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Talk:Beth number

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Difference between aleph numbers and beth numbers

I don't see any difference between anAleph number and a Bet number. Can they be used interchangeably? Then why have both terms? 65.74.59.37 08:41, 21 January 2006 (UTC)Chani

They are different if (and only if) GCH fails. It's not usual to assume GCH, so they can't usually be used interchangeably. --Zundark 11:53, 21 January 2006 (UTC)

Moreover, the definition of the Bet numbers requires the Axiom of Choice, while the definition of the Aleph numbers only uses the Replacement Schema and not the Axiom of Choice. Hence you cannot even use the Bet numbers without assuming AC. Yann Pequignot (talk) 11:01, 13 June 2010 (UTC)

Without AC, it is no longer true that every cardinality includes a well-ordered set, but it is still true that every set has a cardinality. So there isn't any difficulty in defining Beth numbers without AC. — Carl (CBM · talk) 12:25, 13 June 2010 (UTC)
Actually, it's not necessarily true that every set has a "cardinality" (to be precise, in vNBG with Urelements, that there is a (class) function (F) with domain the universe such that F(x) = F(y) if and only if x is equipollent to y). In set theories without urelements, one can take F(x) to be the set of all sets equipollent to x with minimal rank. Even then, we cannot necessarily have that F(x) is equipollent to x.
We note in the article that the von Neumann universes $V_{\omega+\alpha} \!$ have cardinality $\beth_{\alpha} \!$. This appears to be correct (for most definitions of cardinality), even without choice. — Arthur Rubin (talk) 15:46, 13 June 2010 (UTC)
There is a difference between (1) each set having a cardinality, (2) the existence of a uniform selection of representatives for the cardinalities, and (3) each cardinality containing a well-ordered set. I was referring to (1) and you are referring to (2). I think the person I responded to was referring to (3). — Carl (CBM · talk) 15:55, 13 June 2010 (UTC)

Gimel numbers

What's gimel used for? Is it used interchangeably with beth? Because they look similar? Awmorp 23:26, 7 June 2006 (UTC)

I have no idea. I've never seen it outside of Wikipedia. — Arthur Rubin | (talk) 23:59, 7 June 2006 (UTC)
I think that gimel numbers are defined analogously to aleph numbers and beth numbers, but with $\gimel_{\alpha+1}=\gimel_\alpha^{\mathrm{cf}(\gimel_\alpha)}$. However, I'm not aware of any published reference for this, so I've removed the mention of them from the article (which in any case was confusing due to the lack of explanation). --Zundark 08:21, 8 June 2006 (UTC)

Any discussion is taking place at Talk:Beth_two#Merge_into_Beth_Numbers. LambiamTalk 17:54, 2 April 2006 (UTC)

• Merge completed (yesterday), as best I could. If I missed anything relevant that should have been merged, the last version before merge is still there. — Arthur Rubin | (talk) 14:43, 25 July 2006 (UTC)

Clarification needed

"sup" is not defined or linked in any way, as in:

$\beth_{\lambda}=\sup\{ \beth_{\alpha}:\alpha<\lambda \}.$

It would be helpful if it was; I can't be the only reader who finds this complete double-Dutch. --Michael C. Price talk 03:07, 14 February 2007 (UTC)

If the axiom of choice holds, then these cardinals can be identified with their initial ordinals. In that case, the supremum is the least ordinal greater than all those in the set. If the axiom of choice does not hold, then the "obvious" generalization is the direct limit of the sequence of representative sets. So if A is the initial set, then
$\beth_{\lambda} (|A|) = | \bigcup_{\alpha < \lambda} B_{\alpha} |$
where
$\beth_{\alpha} (|A|) = | B_{\alpha} |$
and the Bs are nested. JRSpriggs 08:44, 14 February 2007 (UTC)
Even if the axiom of choice holds, the direct limit of cardinals is not always defined, I'm afraid. The direct limit for a strictly increasing set of cardinals works, which applies here, but the limit of ω1 (the ordinal corresponding to $\aleph_1$) $\aleph_0$'s can be $\aleph_0$ or $\aleph_1$. The cardinality of the union of the specific nested set of sets makes sense, but "direct limit" of cardinals, in general, does not. — Arthur Rubin | (talk) 15:45, 14 February 2007 (UTC)
Also, a description says an equation (I noted with '[clarify]') will use alpha, but alpha is not used.--Dchmelik (talk) 07:35, 1 November 2008 (UTC)
It doesn't say that the equation will use $\alpha$, it says that there is an $\alpha$ such that the equation holds for any $\beta \ge \alpha$. In other words, the equation holds for every sufficiently large ordinal $\beta$. I've reworded it a little, so that it is hopefully clearer. --Zundark (talk) 10:03, 1 November 2008 (UTC)
[Maybe it is because I have not studied much number theory, but] it seems no clearer: α,β,k,μ were re-quantified, so they are not defined as earlier: it seems like $\exists{\alpha}{,}\forall{\beta,k,\mu}{\in}\infty P(\beta,k,\mu.))$ Variables were re-quantified--so maybe not the same, and there is no P(α.)--Dchmelik (talk) 14:59, 3 November 2008 (UTC)
I don't know how you read it that way. The basic form (ignoring the fact that κ and μ are required to be cardinals, and α and β are required to be ordinals) is $\forall\kappa\forall\mu\exists\alpha\forall\beta(\beta\ge\alpha\rightarrow P(\beta,\kappa,\mu))$. --Zundark (talk) 14:09, 6 November 2008 (UTC)

To Arthur Rubin: Given a cardinal $\kappa \,$ and an ordinal $\alpha \,,$ we can define

$\beth_{\alpha} (\kappa) \, = \, \vert W_{\alpha} (A) \vert \,$

where $A \,$ is any set of cardinality $\kappa \,$ and

$W_{\alpha} (A) \, = \, (\{ 0 \} \times A) \cup (\{ 1 \} \times \bigcup_{\beta < \alpha} \mathcal{P} (W_{\beta} (A) ) \, ) \,.$

This is modeled on the alternative definition of the Von Neumann universe. I hope this clears it up. Sorry about the long delay in responding. JRSpriggs (talk) 13:01, 1 November 2008 (UTC)

Better would be
$W_{\alpha} (A) \, = \, (\{ 0 \} \times A) \cup \bigcup_{\beta < \alpha} \{ \langle \beta + 1 , B \rangle \, : \, B \subseteq W_{\beta} (A) \land \vert B \vert \neq 1 \} \,$
which takes out singletons to compensate for the union with previous stages. JRSpriggs (talk) 02:23, 17 February 2010 (UTC)
Well, this is all an interesting conversation, but the core issue still remains: the linked article supremum doesn't go beyond calculus; perhaps the above conversation can be captured in this article or a new article supremum (cardinal numbers) or something like that? linas (talk) 18:11, 21 April 2012 (UTC)
As I said above, there is no problem when the axiom of choice holds. Since almost all of us believe in the axiom of choice, it is not worth the trouble to do more than we have already. As for Arthur Rubin's objection, it does not apply since the cardinal at each stage is larger than at any of the earlier stages. JRSpriggs (talk) 11:47, 22 April 2012 (UTC)

spelling for beth

Who decided to spell the word as "beth", when it's actually pronounced "bet"? Is there a lot of history for this spelling? Do mathematicians pronounce it like the female name beth? Ariel. (talk) 11:01, 22 January 2008 (UTC)

Beth is the traditional English spelling, and is usually pronounced beth in Britain (but, apparently, bayth in America). I would guess that the spelling is based on Tiberian vocalization, but I don't know for sure. It dates from 1650 according to Merriam-Webster - the OED probably has details, if someone feels like checking them. But all this is something that should be covered in the article on the letter, not in this article - mathematicians merely adopted the existing conventions, which long predate the theory of cardinal numbers. --Zundark (talk) 12:49, 22 January 2008 (UTC)
So you are saying mathematicians pronounce it the same way as the hebrew pronunciation? Because in that case it should be noted in the article to pronounce it bet or bais (bays), and not like it looks. Ariel. (talk) 13:12, 22 January 2008 (UTC)
I concur that the pronunciation is as in the Hebrew, but the spelling I've usually seen is Beth, in spite of the fact that I usually refer to the Hebrew letter as Bet. — Arthur Rubin | (talk) 13:56, 22 January 2008 (UTC)
At least in America. — Arthur Rubin | (talk) 14:03, 22 January 2008 (UTC)
No, I wasn't saying that mathematicians pronounce it as in Hebrew, and I don't know how you read my comment that way. Also, the only type of Hebrew I mentioned was Tiberian Hebrew, and in Tiberian Hebrew the pronunciation is beth, so I especially don't see how you read my comment as suggesting bet or bays. I think that the mathematical pronunciation probably varies as much as the non-mathematical pronunciation, and that we therefore can't specify it. But if someone has a good reference for mathematical pronunciation, go ahead and add it to the article. --Zundark (talk) 15:00, 22 January 2008 (UTC)
I've always heard it "beth" as a cardinal, though I say "bet" for the letter in other contexts. CRGreathouse (t | c) 02:53, 23 January 2008 (UTC)
I'm afraid I've heard it pronouced "bayt", but my father speaks Hebrew, as well as working in logic and set theory, so that might be a special case. My parents still spelled it "Beth", though, when we weren't using the symbol. — Arthur Rubin | (talk) 03:01, 23 January 2008 (UTC)
Not to worry, mathematicians butcher eastern european surnames with aplomb. Hardly limited to hebrew. linas (talk) 18:15, 21 April 2012 (UTC)

