Wikipedia:Reference desk/Archives/Mathematics/2009 August 30

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August 30

Principle of mathematical induction

Please solve this problem of mathematical induction

sinx + sin3x + ... +sin(2n-1)x =sin^2nx/sinx —Preceding unsigned comment added by Tipusultan11 (talk • contribs) 07:08, 30 August 2009 (UTC)

If you can tell us where you are getting stuck in solving this problem maybe we can help you with it. What have you tried to do so far? Do you understand induction? Eric. 98.207.86.2 (talk) 07:47, 30 August 2009 (UTC)

Have you to use trig identities? You can do it easily using Euler's formula‎‎ Dmcq (talk) 11:37, 30 August 2009 (UTC)

...and powers thereof. ~~ Dr Dec (Talk) ~~ 11:43, 30 August 2009 (UTC)

As an induction problem it seems simpler than taking powers of the Euler formula, expanding, then comparing terms in the binomial expansion. The base case would be when n = 1, i.e.

$P(1):\ \sin(x)={\frac {\sin ^{2}(x)}{\sin(x)}}\ ,$

and this is clearly true. Next, assume that P(k) is true and prove that this implies that P(k+1) is also true. Well for the case P(k) we have

$P(k):\ \sin(x)+\sin(3x)+\cdots +\sin((2k-1)x)={\frac {\sin ^{2}(kx)}{\sin(x)}}\ .$

We assume that this equality holds. To prove that P(k) implies P(k+1) we simply need to show that

${\frac {\sin ^{2}(kx)}{\sin(x)}}+\sin((2k+1)x)={\frac {\sin ^{2}((k+1)x)}{\sin(x)}}\ ,$

which I won't do because I'm not going to do all of your homework for you ;oP ~~ Dr Dec (Talk) ~~ 12:30, 30 August 2009 (UTC)

...and this last step isn't too hard if you use the formula $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x).$ ~~ Dr Dec (Talk) ~~ 12:42, 30 August 2009 (UTC)

Maybe I'm a stickler for detail, but the conjecture, as written, is false for x=kπ, k integer. --Stephan Schulz (talk) 12:44, 30 August 2009 (UTC)

It's not false, it's true for all x. I assume your problem is that the RHS has a zero denominator for x = mπ (m an integer). Well there's no problem: it's perfectly well defined. The limit of the RHS exists, and is equal to zero. See the article on L'Hôpital's rule. (In fact you don't even need that: just completing the inductive steps that I have outlined above show you that the singularities at x = mπ are removable.) ~~ Dr Dec (Talk) ~~ 12:50, 30 August 2009 (UTC)

Sure, the function can be steadily continued at the undefined points, as limits from both sides agree. But as written, it is not so continued. Even your base case is, strictly, wrong. As I wrote, stickler for details ;-) --Stephan Schulz (talk) 13:15, 30 August 2009 (UTC)

Why is the base case wrong? It has a removable singularity - there is no problem! The base case doesn't require steadily continuing anything. We are canceling functions. I've not tried to evaluate the numerator and denominator before canceling. Would you argue that x/x = 1? I hope not! Well, you're trying to say that there's a problem with the statment that x/x = 1 by saying that the base case isn't well defined. In fact the expression x/x is perfectly well defined; its value at x = 0 is indeterminate, but can be calculated as a limit. Where did I evaluate anything above? ~~ Dr Dec (Talk) ~~ 13:18, 30 August 2009 (UTC)

To make Stephan happy, let's rephrase the problem. Show, using mathematical induction, that for any positive integer n, and any real number y:

$\lim _{x\to y}(\sin(x)+\sin(3x)+\cdots +\sin((2n-1)x))=\lim _{x\to y}{\frac {\sin ^{2}(nx)}{\sin(x)}}\ .$

I always tacitly assume the idea of a limit when I evaluate a function. I'm sure most other mathematicians do too. ~~ Dr Dec (Talk) ~~ 13:30, 30 August 2009 (UTC)

