text:
# Package exercise: change “exercise” to French “exercice”

I am preparing a homework for my students with LaTeX, so I used \usepackage{exercise}. In the .pdf file I got "exercise" and I want it to be in French (exercice) not English.

I put \usepackage[french]{babel} but nothing changed.

Here is my code:

\documentclass[12pt,a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[french]{babel}
\usepackage{exercise}

\begin{document}
\maketitle

\begin{Exercise}[title=...]

\Question{....}

\end{Exercise}

On exercise documentation (§5.2), you can see how the text "constants" (e.g. List of exercises, Exercise, Answer of exercise, ...) are defined. It is done through following macros, respectively:

\def\listexercisename{List of exercises}%
\def\ExerciseName{Exercise}%
...

Thus, you can use following macro to adapt them according to your needs:

\renewcommand{\ExerciseName}{Exercice}

However, for some languages (spanish, italian, dutch... and french), Paul Pichaureau (creator of this package) already did this work. This should be done automatically when you load a language in babel; but how?
You can see (in the same paragraph), that exercise recognise the language loaded via this command: \@ifpackageloaded{babel}{\addto{\captions<LANGUAGE>}{....

In the case of french language, the command is: \addto{\captionsfrenchb}{, thus you should load the frenchb option (and not french) of babel-package to make exercise automatically translate macros in french. → if this works for most languages (at least spanish, dutch, and italian), it now doesn't seem to work with neither french, nor frenchb.

TL;DR: To translate the names in french, you have two one solutions:

• translate each of them manually via \renewcommand{\ExerciseName}{Nouveau nom de mes exercices}-like commands.
• load the frenchb option in babel (and not the french one): already implemented translations will be automatically used;

Edit: With most recent (i.e. post TeXlive 2012) LaTeX distro, neither frenchb nor french options in babel seem to work with automated translation of exercice package (cf. Exercise package with Babel). You should then copy-paste these lines into your .tex files to emulate the translation:

\renewcommand\listexercisename{Liste des exercices}%
\renewcommand\ExerciseName{Exercice}%

The simplest perhaps: line 74 of exercise.sty, replace addto{\captionsfrenchb} with addto{\captionsfrench}.


text:
# Math Help - Algebraic Proofs

1. ## Algebraic Proofs

Using the statement vs reason chart and prove the following equation.

Given: 2x = 12

Prove: x = 6

2. Originally Posted by magentarita
Using the statement vs reason chart and prove the following equation.

Given: 2x = 12

Prove: x = 6
This is very short and simple.

2x = 12 Given

x= 6 Division property of equality

Or you could do it this way:

$2x=12$ Given

$\frac{1}{2}(2)x=\frac{1}{2}(12)$ Multiplication property of equality

$1x=6 \ \$ Multiplicative Inverse and multiplication

$x=6 \ \$Multiplicative Identity

3. ## ok.......

Originally Posted by masters
This is very short and simple.

2x = 12 Given

x= 6 Division property of equality

Or you could do it this way:

$2x=12$ Given

$\frac{1}{2}(2)x=\frac{1}{2}(12)$ Multiplication property of equality

$1x=6 \ \$ Multiplicative Inverse and multiplication

$x=6 \ \$Multiplicative Identity
A very nice step by step reply.

format: plain


text:
# Characterization of stable theories

Let $\kappa$ be an infinite cardinal. A theory $T$ is called $\kappa$-stable if for all model $M\models T$ and all $A\subset M$ with $|A|\leq \kappa$ we have $|S_n^M(A)|\leq \kappa$.

A theory $T$ is called stable if it is $\kappa$-stable for some infinite cardinal $\kappa$.

Question. Is there any characterization for stable theories in terms of non-forking independence relation similar to Kim-Pillay characterization for simple theories?

• What is $S^M_n(A)$? – Taroccoesbrocco Aug 12 '18 at 8:44
• $S_n^M(A)=\{ p : p$ is a complete type over $A \}$ – Lajos Aug 12 '18 at 13:49

Yes. The easy way to see this uses the characterization of stable theories as those simple theories for which non-forking independence satisifies stationarity over models.

Stationarity: For any model $M$ and any set $B$, if $a$ is independent from $B$ over $M$, $a'$ is independent from $B$ over $M$, and $\text{tp}(a/M) = \text{tp}(a'/M)$, then $\text{tp}(a/MB) = \text{tp}(a'/MB)$.

In other words (together with the extension property), any type over $M$ has a unique independent extension to a type over $MB$.

So you can obviously characterize non-forking independence in stable theories by just taking the Kim-Pillay characterization of non-forking independence in simple theories and adding stationarity over models.

Actually, you can get away with fewer axioms than this; in particular, stationarity over models gives you the independence theorem over models for free. So there are various lists of axioms characterizing non-forking in stable theories:

• You can find a stable variant of the Kim-Pillay axioms in Casanovas's book Simple Theories and Hyperimaginaries, Definition 12.2 and Theorem 12.22.
• The fact that non-forking independence in stable theories can be axiomatically characterized may have first been noticed by Harnik and Harrington (long before Kim and Pillay started studying simple theories), see Axioms 0-3 and Theorem 5.8 in their paper Fundamentals of Forking.
• A more modern presentation of the Harnik and Harrington theorem is Theorem 8.5.10 in Tent and Ziegler's book A Course in Model Theory.
• Thank you very much Alex. Is there any characterization for the other unstable classes for instance NSOP1, NTP1, NSOP3 and etc in terms of non-dividing independence? Also thanks for introducing forkinganddividing.com to me. It's a nice website. – Lajos Aug 13 '18 at 15:14
• @Lajos These questions are at the forefront of current research. Kaplan and Ramsey recently made a breakthrough by showing that NSOP1 can be characterized, not by non-forking or non-dividing independence, but by a generalization called Kim independence. See their paper here: arxiv.org/abs/1702.03894 in particular Theorem 9.1. – Alex Kruckman Aug 13 '18 at 15:19
• Thanks a lot for your helpful comment. – Lajos Aug 13 '18 at 15:27
• Does the non-forking independence relation have a property that holds in strongly minimal theories but does not hold in stable theories? – Lajos Aug 14 '18 at 14:31
• @Lajos Well, in a strongly minimal theory, you can characterize forking in terms of algebraic dimension, and that characterization doesn't hold in an arbitrary stable theory. So in that sense yes. But to me that characterization has a different flavor than the usual kind of abstract properties like local character or stationarity. – Alex Kruckman Aug 14 '18 at 14:34


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# Encoding any turing machine to one with 2 symbols

This is a repost from https://stackoverflow.com/questions/56990710/encoding-any-turing-machine-to-one-with-2-symbols since it was brought to my attention that the question would be better suited to this SE than SO.

I am interested in understanding how a Universal Turing Machine would work on a practical level.

I read Rogozhin's paper Small universal Turing machines to understand how a UTM works and how to encode a 2-tag system to act as its input. Next I read Cocke and Minsky's Universality of tag systems with p=2 on how to convert a 2-symbol TM to a 2-tag system.

My literature search is now stuck at the point where I need to convert an arbitrary TM to a 2-symbol TM. 2-symbol means that the TM has an alphabet of 0 and 1 with 0 also acting as a blank, so I strictly have only two symbols at my disposal.

Edit: I am also limited to a single tape.

This question https://stackoverflow.com/questions/4820488/can-a-turing-machine-be-constructed-having-only-two-tape-symbols has a well-written answer that gives the necessary pointers to understand the conversion generally, but it is not explicit enough to know that a possible solution would be a good one.

Can anyone direct me to a paper that shows the state-of-the-art conversion process? I am sure that Cocke and Minsky based their work on proof that any TM can be emulated by a 2-symbol TM.

If there is no literature reference to be had, could anyone help me analyse whether my solution is close enough to optimal? From the post above, it seems like I need to encode the original TM's tape to a fixed-length binary representation. My solution looks like the following, and I would be happy if someone could comment on whether I am way off the mark:

Let's assume that the original TM has 5 states and 8 symbols, so I can use a 3-bit encoding. After reading a word, we would have to encode writing and moving the head like this: 6 states are needed to get back to the beginning of the word that was just read (moving the head once without writing is the equivalent to read 0/write 0 and read 1+write 1, which takes 2 states - repeat 3 times to get to our goal position). Add another 3 states for writing the new word. 6 more states are needed to move back to the beginning of the word and another 6 states to move the head left or right.

The reading operation itself takes 8 states for a 3-bit word (ending up in one of 8 states depending on the result). From here the transition described above is started, so from a given state I can encode all transitions including the read operation in

This encodes all transitions of a single state from the original TM. Since we have 5 original states in total, the complete new TM would need 5*176 = 880 states.

In general terms I would need s*(1+(7*b))*q states, where s is the alphabet size and q the number states of the original TM, and b is the encoding bit depth ceil(log(s)).

This sounds about fair to me, is there something I overlooked? I think writing intermediate state information on the tape is out of the question with only 2 symbols so encoding everything in states seemed the only reasonable option to me.

PS: I also need some pointers in how to get the 2-symbol TM to encode the monadic input to binary. The link above does not give much useful input to this question.

Edit: In order to show that I already tried to get an answer to my second question, here is my specific problem: I have to convert multiple monadic inputs to their binary counterparts using a 2-symbol TM. So the input separation between monadic words would be achieved with a single zero and the end of the tape with two zeros. Now when I start the conversion process, I will have to go back and forth between the monadic inputs and the binary outputs. The binary outputs may naturally contain multiple zeros which would prevent me from finding the end of the tape in the conversion process. I could switch to a different end-of-tape word that cannot occur in the binary numbers, but I am not sure that this is really the most efficient option.

• It looks like you have answered the most of the question. The problem of encoding the input using two symbols is easy for you to solve on your own. I would encourage you to write your own answer, which could become the reference on how to encode any Turing machine to one with 2 symbols. – John L. Jul 11 '19 at 18:52
• encoding everything in states seemed the only [...] option.. Another idea might be to determine the minimum number of bits k needed to encode the tape symbols in binary. Turn the original TM T into a (k+1)-tape TM T' where tapes 1 through k move in sync. Iff in any computation of T, tape cell c holds symbol s at step t, a matching computation of T' cell c on tape i holds the bit #(i-1) of the binary encoding of s at step t. Tape k+1 at any time contains exactly two 1s marking the position of the first unused cell left or under, and right of the head, resp. – collapsar Jul 12 '19 at 0:15
• @Apass.Jack I amended my original question to show my thought process. It does not seem to be easy to solve, unless I am missing something. Any pointers to that easy solution would be appreciated. I think the issue is less trivial than it sounds. – Cerno Jul 12 '19 at 10:12
• @collapsar Thanks a lot for your thoughts.I am sorry that I forgot to mention that I am limited to a 1-tape machine. I will amend the original question. – Cerno Jul 12 '19 at 10:13
• What exactly is your question? It seems like you already know how to do the encoding, but not sure how to do it efficiently? – Dmitri Urbanowicz Jul 12 '19 at 10:31


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## Functions Modeling Change: A Preparation for Calculus, 5th Edition

Because of the $3^x$, it follows that the given function cannot be put in the form $y=kx^p$. Thus, the given function is not a power function.


text:
# Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Clever student:

I know!