Counterfactuals

Arthur Rubin removed [1] the following recent addition to the article:

• The minimum number of possible worlds that might exist (David Lewis, "Counterfactuals")

The addition was so unusual I decided to look it up to see if there was any merit. First, I saw that David Lewis has an article on Wikipedia: David Kellogg Lewis. Second, in looking for any mention of David Lewis, Counterfactuals, and something cardinality-related, I came across a similarly-titled paper which references Dr. Lewis and proves logical (as in mathematical logic) results relating to the cardinality of algebras: see T15 and T16. Donald Nute's Counterfactuals

I have no opinion on the paper or the suitability of either Nute's paper or Lewis' book for this article, but it does seem that they're related in some way.

CRGreathouse (t | c) 13:49, 8 February 2008 (UTC)

Hmmm. Perhaps I was mistaken, but I'm not sure Lewis's definition of "possible worlds" is notable, even if Lewis is notable. Even so, we would need a detailed reference for the statement. Also, under the circumstances, the question of whether Beth 2 means the same thing to Lewis as it does to us seems open. — Arthur Rubin | (talk) 19:06, 8 February 2008 (UTC)
Nute makes explicit reference to 2^(2^stuff), where stuff is a cardinal, so Nute's theorems have beth_2 when stuff is countably infinite. I haven't seen the Lewis paper, but Nute references it so it's not too much of a stretch. Of course this doesn't mean it's notable -- though our list of places that beth_2 appears is limited, so I'd love to keep it as an example if reasonable.
Partially related: have you seen any reasonable examples of beth_3? The article could use one/some if there are any outside of the obvious contrived examples.
CRGreathouse (t | c) 21:10, 8 February 2008 (UTC)

The footnote on page 90 of David Lewis's book "Counterfactuals" states in part that the cardinality of possible worlds in a framework proposed by W.V.O. Quine equals beth 2 which he identifies with the cardinality of the power set of reals. He contrasts this with the cardinality of sentences in a finite language equal to beth 1 which he identifies with the cardinality of reals. Lewis considers propositions as sets of possible worlds. So, the above comports with his comment on page 107 of his book "On the plurality of worlds" that the lowest reasonable estimate of the cardinality of propositions is beth 3. —Preceding unsigned comment added by Kmarkus (talkcontribs) 16:54, 26 August 2010 (UTC)

References

I am Robert Munafo. I just noticed this article acknowledges me as a source. However, it bears so little relation to the material on my web pages (e.g. Aleph-1, Power set, etc.) that I wonder if the citation is still relevant. I wouldn't mind if it gets removed. I don't know much about the legal requirements of GNU Free Documentation License to be able to decide on my own. —Preceding unsigned comment added by Mrob27 (talkcontribs) 02:12, 11 February 2008 (UTC)

It turns out that the reason for this acknowledgement is that this article is partly derived from the old "Beth two" article, and the original version of that article was based very closely on your page 2 to the Power of C. But none of that old text exists in the current article, so the acknowledgement is no longer relevant. (By the way, your page incorrectly says that aleph_2 is the same as 2c only if GCH holds. But, in fact, it can hold without GCH, as it's much weaker.) --Zundark (talk) 09:25, 11 February 2008 (UTC)

Good, then I'l take out the reference to my own web page (smile) (And by the way, thanks for the correction about 2c and GCH) —Preceding unsigned comment added by Mrob27 (talkcontribs) 14:04, 11 February 2008 (UTC)

Now that I've done that, the article has no references. Can someone who knows how, add a tag that says we need to add references? (I'm still new to all the magic Wiki-text stuff.) Robert Munafo (talk) 14:08, 11 February 2008 (UTC)

You can add {{unreferenced}} to the top of the page if you like. It will add a banner to the page asking for citations. (By the way, welcome; I like your page on big numbers a lot.) CRGreathouse (t | c) 15:42, 11 February 2008 (UTC)

Got it, thanks! Robert Munafo (talk) 23:09, 11 February 2008 (UTC)

Comparison to Aleph Numbers

This section says that "Assuming the axiom of choice, infinite cardinalities are linearly ordered; no two cardinalities can fail to be comparable, and so, since by definition no infinite cardinalities are between $\aleph_0$ and $\aleph_1$, the continuum hypothesis can be stated in this notation by saying $\beth_1=\aleph_1$."

If I'm reading this correctly, it states that assuming the axiom choice implies the continuum hypothesis and equivalence of aleph and beth numbers. Since this is clearly incorrect (the continuum hypothesis was proved independant from ZFC), I think it should be reworded in a way that makes it clear that aleph and beth numbers are only equal assuming the continuum hypothesis.Eebster the Great (talk) 18:55, 20 April 2008 (UTC)

It's saying that, assuming AC, the equality $\beth_1=\aleph_1$ is equivalent to CH. It's not saying that this equality (or CH) follows from AC. I've reworded it to make this a little clearer. --Zundark (talk) 22:20, 20 April 2008 (UTC)

I'm very confused. "Assuming the axiom of choice, infinite cardinalities are linearly ordered; no two cardinalities can fail to be comparable, and so, since by definition no infinite cardinalities are between $\aleph_0$ and \aleph_1": By what definition? Isn't the definition of CH that there are no infinite cardinalities between $\aleph_0$ and $\aleph_1$? Isn't this irrelevant and shouldn't the inequality be "$\beth_1 > \aleph_0$"? Then the logic makes sense that CH => $\beth_1 = \aleph_1$. Otherwise CH = "The continuum hypothesis states that there is no cardinal number between the cardinality of the reals and the cardinality of the natural numbers." doesn't seem to support the conclusion.

-- — Preceding unsigned comment added by Schmmd (talkcontribs) 23:16, 26 January 2011 (UTC)

It all makes sense now. --Schmmd (talk) 06:03, 27 January 2011 (UTC)


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# How does a chip interpret different opcodes?

I have learnt in school about opcodes and how it works with the ALU to add numbers and such, now is my question how the computer is able to understand this, because it is codes and the computer needs something to translate that to actually doing that, in't it? (correct me if I'm wrong plz) I also want to, when I understand that, make my own processor so I can learn the computer bottom-up and become a good computer engineer.

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Instructions are often easier to think of as addresses, rather than a commands. These addresses are not a location in memory, but rather a location in hardware to which the parameters of the instruction are sent. This location is a complex set of logic gates designed to fulfill some purpose, e.g. addition, subtraction, etc. –  Polynomial May 15 '12 at 9:28
Back in the old days, you can physically see machines controlled by several levers, switches, buttons, etc. Think of opcodes as settings specifying which lever to pull/push. A certain combination of lever positions will yield to a certain functioning for the machine. –  shimofuri May 15 '12 at 9:40
In this case, the cogs and levers are replaced by bus multiplexers. –  Polynomial May 15 '12 at 10:15
A great example of "cogs and levers" would be arithmometer. It's a type of mechanical calculator. Operator would enter numbers, use levers and switches to select operation and turn winch to provide mechanical energy. Some later models even had electrical motors which made longer calculations easier. –  AndrejaKo Feb 8 '13 at 10:54

At the very lowest level, consider something like microcode. That's what Wouter was talking about when he mentioned Very Long Instruction Word architectures.

A CPU is a collection of busses, registers, memory, and arithmetic logic unit (ALU). Each of these do simple and finite things. Any one higher level instruction is a coordinated set of actions between all the blocks. For example, the low level operations to add a memory location value into the accumulator could be:

1. Enable the operand address onto the memory address bus. Assume memory is always in read mode when not explicitly writing.

2. Enable the accumulator onto the ALU input 1.

3. Enable the memory data bus onto the ALU input 2.

4. Set the ALU operation to addition.

5. Wait the minimum number of clock ticks so that you know the output of the ALU has settled. In this case it includes the memory read time, the ALU propagation time, and any intermediate data path propagation times.

6. Latch the ALU output into the accumulator.

When you break it down into the basic hardare operations, you note that orchestrating a instruction is mostly routing data to the right places in the right sequence. If this were implemented with descrete logic, #1 would be asserting a single output enable line of a tri-state buffer that drives the memory address bus. #2 and #3 likewise require asserting a single line. #4 is a little different in that the ALU is usually a canned logic block itself and often has a set of lines that code the operation. For example, 000 might be pass input 1 to output, 001 add both inputs to the output, 010 logical AND both inputs to the output, etc.