I suspect you're missing a negation somewhere above. But yes, for me (the function defined over the reals by the formula) x/x is different from (the function defined over the reals by the constant formula) 1. In set notation, 1 is $(R\times \{1\})$, and x/x is $(R\backslash \{0\}\times {1})$. The one is a proper subset of the other (and so can of course be extended), and hence they are not the same. --Stephan Schulz (talk) 13:54, 30 August 2009 (UTC)

...and this helps to answer Tipusultan11's question, how? (p.s. So what if I haven't not missed no negatives?) ~~ Dr Dec (Talk) ~~ 14:02, 30 August 2009 (UTC)

Stephan, let's stop this now. You can carry on worrying about the validity of expressions like x/x, so that I can try to help answer people's questions here on the RD. Maybe Tipusultan11 should just hand his homework in with the solution: Question not well defined? ~~ Dr Dec (Talk) ~~ 14:08, 30 August 2009 (UTC)

For OP's interest, it will probably help him get full marks if he puts sin x != 0 somewhere. --91.145.89.58 (talk) 16:28, 30 August 2009 (UTC)

...meaning? ~~ Dr Dec (Talk) ~~ 16:39, 30 August 2009 (UTC)

In the case OP has to compose math test answers, it will steer him away from all trouble if he mentions that there's no equality when sin x = 0; x = pi * n. Forgive if my english is bad. --91.145.89.58 (talk) 16:52, 30 August 2009 (UTC)

Dr Dec. You write: "I always tacitly assume the idea of a limit when I evaluate a function. I'm sure most other mathematicians do too". How can you be that sure after meeting a mathematician who do not tacitly assume the idea of a limit? Your tacit assumption may lead you into trouble. Stephan's objection is perfectly valid. Bo Jacoby (talk) 17:29, 30 August 2009 (UTC).

Even if Stephan were a mathematician, the statement "most other" does not mean "all". Besides that, Stephan is a computer scientist and not a mathematician and as such does not count when it comes to qualifying the valididity of statements involving mathematicians (exhaustive or not). One is not going to go very far with one's mathematical thinking if one stops at the first problem. If one were to be presented with a problem that involved indeterminate forms, then one may stop if one chose. However, I might prefer to proceed with common sense. ~~ Dr Dec (Talk) ~~ 19:01, 30 August 2009 (UTC)

When composing functions with restricted domains, the domain of the composition is the largest set on which all the intermediate functions are defined. For example, working over the real numbers, the function given by f(x) = (√x)² is only defined for x≥0, even though it can be extended to a function defined for all x. (Such an extension must of course be non-unique, which is why we cannot ask what f(-2) is.) If a student claimed that (sin x)/x was defined at 0, I would mark it as incorrect. And I don't mean this pedantically; I don't believe I've ever said "And since (sin x)/x is 1 at 0..."; instead I would only say "And since (sin x)/x goes to 1 at 0..." The distinction between a function and its continuation, completion, or closure is an important one which becomes more important the deeper you go in mathematics; in my experience a practicing mathematician would be unlikely to blur this line, even in informal discussions. Tesseran (talk) 05:53, 31 August 2009 (UTC) [Edit: fixed typo, thanks Pt.] Tesseran (talk) 18:51, 31 August 2009 (UTC)

As functions on R, sin²x/sinx is definitely not equal to sinx. The OP's teacher may or may not care if he/she attends to that kind of detail, but if for instance this were an analysis class that would not fly. 67.100.146.151 (talk) 08:16, 31 August 2009 (UTC)

By the way, (sin x)/x goes to 1 and not 0, as x→0. — Pt _(T) 10:35, 31 August 2009 (UTC)

Program the function f by f(x) = x/x. Then f(x) evaluates to one when x is not equal to zero, and gives either an "zero divide" error message, or the value 0 (because of the zero in the numerator) when x = 0. Which programming languages return f(0) = 1 by evaluating lim_x→0f(x) ? Bo Jacoby (talk) 22:07, 1 September 2009 (UTC).