$x^{0}$ =  $x^{1-1}$ = $x^{1} x^{-1}$ = $\frac{x}{x}$ = $1$.

Now we just plug in x=0, and we see that zero to the zero is one!

Cleverer student:

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

$0^{x}$$0^{1+x-1}$$0^{1} \times 0^{x-1}$$0 \times 0^{x-1}$$0$

which is true since anything times 0 is 0. That means that

$0^{0}$ = $0$.

Cleverest student :

That doesn’t work either, because if $x=0$ then

$0^{x-1}$ is $0^{-1} = \frac{1}{0}$

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function $x^{x}$ and see what happens as x>0 gets small. We have:

$\lim_{x \to 0^{+}} x^{x}$ = $\lim_{x \to 0^{+}} \exp(\log(x^{x}))$

= $\lim_{x \to 0^{+}} \exp(x \log(x))$

= $\exp( \lim_{x \to 0^{+} } x \log(x) )$

= $\exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )$

= $\exp( \lim_{x \to 0^{+} } -x )$

= $\exp( 0)$

= $1$

So, since  $\lim_{x \to 0^{+}} x^{x}$ = 1, that means that $0^{0}$ = 1.

High School Teacher:

Showing that $x^{x}$ approaches 1 as the positive value x gets arbitrarily close to zero does not prove that $0^{0} = 1$. The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that $0^{0}$ is undefined. $0^{0}$ does not have a value.

Calculus Teacher:

For all $x>0$, we have

$0^{x} = 0$.

Hence,

$\lim_{x \to 0^{+}} 0^{x} = 0$

That is, as x gets arbitrarily close to $0$ (but remains positive), $0^{x}$ stays at $0$.

On the other hand, for real numbers y such that $y \ne 0$, we have that

$y^{0} = 1$.

Hence,

$\lim_{y \to 0} y^{0} = 1$

That is, as y gets arbitrarily close to $0$, $y^{0}$ stays at $1$.

Therefore, we see that the function $f(x,y) = y^{x}$ has a discontinuity at the point $(x,y) = (0,0)$. In particular, when we approach (0,0) along the line with x=0 we get

$\lim_{y \to 0} f(0,y) = 1$

but when we approach (0,0) along the line segment with y=0 and x>0 we get

$\lim_{x \to 0^{+}} f(x,0) = 0$.

Therefore, the value of $\lim_{(x,y) \to (0,0)} y^{x}$ is going to depend on the direction that we take the limit. This means that there is no way to define $0^{0}$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$.

Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function $f(x,y) = y^x$ for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

$y^x$ := $1 \times y \times y \cdots \times y$

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

$y^{x}$ = $1 \times y$.

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

$y^{0}$ = $1$

which holds for any y. Hence, when y is zero, we have

$0^0 = 1$.

Look, we’ve just proved that $0^0 = 1$! But this is only for one possible definition of $y^x$. What if we used another definition? For example, suppose that we decide to define $y^x$ as

$y^x$ := $\lim_{z \to x^{+}} y^{z}$.

In words, that means that the value of $y^x$ is whatever $y^z$ approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

[Clarification: a reader asked how it is possible that we can use $y^z$ in our definition of $y^x$, which seems to be recursive. The reason it is okay is because we are working here only with $z>0$, and everyone agrees about what $y^z$ equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

$0^0$ = $\lim_{x \to 0^{+}} 0^{x}$ = $\lim_{x \to 0^{+}} 0$ = $0$

Hence, we would find that $0^0 = 0$ rather than $0^0 = 1$. Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what $y^x$ means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is $0^0$ really? Well, for x>0 and y>0 we know what we mean by $y^x$. But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of $y^x$ is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what $y^x$ means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that $0^0=1$? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use $0^0=0$ or if we say that $0^0$ is undefined. For example, consider the binomial theorem, which says that:

$(a+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}$

where $\binom{x}{k}$ means the binomial coefficients.

Now, setting a=0 on both sides and assuming $b \ne 0$ we get

$b^x$

= $(0+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} 0^k b^{x-k}$

= $\binom{x}{0} 0^0 b^{x} + \binom{x}{1} 0^1 b^{x-1} + \binom{x}{2} 0^2 b^{x-2} + \hdots$

= $\binom{x}{0} 0^0 b^{x}$

= $0^0 b^{x}$

where, I’ve used that $0^k = 0$ for k>0, and that  $\binom{x}{0} = 1$. Now, it so happens that the right hand side has the magical factor $0^0$. Hence, if we do not use $0^0 = 1$ then the binomial theorem (as written) does not hold when a=0 because then $b^x$ does not equal $0^0 b^{x}$.

If mathematicians were to use $0^0 = 0$, or to say that $0^0$ is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using $0^0 = 1$.

There are some further reasons why using $0^0 = 1$ is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.

This entry was posted in -- By the Mathematician, Math, Philosophical. Bookmark the permalink.

### 1,144 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

1. Stephan H. says:

The upper limit of the index k in the binomial theorem is n, not infinity. But the argument still holds. Thank you for convincing me: 0^0 is NOT (or rather, should not be allowed to remain) undefined, even though it is a singularity in z = x^y.

2. Alan Feldman says:

If one insists on y^x being a continuous function, then you run into multiple answers, depending on how you take the limit. But if you drop that requirement there are many reasons to define 0^0 as 1.

3. VDThinh says:

It is a mistake when you write 0^0=0^1. 0^{x-1} because 0^{x-1} don’t exist if x<1.

4. RustyFoxVevo says:

Most scholar mathematician:- By laws of indices, the high school teacher is right.

a^0=a^m/a^m which is 1 for (a,m)>1. But 0^0=0, for which the answer terms down to 0/0 which is not defined or imdetermined.

5. Anonymous Mathmatician says:

All of these could be possible for “anything” to the zero power is one. However, zero times itself is still zero. The question remains. Is this one, or zero? I believe that it is a problem that can not be solved in our particular dimension. Yet, in the fourth dimension it could be solvable because of it’s the warped structure making other numbers possible.

6. Banafsheh says:

Very interesting.
Thank you professor.

7. Stephan H. says:

Pascal says:
(a+b)^n= Σ(i=0:n) nCi a^i b^(n-i)
Example 1:
(1+2)^3=
1•1^0•2^3+
3•1^1•2^2+
3•1^2•2^1+
1•1^3•2^0
= 8+12+6+1 = 27
Example 2 (“proof” that 0^0 is 1):
(0+1)^2=
1•0^0•1^2+
2•0^1•1^1+
1•0^2•1^0
= (0^0) = 1

8. Stephan H. says:

The binomial formula nails down the answer.
Most directly:
1 =
(0+1)^1 =
Σ(i=0:1) 1!/(i!(1-i)!) • 0^i•1^(1-i) =
1•0^0•1^1 + 1•0^1•1^0
= 0^0
Q.E.D.

9. The answer can be demonstrated quite simply and it is undefined:

Consider:
x^0 = x^1 * x^-1

When we set x to 0, the equation becomes:

0^0 = 0 * (1/0) = undefined

As we know 1/0 is undefined, then we know that 0^0 is undefined.

10. Alan Feldman says:

Your first equation is not valid for x = 0, but you use it as if it is. So this does not prove anything.

11. HK says:

This “problem” is of the same type (though a little more involved) as determining 0! (zero factorial).
Given the natural definition for n-factorial (all natural numbers multiplied together up to n) and natural definition of a^n (a multiplied by itself n times), neither 0! nor 0^0 has any true conceptual meaning. They are mathematical objects of the mind that pop out because of our algebraic fiddling – that is to say, our use of notation.
A good example of this is the first term of the Taylor Series for e^x.
e^x = 1 + x + x^2/2 + x^3/6 + x^4/24 + …
We choose for notational convenience to rewrite this as
e^x = Σ(k=0:inf) [ x^k / k! ]
Now, looking carefully at the first term of our infinite sum, we find we have x^0 / 0!
Now, this first term equals 1. It equaled 1 before we chose to use infinite sum notation to represent it. Given that the first term in the Taylor Series is 1, and our convenient notation pops out x^0/0!, we must decide quickly that 0! must equal 1 to preserve the notation.
Now, what is e^0. Before we started using our summation notation, we knew it to equal 1. Now, for x=0, our convenient summation notation gives us a first term of 0^0/0!
Now we need to accept 0^0 = 1 to preserve out notation.
The only alternative here is to go back and rewrite our Taylor Series as
e^x = 1 + Σ(k=1:inf) [ x^k / k! ]
Neither 0^0 nor 0! have any sensible “meaning” related to our natural definitions of powers and factorials, but for the convenience of our most algebraically useful notation, we choose to say 0!=1 and 0^0 =1. We accept these values and carry on.

12. fatih says:

use polynom 0°= 1

13. STEPHAN HEUMANN says:

HK nailed it, both conceptually and formally.. we accept as a matter of definition that 0! = 1. Why is there any controversy that 0^0 = 1 as well? I think it’s because we think about 0*0 as a factor of 0^0, and trust intuition.

Fatih, what are you talking about? It sounds like you’re regarding a function polynom() in some math toolkit as the ultimate authority on math theory. (“Just google it!”) Or…?

14. Mathematical enthusiast says:

Actually 0^0 is one of the indeterminate forms according to limits.

If f(x)=0,g(x)=0. Lt x tends to a f(x)^g(x)=e^Lt tends to a f(x)g(x)

15. Alan Feldman says:

0^0 is not a limit. Neither is 1^oo, which is equal to 1. But 1^oo is often used to _represent_ a particular _form_ of limits. For example, lim(h->oo) (1+1/h)^h = e, not 1. Yet the base approaches 1 while the exponent grows without limit. Similarly, 0^0 is often used to represent the form of limits where the base and exponent both approach zero.