The point is that at the level described above, each instruction is just asserting a certain set of control lines in sequence, possibly with minimum wait times between some actions. A stripped down CPU could simply tie each bit in the instruction word to one of these control lines. That would be simple and highly flexible, but one drawback is that the instruction word would need to be quite wide to contain all the necessary control lines and a operand address field. The instruction memory is used very inefficiently.

A bunch of years ago, there were machines with microcoded architecture. They worked as I described at the low level, but these microinstructions weren't what you wrote when you programmed the machine. Each user level instruction essentially kicked off a small microcode routine. The user instructions would now be more compact with less redundant information in them. This was good because there could be many many of them and memory was limited. But the actual low level control of the hardware was done from microcode. This was stored in a special wide and fast and therefore expensive memory, but it didn't need to be very big because there were only a few microcode instruction for each user level opcode.

Nowadays, relatively simple machines like microcontrollers don't really have microcode. The instruction set has been made a little simpler and more regular so that it can be decoded directly by hardware, although that hardware may have a sort of sequencer or state machine that isn't exactly a microcode engine but sortof does that job with combinatorial logic with pipeline stages where things get held up waiting on clock edges. This is one reason, for example, that smaller PICs have a CPU clock that is 4x the instruction clock. That allows 4 clock edges per instruction, which is useful for managing propagation delays thru the logic. Some of the PIC documentation even tells you at what Q edges what operations are performed.

So if you want to get something very basic up and running, try implementing just a microcode machine. You may need a 24 bit or wider instruction word, but for demonstration and learning that is fine. Each bit controls a single flip flop clock, enable line, or whatever. You will also need some way of sequencing thru the instructions and doing conditional branching. A brute force way is to put a new address for possible use depending on the result of some conditional operation right into the microcode word.

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In a VLIW architecture there is no microcode, the bits is the (ISA = Instruction Set Architecture = assembly level) instructions directly control the internal logic. –  Wouter van Ooijen May 15 '12 at 15:18
@Wouter: Or you can look at it like there is only microcode that the programmer programs directly. It depends on whether you consider microcode as instruction bits directly controlling hardware or a layer of execution below the user-visible instruction level. –  Olin Lathrop May 15 '12 at 17:36
One may simplify branching logic considerably by having some bits of the "program counter" loaded unconditionally from bits in the microcode word, while others are loaded based upon various conditions. If there are four conditions, one might use two bits in the microcode word to select one of the bits to feed a bit of the PC address, or one might simply feed all four conditions into PC address bits. The former approach would reduce the ROM size, while the latter would reduce the amount of other circuitry. –  supercat Aug 22 '12 at 21:36
The Apple ][ floppy drive controller "Woz Machine" ran at 2MHz, and used a 256x8 PROM to convert four inputs (read-head data, bit 7 of a shifter, and two 6502-controlled latches), plus four bits of state, into four outputs (two shifter-mode outputs, shifter-clear, and a shifter-data output). In many states, the state machine wouldn't care about all of the inputs, but the PROM contents would simply be duplicated as needed in the addresses where the "don't care" state was high and where it was low. –  supercat Aug 22 '12 at 21:49

Any translation from one set of codes to another set of codes can be done by a (conceptually) simple piece of logic. A ROM for instance will do. A simple (not pipelined or microprogrammed) processor translates (parts of) the instruction to bits that control the various hardware parts inside the CPU, like the ALU function, register selection, and control lines for buffers, multiplexers, read/write selects, etc.

An interesting case is te Very Long Instruction Word type of architectures, where there is no such translation: the bits in the instruction directly control the functions inside the CPU.

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To keep things simple, imagine that an opcode represents the actual control signals inside a chip. For an imaginary processor, an opcode for ADD, R1, R2, R3 may be:

1111 0001 0010 0011

The first '1111' may represents the ADD operation and might be used as a direct control signal to the ALU to select the addition operation. For this imaginary processor, it's 1111 but real-world processors will need to refer to the Instruction Set Architecture (ISA) document for the actual bits.

The other three represent the operands: R1, R2 and R3. All these signals can be connected directly to the internal function units of the processor. If a translation is needed between opcode and control signals, something as simple as a ROM will do.

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# Ncert Solutions For Class 10 Maths Ex 3.6

## Ncert Solutions For Class 10 Maths Chapter 3 Ex 3.6

Solve the following pairs of equations by reducing them to a pair of linear equations:

Question 1:

12a+13b=2$\frac{1}{2a}+\frac{1}{3b}=2$

13a+12b=136$\frac{1}{3a}+\frac{1}{2b}=\frac{13}{6}$

Solution:

Let us assume 1a=mand1b=n$\frac{1}{a}=m\:and\:\frac{1}{b}=n$, then the equation will change as follows.

m2+n3=23m+2n12=0$\frac{m}{2}+\frac{n}{3}=2\:\Rightarrow 3m+2n-12=0$   …….(1)

m3+n2=1362m+3n13=0$\frac{m}{3}+\frac{n}{2}=\frac{13}{6}\:\Rightarrow 2m+3n-13=0$   …….(2)

Now, Using cross-multiplication method, we get

m26(36)=n24(39)=194$\frac{m}{-26-\left ( -36 \right )}=\frac{n}{-24-\left ( -39 \right )}=\frac{1}{9-4}$ m10=q15=15$\frac{m}{10}=\frac{q}{15}=\frac{1}{5}$ m10=15andq15=15$\frac{m}{10}=\frac{1}{5}\:and\:\frac{q}{15}=\frac{1}{5}$

m = 2 and n = 3

1a=2and1b=3$\frac{1}{a}=2\:and\:\frac{1}{b}=3$ a=12andb=13$a=\frac{1}{2}\:and\:b=\frac{1}{3}$

Question 2:

4a+3b=14$\frac{4}{a}+3b=14$

4a4b=23$\frac{4}{a}-4b=23$

Solution:

Putting 1a=m$\frac{1}{a}=m$ in the given equation we get,

So, 4m + 3b = 14     => 4m + 3b – 14 = 0  ………..(1)

3m – 4b = 23     => 3m – 4b – 23 = 0  ………..(2)

By cross-multiplication, we get,

m6956=b42(92)=1169$\frac{m}{-69-56}=\frac{b}{-42-\left ( -92 \right )}=\frac{1}{-16-9}$ m125=b50=125$\frac{m}{-125}=\frac{b}{50}=\frac{-1}{25}$ m125=125andb50=125$\frac{m}{-125}=\frac{-1}{25}\:and\:\frac{b}{50}=\frac{-1}{25}$

m = 5 and b = -2

m = 1a$\frac{1}{a}$ = 5

a = 15$\frac{1}{5}$

b = -2

Question 3:

7a2bab=5$\frac{7a-2b}{ab}=5$

7b2a=5$\frac{7}{b}-\frac{2}{a}=5$ …..(i)

8a+7bab=15$\frac{8a+7b}{ab}=15$

8b+7a=15$\frac{8}{b}+\frac{7}{a}=15$ …..(ii)

Solution:

Substituting 1a=m$\frac{1}{a}=m$ in the given equation we get,

– 2m + 7n = 5     => -2 + 7n – 5 = 0  ……..(iii)

7m + 8n = 15     => 7m + 8n – 15 = 0 ……(iv)

By cross-multiplication method, we get,

m105(40)=n3530=11649$\frac{m}{-105-\left ( -40 \right )}=\frac{n}{-35-30}=\frac{1}{-16-49}$ m65=n65=165$\frac{m}{-65}=\frac{n}{-65}=\frac{1}{-65}$ m65=165andb65=165$\frac{m}{-65}=\frac{1}{-65}\:and\:\frac{b}{-65}=\frac{1}{-65}$

m = 1 and n = 1

m = 1a$\frac{1}{a}$ = 1        n = 1a$\frac{1}{a}$ = 1

a = 1    b = 1

Question 4:

10a+b+2ab=4$\frac{10}{a+b}+\frac{2}{a-b}=4$

15a+b5ab=2$\frac{15}{a+b}-\frac{5}{a-b}=-2$

Solution:

Substituting 1a+b=m$\frac{1}{a+b}=m$ and 1ab=n$\frac{1}{a-b}=n$ in the given equations, we get,

10m + 2n = 4      =>  10m + 2n – 4 = 0      …..(i)

15m – 5n = -2     =>   15m – 5n + 2 = 0    …..(ii)

Using cross-multiplication method, we get,

m420=n60(20)=15030$\frac{m}{4-20 }=\frac{n}{-60-\left ( -20 \right )}=\frac{1}{-50-30}$ m16=n80=180$\frac{m}{-16}=\frac{n}{-80}=\frac{1}{-80}$ m16=180andb80=180$\frac{m}{-16}=\frac{1}{-80}\:and\:\frac{b}{-80}=\frac{1}{-80}$

m = 15$\frac{1}{5}$ and n = 1

m = 1a+b=15$\frac{1}{a+b}=\frac{1}{5}$ and n = 1ab=1$\frac{1}{a-b}=1$

a + b = 5  …..(iii)

and a – b = 1 …..(iv)