Big O of sin x

$\sin(x)={\mathcal {O}}(f(x)){\mbox{ as }}x\to \infty$

Can anything meaningful be said about f(x)? SpinningSpark 13:13, 30 August 2009 (UTC)

Could you please qualify your notation. I know ${\mathcal {O}}$ as a function ring. Do you mean O(f)? ~~ Dr Dec (Talk) ~~ 13:19, 30 August 2009 (UTC)

Sorry, I meant Big O notation. I copied the style from sorting algorithm so someone should take a look at that article if it's wrong. SpinningSpark 13:34, 30 August 2009 (UTC)

Well the notation means that, for sufficiently large x, $|\sin(x)|\leq k|f(x)|$ for some fixed positive constant k. This isn't very illuminating. The Big O notation is usually used to describe growth, but since the sine function doesn't grow it's a bit of a non-starter. For example, if $g(x)=O(f(x))$ then we know that the infimum of $|f(x)|$, for very large x, grows like the infimum of $|g(x)|$. But the infimum of $|\sin(x)|$ is zero: as small as it gets; so this doesn't tell us anything. Why did you ask the question? Can you give us more information and context? ~~ Dr Dec (Talk) ~~ 13:58, 30 August 2009 (UTC)

(ec) Saying sin(x) = O(f(x)) means that there is some k such that for x sufficiently large we have |sin(x)| ≤ k|f(x)|. This will always be satisfied provided that 1 ≤ k|f(x)|, which is the same as 1/k ≤ |f(x)|. So sin(x) = O(f(x)) means that the values of f(x) will eventually stay outside of some interval around 0. That is, there is some ε such that ε ≤ |f(x)| for x sufficiently large. It's not hard to see that this condition is equivalent to sin(x) = O(f(x)). Another way to say this is that the lim inf of f(x) as x goes to infinity is not zero. Staecker (talk) 14:05, 30 August 2009 (UTC)

Is this true? $|\sin(x)|$ will become zero infinitly many times, not matter how large x is. For example, try $f(x)=2\sin(x).\,$ Clearly $|\sin(x)|\leq |2\sin(x)|$ for all x (where k = 1), but f is zero infinitly many times and does not have a well-defined limit. I guess that the statement might be that If a limit exists then it will not be zero. ~~ Dr Dec (Talk) ~~ 14:18, 30 August 2009 (UTC)

While $\scriptstyle \lim _{x\rightarrow \infty }2\sin x$ does not exist, $\scriptstyle \liminf _{x\rightarrow \infty }2\sin x$ certainly does; it is –2. But shouldn't it be $\scriptstyle \limsup _{x\rightarrow \infty }|f(x)|$ (which is 2 in your example) that should not be 0? —JAO • T • C 14:30, 30 August 2009 (UTC)

Good point! I missed the inf. In that case then

$\lim _{x\to \infty }\inf |f(x)|\neq 0$

as Staecker quite rightly says. ~~ Dr Dec (Talk) ~~ 14:42, 30 August 2009 (UTC)

I think Dec's objection is valid- |f(x)| = |2sinx| does have lim inf equal to zero, but it is certainly the case that sin(x) = O(f(x)). So there are more possibilities for f. Perhaps the correct weaker condition is that: $\liminf _{x\to \infty }|f(x)|\neq 0$ or $\liminf _{x\to a}|f(x)|=0$ for all a being an integer multiple of π. (Or something like that.) Staecker (talk) 17:38, 30 August 2009 (UTC)

Sorry Dr. Dec, there is no context, it just arose out of some idle scribblings on scrap paper. SpinningSpark 14:38, 30 August 2009 (UTC)

Just to check that I am following this, would it be right to say,

$\sin(x)=\mathrm {O} (1)){\mbox{ for }}k\geq 1$

$\because |\sin(x)|\leq k|1|$

SpinningSpark 15:29, 30 August 2009 (UTC)