Various expressions fall into the 1^oo category. In general their limits don’t equal 1 or even each other. The same goes for 0^0. So what is 0^0? If you insist on 0^x being continuous, then you need limits. And in this case they do not give a unique answer. But why does it have to be continuous? If you want Sum(n=0->oo) x^n/n! to equal e^x for _all_ x, 0^0 must be 1. If you insist that 0^0 is undefined, then so is the preceding formula for x = 0. You can’t have it both ways. I choose 0^0 = 1 because I find it rather ugly to have an exception for the e^x series. In addition, every math book I have ever read which contains the series for e in summation form says that it is good for _all_ x. All x includes zero. Therefore, 0^0 = 1.

Bottom line, you have to choose between 0^0 = 1 and the summation form of the series for e^x being undefined for x = 0. Since there are many other situations where 0^0 = 1 is a very useful thing, it is wise to define it sas such.

16. Alan Feldman says:

I meant if you want y^x to be continuous at (0,0) you need limits. Sorry for the goof.

17. Richard says:

*Strictly speaking* zero to the zero power is not defined. However there are many situations where it is convenient to *assign* it a particular value. And for the vast majority of these situations, the value that works is 1.

18. Alan Feldman says:

@Richard. If you assign 1 to 0^0, you’ve defined 0^0 to be 1.

Assign and define are pretty much the same thing, but in different orders.

It is convenient to define y^0 to be 1.

It is convenient to define y^(-x) as 1/y^x for y != 0.

It is convenient to assign the nth roots of a number to the 1/nth power of a that number, using the positive root if n is even.

For real x and positive real y it is convenient to define y^x as e^(x ln y), where ln y is defined by the usual integral.

For complex numbers it is convenient to extend y^x using Euler’s formula.

For 0^0 it is convenient to assign it a value of 1.

What is the difference between these? Why are any of them “strictly speaking” more “correct” than the others?

And the clincher: According to you, Sum(n=0->oo) x^n/n! is e^x except at x = 0, where it is “strictly speaking” not defined. But it’s “convenient” to make it 1. Yet every math book I have ever seen that gives this summation formula for e^x says it is good for _all_ x. _All_ includes _0_.

Still further even also yet again: You have a choice. Accept that 0^0 = 1 (which it does), or add an exception to e^x = Sum(n=0->oo) x^n/n! for x = 0. I have yet to see any reference declare this exception, yet so many refuse to accept the inescapable consequence that 0^0 = 1.

And there’s more! It is convenient to define 0! as 1 (and, using a suitable argument [or operand], agrees with the Gamma function, which is a generalization or an extension of the factorial function).

And the same for nC0 = 1 (unless you use the factorial definition of nCk).

Why do we have all these convenient definitions and yet only 0^0 = 1 meets so much resistance from so many?

19. Richard says:

@Alan Feldman
I certainly did not mean to open a can of worms.  Perhaps my choice of the words “strictly speaking” and “assign” were not the best.  But I largely stand by my assertion that, applying the most rigorous standard, 0^0 is not defined.  This is *not* my own proclamation, but rather what I was unequivocally taught by every math professor I had at Villanova (most of them PhD’s) in the course of obtaining my B.S. degree in mathematics (admittedly forty years ago).  I agree that, when it comes to 0^0, there are times when it’s best to  consider it as representing 1.    But if you insist that 0^0 is formally, rigorously mathematically defined, then (unless things have changed in the last few decades) your argument is not so much with me but rather with the whole academic community.
As to why “0^0 = 1” is so controversial, I really don’t know.

20. Bobbie says:

Simple answer for a GCSE Student? Please no complicated things, just simple yes no, 0 or 1, thanks.

21. Richard says:

@Bobbie
I’m afraid that any simple yes no 0 or 1 answer would be inadequate.  If you’re getting your information from a college math professor or a math textbook, you will most likely be told that, because of certain fundamental technical issues, 0^0 is mathematically undefined.  Having said that, there are situations where 0^0 does come up, and in most of those cases it represents a 1.   Some but not all math buffs take that as proof that 0^0 is fundamentally intrinsically 1; but that opinion is controversial.  I realize that this is not the simple uncomplicated answer you’re looking for.  But I hope you find it at least somewhat helpful.

22. Alan Feldman says:

@Richard

What is your “rigorous standard”? How is 0^0 = 1 any less valid than any of the other generalizations of y^x?

Those math professors are wrong.

Again, you cannot have e^x = Sum(n=0->oo) x^n/n! work for all numbers, but I can virtually guarantee those math professors would say the formula is good for _all_ x, and _all_ x includes _0_. So they contradict themselves. You cannot have this e^x formula work for all numbers unless 0^0 = 1. Can you please address this and my other points? You have to choose: Accept that 0^0 = 1, or add an exception to the e^x series. Please tell me what you choose.

BTW, software is drifting in the direction of 0^0 = 1. And it certainly makes programming series easier!

23. Alan Feldman says:

@Bobbie

0^0 = 1

I don’t know what the test graders will assume. Just pray it doesn’t come up. I think it’s rather unlikely. But if they insist it’s undefined, they’re wrong. (^_^)

24. Richard says:

@Alan Feldman
I am not a mathematician, so I must rely on what the experts say.  I’m not going to dismiss the teachings of a dozen university professors just because someone I don’t even know says they are all wrong.  Please give me a few days; I will see if I can contact some of them and ask them about this.  They are PhD’s who have spent their lives studying and teaching math.  It’s their job to know this stuff.  I trust their judgement more than I trust what I read in a discussion thread on the Internet.  In the meantime, with all due respect, exactly what are *your* credentials?

25. Alan Feldman says:

Look. There are two choices:

1) 0^0 is undefined, in which case e^x = Sum(n=0->oo) x^n/n! is undefined for x = 0, and is therefore not valid for all x, or

2) 0^0 = 1, in which case e^x = Sum(n=0->oo) x^n/n! is valid for all x, including 0.

You can’t take one from column 1 and one from column 2. You must choose between 1 and 2. I keep making this point, and you keep ignoring it, as do all the other “undefiners”. Ask these professors which of the two statements is true. Let me know what they say.

This is a simple argument. If you want the series for work for x=0 you have to have 0^0=1. It cannot be otherwise, regardless of anyone’s credentials.

The mathematician here says it’s 1.

Hey, I have a reference book that says 0^a = 0 for a != 0. I don’t believe it, and never have. If a is negative, you’re dividing by 0 and that’s a big no-no.

Again, I want someone to answer my challenge of statement 1 vs 2 above. You can’t have it both ways.

Bottom line: Either 0^0 = 1, or e^x = Sum(n=0->oo) x^n/n! has an exception at zero.

You can’t have both 0^0 undefined and the series work for all x. You can’t! Ask your professors.

If you want credentials, take Euler. He said 0^0 = 1. It’s hard to outdo his credentials!

26. The Physicist says:

@Alan Feldman
I hate to step into the middle of this debate. On behalf of the other readers, I genuinely appreciate that you’re having it.
The Taylor expansion of e^x is $e^x=1+x+x^2/2+\ldots$ To save space we typically write that as $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$ with the understanding that we’re using $\frac{x^0}{0!}=1$, even when x=0, because the first term in the expansion is always 1. Here 0^0=1 is just a matter of notational convenience, not something particularly profound.

27. Richard says:

@The Physicist
@Alan Feldman
I very much like the idea of describing 0^0=1 as “a matter of notational convenience, not something particularly profound.”    I would even go one step further and call it “a matter of notational convenience, not something rigorously and fundamentally proven.”    I was reading some old postings in this discussion, and I think Brian Wynne summed it up best of all in his posting of May 31, 2016, which begins: “Declaring or defining a value for a troublesome expression is not proof. How you define the value of 0^0……….”    (I hope you’ll read his whole posting; I think he makes some excellent points.)   My personal opinion is that there is nothing inconsistent about considering 0^0 to be undefined/indeterminate in the formal sense, and still defining it as 1 when doing so simplifies notation and/or avoids having to have special cases.

28. Ali says:

a^0 = 1, provided that “a not equal to 0”
Proof:
a ^ 0 = a ^ (1-1)
= a ^ 1 x a ^ – 1
= a / a = 1
But we know that a / a = 1 is true when “a not equal to 0” since 0 / 0 is undefined.

29. Alan Feldman says:

@Ali,

Howard Ludwig and I have explained why your argument is not definitive. In brief, you could just as well say that 0^2 is undefined, because 0^2 = 0^3 / 0^1, which is division by zero, which is undefined. Hell, you could even say 0 itself is undefined by the same argument! There are several other ways to show that 0^0 = 1. Please see posts by Howard and me.

30. Jody says:

a^k = 1·a^k
So…
a^3 = 1·a^3 = 1·a·a·a
a^2 = 1·a^2 = 1·a·a
a^1 = 1·a^1 = 1·a
a^0 = 1·a^0 = 1
Let a = 0
0^3 = 1·0^3 = 1·0·0·0
0^2 = 1·0^2 = 1·0·0
0^1 = 1·0^1 = 1·0
0^0 = 1·0^0 = 1

Maybe I’m over simplifying everything, but an exponent is an operation defined as repeated multiplication. So when 0 is the exponent, the base is never multiplied and 1 is the only factor left. This is the case for any number as well as zero.

31. Alan Feldman says:

Richard writes: “I very much like the idea of describing 0^0=1 as “a matter of notational convenience, not something particularly profound.” I would even go one step further and call it “a matter of notational convenience, not something rigorously and fundamentally proven.” ”

Not proven? Not proven based on what? How is it proven that y^x = e^(x ln y)? (Assuming positive real y and real x, of course.)

Please read my posts, and those of betaneptune and Howard Ludwig. We have addressed all these points (including Brian Wynne’s argument, which actually disproves itself!), and then some. Please tell us why our points are wrong, instead of repeating points we have addressed.

Again: We have addressed all these points. Please stop repeating them and tell us why our points are wrong.

32. Alan Feldman says:

@The Physicist

When we write the e^x formula in summation form, we are actually being more precise. There is no need to infer the pattern of succeeding terms, because it is specified explicitly.

You say then that 0^0 = 1 is simply notational convenience and not something profound. One could say the same about 0! = 1, and nC0 = 1 (the combination bit).