Adding equation (iii) and (iv), we get

2a = 6   => a = 3 …….(v)

Putting the value of a = 3 in equation (3), we get

b = 2

Hence, a = 3 and b = 2

Question 5:

2a+3b=2$\frac{2}{\sqrt{a}}+\frac{3}{\sqrt{b}}=2$

4a9b=1$\frac{4}{\sqrt{a}}-\frac{9}{\sqrt{b}}=-1$

Solution:

Substituting 1a=m$\frac{1}{\sqrt{a}}=m$ and 1b=n$\frac{1}{\sqrt{b}}=n$ in the given equations, we get

2m + 3n = 2 …….(i)

4m – 9n = -1 ……(ii)

Multiplying equation (i) by 3, we get

6m + 9n = 6 ………..(iii)

Adding equation (ii) and (iii), we get

10m = 5

m = 12$\frac{1}{2}$ ……(iv)

Now by putting the value of ‘m’ in equation (i), we get

2×12+2n=2$2\times \frac{1}{2}+2n = 2$

3n = 1

n = 13$\frac{1}{3}$

m = 1a=12$\frac{1}{\sqrt{a}}=\frac{1}{2}$

a=2$\sqrt{a}=2$

a = 4

n = 1b=13$\frac{1}{\sqrt{b}}=\frac{1}{3}$

b=3$\sqrt{b}=3$

b = 9

Hence, a = 4 and b = 9

Question 6:

5a1+1b2=2$\frac{5}{a-1}+\frac{1}{b-2}=2$

6a13b2=1$\frac{6}{a-1}-\frac{3}{b-2}=1$

Solution:

Substituting 1a1=m$\frac{1}{a-1}=m$ and 1b2=n$\frac{1}{b-2}=n$ in the given equations, we get,

5m + n = 2 ……(i)

6m – 3n = 1 …..(ii)

Multiplying equation (i) by 3, we get

15m + 3n = 6 ……..(iii)

Adding (ii) and (iii) we get

21m = 7

m = 13$\frac{1}{3}$

Putting this value in equation (i), we get

5×13+n=2$5\times \frac{1}{3}+n=2$ n=253=13$n=2-\frac{5}{3}=\frac{1}{3}$ m=1a1=13a1=3a=4$m=\frac{1}{a-1}=\frac{1}{3}\:\:\Rightarrow a-1=3\:\:\Rightarrow a=4$ n=1b2=13b2=3b=5$n=\frac{1}{b-2}=\frac{1}{3}\:\:\Rightarrow b-2=3\:\:\Rightarrow b=5$

Hence, a = 4 and b = 5

Question 7:

6a + 3b = 6ab

=> 6b+3a=6$\frac{6}{b}+\frac{3}{a}=6$ …..(i)

2a + 4b = 5ab

=> 2b+4a=5$\frac{2}{b}+\frac{4}{a}=5$ …….(ii)

Substituting 1a=m$\frac{1}{a}=m$ and 1b=n$\frac{1}{b}=n$
3m + 6n – 6 = 0

4m + 2n – 5 = 0

By cross-multiplication method, we get

m30(12)=n24(15)=1624$\frac{m}{-30-\left ( -12 \right )}=\frac{n}{-24-\left ( -15 \right )}=\frac{1}{6-24}$ m18=n9=118$\frac{m}{-18}=\frac{n}{-9}=\frac{1}{-18}$ m18=118andn9=118$\frac{m}{-18}=\frac{1}{-18}\:and\:\frac{n}{-9}=\frac{1}{-18}$

m = 1 and n = 1/2

m = 1/a = 1 and n = 1/b = 1/2

a = 1                              b = 2

Hence, a = 1 and b = 2

Question 8:

13a+b+13ab=34$\frac{1}{3a+b}+\frac{1}{3a-b}=\frac{3}{4}$

12(3a+b)12(3ab)=18$\frac{1}{2\left ( 3a+b \right )}-\frac{1}{2\left ( 3a-b \right )}=\frac{-1}{8}$

Solution:

substituting 13a+b=mand13ab=n$\frac{1}{3a+b}=m\:and\:\frac{1}{3a-b}=n$ in the given equations, we get

m + n = 3/4  ……(1)

m/2 – n/2 = -1/8

m – n = -1/4  ……(2)

Adding (1) and (2), we get

2m = 3/4 – 1/4

2m = 1/2

Putting in (2), we get

1/4 – n = -1/4

n = 1/4 + 1/4 = 1/2

m = 13a+b=14$\frac{1}{3a+b}=\frac{1}{4}$

3a + b = 4  …….(3)

n = 13ab=12$\frac{1}{3a-b}=\frac{1}{2}$

3a – b = 2 ………(4)

Adding equations (3) and (4), we get

6a = 6

a = 1 ……..(5)

Putting in (3), we get

3(1) + b = 4

b = 1

Hence, a = 1 and b = 1


text:
# Estimating Partition functions

I have a finite state ensemble with an energy functional (you can think of it as an ferromagnetic Ising model if you like), and I need very careful estimates of the partition function. What methods are available to me to get reasonable estimates (in reasonable temperature regimes) of $Z(\beta)$?

• Why do you need the numerical value of the partition function? It's just a normalization factor -- there's a number of ways at getting at any sort of observable without explicit reference to the value of Z.
– wsc
Jan 18, 2011 at 23:56

Good question! The reponse really depends on what you 'need': if you're sampling a small system and you're lucky enough to actually be able to count (or estimate...) the number of states $\mathcal{N}(E)$ for a given energy $E$, then you can explicitly work out $Z$. Such algorithms exist for the Ising model, but in the generic case you're not this lucky.
There is the possibility of 'naively sampling' the integral/sum you're doing: you can use any Monte Carlo algorithm (especially for Ising-like systems the literature is very rich) to generate a sequence of states and you use this representative sample to calculate $Z$ for any range of temperatures. (It's important though that you do not change $\beta$ during an MC run, since this tends to break detailed balance.)


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 4.1: Ratios and Rates

Difficulty Level: At Grade Created by: CK-12

## Introduction

The Top of the World - Mount Everest

Josh sat at the breakfast table with his nose in a book. His Mom poured some cereal into a bowl and put it down in front of him. Karen, Josh’s sister came into the kitchen and sat down next to him.

“I am reading a book about Mount Everest,” Josh said without looking up.

“Alright Josh, now put the book down and eat something,” their Mom instructed.

“It is terrific. Do you know that during the best year on Everest, and best meaning the year that the least number of people died, that 129 people summitted and only 8 died,” Josh said smiling.

“That hardly seems like something to smile about,” Mom said sipping her coffee.

“Yeah, how morbid,” Karen chimed in.

“Listen it may be morbid, but it is a fact. During the worst year, only 98 people summitted and 15 died. That is a lot of people. I mean we can compare the number of people who summitted and the number of people who didn’t make it,” Josh said.

“Well, it proves that Everest is a dangerous place and not a trip to be taken lightly,” Mom said.

“Yes, but think about how amazing it would be to stand on the top of the world!” Josh said.

“How do they figure out that the best year is the best and the worst year is the worst based on those numbers?” Karen asked.

“They compare the ratios,” Josh explained. “Let me tell you how.”

Do you know how? Given this information, Josh is going to show Karen how to simplify and compare ratios. In this lesson, you will learn how to write ratios and how to understand them better. Pay close attention and you can work with the ratios of the best and worst years on Everest at the end of the lesson.

What You Will Learn

In this lesson, you will learn how to execute the following skills.

• Write, compare and order ratios.
• Find and apply unit rates.
• Find and apply equivalent rates.
• Solve real-world problems involving ratios and rates.

Teaching Time

I. Write, Compare and Order Ratios

A ratio is a comparison of two things or two quantities. The key thing to notice about ratios is that we are comparing. There are many different ways to compare things in mathematics. For example, imagine that there are 25 students in a class; 12 boys and 13 girls. You could say that the ratio of boys to girls is 12 to 13. You could also write this ratio as 12 : 13 or 1213\begin{align*}\frac{12}{13}\end{align*}. The ratio of boys to total students can be written as 12 to 25, 12 : 25, or 1225\begin{align*}\frac{12}{25}\end{align*}. The ratio of girls to total students can be written as 13 to 25, 13 : 25, or 1325\begin{align*}\frac{13}{25}\end{align*}. You will also see that the order that you compare makes a difference. If you are comparing boys to total students, then the number of boys comes first or if you are using fraction form, it is the numerator.