That's correct. To say that a function f is such that f(x) = O(1) means that, for sufficiently large x, f is bounded, i.e. |f(x)| ≤ k for all sufficiently large x. Like I said: the big O notation is used to describe the growth of a function. Say that f(x) = O(g(x)) then what does that mean? Well consider the cone given by y = ±g(x). Then f(x) will, for sufficiently large x, always stay inside a cone of that shape. So, e.g., if f(x) = O(x) then, for sufficiently large x, f(x) will always lie inside a cone given by the two straight lines y = ±kx, for some k (i.e. f(x) ≤ k|x|). ~~ Dr Dec (Talk) ~~ 15:50, 30 August 2009 (UTC)

The usefulness of sine and cosine

So, let's assume that sine and cosine haven't been invented. We want to try to express, in big O notation, the statement that the limit of the infimum of some smooth function f is non-zero (whilst also, theoretically, being able to take any other real value). How could that be done if we didn't know about sine or cosine? (If we don't know about sine or cosine then we couldn't use a power series for them either!) ~~ Dr Dec (Talk) ~~ 14:51, 30 August 2009 (UTC)

That's $\liminf f>0$. The big-Oh notation says something about absolute size. You cannot separate statements about the liminf and limsup. Besides, $O(\cdot )$ only implies an upper bound. There is a corresponding notation for lower bounds ($\Omega$, see Big Oh notation), although it's not as commonly seen. The definition also has absolute values. I don't think it is worthwhile trying to express non-quantitative statements in big-Oh notation. Phils 16:16, 30 August 2009 (UTC)

After reading the above more carefully, I realize that you are referring to the fact that $\sin x=O(f(x))$ implies $\liminf |f|>0$, i.e. the function to tested is in the $O(\cdot )$. In that case $1=O(f(x))$ works just as well. My remark about absolute value still holds. Phils 16:26, 30 August 2009 (UTC)

In computer science when we want to be precise, for a given function f, we define O(f) as the set of functions h for which there exists a constant K such that for sufficiently large x, $h(x)/f(x)<K$. So we would then say $\sin \in O(1)$ where we are abusing notation and writing "1" to denote the constant function $x\mapsto 1$. Re Declan: big O notation in the treatment I'm accustomed to is insensitive to additive constants. I don't think it would be used in the way you are asking. 67.122.211.205 (talk) 01:52, 31 August 2009 (UTC)

What mathematics things are vs. what mathematical things do

Is only the latter important? It seems to me like defining 'dog' as something that barks, bites, etc. But not telling what a 'dog' is. On the other hand, the only way of defining 'division' I can thing of is over its output.Quest09 (talk) 16:50, 30 August 2009 (UTC)

...so are you suggesting we define a dog by its output? YUK! :oP ~~ Dr Dec (Talk) ~~ 17:01, 30 August 2009 (UTC)

But surely all objects are defined by their interaction with obsevers anyway? The "are" is only assessed by interaction with the world. --Leon (talk) 19:18, 30 August 2009 (UTC)

Here's an interesting example of this problem (borrowed from a book, but I've forgotten which one EDIT maybe the one in the OUP very short intro series by Gowers). What "is" the king in chess? It's not a particular piece in your chess set, because if you've lost your king you can use any old bit of junk lying around. Is a king anything more than "something which can move only a single square in one direction except for castling, and...."? 87.194.213.98 (talk) 19:25, 30 August 2009 (UTC)

If you can find a difference between the two then perhaps you can find a scientific explanation of Transubstantiation ;-) Dmcq (talk) 19:53, 30 August 2009 (UTC)

Mathematicians are concerned with the fact that the reals form a complete ordered field (what they do), and often impatient with things like the fact that the reals can be regarded as equivalence classes of Cauchy sequences or as Dedekind cuts, etc. (what they (supposedly?) are).