I suppose that by saying it’s not “profound” you mean it can’t be derived from the core definition of exponentiation, y^n = n factors of y multiplied together. So how do you derive y^(-n) = 1/y^n for n a positive integer? (And y^0, too.) You don’t and you can’t. How can you multiply something by itself a negative number of times? You can’t. So you define it to be that because it “makes sense”. It is useful and consistent with the basic laws of exponents. You do it by continuing the pattern using division. You have _generalized_ the concept of exponentiation, not derived anything. The same goes for y^x = e(x ln y). And in that case the typical derivation in calculus is to choose certain integral in a totally ad hoc manner and show it has all the right properties. You can, as Howard Ludwig has in one or more of his posts, that 0^0 has to equal 0 or 1 to satisfy the exponent laws. Choosing 0 causes an ugly exception to y^0 = 1 and is not useful for anything anyway. So it’s obviously 1.

It seems to me that by making all these definitions we are simply generalizing the idea of exponentiation in the most sensibly possible way. I mean, take 2^3.7. What does it mean to multiply 3.7 factors of 2? It makes no sense. We have done is to extend or generalize the definition of exponentiation in a useful way. One can then generalize even further to complex numbers by using Euler’s formula.

Why is 0^0 any less “profound” than the other generalizations? These other generalizations are simply ways that satisfy the exponent laws when the exponent is not a positive integer. I suppose you can say those are derivations, but it is the same as what we do to determine 0^0. And there are actually even more situations showing that 0^0 = 1. Howard Ludwig has explained this in great detail in his posts in this thread.

Please check my other posts, and those of betaneptune and Howard Ludwig, and tell us where we went astray.

33. The Physicist says:

@Alan Feldman
My concern was that you may have been making an unwarranted appeal to authority, by claiming that e^x is defined as $e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$ as opposed to $e^x=1+x+\frac{x^2}{2}+\ldots$. The second form is the Taylor polynomial and is exactly the same as e^x. The first is the same if you (fairly) define 0^0=0!=1.

I would say that 0!=1 is more convenient and that’s why we use it. And you’re right, defining y^(-n) to be the multiplicative inverse of y^n in order to be consistent with the exponential rules for positive numbers is a reasonable thing to do, but that doesn’t make it the “right” thing to do. It’s reasonable because we find that we can apply the exponential rules generally (not just to positive integers), when y^(-n) is properly defined. In fact, that is why it is so defined.
The particular issue with 0^0, is that there’s no one immediate “correct” definition. It’s been pointed out in this thread that you can’t “do the math” and find the value of 0^0 directly the way you can with, say, 2^3. Instead, when we bother to give it a value, it has to be in some context. For 0! there is a single defining principle (that n!=n(n-1)! and n!=1*2*3…n for n≥1) that implies that 0!=1. We need to follow a pattern and plug in a definition that works, just like with negative exponents. But that doesn’t mean that 0!=1 is “true”.
In order to define the value of something that doesn’t intrinsically have a value, we extend patterns. In this case, that generally means using limits. Regardless of how you do the limits, $\lim_{x\to2,y\to3}x^y=8$. Therefore, it is fair to define 2^3=8 (assuming that, for some reason, that was difficult to figure out directly) without any context. However, if you do the same thing for $\lim_{x\to0,y\to0}x^y$ you find that you suddenly need to be very careful about how you do the limit. In particular, if you do the x limit first you’ll get 0 and if you do the y limit first you’ll get 1.
This is what x^y looks like near 0^0; it’s not pretty.
That said, it is definitely reasonable, in most reasonable situations, to declare that 0^0=1.

34. Alan Feldman says:

@The Physicist

The two forms of the Taylor series for e^x are the same. Every math book I’ve ever read that gives the summation version says that it is valid for all x. “All x” includes zero. Therefore it is valid for zero. And that can only true if 0^0 = 1. Therefore, either 0^0 = 1, or all the math books are wrong and the summation version is not valid for x = 0. You can’t have it both ways.

In _Introduction to Analysis_ by Rosenlicht there is a section on power series. His proofs work only if 0^0 is defined. In one instance he talks about the series Sum(n=0->oo) c_n (x-a)^n. In some cases it converges only if x = a. So if 0^0 is undefined, such series don’t converge and the proof falls apart. Basically the radius of convergence is defined assuming the power series works at least for x-a = 0. And that can only be true if 0^0 is defined. And in the case of many series, like e^x, cos x, the binomial theorem — they only work if 0^0 = 1. So it’s 1. I’m not aware of any exceptions.

0! = 1 is true. It’s a generalization of the factorial, if you wish, as is the Gamma function! Hey, you can define n! as 1*1*2*…*n and 0! is 1 without any qualms.

You can’t “directly find the value” of the other generalizations of y^x. Why is this one any different?

Limit arguments only apply if you insist that the function be continuous. Why does it need to be continuous?

Using the exponent laws alone you can narrow down 0^0 to either 0 or 1, as Howard has shown. Series written in summation form that start with a constant require 0^0 = 1. So it’s 1.

It’s the null or empty product. It’s the number of mappings or functions from the empty set to the empty set. It works for the derivative of x^n. It works for y^0 = 1. Howard Ludwig discusses at least some of these and others in his posts. He also makes the point that why say, paraphrasing, ~”It’s undefined, but wherever it shows up you need to substitute 1. What sense does that make?~”

If 0^0 = 1 is only “definitely reasonable”, then why can’t the same be said for y^(-n), n a positive integer, y^0 = 1, and y^x = e^(x ln y)?

Please read my previous posts, and those of betaneptune and Howard Ludwig, and tell us where we went wrong.

35. The Physicist says:

@Alan Feldman
In the books you’re talking about, do they explicitly say “0^0=1 in all cases”, or is that only implied because it’s used? I use that convention all the time, but it’s important to be careful when you do because, to be clear, 0^0 is a known and established indeterminate form. If any book says that it has a single, definite value in all cases, then that book is definitely wrong.

36. Alan Feldman says:

No. But it has to be true for the summation formulas for power series to be valid at zero.

Indeterminate forms apply to categories of limits. Best example I can think of is 1^oo. Clearly this is one, but you can make it any positive value you want using lim(h->oo) (1+x/h)^h, which gives e^x. But clearly this doesn’t mean that 1^oo = e^x (!). 0^0 as an indeterminate form applies to an expression where both the base and the exponent go to zero as the limit is taken. It doesn’t mean that 0^0 itself is indeterminate any more than lim(h->oo) (1+x/h)^h implies that 1^oo is indeterminate. 1^oo = 1, as does 0^0.

37. Brian Wynne says:

Why do you think that, if 0^0 is so obviously equal to 1, the entire math community has not wholeheartedly embraced it and defined it to be so? Yet, there is still something inherently unsettling about defining it for the sake of implication. Perhaps, the resistance is due to the fact that 0^0 = 1 does not hold for ALL of the different implications that arise FROM IT. You have eloquently shown a one-sided argument by clearly demonstrating so many instances that lead TO IT. But a good definition is biconditional. If your definition holds, then a^0 = 1, for ALL a, including zero. This implies that (a^0) / (0^0) = 1, for all a.

(a^0) / (0^0) ?= (a/0)^0, for all a.

38. Richard says:

@Alan Feldman
@The Physicist
@Brian Wynne
Alan Feldman writes:
“You can, as Howard Ludwig has in one or more of his posts, that 0^0 has to equal 0 or 1 to satisfy the exponent laws. Choosing 0 causes an ugly exception to y^0 = 1 and is not useful for anything anyway. So it’s obviously 1.”
This seems to suggest that difficult mathematical issues can be rigorously and formally resolved by simply rejecting what we think is “ugly” and accepting as true what we see as “useful.”
(And for that matter, what about the “ugly exception” to 0^y=0 caused by choosing 1?)
I think most would agree that pragmatically defining 0^0 to be 1 in most instances is useful and logical.  But that’s not the same thing as establishing by rigorous proof that 0^0 is fundamentally, intrinsically, and exclusively equal to 1.   The sigma notation for the Taylor series for e^x (and for any Taylor series for that matter) does not prove that 0^0=1.   The reason for assigning the value 1 to 0^0 in the first place is to make it “work” for that (and similar) notation.  Using that notation as proof is begging the question.

39. STEPHAN H. says:

It’s time to close this discussion.

Those high-school math teachers] (if they insist that 0^0 is 0 or undefined) are just plain wrong. They do correctly assert that x^0 = 1 for x > 0, so they will concede that e^0 = 1 (where e=2.71828… is the base of natural logarithms).
The definition of the fundamental exponential function exp(x) = e^x , is:
exp(x) ≡ Σ(k=0, 1, … ∞) x^k/k!
In the simplest case, exp(0) = e^0 = 1 , all but the first term in the series vanish because for k>0, 0^k = 0 and k! > 0 (so 0^k/k! = 0). This leaves
exp(0) = 1 = 0^0 / 0!
and since 0! = 1 (a settled fact),
0^0 = 1
QED.

It is time to close this discussion.

40. Alan Feldman says:

@Brian Wynne

(a^0) / (0^0) ?= (a/0)^0, for all a.

No. This doesn’t work for any a. And the reason it doesn’t work is that the exponent laws are not valid when there is division by zero. But it is still true that 0^0 = 1.

The fact that you can’t express the LHS in the form of the RHS means nothing as to the value of the LHS.

Nonsense from nothing:

0 = 0
a^2-a^2 = a^2-a^2
a^2-a^2 = a(a-a)
a^2-a^2 = (a+a) (a-a)
(a+a)(a-a) = a(a-a)
a+a = a
2 = 1

The above also doesn’t work when you allow division by zero, which is what you’re doing when you cancel the (a-a)’s. This doesn’t mean that and of the equalities before canceling out the (a-a)’s aren’t true.

41. Alan Feldman says:

@Richard

I’m still waiting for the “rigorous proof” that y^x = e^(x ln y). And when you prove it you have to state your assumptions. The only “pure” case is the basic exponent laws with positive integral exponents. The rest is generalizing via definitions.

And you clearly didn’t understand my points. Let’s try again:

There are times when multiple solutions come up. You are allowed to reject the nonsensical ones. What is the length of a side of a square with an area of 4? Well, we know that the area is the length squared, so we can say that L^2=4. This gives L=+/-2. We reject the -2, as it is meaningless, and have rigorously proven that the side of the square is 2. We do similar things in physics. The integral of the wave function over all space is not allowed to become infinite. So any wave function that grows without limit as x increases is properly rejected, even though it may be a solution of the Schroedinger equation. So by using the basic exponent laws we can easily show that 0^0 has to be zero or 1. Everything else points to 1. So it’s 1. We reject the bad answer just as we do in many other cases. Remember, the exponent laws are not enough to determine an an answer, but it offers a solution that is consistent with every other method of determining 0^0. You are focusing on the exponent-law argument to the exclusion of everything else. You are “cherry picking”, or whatever its opposite is. Everything I have said in totality clearly points to 0^0 = 1.