That is a great question. There are three different ways to write ratios. You can write them with a colon between the two values that you are comparing, you can write them using the word “to”, and you can write them by putting the values in fraction form. You can choose which way you want to write a ratio and these ways are interchangeable too.

Take a minute to write the definition of a ratio and the three ways to write a ratio in your notebook.

We just looked at the three ways to write a ratio when you are making a simple comparison. Remember to read the information carefully so that you are clear what is being compared. Now let’s look at an example where you might need to figure something out to write the ratio.

Example

There are 32 red and yellow candies in a bag. There are 10 yellow candies. What is the ratio of red candies to total candies in the bag?

We need the ratio of red candies to total candies. We know that there are 32 total candies. We need to find the number of red candies in the bag before writing the ratio.

3210=22\begin{align*}32-10 = 22\end{align*}

Now write the ratio of red candies to total candies. Because a ratio is a comparison, it can be simplified. Make sure to reduce the ratio to lowest terms. This makes it easier to understand the quantities that are being compared.

red candiestotal candies=2232=1116\begin{align*}\frac{\text{red candies}}{\text{total candies}} = \frac{22}{32} = \frac{11}{16}\end{align*}

We also could have written this ratio in two other ways.

22 to 32 then simplified to 11 to 16

OR

22 : 32 then simplified to 11 : 16

All of these answers would have been correct. Remember that you can interchange the form that you choose to write a ratio.

We can also compare ratios. This is when we have two or more ratios and we want to figure out which ones are larger and which ones are smaller. Let’s look at an example.

Example

Mr. Collison’s class has 30 total students. Of these, 12 are boys. Mrs. Peterson’s class has 25 students. Of these, 11 are boys. Which class has a higher ratio of boys to total students?

First, find the ratio of boys to total students for both classes. You need to do this first because these are the quantities that you are comparing.

Mr. Collison’s class: 1230=25\begin{align*}\frac{12}{30} = \frac{2}{5}\end{align*}

Mrs. Peterson’s class: 1125\begin{align*}\frac{11}{25}\end{align*}

Now compare the ratios the same way you compare fractions. Find a common denominator and compare the numerators.

The least common denominator is 25.

25(55)=1025\begin{align*}\frac{2}{5} \left(\frac{5}{5} \right) = \frac{10}{25}\end{align*}

1025<1125\begin{align*}\frac{10}{25} < \frac{11}{25}\end{align*}, since 10 is less than 11.

Mrs. Peterson’s class has a higher ratio of boys to total students.

Note: In Example 2, even though Mr. Collison’s class had a larger number of boys in the class, Mrs. Peterson’s class had a larger ratio of boys to total students.

We can also order ratios. When you have more than two ratios, you can write them in order from least to greatest or from greatest to least. Let’s look at an example.

Example

Order the following ratios from least to greatest: 10 to 15, 1636\begin{align*}\frac{16}{36}\end{align*}, 12 : 48

The first thing to notice is that these ratios are all in different forms. Let’s write them in the same form first of all. Let’s work with fraction form so that we can apply what we know about comparing and ordering fractions.

10 to 15=1015\begin{align*}10 \ \text{to} \ 15 = \frac{10}{15}\end{align*}

1636\begin{align*}\frac{16}{36}\end{align*}

12:48=1248\begin{align*}12:48 = \frac{12}{48}\end{align*}

Now notice that none of these fractions are in simplest form. We can simplify them and that will make it much easier to order them.

101516361248=23=49=14\begin{align*}\frac{10}{15} &= \frac{2}{3}\\ \frac{16}{36} &= \frac{4}{9}\\ \frac{12}{48} &= \frac{1}{4}\end{align*}

If you understand fractions, we can simply order them right now. We know that one-fourth is the smallest part. That four is almost half of nine, so that would be the middle value, and that two-third would be the greatest part.

The answer is 14,49,23\begin{align*}\frac{1}{4}, \frac{4}{9}, \frac{2}{3}\end{align*}.

Well in that case, you could rewrite them all using a common denominator. Then you will be able to order the numerators. Let’s take a look at that.

144923=936=1636=2436\begin{align*}\frac{1}{4} &= \frac{9}{36}\\ \frac{4}{9} &= \frac{16}{36}\\ \frac{2}{3} &= \frac{24}{36}\end{align*}

You can see that our original work was correct.

II. Find and Apply Unit Rates

A unit rate is a special kind of ratio, where the second number, or the denominator, is equal to one. With a unit rate, you are comparing a quantity to one. Some common examples of unit rates are miles per gallon, price per pound, and pay rate per hour.

To find a unit rate, simplify the ratio so that you have a 1 in the denominator. You can simply divide the first number in the ratio by the second. Make sure you keep track of the units.

Write this information on unit rates down in your notebook. Then continue with the example.

Example

Kayla bought 5.5 pounds of apples. She paid a total of $7.15. What was the unit rate of the apples per pound? A key word is here “per pound”. When you see the word “per”, you should know that you are working with unit rates. You need to find the price per pound. We can write the ratio of price to pounds using the information in the problem. That is what we are comparing, so that is how we write the ratio. Then we can fill in the given information. pricepounds=$7.155.5 pounds\begin{align*}\frac{price}{pounds} = \frac{\ 7.15}{5.5 \ pounds}\end{align*}

Now divide to find the price per pound. We divide the price by the number of pounds that she bought.

5.5)7.15¯¯¯¯¯¯¯¯¯¯¯¯\begin{align*}{5.5 \overline{){7.15\; }}}\end{align*}

$7.155.5 pounds=$1.301 pound\begin{align*}\frac{\7.15}{5.5 \ pounds} = \frac{\ 1.30}{1 \ pound}\end{align*}

The unit rate is $1.30 per pound. Example Brian worked for 8 hours yesterday and made a total of$86. What is his pay rate?

Write a ratio comparing pay rate to hours worked. The rate that we are looking for is what Brian made “per” hour. Even though the problem doesn’t use the word “per” a pay rate is per hour.

pay ratehours worked=868 hours\begin{align*}\frac{\text{pay rate}}{\text{hours worked}} = \frac{\ 86}{8 \ hours}\end{align*} Now divide to find the unit rate.868 hours=10.751 hour\begin{align*}\frac{\86}{8 \ hours} = \frac{\ 10.75}{1 \ hour}\end{align*} Brian’s pay rate is10.75 per hour.

You may also see rates that are equal. Let’s look at equivalent rates.

III. Find and Apply Equivalent Rates

An equivalent rate is a rate that is equal but is written in a different form.'

To find an equivalent rate, you need to write a fraction that is equivalent to the given fraction, or rate.

Example

Find an equivalent rate for this comparison.

2.001=?8\begin{align*}\frac{\2.00}{1} = \frac{?}{8}\end{align*} Notice that we have a unit rate here. We know that the unit rate is two dollars for every one thing. We want eight. We can figure out how to work on this problem by thinking mathematically. 1×8=8\begin{align*}1 \times 8 = 8\end{align*} Just like when we worked with fractions, whichever operation we perform with the denominator must be performed with the numerator too. We multiplied by eight, so we need to do that with the numerator too.2.001=16.008\begin{align*}\frac{\2.00}{1} = \frac{\ 16.00}{8}\end{align*} These two rates are equivalent. As long as you multiply the numerator and the denominator by the same value, you will always create an equivalent rate! IV. Solve Real – World Problems Involving Ratios and Rates Now that you understand ratios and rates, we can look at applying this information when working with real – world problems. Let’s look at some examples. Example Myra’s team scored 10 goals in the last 3 games. At this rate, how many goals will Myra’s team score in 6 games? First, write the ratio to show the team’s scoring rate. goalsgames=10 goals3 games\begin{align*}\frac{goals}{games} = \frac{10 \ goals}{3 \ games}\end{align*} You need to know the equivalent rate for 6 games. Notice that the second ratio has the games in the same spot of the denominator. Be sure that you write the ratios so that the same quantities are being compared. If you mix them up, you get a different result. 10 goals3 games=? goals6 games\begin{align*}\frac{10 \ goals}{3 \ games} = \frac{? \ goals}{6 \ games}\end{align*} Look at the two fractions. The denominator is doubled in the second fraction. So, multiply the first fraction by an equivalent of 1 in order to get the second fraction. 10 goals3 games(22)=20 goals6 games\begin{align*}\frac{10 \ goals}{3 \ games} \left(\frac{2}{2}\right) = \frac{20 \ goals}{6 \ games}\end{align*} An equivalent rate is 20 goals in 6 games. So at this rate, Myra’s team will score 20 goals. Example A store sells salmon for6.99 per pound. What is the rate for 6 pounds of salmon?

You know the rate per pound. You need to find the rate for 6 pounds. So you can multiply the unit rate by 6 to find the equivalent ratio.