The analogy to the king in chess is on the mark.

But this might be argued about philosophically. Michael Hardy (talk) 01:35, 31 August 2009 (UTC)

The fact that you have mentioned two completely different constructions of the reals that are both very popular shows how unimportant constructions are to the vast majority of mathematics. "The reals are the completion of the field of fractions of a set with at least one element and a successor function." is the most useful definition (albeit lacking a few key details for the sake of conciseness). --Tango (talk) 01:43, 31 August 2009 (UTC)

That's exactly my point in mentioning the two examples. Michael Hardy (talk) 21:05, 31 August 2009 (UTC)

The issue you're up against is the endless quest for syntactic proofs of theorems, since who can really say anything about the semantics of uncountable entities like real numbers, to say nothing of (e.g.) Hilbert spaces (the substrate of quantum theory)? Jean-Yves Girard has some pretty good rants along these lines, e.g. [1] 67.122.211.205 (talk) 03:11, 31 August 2009 (UTC)

PUZZLING PUZZLE

Guys I have been trying to solve this for a long time but I am not able to..please help me with this puzzle.(Thot it would fit in here better rather than the misc. desk)

The diagram shows a 4 x 5 grid with some filled cells. Find the numbers in the remaining cells according to the following rules: a) Each number can only take values from 1 to 5. There are 4 such full sets(1-5). b) Sum of each row is same. c) Sum of each column is same. d) Adjacent numbers cannot be the same.

-	-	-	1	5
-	-	-	-	1
2	4	-	-	-
4	-	-	-	-

thanks —Preceding unsigned comment added by 117.193.136.45 (talk) 16:53, 30 August 2009 (UTC)

An impractical but possible way to solve any such problem is to formulate and solve it as an Integer Programming Problem. There are softwares which automatically solve IPP's and you could use those.--Shahab (talk) 17:46, 30 August 2009 (UTC)

But, I would hope, 117.193.136.45 wants some kind of mathematical algorithm to solve the problem. I know that computers use algorithms, but that's just no fun! I was trying to find a connexion with sudoku. ~~ Dr Dec (Talk) ~~ 18:28, 30 August 2009 (UTC)

1	3	5	1	5
5	2	4	3	1
2	4	2	5	2
4	3	1	3	4

Just used good old fashioned trial and error. It's clear that the row and column sums must be 15 and 12 respectively.--RDBury (talk) 19:33, 30 August 2009 (UTC)

My solution was

2	4	3	1	5
4	2	3	5	1
2	4	3	1	5
4	2	3	5	1

which has a more obvious pattern. I can't see how it could have taken any length of time at all. Dmcq (talk) 19:41, 30 August 2009 (UTC)

Umm Dmcq, what about the constraint: d) Adjacent numbers cannot be the same? -hydnjo (talk) 19:49, 30 August 2009 (UTC)

Oops sorry, I should have read through to the end. Silly me Dmcq (talk) 20:09, 30 August 2009 (UTC)

I found the same solution as RDBury, also by trial and error. Note that since the column sums must be 12, there are only three ways to complete column 1 and three ways to complete column 5. That gives you 9 starting points, most of which quickly lead to dead ends. Gandalf61 (talk) 19:58, 30 August 2009 (UTC)

Damn. I followed this from Science to Miscellaneous, but didn't realize (didn't read) that is had also been moved here. However, under the premise that it is better the teach a man to fish than to give him a fish, I'll copy my answer here. -- Tcncv (talk) 23:59, 30 August 2009 (UTC)

There is a solution (and only one), but it would spoil the fun for me to give it outright. One way to find it is to use a backtracking approach, which is a way to solve many logic problems like this. (See also the Ariadne's thread article.) First, make a choice and see where that takes you. Fill in other cells that can be derived from that choice. When you cannot fill in any more, make another choice. If you come to a dead end, backtrack to your most recent choice and select another. If you run out of options at that point back up further and select another option for the previous choice.