As for my summation format argument, again you missed the point. Every book that gives the formula for e^x in summation form also says it is valid for all x. All x includes 0. There are no asterisks with side conditions or what not. All x means all x. And for the formula to be true for 0, 0^0 has to be 1. As I’ve said many, many times, you have two choices:

1) 0^0 = 1 and e^x = Sum(0->oo) x^n/n!

-XOR-

2) 0^0 is undefined, and therefore we have an exception in the e^x formula for x = 0, meaning that the formula is not true for all x.

Please choose 1 or 2. Don’t sidetrack into other arguments as you’ve just done. Don’t tell me it doesn’t prove anything. Don’t go off on any other tangents. This is a simple question: Is it 1, or is it 2? You can’t have it both ways.

No one has answered this question. You can be the first.

42. Alan Feldman says:

Wow, I spoke too soon! STEPHAN H. has answered my question. And the answer is 1. Thank you STEPHAN H.

OK.

43. Richard says:

@STEPHAN H.
If you feel the discussion should close, then I don’t think it’s such a good idea to make a posting declaring that you are right and proclaiming to all who disagree with you that we are wrong.  That’s like sending us engraved invitations to continue presenting *our* arguments.
I actually agree with you that it’s getting to be time to end the discussion, for the simple reason that it’s just not getting anywhere.  No one is changing anyone’s mind about anything.
I feel a good way to move in that direction is to see if there is something, no matter how small or trivial, that we can all agree on.  Something like:
One of the most debated controversies in mathematics is the question of zero to the zero power.  This issue has been argued for centuries, and the disagreement is not likely to end.
I doubt anyone would disagree with that.  And I suspect it’s probably the only thing we’ll all be able to agree on.

44. Alan Feldman says:

@Richard, who writes:

“Alan Feldman writes:
“You can, as Howard Ludwig has in one or more of his posts, that 0^0 has to equal 0 or 1 to satisfy the exponent laws. Choosing 0 causes an ugly exception to y^0 = 1 and is not useful for anything anyway. So it’s obviously 1.”
This seems to suggest that difficult mathematical issues can be rigorously and formally resolved by simply rejecting what we think is “ugly” and accepting as true what we see as “useful.”
(And for that matter, what about the “ugly exception” to 0^y=0 caused by choosing 1?)”

Sorry, but I have more to add.

There’s another ugly thing and it occurs with y^x = e^(x ln y).

We have the exponent law y^m y^n = y^(m+n). Clearly this works for positive integral exponents. But when we try to extend to rational exponents, we run into a problem.

What should 4^(1/2) be? Well, we want it to satisfy the exponent laws. And with this one we have
(+/-2)^2 = 4. But we want a unique value. So we throw out the negative one. We need to do this for all even roots. But the above formula does it for us. So even with this accepted formula we are throwing out ugly values, in no small part because it’s useful to do so.

Now I completely forgot to comment on 0^y = 0. I don’t think it’s ugly to choose 0^0 = 1. Zero is a special case. E.g., you can’t take 0 to a negative exponent (and probably not an imaginary one, either). Notice that we set 0^y for positive real y to be 0. For positive integral exponents this is obvious. But we do it for the in-between values because it makes the exponent laws work. And notice that we can’t use the log definition! Additionally y^0 is far more important than 0^y. Maybe instead of ugly I should have said impractical. But remember it’s much more than this that shows that 0^0 = 1. Much more. Howard Ludwig, betaneptune, and I have written much about that in our other posts.

OK.


text:
# Imaginary line (mathematics)

In geometry, an imaginary line is a straight line that only contains one real point. It can be proven that this point is the intersection point with the conjugated line.

It is a special case of an imaginary curve.

It can be proven that there exists no equation of the form $ax+by+cz=0$ in which a, b and c are all real coefficients. However there do exist equations of the form $ax+by+cz=0$, but at least one of the coefficients need be nonreal.

As follows, it can be proven that, if an equation of the form $ax+by+cz=0$ in which a, b and c are all real coefficients, exist, the straight line is a real line, and it shall contain an infinite number of real points.

This property of straight lines in the complex projective plane is a direct consequence of the duality principle in projective geometry.

In the complex plane (Argand Plane), we have a term called "imaginary axis".In Argand plane, y-axis is imaginary axis. All numbers in this axis are in form of 0+ib form.

## Argument

An argument is the angle or projection of any complex number in the Argand plane on the real axis (x-axis), denoted Arg(z). The argument can be easily found by following procedure:

```      If a+ib is any complex number foming angle A on real axis then,
cosA = a/√a^2+b^2 sinA= b/√a^2+b^2 tanA=b/a
```

arg(z)=A

## Properties of argument

• arg(AxB)=arg(A) + arg(B)
• arg(A/B)=arg(A) - arg(B)
• arg(z)=0 if and only if z lies in +ve real axis
• arg(z)=180 if and only if z lies in -ve real axis
• arg(z)=90 if and only if z lies in +ve imaginary axis
• arg(z)=-90 if and only if z lies in -ve imaginary line

Domain of argument = R Range = (-180,180)

## Modulus

Modulus of any complex no. a+ib is

mod(z)=√a^2+b^2

In Argand plane, modulus denotes distance between a complex number and the origin (0,0).

Example: mod(z)=2 denotes locus of all complex numbers z lying in circle of radius 2 at centre (0,0)


text:
# Associated bundles

If two $${G}$$-bundles $${(E,M,F)}$$ and $${(E^{\prime},M,F^{\prime})}$$, with the same base space and structure group, also share the same trivializing neighborhoods and transition functions, then they are each called an associated bundle with regard to the other. It is possible to construct (up to isomorphism) a unique principal $${G}$$-bundle associated to a given $${G}$$-bundle; going in the other direction, given a principal $${G}$$-bundle and a left action of $${G}$$ on a fiber $${F}$$, we can construct a unique associated $${G}$$-bundle with fiber $${F}$$. In particular, given a principal bundle $${(P,M,G)}$$, the rep of $${G}$$ on itself by inner automorphisms defines an associated bundle $${(\mathrm{Inn}P,M,G)}$$, and the adjoint rep of $${G}$$ on $${\mathfrak{g}}$$ defines an associated bundle $${(\mathrm{Ad}P,M,\mathfrak{g})}$$. If $${G}$$ has a linear rep on a vector space $${\mathbb{K}^{n}}$$, this rep defines an associated bundle $${(E,M,\mathbb{K}^{n})}$$, which we explore next.

The above depicts how given a principal bundle, we can construct an associated bundle for the action of $${G}$$ on a vector space $${\mathbb{K}^{n}}$$ by a linear rep, on itself by inner automorphisms, and on its Lie algebra $${\mathfrak{g}}$$ by the adjoint rep. The action of the structure group is shown in general and for the case in which $${G}$$ is a matrix group, with matrix multiplication denoted as juxtaposition. Although denoted identically, the $${f_{i}}$$ are those corresponding to each bundle.

 Δ The $${G}$$-bundle $${E}$$ with fiber $${F}$$ associated to a principal bundle $${P}$$ is sometimes written $${E=P\times_{G}F\equiv(P\times F)/G}$$, where the quotient space collapses all points in the product space which are related by the right action of some $${g\in G}$$ on $${P}$$ and the right action of $${g^{-1}}$$ on $${F}$$.


text:
06 325 750

# Station  DMore Machines

Play with the $1 \leftarrow 4$ machine and the $1 \leftarrow 5$ machine.

### Question 1

What is the $1\leftarrow 4$ code for the number twenty-nine?

### Question 2

What is the $1\leftarrow 5$ code for the number twenty-nine?

### Question 3

Can you work backwards?
What number has $132$ as its code in a $1\leftarrow 4$ machine?

### Question 4

What number has $132$ as its code in a $1 \leftarrow 5$ machine?

Well done! Now you can either go to the next island or perhaps go to the Wild Exploration Station below.

Register NOW and unlock all islands!


text:
## Problem 1¶

(a) Write a trapezoid(y,h) function that approximate definite integrals using the Trapezoid Rule

\begin{equation*} \int_a^b f(x) dx = h \left(\frac{1}{2} y_0 + \sum_{i=1}^{N-1} y_i + \frac{1}{2} y_N \right) + O(h^2) \end{equation*}

In this formula, the $y_i$ are values $y_i = f(x_i)$ on $N+1$ evenly spaced gridpoints between $a$ and $b$, where $x_i = a + i h$, and $h = (b-a)/N$.

Note that the natural mathematical notation for the Trapezoid rule uses 0-based indexing, but Julia uses 1-based indexing for its vectors. You will have to negotiate this difference when writing your code (i.e. adjust the formula to use 1-based indices).

In [ ]:

(b) Test your trapezoid(y,h) function on the integral $\int_0^1 x \, dx = 1/2$. To do this, choose a smallish value of $N$ (perhaps 10), construct a vector $y$ of length $N+1$, fill it with evelny spaced values of $y=f(x)$ between 0 and 1, and then run trapezoid(y,h) using the appropriate value of $h$. If you don't get 0.5 as an answer, something's wrong!

In [ ]:

## Problem 2¶

Write a wrapper function trapezoid(f, a, b, N) that takes a function $f(x)$, and interval $a,b$ and a discretization number $N$, uses those arguments to construct a vector $y$ and a gridspacing $h$, and then calls trapezoid(y,h). Test it on $f(x) = x, a=0, b=1, N=10$.

In [ ]:

## Problem 3¶

Use the trapezoid(f,a,b,N) function to approximate the integral

\begin{equation*} \int_0^1 x\, e^{2x} \, dx \end{equation*}

using $N=16$ and $N=32$. The exact value of the integral is $(e^2-1)/4$. What is the error in both cases? Does the error scale as expected, $O(h^2)$?

In [ ]:

## Problem 4¶

(a) Write a simpson(y,h) function that approximates definite integrals using Simpson's rule

\begin{equation*} \int_a^b f(x) \, dx = \frac{h}{3} \left(y_0 + 4 \sum_{odd \, i=1}^{N-1} y_i + 2 \sum_{even \, i=2}^{N-2} y_i + y_N\right) + O(h^4) \end{equation*}

As in the Trapezoid Rule, the $y_i$ are values $y_i = f(x_i)$ on $N+1$ evenly spaced gridpoints between $a$ and $b$, where $x_i = a + i h$, and $h = (b-a)/N$.