$6.991 pound(66)=$41.946 pounds\begin{align*}\frac{\6.99}{1 \ pound} \left(\frac{6}{6} \right) = \frac{\41.94}{6 \ pounds}\end{align*}

The rate for 6 pounds of salmon is 41.94. Now let’s go back to the problem from the introduction and work on solving that problem. ## Real-Life Example Completed The Top of the World - Mount Everest Here is the original problem once again. First, reread it. Then you will need to write and simplify two ratios. One ratio will represent the numbers from the best year on Everest and one ratio will represent the numbers from the worst year on Everest. There are two parts to your answer. Josh sat at the breakfast table with his nose in a book. His Mom poured some cereal into a bowl and put it down in front of him. Karen, Josh’s sister came into the kitchen and sat down next to him. “What are you reading?” she asked. “I am reading a book about Mount Everest,” Josh said without looking up. “Alright Josh, now put the book down and eat something,” their Mom instructed. “It is terrific. Do you know that during the best year on Everest, and best meaning the year that the least number of people died, that 129 people summitted and only 8 died,” Josh said smiling. “That hardly seems like something to smile about,” Mom said sipping her coffee. “Yeah, how morbid,” Karen chimed in. “Listen it may be morbid, but it is a fact. During the worst year, only 98 people summitted and 15 died. That is a lot of people. I mean we can compare the number of people who summitted and the number of people who didn’t make it,” Josh said. “Well, it proves that Everest is a dangerous place and not a trip to be taken lightly,” Mom said. “Yes, but think about how amazing it would be to stand on the top of the world!” Josh said. “How do they figure out that the best year is the best and the worst year is the worst based on those numbers?” Karen asked. “They compare the ratios,” Josh explained. “Let me tell you how.” Now write your ratios. Remember to simplify them. There are two parts to your answer. Solution to Real – Life Example We are going to write two ratios that we can compare. One ratio will represent the best year on Everest and one will represent the worst year on Everest. First, let’s look at the numbers from the best year on Everest. 129 people summitted 8 died The ratio is 129 : 8. However, we want to simplify this ratio to get a better idea of the size of the ratio. Let’s rewrite it in fraction form and simplify it. 1298=161\begin{align*}\frac{129}{8}= \frac{16}{1}\end{align*} This means that for every 16 people who summitted one did not make it. Now let’s look at the numbers for the worst year on Everest. 98 people summitted 16 people died 9816=6.5 to 1\begin{align*}\frac{98}{16} = 6.5 \ to \ 1\end{align*} This means that for every 612\begin{align*}6 \frac{1}{2}\end{align*} people who summitted, one did not make it. From these ratios, you can see how the best year compares to the worst year on Everest. ## Vocabulary Here are the vocabulary words that are found in this lesson. Ratio a way of comparing two numbers or quantities. Ratios can be written in fraction form, with a colon or by using the word “to”. Unit Rate a ratio that is comparing a quantity to one. The word “per” is a key word with unit rates. Equivalent Rate two rates that are equal although different values are being used to represent the same quantities. Resources www.mnteverest.net ## Time to Practice Directions: Look at each ratio. Then write it in the other two ways that it can be written. 1. 16 to 3 2. 4 to 5 3. 1 : 4 4. 121\begin{align*}\frac{12}{1}\end{align*} 5. 6 : 11 6. 33 to 100 7. 49\begin{align*}\frac{4}{9}\end{align*} 8. 3 to 4 9. 45 to 12 10. 12 : 12 Directions: Simplify each ratio and write your answer in fraction form. 1. 4 to 12 2. 5 : 20 3. 36 to 6 4. 18 : 36 5. 20 to 100 Directions: Use what you have learned about ratios to solve each problem. In Kyle’s drawer, there are 14 pairs of white socks and 8 pairs of black socks 1. Write the ratio of black socks to white socks. 2. Write the ratio of black socks to total socks. 3. Write the ratio of white socks to total socks. 4. There are 150 apartments in the Gray building. Of these, 60 are rented and the rest are owned. There are 65 apartments in the Black building. Of these, 45 are rented and the rest are owned. Which building has a larger ratio of owned apartments to total apartments, and by how much? 5. Susan drove a total of 182 miles in three and a quarter hours. If she drove the same speed the whole drive, what was her rate in miles per hour? 6. Holly works at a library reshelving books. She reshelved 960 books in 4 hours. What is Holly’s rate of reshelving in books per hour? 7. Sam bought 9.5 pounds of peaches to make a pie. The peaches cost15.39. What was the unit rate of the peaches?
8. A vacation house costs \$1,100 to rent per month. If there are 30 days in the month, what is the daily rate?
9. Julie can type 55 words per minute. What is the equivalent rate of typing 15 minutes?
10. Don can wrap 8 presents in an hour. What is Don’s rate for 12 presents?

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# Is $T^2$ equivalent to 1 Way MANOVA when we have just two populations?

When we're in the presence of two populations, for the same assumptions (random samples, multivariate distribution, same covariance matrix), and we want to test the mean vectors, we have two options:

• $T^2$ statistic (a Mahalanobis distance function)
• A statistic,function of the Wilk's lambda (MANOVA)

Both statistics have the same distribution $F_{p,n-p-1}$.

I was thinking that both should be equivalent. However, I don't see how the MANOVA statistic which involves a fraction of determinants can be made to be equal to the $T^2$ statistic.

Any help would be appreciated.

• The answer to the title question is: Yes. – amoeba says Reinstate Monica Apr 25 '17 at 15:03
• Regarding the math, MANOVA looks at $\Sigma_W^{-1}\Sigma_B$, but for two classes $\Sigma_B$ is rank-one, so this whole expression reduces to T2 which is $(\mu_1-\mu_2)^\top\Sigma_W^{-1}(\mu_1-\mu_2)$. Or something along these lines. – amoeba says Reinstate Monica Apr 25 '17 at 15:06


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# Why does $\arctan(x)= \frac{1}{2i}\log \left( \frac{x-i}{x+i}\right)+k$?

Letting $x=\tan(u)$, $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{1+\tan(u)^2}\sec^2(u) \, du=u+k=\arctan(x)+k$$ Also, $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{(x-i)(x+i)} \, dx=\frac{1}{2i}\int\frac{1}{(x-i)}-\frac{1}{(x+i)} \, dx$$ $$=\frac{1}{2i}\left(\ln(x-i)-\ln(x+i)\right)+c$$ $$=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+c$$ Giving $$\arctan(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+q$$

Why is this correct? What is the nature of $q$ (is it 'flexible' so the equality doesn't really mean much)?

I think it probably has something to do with the relationship between $\log(z)$ and $\arg(z)$, but $\arg(z\pm i)$ is hard to calculate neatly.

• $e^{iz} = \cos z+i\sin z$. I've posted a fairly short argument based on that. Jun 7, 2013 at 22:29
• It would be a nice exercise to plug in x = 1 on both sides of that equation. arctan(1) is ofcourse a well known value. Try to get that value from the right side too... Jun 7, 2013 at 23:16
• Are there typos in the u-substitution? Jun 7, 2013 at 23:52
• Don't think so, what would be incorrect? Jun 8, 2013 at 0:15
• NB: the constant q above is $\pi/2$. To connect with the formula from @MichaelHardy, use $\ln(-z) = \ln(e^{i\pi}z) = \ln(e^{i\pi})+\ln(z)=i\pi+\ln(z)$. Aug 12, 2021 at 0:43

The basic identity used here was discovered in the 18th century by Leonhard Euler: $$e^{iz} = \cos z+i\sin z$$ where of course the cosine and the sine are of $$x$$ in radians.

It follows that $$\cos z = \dfrac{e^{iz}+e^{-iz}}{2}$$ and $$\sin z = \dfrac{e^{iz}-e^{-iz}}{2i}$$.

Therefore $$a= \tan z = \frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})} = -i\frac{e^{2iz}-1}{e^{2iz}+1} = -i\frac{b-1}{b+1}.$$ \begin{align} a & = -i\frac{b-1}{b+1} \\[10pt] (b+1)a & = -i(b-1) \\[10pt] b(i+a) & = i-a \\[10pt] b & = \frac{i-a}{i+a} \\[10pt] e^{2iz} & = \frac{i-a}{i+a} \\[10pt] 2iz & = \log \frac{i-a}{i+a} \end{align} The logarithm, like the arctangent, is "multiple-valued".

• Thank you, it's nice to have things explicitly stated from the basics.
– Meow
Jun 7, 2013 at 22:33

We have that $$i\sin z=\sinh iz$$ $$\cos z=\cosh iz$$

This means that $$i\tan z=\tanh iz$$

But $$\tanh^{-1}z=\frac 1 2\log\left(\frac{1+z}{1-z}\right)$$

• Does that definition for $\tanh^{-1}(z)$ come from $\tanh(z)= \frac{e^z-e^{-z}}{e^z+e^{-z}}$? That's the problem I was having.
– Meow
Jun 7, 2013 at 22:10
• @Alyosha Yes. Actually, it is easier yo use $$\tanh z=\frac{e^{2z}-1}{e^{2z}+1}$$ to find the inverse.
– Pedro
Jun 7, 2013 at 22:11

This is correct, and by letting $x$ tend to $+\infty$ you can find the constant term: $$\arctan(x) = \frac{1}{2i} \log(\frac{x-i}{x+i}) + \frac{\pi}{2},$$ since $\log(\frac{x-i}{x+i})$ tends to $\log(1) = 0$ and $\arctan(x)$ tends to $\pi / 2$.