This puzzle has a couple of good places to start. For ease of reference, I'll refer to the cells by row and column – (1,1) through (4,5). First you need to figure out the needed row and column sums. Since the sum of all of the available numbers is 4 × (1 + 2 + 3 + 4 + 5) = 60, the rows must each total 15 and the columns 12. Take a look at column 1. You have two numbers already whose sum is 6, so the remaining two numbers in cells (1,1) and (2,1) must also equal 6 (giving a column total of 12). Your options are 1 & 5, 2 & 4, and 5 & 1. The combination 3 & 3 would be an immediate adjacency violation; so would 4 & 2, given the existing value in cell (3,1). Tentatively select one pair and fill them in, keeping track of the other options in case you have to backtrack.

Now take a look at row 1. You already have three of five values, so you can apply the same logic. After that, look at column 2, then row 2, then column 3 etc. As you consider numbers, you can immediately eliminate any that cause adjacency violations. You can also watch for cases where the same number is used more than the allotted four times. That would also prompt you to back track.

Good luck. -- Tcncv (talk) 23:20, 30 August 2009 (UTC)

Sudoku boards and groups with 9 elements

Think about a completed sudoku board: it's a 9 × 9 board where we must fill the board with the numbers {1,…,9} in such a way that the same number can't appear in the same colomn, the same row, or in any of the 3 × 3 sub-squares that make up the 9 × 9 total board. Now, for me, the first two conditions (i.e. the same number can't appear in the same colomn or the same row) remind me of a group table for a group, say G, with nine elements. Is there a correspondence between completed Sudoku boards and groups with nine elements? It's been about five years since I did any finite group theory or number theory, so please forgive my ignorance. But I seem to rememeber subgroups forming little blocks in the table like the 3 × 3 blocks we have on sudoku boards. Is there a 1-1 correspondence between group tables and sudoku boards (I doubt it, but I'm so rusty that I forget). Although, I seem to remember some enumeration results from representation theory of finite groups, but again; it's so long ago that they're nothing more than quiet little bells ringing ~~ Dr Dec (Talk) ~~ 18:51, 30 August 2009 (UTC)

There are only two groups of order 9: the cyclic group and the direct product of two groups of order 3, so no such correspondence exists. If you draw the Cayley table of a group, that is, a grid whose rows and cols correspond to the group elements and whose ij entry is the product of the corresponding elements, a subgroup doesn't look like a "sub-square" in which every element (of the big group) occurs once, so this is also not like the small squares in a sudoku grid. You might be interested in Latin square, a correct sudoku grid is a special kind of Latin square. As for rep thy, you might have character tables in mind perhaps.87.194.213.98 (talk) 19:21, 30 August 2009 (UTC)

I already know about Latin squares, but I thought a group theoretical approach might be interesting. I know, from Lagrange's theorem, that the order of any subgroup must divide the order of the group, so subgroups of order 1, 3 and 9 are all possible. Could you please say a few words as to why there aren't any order 9 groups with subgroups of order 3? ~~ Dr Dec (Talk) ~~ 19:29, 30 August 2009 (UTC)

There are, all groups of order 9 have subgroups of order 3. If <g> is cyclic of order 9 then g^3 generates a subgroup of order 3, and if a group G is the direct product of two groups of order 3 then it has several subgroups of order 3. Have a look at the Mathematics_of_sudoku article, they mention that not-Burnside's lemma can be used to enumerate solutions up to a notion of equivalence. That article also talks about Cayley tables and sudoku grids.87.194.213.98 (talk) 19:40, 30 August 2009 (UTC)

Excellent! I'd seen the maths of sudoku section in the sudoku article, but had some how managed to miss the fact that there was an entire article. I know what I'll be doing for the next half hour... ~~ Dr Dec (Talk) ~~ 21:15, 30 August 2009 (UTC)