Remember that for Simpson's Rule $N$ must be odd! And be careful, as you implement this formula written with 0-based indices using 1-based Julia indices, even and odd switch places!

In [ ]:

(b) Test your simpson(y,h) function on the definite integral $\int_0^1 x^2 \, dx = 1/3$ using a smallish value of $N$.

In [ ]:

## Problem 5¶

As in problem 2, write a wrapper function simpson(f, a, b, N) that translates its arguments into appropriate values y,h and then calls simpson(y,h). Make sure it gets the right answer to $\int_0^1 x^2 \, dx = 1/3$.

In [ ]:

## Problem 6¶

Use the simpson(f,a,b,N) function to approximate the integral

\begin{equation*} \int_0^1 x\, e^{2x} \, dx \end{equation*}

using $N=16$ and $N=32$. The exact value of the integral is $(e^2-1)/4$. What is the error in both cases? How do those errors compare to those of the Trapezoid Rule? Does the error scale as expected, $O(h^4)$?

In [2]:

Out[2]:
simpson (generic function with 2 methods)

## Problem 7¶

Make a log-log plot of error versus $N$ for the integral of problem 6, for $N=2^n+1$ for $n=2$ through $10$. Show the error for Trapezoid in red and the error for Simpson in blue. Can you relate the slope of the log-log error lines to the expected errors $O(h^2)$ and $O(h^4)$?

In [ ]:


text:
• 论文 •

### 耐受偏二甲肼的芦苇变异株系的筛选

1. 1 中国科学院植物研究所光合作用与环境分子生理学重点实验室,北京 100093
2 北京特种工程设计院,北京 100028
• 收稿日期:2004-05-26 接受日期:2005-06-17 出版日期:2005-05-26 发布日期:2005-08-30
• 通讯作者: 麻密
• 基金资助:
全军重点污染治理项目(SGS/2001-07)

### SELECTION OF UDMH TOLERANT VARIANT LINES OF AN AQUATIC REED

YANG Shu-De1, XU Wen-Zhong1, LIU Shi-Rui2, ZHANG Tong2, MA Mi1,*()

1. 1 Key Laboratory of Photosynthesis and Environmental Molecular Physiology, Institute of Botany, Chinese Academy of Sciences, Beijing 100093, China
2 Beijing Special Engineering Design and Research Institution, Beijing 100028, China
• Received:2004-05-26 Accepted:2005-06-17 Online:2005-05-26 Published:2005-08-30
• Contact: MA Mi
• About author:* E-mail: mami@ibcas.ac.cn

## 被引次数

3 | 1

Abstract:

Unsymmetrical dimethyl-hydrazine (UDMH) is a highly volatile and reactive compound widely used as a liquid rocket propellant fuel in manned and unmanned space and missile systems. Exposure to UDMH is an important occupational health issue, which will increase in significance as space applications increase. Several methods have been tested for removing UDMH from the environment, such as catalytic oxidation, active carbon absorption and ion exchange resins. However, little attention has been given to the application of constructed wetlands to remove UDMH from groundwater even though constructed wetlands are a promising measure for wastewater treatment.
In expectation of constructing of an effective wetland by the variant reed lines to deal with UDMH polluted groundwater, attempts were made to screen for UDMH tolerant variant lines of reeds using cellular engineering technology in this study.
Seeds of the reed (Phragmites communis) collected from the river banks near by a satellite-launching basement were sterilized with 70% ethanol and 2% hypochlorous acid and then transferred to a MS medium for germination. After germination, hypocotyl seedling segments were used to induce callus on the MS medium supplemented with 1 mg·L-1 2,4-dichlorophenoxyacetic acid. The UDMH-tolerant cell lines were obtained by continuous selection on the UDMH containing medium supplemented with 0.5 mg·L-1 2,4-dichlorophenoxyacetic acid. Three levels of UDMH tolerant cell lines were screened for that contained either 1.63, 3.26, or 8.15 mmol·L-1 UDMH after 3-6 times of subculture. Relative growth, measured at the end of a 23-day incubation, was 90.4%, 84.3%, and 43.4% of the control in the three treatments, respectively, and, at the end of a 43-day incubation, was 95.6%, 91.7%, and 46.8% of the control, respectively. These results showed that UDMH could inhibit the growth of the callus at 8.15 mmol·L-1. Regeneration of plantlets from the selected cell lines was conducted on UDMH containing MS medium supplemented with 0.1 g·L-1 6-Benzylaminopurine and 0.01 g·L-1 1-naphthyl acetic acid. However, only the cell lines grown in the lower two treatments could be successfully differentiated, the 8.15 mmol·L-1 UDMH failed to differentiate. These results indicated that the suppressive effect of UDMH on callus differentiation was greater than on callus growth. Plantlets screened under UDMH stress were more tolerant to UDMH than the control, which were plantlets regenerated from callus without UDMH stress. Root growth of the differentiated plantlets was enhanced by transferring them to a hormone-free MS medium supplemented with UDMH for 35 days prior to successful transplantation of the plantlets into the soil. Finally, more than 70% of regenerated plantlets could be successfully transferred to the greenhouse. Further experiments estimating the growing status and remediation efficiency in UDMH polluted water of these variant seed lines were undertaken.


text:
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# Questions

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### Strong Law of Large Numbers for weakly dependent random variables

oct 31 09 at 2:50 michael lacey 2264

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jul 28 11 at 18:57 Adrien Hardy 766211

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may 25 at 19:05 Kjetil B Halvorsen 91648

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mar 21 10 at 1:31 Dan Lee 43315

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text:
Open Access Publications from the University of California

## Precision Black Hole Mass Measurement in Luminous Early-Type Galaxies

• Author(s): Boizelle, Benjamin David
The supermassive black hole (BH) census remains very incomplete at the highest masses ($\gtrsim10^9$ $M_\sun$), limiting our understanding of BH/host galaxy co-evolution for the most massive galaxies. I present analysis of circumnuclear ionized atomic and both warm and cold molecular gas kinematics, as well as results of detailed gas-dynamical modeling, for a sample of nearby, luminous early-type galaxies (ETGs). Keck OSIRIS integral-field unit and \textit{Hubble Space Telescope} (\textit{HST}) STIS spectroscopy reveal H$_2$ 1$-$0 S(1) and H$\alpha$ in approximately Keplerian rotation within \mbox{NGC 1275}. We build and optimize models of dynamically-warm disk rotation to better constrain its BH mass $M_{\rm BH}=(1.20^{+0.32}_{-0.44})\times10^9$ $M_\sun$, although significant turbulent velocity dispersion complicates model fitting.
Dynamically cold molecular gas emission is free from nearly all of the systemics that plague $M_{\rm BH}$ determination when using dynamically warm tracers. I present Atacama Large Millimeter/submillimeter Array (ALMA) $\sim0\farcs3-$resolution CO(2$-$1) and continuum imaging of a sample of early-type galaxies (ETGs) that host circumnuclear dusty disks. Given their incredible data sensitivities, these dynamically cold disks yield the most sensitive possible tracers of the central gravitational potential. In several targets we detect central velocity upturns that suggest unresolved Keplerian rotation. We find these disks are formally stabilized against fragmentation and gravitational collapse. The continuum emission is in all cases dominated by a central, unresolved source and appears to be consistent with hot accretion.
I present follow-up, $0\farcs1-$resolution CO(2$-$1) imaging of one of these promising ETGs -- NGC 3258 -- to highly resolve its BH sphere of influence and map out the molecular kinematics in exquisite detail. Its very regular, nearly symmetric rotation enables the most detailed thin disk modeling to date. Moderate kinematic twists are incorporated using tilted-ring formalism, and the extended mass profile is amended to allow for a variable stellar mass density profile. Preliminary model optimization returns $M_{\rm BH}=2.23\times10^9$ $M_\sun$ with an anticipated total uncertainty of around 5\%.


text:
# What's the difference between Matplotlib.pyplot and Matplotlib.figure?

## matplotlib.pyplot

The matplotlib.pyplot is a collection of functions that make matplotlib work like MATLAB. Each pyplot function makes some change to a figure: e.g., creates a figure, creates a plotting area in a figure, plots some lines in a plotting area, decorates the plot with labels, etc.

In matplotlib.pyplot, various states are preserved across function calls, so that it keeps track of things like the current figure and plotting area, and the plotting functions are directed to the current axes

## matplotlib.figure

The figure keeps track of all the child Axes, a smattering of 'special' artists (titles, figure legends,etc), and the canvas. A figure can contain any number of Axes, but will typically have at least one.

## Example

from matplotlib import pyplot as plt

plt.rcParams["figure.figsize"] = [7.50, 3.50]
plt.rcParams["figure.autolayout"] = True

fig = plt.figure("I am figure window")

plt.show()


text:
# What is the "Carbonic acid-bicarbonate" buffer system?

Jul 18, 2015

It's a buffer system that consists of a weak acid and its conjugate base.

#### Explanation:

The carbonic acid - bicarbonate buffer system consists of carbonic acid, a weak acid, and the bicarbonate anion, its conjugate base.

The important thing to realize here is that carbonic acid, ${H}_{2} C {O}_{3}$, is actually formed when carbon dioxide, $C {O}_{2}$, is dissolved in water.

After carbon dioxide is dissolved, it combines with the water molecules to form carbonic acid according to the following equilibrium reaction

$C {O}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{2} C {O}_{3 \left(a q\right)}$

Being a weak acid, carbonic acid can then go on and donate its protons in two steps to form the bicarbonate, HCO_3""^(-), and the carbonate anions, $C {O}_{3}^{2 -}$.

The equilibrium of interest will be

${\underbrace{{H}_{2} C {O}_{3 \left(a q\right)}}}_{\textcolor{g r e e n}{\text{weak acid")) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + underbrace(HCO_(3(aq))^(-))_(color(blue)("conjugate base}}}$

In essence, if a disturbance occurs in the concentration of the species involved in this equilibrium reaction, the equilibrium will shift in a direction that will compensate that disturbance - think Le Chatelier's Principle.

If a strong acid is introduced in the system, which is equivalent to having an increased concentration of hydronium ions, it will react with the bicarbonate anion and form carbonic acid, a weak acid.