This is valid for $x > 0$. Unfortunately, there is a difficulty with the $\log$ function at $-1$, and for $x < 0$ we get the formula $$\arctan(x) = \frac{1}{2i} \log(\frac{x-i}{x+i}) - \frac{\pi}{2}$$ by a similar argument.

That the logarithm and inverse trigonometric functions are related shouldn't be surprising, though. Remember Euler's formula: $e^{ix} = \cos(x) + i \sin(x)$ that relates the exponential to the trigonometric functions.

• Surely the logarithmic term will be multivalued? Or this covered by the fact the inverse tangent is also multivalued? (Infact, now I think about it, it is indeed) Jun 7, 2013 at 22:04
• @AndrewD You're right, both are multivalued, but unfortunately not in quite the same way Jun 7, 2013 at 22:17

The other answers are fine. However, to my taste, there is a more conceptual explanation involving covering space theory.

In summary, the above relationship between inverse tangent and complex logarithm holds since the tangent function and exponential function are "essentially the same" universal topological covering map. In even more informal terms, the exponential map and the tangent are (up to isomorphism) both a solution to the same optimization problem (if one views a universal cover, or more generally a universal object, as a solution to an optimization problem, as done here).

Let me explain. Firstly, both the exponential function $$\operatorname{exp}\colon \mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$$ and the tangent function $$\operatorname{tan}\colon \mathbb{C}\rightarrow \mathbb{P}^1\setminus\{\pm i\}$$ are universal covers. Here, $$\mathbb{P}^1$$ denotes the Riemann sphere.

Now, take a homeomorphism $$f\colon \mathbb{C}\setminus\{0\}\rightarrow \mathbb{P}^1\setminus\{\pm i\}$$. As an example, consider the following (restricted) Möbius transformation $$f(z):=-i\cdot\frac{z-1}{z+1}.$$ This map is induced from the unique autmorphism $$\hat{f}$$ of the Riemann sphere with $$\hat{f}(0)=i, \hat{f}(\infty)=-i$$ and $$\hat{f}(1)=0.$$ Its inverse is given by $$f^{-1}(z)=\frac{i-z}{i+z}.$$ Next, consider the following diagram of covering maps $$\require{AMScd} \begin{CD} \mathbb{C} @.\mathbb{C}\\ @V\operatorname{exp} VV @VV\operatorname{tan}V\\ \mathbb{C}\setminus\{0\}@>>f> \mathbb{P}^1\setminus\{\pm i\} \end{CD}$$

Since $$\operatorname{exp}\colon \mathbb{C}\rightarrow \mathbb{C}\setminus\{0\}$$ is a universal cover and $$f$$ is a biholomorphism (in particular a homeomorphism), the composition $$f\circ \operatorname{exp}\colon \mathbb{C}\rightarrow \mathbb{P}^1\setminus\{\pm i\}$$ is a universal cover. Furthermore, we have $$f(\operatorname{exp}(0))=0=\operatorname{tan}(0)$$.

In other words, the covering maps $$f\circ\operatorname{exp}\colon (\mathbb{C},0)\rightarrow (\mathbb{P}^1\setminus\{\pm i\},0)$$ and $$\operatorname{tan}\colon (\mathbb{C},0)\rightarrow (\mathbb{P}^1\setminus\{\pm i\},0)$$ are both initial objects in the category of (pointed) coverings of the pointed topological space $$(\mathbb{P}^1\setminus\{\pm i\},0)$$. The objects of this category are covering maps $$p\colon(X,x)\rightarrow (\mathbb{P}^1\setminus\{\pm i\},0)$$ with $$p(x)=0$$. Here, $$(X,x)$$ is any topological space $$X$$ with base point $$x$$. The morphisms are fiber-preserving continuous maps between coverings of $$(\mathbb{P}^1\setminus\{\pm i\},0)$$ that preserve the base point.

Now, initial objects are unique up to unique isomorphism. Said differently, there exists a unique homeomorphism $$g\colon \mathbb{C}\rightarrow \mathbb{C}$$ with $$g(0)=0$$ such that the following diagram commutes $$\require{AMScd} \begin{CD} \mathbb{C} @>g>>\mathbb{C}\\ @V\operatorname{exp} VV @VV\operatorname{tan}V\\ \mathbb{C}\setminus\{0\}@>>f> \mathbb{P}^1\setminus\{\pm i\} \end{CD}$$ This mapping reads as $$g(z)=\frac{1}{2i}z$$. It should not come as a suprise that $$g$$ has to be an automorphism of the complex plane (and hence of the form $$g(z)=az$$ with $$a\in \mathbb{C}\setminus\{0\}$$): A lift of a holomorphic mapping between Riemann surfaces along a unbranched holomorphic covering map is necessarily holomorphic (see Theorem 4.9 in Forster's Lectures on Riemann surfaces).

Anyways, by reversing arrows (and choosing branches of complex logarithm and tangent) we obtain the formula given in Michael Hardy's answer

$$\operatorname{arctan}(z)=(g\circ \operatorname{log}\circ f^{-1})(z)=\frac{1}{2i}\operatorname{log}\big(\frac{i-z}{i+z}\big).$$


text:
My Math Forum Non-computable Reals

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 June 12th, 2017, 04:03 PM #1 Senior Member   Joined: Jun 2014 From: USA Posts: 363 Thanks: 26 Non-computable Reals Question 1: Can a non-computable real $x$ be created using the following method? Let $s \in ( \,0,1) \,$ be a computable number that is irrational. A turing machine can therefore encode the binary expansion of $s$ to any desired number of bits: $$s = s_1s_2s_3s_4\dots\text{, where each } s_i \in \{0, 1\}$$ Let $t_1, t_2, t_3, \dots$ be a listing of binary strings $t_i$ created using each bit in $s$ as follows: $$\begin{matrix} t_1 \text{ }| & t_{1,1} = s_1 & \longrightarrow & t_{1,2} = s_2 & \text{ } & t_{1,3} = s_6 & \longrightarrow & \dots \\ \text{ } & \text{ } & \swarrow & \text{ } & \nearrow & \text{ } & \swarrow\\ t_2 \text{ }| & t_{2,1} = s_3 & \text{ } & t_{2,2} = s_5 & \text{ } & t_{2,3} = s_8 & \text{ } & \dots \\ \text{ } & \downarrow & \nearrow & \text{ } & \swarrow & \text{ } & \nearrow \\ t_3 \text{ }| & t_{3,1} = s_4 & \text{ } & t_{3,2} = s_9 & \text{ } & t_{3,3} = s_{13} & \text{ } & \dots \\ \text{ } & \text{ } & \swarrow & \text{ } & \nearrow & \text{ } & \swarrow \\ t_4 \text{ }| & t_{4,1} = s_{10} & \text{ } & t_{4,2} = s_{12} & \text{ } & t_{4,3} = s_{19} & \text{ } & \dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix}$$ Let $x'$ be Cantor's diagonal sequence with respect to the listing of binary strings $t_i$: $$x' = x_1x_2x_3\dots \text{, where each } x_i = \begin{cases} 0 & \text{if } t_{i,i} = 1 \\ 1 & \text{if } t_{i,i} = 0 \end{cases}$$ Let $x \in ( \,0,1) \,$ be created using the bits $x_i$ of $x'$ as follows: $$x = 0.x_1x_2x_3\dots$$ Comments: Where $s$ is computable and $x$ is computed from $s$, I believe that we must consider $x$ to be computable as well. Perhaps I am mistaken, thus my initial question. Question 2: If $x$ is in fact computable because $s$ was computable, then is it not possible to find a suitable $s$ so as to compute any given $x$? That is, I am tempted to speculate that $s$ can be computed in a manner such that the bits of the diagonal $x'$ of the listed strings $t_i$ can in turn result in any given $x$. Where the computable numbers are countable, I believe my speculation must be incorrect, however. Trivially, any desired bit $x_i$ can result from a given $s$. Showing that a suitable $s$ exists to compute all desired bits $x_i$ for a given $x$ is not trivial, however. My initial thought was that a suitable $s$ for any given $x$ could be constructed from a computable sequence of computable numbers using the "interleaving" technique shown in the accepted answer to this stackexchange thread (that is, create $s$ by interleaving a computable sequence of computable numbers $y_1, y_2, y_3\dots$), but if doing that, we might as well just interleave the computable sequence of computable numbers to directly create $x$ itself, so I am stalled out at the moment: https://math.stackexchange.com/quest.../183383#183383 Chances are, I just need to step back from this for a little bit. Getting an answer to my initial question would be a nice start. Thank you.
 June 12th, 2017, 04:13 PM #2 Senior Member   Joined: Aug 2012 Posts: 1,858 Thanks: 513 I didn't work through your post yet but you can create a noncomputable real by choosing a programming language, enumerating all the programs (by program we mean the kind that either halts or not, as in Turing machines), and writing a 1-bit or a 0-bit in the n-th binary position depending on whether program n halts or not. This is definable but can't be computable else we could use the bitstring to solve the halting problem. This method is similar but not identical to Chaitin's Omega. ps -- I read through your post. Your method can't do what you think it does since you're claiming something false. However can you step back and agree or disagree that you understand why if we simply list all the computable reals (of which there are only countably many) and take the anti-diagonal, the anti-diag can NOT be computable. That will give us a common base of understanding here. And reading through your post a second time makes me think this is the heart of the matter. Let's just handle that case first, what do you think? Last edited by Maschke; June 12th, 2017 at 04:21 PM.
June 12th, 2017, 06:20 PM   #3
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Quote:
 Originally Posted by Maschke ps -- I read through your post. Your method can't do what you think it does since you're claiming something false.
What do I think my method does? I speculated that $x$ is computable because some 'simple' tweaks to a program for $s$ would seemingly spit out $x$ instead. I'm literally asking though, so I don't think my method does anything at this point quite frankly.

