not a number in OP's incorrectly transcribed version of Gallian's $\Bbb{Z}/n\Bbb{Z}$.
Eric Towers
Jan 20, 2022 14:54
@BillDubuque : As already commented: "If $\{0, \dots, n−1\}$ is a subset of $\Bbb{Z}$, as OP has written, then $n$ isn't a number in $\Bbb{Z}_n$. In short: my two comments are related."
Eric Towers
Jan 20, 2022 14:54
@BillDubuque : As already commented: "Gallian's 7th edition, p. 42 does not have the "${}\subset \Bbb{Z}$". Are you sure the "${}\subset \Bbb{Z}$" appears where you have it?"
Eric Towers
Jan 20, 2022 14:54
@BillDubuque : I cannot possibly be adding to the confusion caused by OP's writing of wrong things.
Eric Towers
Jan 20, 2022 14:54
@BillDubuque : No, I am not. I am attempting to get OP to actually write Gallian's definition, which has not yet happened.
Eric Towers
Jan 20, 2022 14:54
@BillDubuque : No. $n$ lives in an equivalence class in Hungerford's definition, so Hungerford's definition can make sense of it (modulo notation intended to help the reader).
Eric Towers
Jan 20, 2022 14:54
@BillDubuque : If $\{0, \dots n-1\}$ is a subset of $\Bbb{Z}$, as OP has written, then $n$ isn't a number in $\Bbb{Z}_n$. In short: my two comments are related.
Eric Towers
Jan 20, 2022 14:54
Gallian's 7th edition, p. 42 does not have the "${} \subset \Bbb{Z}$". Are you sure the "${} \subset \Bbb{Z}$" appears where you have it?
Eric Towers
Jan 20, 2022 14:54
As you've presented it, Gallian's definition cannot make sense of "$n+n$" in $\Bbb{Z}/n\Bbb{Z}$, but Hungerford's can.
 

 Discussion on question by Lance: Is i…

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Eric Towers
Jan 12, 2022 16:36
... and its running at least 100-times slower than your laptop's general purpose CPU ... this is likely a wash (meaning "essentially cancels out").
Eric Towers
Jan 12, 2022 16:36
From the comment block at the code you reference: "This program demonstrates concept behind AKS algorithm and doesn't implement the actual algorithm (This works only till n = 64)" So we should expect this implementation to misbehave in inputs bigger than $64$.
Eric Towers
Jan 12, 2022 16:36
No. The first number there (partially) factors as $491 \cdot 7237 \cdot \text{big number} = 662\,635\cdots{}611\,007$.
Eric Towers
Jan 12, 2022 16:36
You might also read Clavier et al. 2012 describes implementing probable and provable 512 to 2048 bit (154 to 616 digit) prime generation on a 30 MHz AT90SC in sub-second to several seconds time.
Eric Towers
Jan 12, 2022 16:36
It might help to read up on probable prime testing -- most such tests make no attempt to factor the number. There are prime proving algorithms, e.g. Pocklington and AKS that do not attempt to find factors of the given number. AKS implementation.
 

 Discussion on question by Burzin: How…

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Eric Towers
Dec 19, 2021 10:35
A method to recognize algebraics is to generate powers of the number and search for a $\Bbb{Z}$-linear relationship among the powers. A small instance of this is described using the LLL algorithm. Numbers that aren't "obviously algebraic" don't come with a hint for how large the degree could be, so this isn't really an algorithm as much as it is a search over "low" degrees (up to a few hundred or a few thousand, depending on your tolerance for long computation times).
 

 Discussion on question by user264745:…

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Eric Towers
Dec 13, 2021 20:07
No. The above is definitely a definition. You want to know if using this new definition gives results consistent with the old definition for functions to which both can simultaneously apply.
Eric Towers
Dec 13, 2021 20:07
Your (2) works for me. Now my counter-question : if you read the exercise with fresh eyes, do you see that this is the question?
Eric Towers
Dec 13, 2021 20:07
(1) You shouldn't think about the previous definition (def 6.1) because it only defines the bundle of symbols on the right-hand side of the equals sign; you're being given the definition of the left-hand side in terms of a limit of a bundle of symbols defined by that previous definition. All definitions work this way: the new, currently undefined, thing is defined in terms of undefined atomic things and previously defined things.
Eric Towers
Dec 13, 2021 20:07
"there there" \mapsto "then there". Why do you imagine my statement must agree with anyone else's? Additionally, you are responsible for synthesizing these various comments into your understanding.
Eric Towers
Dec 13, 2021 20:07
I did not reference ArcticChar. You did. It is entirely baffling that you keep trying to draw some sort of relationship between comments that does not exist.
Eric Towers
Dec 13, 2021 20:07
(1): Literally in the text you quote "Define $\int_0^1 \dots$". A definition for $\int_0^1 f(x)\,\mathrm{d}x$ for $f$ defined on $(0,1]$ is literally given to you in the question. (2): If $f$ is defined on $[0,1]$, there there is a restriction, $f|_{(0,1]}$ defined only on $(0,1]$. Does $\int_0^1 f(x)\,\mathrm{d}x = \int_0^1 f|_{(0,1]} \,\mathrm{d}x$, where the latter integral is defined by the statement of the problem.
 

 Discussion between user264745 and Eri…

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Eric Towers
Dec 13, 2021 18:38
No. The above is definitely a definition. You want to know if using this new definition gives results consistent with the old definition for functions to which both can simultaneously apply.
Eric Towers
Dec 13, 2021 18:38
Your (2) works for me. Now my counter-question : if you read the exercise with fresh eyes, do you see that this is the question?
Eric Towers
Dec 13, 2021 18:38
I did not reference ArcticChar. You did. It is entirely baffling that you keep trying to draw some sort of relationship between comments that does not exist.
Eric Towers
Dec 13, 2021 18:38
(1) You shouldn't think about the previous definition (def 6.1) because it only defines the bundle of symbols on the right-hand side of the equals sign; you're being given the definition of the left-hand side in terms of a limit of a bundle of symbols defined by that previous definition. All definitions work this way: the new, currently undefined, thing is defined in terms of undefined atomic things and previously defined things.
Eric Towers
Dec 13, 2021 18:38
"there there" \mapsto "then there". Why do you imagine my statement must agree with anyone else's? Additionally, you are responsible for synthesizing these various comments into your understanding.
Eric Towers
Dec 13, 2021 18:38
(1): Literally in the text you quote "Define $\int_0^1 \dots$". A definition for $\int_0^1 f(x)\,\mathrm{d}x$ for $f$ defined on $(0,1]$ is literally given to you in the question. (2): If $f$ is defined on $[0,1]$, there there is a restriction, $f|_{(0,1]}$ defined only on $(0,1]$. Does $\int_0^1 f(x)\,\mathrm{d}x = \int_0^1 f|_{(0,1]} \,\mathrm{d}x$, where the latter integral is defined by the statement of the problem.
 

 Discussion on question by SiNui: Krus…

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Eric Towers
Dec 7, 2021 20:45
Works for me...