The equilibrium will thus shift to the left.

The fact that a strong acid is converted to a weak one will prevent to acidity of the solution to increase significantly.

Likewise, if a strong base is introduced, it will react with the carbonic acid to form the bicarbonate anion, thus reducing the potential increase in pH.

The equilibrium will shift right.

This buffer is actually used by the body to regulate blood acidity.

Check out this really cool demonstration by Professor Diane O'Dowd of how this buffer works

https://vimeo.com/20298747


text:
# LightPhysical#

(Physical)

No corresponding User page.

## Description#

An abstract physical light.

## Public Attributes#

Type Name Visual Hint Description
double[3] color VISUAL_HINT_COLOR Emission color of the light.
double exposure VISUAL_HINT_DEFAULT Exposure of the light.
double intensity VISUAL_HINT_DEFAULT The gain factor of the light.
double[3] contribution VISUAL_HINT_COLOR Contribution of the light on the surface.
long mode VISUAL_HINT_DEFAULT Set the mode to use when computing the illumination. The physical mode takes in account the surface of the light to compute its intensity.
double saturation VISUAL_HINT_DEFAULT Saturation of the light.
long sample_count VISUAL_HINT_SAMPLE_PER_PIXEL Set the number of samples per pixel to use to evaluate the light.
double low_light_threshold VISUAL_HINT_DEFAULT Unshadowed illumination below which a light sample is readily assumed to be occluded, hence bypassing the calculation of its actual visibility to a given shading point so as to increase rendering speed.
double diffuse_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the diffuse lobes.
double reflection_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the reflection lobes.
double transmission_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the transmission lobes.
double subsurface_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the subsurface lobes.
double volume_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the volumes.
double indirect_diffuse_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the diffuse lobes.
double indirect_reflection_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the reflection lobes.
double indirect_transmission_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the transmission lobes.
double indirect_subsurface_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the subsurface lobes.
double indirect_volume_weight VISUAL_HINT_PERCENTAGE Multiply the light contribution on the volumes.
bool unseen_by_camera VISUAL_HINT_DEFAULT Set whether the object is hidden from the camera (primary rays) or not.
bool enable_light_occlusion VISUAL_HINT_DEFAULT Set whether the light is occluder or not.
double[3] shadow_color VISUAL_HINT_COLOR Shadow color of the light.
reference (Group) shadow_group VISUAL_HINT_GROUP Group of objects that have to be considered when casting a ray for shadows or global illumination. If not set, the corresponding groups in the layer are taken by default.
long diffuse_max_depth VISUAL_HINT_DEFAULT Set whether the light is hidden from the diffuse reflections and transmissions after a number of diffuse bounces.
long reflections_max_depth VISUAL_HINT_DEFAULT Set whether the light is hidden from the specular and glossy reflections after a number of reflections bounces.
long transmissions_max_depth VISUAL_HINT_DEFAULT Set whether the light is hidden from the specular and glossy transmissions after a number of transmissions bounces.
long volume_max_depth VISUAL_HINT_DEFAULT Set whether the light is hidden from volume scattering after a number of scattering bounces.
long shadows_max_depth VISUAL_HINT_DEFAULT Set whether the light casts shadows after a number of bounces.
long total_max_depth VISUAL_HINT_DEFAULT Set whether the light is hidden from reflections and transmissions after a total number of indirect bounces.
long attenuation_mode VISUAL_HINT_DEFAULT Enable the light intensity attenuation that multiplies the falloff to get the final light intensity decay.
double attenuation_near_start VISUAL_HINT_DISTANCE Set the distance from the light at which near attenuation starts. Any object whose distance is shorter than the specified value will not receive any illumination from the light.
double attenuation_near_end VISUAL_HINT_DISTANCE Set the distance from the light at which near attenuation ends. Any object whose distance is longer than the specified value will receive the full illumination from the light.
double attenuation_far_start VISUAL_HINT_DISTANCE Set the distance from the light at which far attenuation starts. Any object whose distance is shorter than the specified value will receive the full illumination from the light.
double attenuation_far_end VISUAL_HINT_DISTANCE Set the distance from the light at which far attenuation ends. Any object whose distance is longer than the specified value will not receive any illumination from the light.
curve attenuation_curve VISUAL_HINT_DEFAULT Multiplies the light intensity profile. The specified curve uses the distance of the object to the light as input.
string light_path_expression_label VISUAL_HINT_DEFAULT Set the light label in light path expressions.
bool aov_exclusive VISUAL_HINT_DEFAULT If turned on, this light will not contribute to the beauty but only to Light Path Expression AOVs, and under the 'X' symbol instead of the usual 'L'. When enabled, this light will not contribute to the unoccluded results ('U' symbol in LPEs) either and will not be seen by the camera.

## Inherited Public Attributes#

Type Name Visual Hint Description
double[3] translate VISUAL_HINT_DISTANCE Set the XYZ position of the item relative to its parent item.
double[3] rotate VISUAL_HINT_ANGLE Set the XYZ rotation of the item relative to its parent item.
double[3] scale VISUAL_HINT_SCALE Set the XYZ scaling of the item relative to its parent item.
double[3] shear VISUAL_HINT_SCALE Set the XYZ shear of the item relative to its parent item.
long rotation_order VISUAL_HINT_DEFAULT Specify on which order the Euler rotations are composed. By default Clarisse's uses YXZ
reference (SceneItem) parent VISUAL_HINT_DEFAULT Set the parent item of the item.
bool inherit_transform VISUAL_HINT_DEFAULT Does the current object inherits its parent transformation ?
long reference_frame VISUAL_HINT_FRAME Set the reference frame that defines the base transform and deformation state of the item. This reference is used when computing texture projections or when generating point cloud (in Global Base) on the base geometry.
object[] (Constraint) constraints VISUAL_HINT_DEFAULT Stack of constraint operators. Constraints like Target, Point, Orient... are applied to a Scene Item in order to drive the transformation of the item.
double[3] scale_pivot VISUAL_HINT_DISTANCE Set the coordinates of the scale pivot.
double[3] rotate_pivot VISUAL_HINT_DISTANCE Set the coordinates of the rotation pivot.
double[3] translate_offset VISUAL_HINT_DISTANCE Set the XYZ position of the pivot point of the item.
double[3] rotate_offset VISUAL_HINT_ANGLE Set the XYZ rotation of the pivot point of the item.
double[3] scale_offset VISUAL_HINT_SCALE Set the XYZ scaling of the pivot point of the item.
long rotation_offset_order VISUAL_HINT_DEFAULT Set the rotation axis mode.
bool display_visible VISUAL_HINT_DEFAULT Set whether the item is shown in the 3D View or not.
bool display_pickable VISUAL_HINT_DEFAULT Set whether the item is pickable in the 3D View or in the Image View.
long display_color VISUAL_HINT_DEFAULT Set the item color for constant or simple shading display in the 3D View.
double[3] display_custom_color VISUAL_HINT_COLOR Set the item custom color used for display in the 3D View when the Display Color attribute is set to Custom.
bool enable_motion_blur VISUAL_HINT_DEFAULT Set whether the renderer will do motion blur on the item or not.
bool enable VISUAL_HINT_DEFAULT Enable the illumination of the light during the computation of the shading for all objects.