There are 2 non-isomorphic groups of order 9. The actual number of Cayley tables that form groups is a lot more than that. The problem is that groups require associativity which puts way more structure on the array than is applicable for Sudoko squares.--RDBury (talk) 19:51, 30 August 2009 (UTC)

If we restrict ourselves to the set {1,...,9} then the only isomorphisms are permutations of the symbols and you can permute the symbols in a completed Sudoko board without breaking it, so I don't think that makes much difference (although rigour requires we mention it). What if we consider quasigroups instead? According to the article there is (modulo permutation of symbols) a 1-1 correspondence between quasigroups and Latin squares. Is there anything interesting to say about those quasigroups that correspond to a Suduko board? --Tango (talk) 01:35, 31 August 2009 (UTC)

As an aside note that all groups of order p² (p prime) are abelian and the converse of Lagrange's theorem holds for all abelian groups.--Shahab (talk) 19:55, 30 August 2009 (UTC)

If G is a group of order p² where p is prime, it must equal to its center. To see this, observe that the center of G, Z(G), is either trivial or equals G, since G/Z(G) cannot be cyclic (here we are using the hypothesis that p is prime). Now note the fact that the center of a p-group (a group of prime-power order) can never be trivial, and therefore G is abelian (the fact that the center of a p-group is non-trivial is often proven using a simple argument with orbits and group actions). --PST 12:13, 31 August 2009 (UTC)

Random trig question

I know that it is possible to perform inverse trigonometric functions with a calculator. I have been told that it is not possible by hand. However, trigonometry existed centuries before calculators were invented. So, is it possible to perform these function by hand? If not, how did the ancient mathematicians do it? Intelligentsium 23:21, 30 August 2009 (UTC)

It is possible, it just depends how much time you have on your hands. For example, in the olden days people used log tables to evaluate logarithms. These were basic grids that once you knew the base of the log (say n) and the value of the variable (say x), would give you the answer (say log_n(x)). Well, in that case, finding a number in the middle of the table that was close to the number you wanted to find the inverse of would tell you to which base, and of which value, it was the inverse of. I assume that trig' functions were the same. It's just like a group table: find the number in the group table and you know what's the inverse of what. ~~ Dr Dec (Talk) ~~ 00:13, 31 August 2009 (UTC)

Well, obviously, when you see a table of values of sine and cosine in a book published in 1908, or 1650, that tells you something. Maybe what was meant was that the methods needed in order to do it efficiently by hand cannot be taught at the very elementary level at which instruction in trigonometry takes place today.

As for log tables, where you looked up log_n(x) if you knew n and x, that's not how it was done because it's not an efficient way to do it. Base-10 log tables were commonplace. If you wanted log₂(3), you used a base-10 log table and found (log₁₀(3))/log₁₀(2). Tables of base-e logarithms also existed, but were not in an appendix to every book. Michael Hardy (talk) 01:28, 31 August 2009 (UTC)

Of course you can compute inverse trig functions by hand, for example with Padé approximants. Stupendous amounts of research went into developing methods to minimize the amount of labor required by those computations. In practice you would look your number up in a table and use linear or quadratic interpolation on the nearest values you could find to the one you wanted. But somebody had to prepare the tables, and they did it without computers... 67.122.211.205 (talk) 01:42, 31 August 2009 (UTC)

And the books of tables all had little errors in them which is why Charles Babbage designed his Difference engine. Dmcq (talk) 07:47, 31 August 2009 (UTC)

Maybe we should add that if it were impossible to compute values of inverse trigonometric functions by hand, then it would be impossible to do so via computers as well. Michael Hardy (talk) 21:07, 31 August 2009 (UTC)

I wouldn't say that. I mean, sure, they are Turing equivalent, but some amounts of hand calculation are just impractical. You can calculate pi to 100's of digits by hand, maybe even 1000's of digits, but not millions of digits. But it can be done with computers. 70.90.174.101 (talk) 00:59, 1 September 2009 (UTC)