Quote:
 Originally Posted by Maschke However can you step back and agree or disagree that you understand why if we simply list all the computable reals (of which there are only countably many) and take the anti-diagonal, the anti-diag can NOT be computable. That will give us a common base of understanding here. And reading through your post a second time makes me think this is the heart of the matter. Let's just handle that case first, what do you think?
I agree. Namely, if $C$ is a listing of all computable reals, Cantor's diagonal would produce a real not in the list, so such an anti-diagonal could not be computable. I think this next question highlights my (mis)understanding if you'll put up with one more quick definition so I can ask it:

If $C^I = c_1, c_2, c_3\dots$ is a listing of all computable reals in (0, 1) that are irrational, then there exists a real number $s^C \in ( \,0,1) \,$ such that the binary sequences $t_i$ produced using my above method would correlate with $C^I$ such that:

$c_1 = 0.t_{1,1}t_{1,2}t_{1,3}\dots$
$c_2 = 0.t_{2,1}t_{2,2}t_{2,3} \dots$
$c_3 = 0.t_{3,1}t_{3,2}t_{3,3}\dots$
.
.
.

In this case, $s^C$ cannot be computable either, right? I say this because I am still speculating that "$s$ is computable implies $x$ is computable" and similarly "$s^C$ is computable implies $x^C$ is computable" too. But, by definition of the list, $x^C$ could not be computable because $x^C$ is not in the list, so $s^C$ cannot be computable either.

June 12th, 2017, 08:22 PM   #4
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 Originally Posted by AplanisTophet If $C^I = c_1, c_2, c_3\dots$ is a listing of all computable reals in (0, 1) that are irrational
Should just be "If $C^I = c_1, c_2, c_3\dots$ is a listing of all computable reals in [0, 1]." No need to restrict to irrationals, but infinite binary expansions should be used in the case of dyadic rationals where also 1 should be interpreted as 0.111... and 0 should be interpreted as 0.000..., otherwise there's nothing to stop $x^C$ from being equal to 0 or 1 and therefore both computable and not in the list.

Sorry.

Last edited by AplanisTophet; June 12th, 2017 at 08:39 PM.

June 12th, 2017, 08:40 PM   #5
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Quote:
 Originally Posted by AplanisTophet I agree. Namely, if $C$ is a listing of all computable reals, Cantor's diagonal would produce a real not in the list, so such an anti-diagonal could not be computable.
Didn't work through your argument yet, just trying to sort this out myself.

Yes that's the problem. We have a list of all the computable numbers, so the anti-diag can't be computable. But if each row of the list contains digits cranked out by an algorithm, and the anti-diag rule "choose something other than the n-th digit of the n-th row" is clearly an algorithm, so why isn't the anti-diag computable?

The answer seems to be that the enumeration itself is not computable. I found this stackexchange thread. See Carl Mummert's answer. I still find this a bit murky -- more murky than I remember it being. I want to give this some more thought. I seem to recall understanding this at one point in the past but now it seems elusive. It's getting late, I'll sort this out tomorrow.

In any event, does this have any bearing on your original question? You mentioned Cantor diagonalization.

June 12th, 2017, 08:47 PM   #6
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Quote:
 Originally Posted by Maschke Didn't work through your argument yet, just trying to sort this out myself. Yes that's the problem. We have a list of all the computable numbers, so the anti-diag can't be computable. But if each row of the list contains digits cranked out by an algorithm, and the anti-diag rule "choose something other than the n-th digit of the n-th row" is clearly an algorithm, so why isn't the anti-diag computable? The answer seems to be that the enumeration itself is not computable. I found this stackexchange thread. See Carl Mummert's answer. I still find this a bit murky -- more murky than I remember it being. I want to give this some more thought. I seem to recall understanding this at one point in the past but now it seems elusive. It's getting late, I'll sort this out tomorrow. In any event, does this have any bearing on your original question? You mentioned Cantor diagonalization.
Thank you.

Again, sorry, see my revised definition of $C^I$ so that I can ensure the anti-diagonal isn't in the list. This is perhaps easier with binary sequences instead of the reals because of potentially tripping over the two different expansions for dyadic rationals. I think I patched it up, but essentially, that's what I arrive at is that $s^C$ cannot be computable again assuming if it were to be, then $x^C$ must be also which is impossible.

June 18th, 2017, 09:07 PM   #7
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Quote:
 Originally Posted by AplanisTophet Comments: Where $s$ is computable and $x$ is computed from $s$, I believe that we must consider $x$ to be computable as well. Perhaps I am mistaken, thus my initial question.
Yes of course. Now that I've looked at this, the zigzag is a computable function of the digits of $s$. If $s$ is computable so is the zigzag and vice versa.

Quote:
 Originally Posted by AplanisTophet If $x$ is in fact computable because $s$ was computable, then is it not possible to find a suitable $s$ so as to compute any given $x$? That is, I am tempted to speculate that $s$ can be computed in a manner such that the bits of the diagonal $x'$ of the listed strings $t_i$ can in turn result in any given $x$. Where the computable numbers are countable, I believe my speculation must be incorrect, however.
I don't see how that works at all. To produce a noncomputable zigzag you have to start with a noncomputable $s$.

Last edited by Maschke; June 18th, 2017 at 09:14 PM.

June 18th, 2017, 10:26 PM   #8
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Quote:
 Originally Posted by Maschke I don't see how that works at all. To produce a noncomputable zigzag you have to start with a noncomputable $s$.

Quote:
 Originally Posted by AplanisTophet ...but essentially, that's what I arrive at is that $s^C$ cannot be computable again assuming if it were to be, then $x^C$ must be also which is impossible.
In other words, we can of course find a $s$ for any given $x$, but where we can prove that $x$ is not computable, then $s$ cannot be either. In this case, we can compare $s$ to the function $f( \,i,n) \,$ in Carl Mummert's answer (as shown in the stackexchange thread you found here https://math.stackexchange.com/quest...utable-numbers ). The function $f$ is not computable in that answer, just as I cannot find my computable $s$.

June 19th, 2017, 08:28 AM   #9
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Quote:
 Originally Posted by AplanisTophet Sorry, I didn't realize you were still looking into this. You had already found the answer.
I've been trying to understand the fundamental reason that an enumeration of computable reals must itself be noncomputable. It's because a computable enumeration would violate the noncomputability of the halting problem. If you had a computable function that returns a halting algorithm for each input $n$, that would amount to knowing which algorithms halt, which is impossible.

However I have a couple of questions in my mind that I'm still working through.

In any event, this gives us a constructive Cantor proof in a sense. A constructivist who denies the existence of anything that's not computable would not regard the computable numbers as being countable, because there is no computable enumeration. Meaning (to a constructivist) that there is no enumeration at all.

Last edited by Maschke; June 19th, 2017 at 08:40 AM.

June 19th, 2017, 07:32 PM   #10
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 Originally Posted by Maschke I've been trying to understand the fundamental reason that an enumeration of computable reals must itself be noncomputable.
I assume you mean an enumeration of all computable reals. If $s$ is computable, then $\{t_i\}, i \in \mathbb{N}$ is computable using my method from the OP. If $s$ is not computable, then $\{t_i\}, i \in \mathbb{N}$ is not computable either.

We do have $\{ x \in ( \,0,1) \, : \text{the sequence }\{0.t_i\}, i \in \mathbb{N} \text{ lists all computable reals in } [ \,0,1] \, \}$. I find that somewhat interesting, noting that each $x$ in the set is not computable.

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