## CID#

class "LightPhysical" "Light" {
abstract yes
#version 0.9
category "/Light/Physical"
doc "An abstract physical light."
attribute_group "lighting" {
color "color" {
doc "Emission color of the light."
texturable yes
animatable yes
value 1 1 1
}
double "exposure" {
doc "Exposure of the light."
animatable yes
numeric_range yes -100 100
ui_range yes -5 20
value 0.0
}
long "intensity_unit" {
doc "The photometric unit used to determine the luminous intensity of the light."
hidden yes
preset "Lumen (lm)" "0"
preset "Lux (lx)" "2"
preset "Nit (nt)" "1"
value 1
}
double "intensity" {
doc "The gain factor of the light."
animatable yes
numeric_range yes 0.0 1.79769313486232e+308
value 1
}
color "contribution" {
doc "Contribution of the light on the surface."
texturable yes
animatable yes
value 1 1 1
}
long "mode" {
doc "Set the mode to use when computing the illumination. The physical mode takes in account the surface of the light to compute its intensity."
preset "Normalized" "0"
preset "Physical" "1"
value 0
}
double "saturation" {
doc "Saturation of the light."
animatable yes
slider yes
numeric_range yes 0.0 100
ui_range yes 0.0 5
value 1
}
}
attribute_group "lighting>sampling" {
sample_per_pixel "sample_count" {
doc "Set the number of samples per pixel to use to evaluate the light."
numeric_range yes 0.0 65536
ui_range yes 0.0 256
value 16
}
double "low_light_threshold" {
doc "Unshadowed illumination below which a light sample is readily assumed to be occluded, hence bypassing the calculation of its actual visibility to a given shading point so as to increase rendering speed."
null_label "Use Renderer Settings"
null_value "-1.0"
numeric_range yes 0.0 1.79769313486232e+308
ui_range yes 0.0 1.79769313486232e+308
value -1
}
}
attribute_group "lighting>visibility>weights>direct" {
collapsed yes
percentage "diffuse_weight" {
doc "Multiply the light contribution on the diffuse lobes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
percentage "reflection_weight" {
doc "Multiply the light contribution on the reflection lobes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
percentage "transmission_weight" {
doc "Multiply the light contribution on the transmission lobes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
percentage "subsurface_weight" {
doc "Multiply the light contribution on the subsurface lobes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
percentage "volume_weight" {
doc "Multiply the light contribution on the volumes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
}
attribute_group "lighting>visibility>weights>indirect" {
collapsed yes
percentage "indirect_diffuse_weight" {
doc "Multiply the light contribution on the diffuse lobes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
percentage "indirect_reflection_weight" {
doc "Multiply the light contribution on the reflection lobes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
percentage "indirect_transmission_weight" {
doc "Multiply the light contribution on the transmission lobes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
percentage "indirect_subsurface_weight" {
doc "Multiply the light contribution on the subsurface lobes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
percentage "indirect_volume_weight" {
doc "Multiply the light contribution on the volumes."
slider yes
numeric_range yes 0.0 65536
ui_range yes 0.0 1
value 1
}
}
attribute_group "lighting>visibility" {
collapsed yes
bool "unseen_by_camera" {
doc "Set whether the object is hidden from the camera (primary rays) or not."
value yes
}
bool "enable_light_occlusion" {
doc "Set whether the light is occluder or not."
value no
}
}
collapsed yes
doc "Shadow color of the light."
animatable yes
value 0.0 0.0 0.0
}
doc "Group of objects that have to be considered when casting a ray for shadows or global illumination. If not set, the corresponding groups in the layer are taken by default."
filter "SceneObject"
null_label "Use 3D Layer"
input "motion" "geometry"
dg_cyclic yes
value ""
}
}
attribute_group "lighting>visibility>path_depth" {
collapsed yes
long "diffuse_max_depth" {
doc "Set whether the light is hidden from the diffuse reflections and transmissions after a number of diffuse bounces."
null_label "Infinite"
null_value "-1.0"
numeric_range yes 0.0 65536
ui_range yes 0.0 16
value -1
}
long "reflections_max_depth" {
doc "Set whether the light is hidden from the specular and glossy reflections after a number of reflections bounces."
null_label "Infinite"
null_value "-1.0"
numeric_range yes 0.0 65536
ui_range yes 0.0 16
value -1
}
long "transmissions_max_depth" {
doc "Set whether the light is hidden from the specular and glossy transmissions after a number of transmissions bounces."
null_label "Infinite"
null_value "-1.0"
numeric_range yes 0.0 65536
ui_range yes 0.0 16
value -1
}
long "volume_max_depth" {
doc "Set whether the light is hidden from volume scattering after a number of scattering bounces."
null_label "Infinite"
null_value "-1.0"
numeric_range yes 0.0 65536
ui_range yes 0.0 16
value -1
}
doc "Set whether the light casts shadows after a number of bounces."
null_label "Infinite"
null_value "-1.0"
numeric_range yes 0.0 65536
ui_range yes 0.0 16
value -1
}
long "total_max_depth" {
doc "Set whether the light is hidden from reflections and transmissions after a total number of indirect bounces."
null_label "Infinite"
null_value "-1.0"
numeric_range yes 0.0 65536
ui_range yes 0.0 32
value -1
}
}
attribute_group "lighting>attenuation" {
collapsed yes
long "attenuation_mode" {
doc "Enable the light intensity attenuation that multiplies the falloff to get the final light intensity decay."
preset "None" "0"
preset "Near Only" "1"
preset "Far Only" "2"
preset "Near & Far" "3"
preset "Curve" "4"
value 0
}
distance "attenuation_near_start" {
doc "Set the distance from the light at which near attenuation starts. Any object whose distance is shorter than the specified value will not receive any illumination from the light."
animatable yes
numeric_range yes 0.0 1000000000
ui_range yes 0.0 100
value 0.0
}
distance "attenuation_near_end" {
doc "Set the distance from the light at which near attenuation ends. Any object whose distance is longer than the specified value will receive the full illumination from the light."
animatable yes
numeric_range yes 0.0 1000000000
ui_range yes 0.0 100
value 0.1
}
distance "attenuation_far_start" {
doc "Set the distance from the light at which far attenuation starts. Any object whose distance is shorter than the specified value will receive the full illumination from the light."
animatable yes
numeric_range yes 0.0 1000000000
ui_range yes 0.0 100
value 10
}
distance "attenuation_far_end" {
doc "Set the distance from the light at which far attenuation ends. Any object whose distance is longer than the specified value will not receive any illumination from the light."
animatable yes
numeric_range yes 0.0 1000000000
ui_range yes 0.0 100
value 10.1
}
curve "attenuation_curve" {
doc "Multiplies the light intensity profile. The specified curve uses the distance of the object to the light as input."
value[] {
key 4 0.0 1 0.0 0.0 0.0 -0.3333 0.0 0.3333 0.0 yes yes no yes
key 4 1 0.0 0.0 0.0 0.0 -0.3333 0.0 0.3333 0.0 yes yes no yes
}
}
}
attribute_group "AOV" {
collapsed yes
string "light_path_expression_label" {
doc "Set the light label in light path expressions."
null_label "lights"
null_value "lights"
value "lights"
}
bool "aov_exclusive" {
doc "If turned on, this light will not contribute to the beauty but only to Light Path Expression AOVs, and under the \'X\' symbol instead of the usual \'L\'. When enabled, this light will not contribute to the unoccluded results (\'U\' symbol in LPEs) either and will not be seen by the camera."
value no
}
}
}


text:
## Part 1 Summary

To complete this part, you must login to your account on the Math Department Sage server and demonstrate that you know how to input a matrix, augment a matrix with a vector, and compute the echelon form of an (augmented) matrix.

Completing Part 1 is worth 1 point. If you are not interested in completing Part 2 of the assignment, you may turn in your paper and leave the lab after completing Part 1.

Make sure your name is on your paper and it was signed by the instructor.

## Part 2 Summary

To complete Part 2, you must first download the Sage worksheet file Math317-Lab2.sws from either Blackboard or GitHub, load this worksheet into your Sage session and follow the instructions provided. You must then save your completed Sage worksheet to a file and upload the resulting sws file to Blackboard.

## Finishing

When you have finished working or it is 2pm (whichever comes first):

1. stop your Sage worksheet (Action $\rightarrow$ Save and quit worksheet; or use Stop button),

2. sign out of Sage,

5. hand it in to the instructor.

## Part 1 Details

In Homework 3, we proved that the vector ${{\mathbf{b}}}= (1, -1, 1, -1)$ is as a linear combination of the vectors ${{\mathbf{v}}}_1 = (1 , 0 , 1 , -2)$, ${{\mathbf{v}}}_2 = (0 , -1 , 0 , 1)$, and ${{\mathbf{v}}}_3 = (1, -2 , 1 , 0)$.

Do you remember how we did this?

That's right! We recognized that if the matrix $A$ has first column ${{\mathbf{v}}}_1$, second column ${{\mathbf{v}}}_2$, and third column ${{\mathbf{v}}}_3$, then writing ${{\mathbf{b}}}$ as a linear combination of the given vectors is the same as solving the system $A{{\mathbf{x}}}= {{\mathbf{b}}}$. That is, if we find ${{\mathbf{x}}}= (x_1, x_2, x_3)$ that solves this system, then ${{\mathbf{b}}}= x_1{{\mathbf{v}}}_1+x_2{{\mathbf{v}}}_2+x_3{{\mathbf{v}}}_3$ is the desired linear combination.

In this first part of this lab assignment, we will simply re-solve the above problem, but this time we will make Sage do the tedious work. Carry out the following steps:

1. Login to any machine in Carver 449, open up a browser (preferably Chrome) and navigate to https://sage.math.iastate.edu/

Login to your account on the Math Department Sage server using the login information you were given for Lab 1. Once you are logged in, create a new worksheet and name it Lab2.

2. In the Lab2 worksheet, click the first cell and type the following:

b = vector([1, -1, 1, -1]); print b

Then type Shift+Enter or click evaluate.

Did Sage print what you expect? If so, move on to number 3.

If not, try again and/or ask the instructor for help.

3. Click somewhere inside the next cell and enter the following expression:

 A = matrix(4, 3, [1, 0, 1, 0, -1, -2, 1, 0, 1, -2, 1, 0]) print A

What did Sage print?

Is it a matrix with three columns, $\mathbf{v}_1$, $\mathbf{v}_2$, $\mathbf{v}_3$? If not, try again and/or your neighbor for help.

Can you decipher the syntax we used to input this matrix $A \in \mathbb{R}^{4\times 3}$? If so, move on to number 4.

If not, think a litter harder and/or discuss it with your neighbor.

4. Next, have Sage augment the matrix $A$ with the vector $b$ by entering the following into the next worksheet cell:

Ab = A.augment(b, subdivide=True) print Ab

5. Ask Sage to put your augmented matrix in reduced-row echelon form:

Ab.rref()

You should now see two pivots (both equal to 1). Let the free variable $x_3 = s$ and write the vector $\mathbf{b}$ as a linear combination of the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, $\mathbf{v}_3$ involving $s$.

$$\mathbf{b}= \begin{bmatrix} 1 \\ -1 \\ 1 \\ -1\end{bmatrix}= \underline{\phantom{XXX}}\mathbf{v}_1 + \underline{\phantom{XXX}} \mathbf{v}_2+ \underline{\phantom{XXX}} \mathbf{v}_3.$$

If you want to stop here, ask the instructor to check your work and sign below then save and quit. Otherwise, move on to Part 2

Instructor signature: (1 point)

## Part 2 Details

Load the Math317-Lab2.sws file into your Sage session and complete the exercises.


text:
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I'm looking for something like a Grassmannian, but which parameterizes the submodules of a module rather than the subspaces of a vector space. Most specifically, I'm looking for something which parameterizes the submodules of specifically $Z^n$. \mathbb{Z}^n$. So another way to say it is that I'm looking for a space parameterizing for the subgroups of a free abelian group. (A moduli space?) I've seen some references to the concept of a "Grassmannian of submodules" here and there (like the papers on the first page of https://www.google.com/search?q=%22grassmannian+of+submodules%22) but can't figure out if this handles modules like$Z^n$\mathbb{Z}^n$ or not.
EDIT: to give a bit more info, the only specific application I really care about is parameterizing the free subgroups of a free abelian group, or the "lattices" in the Z-module $Z^n$, \mathbb{Z}$-module$\mathbb{Z}^n$, etc. A solution which works only for that, but which doesn't handle more exotic modules would be fine for my purposes. (And if it doesn't work out for all free abelian groups in general, then having a solution for at least finitely generated free abelian groups would even be a great start.) I framed the question in terms of the "submodules of a module" in general just because I saw some references to there being a "Grassmannian of submodules" before, so I thought such a construction might be widely known. 2 clarification I'm looking for something like a Grassmannian, but which parameterizes the submodules of a module rather than the subspaces of a vector space. Most specifically, I'm looking for something which parameterizes the submodules of specifically Z^n.$Z^n$. So another way to say it is that I'm looking for a space parameterizing for the subgroups of a free abelian group. (A moduli space?) I've seen some references to the concept of a "Grassmannian of submodules" here and there (like the papers on the first page of https://www.google.com/search?q=%22grassmannian+of+submodules%22) but can't figure out if this handles modules like Z^n$Z^n$or not. Does anyone know if such an object exists and if so, how to construct it? Where I can get more information on this? EDIT: to give a bit more info, the only specific application I really care about is parameterizing the free subgroups of a free abelian group, or the "lattices" in the Z-module$Z^n\$, etc. A solution which works only for that, but which doesn't handle more exotic modules would be fine for my purposes.