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MIF, seems to be a forum bug, if I try to move the link "many others" up to the preceding paragraph, I get the error "BBC code cannot be nested in url tags".
Any (elementary) number theory, discrete math, or intro to proofs book will contain material on modular arithmetic. If you want to just learn modular arithmetic, I would go with sources on the web. Rutguers has an introduction, but there are
Avon, we're are talking about functions that take on complex values. Saying f(x) > 0 makes no sense.
does solving |g(k)|'=0 and g'(k)=0 give the same critical points?
Sorry, but once again I have to answer you with a question.
given a function f(k) defined on the reals
Does it map into the reals? Or into the complex numbers?
Yes.
George, you keep posting about this "element at infinity". Prove it exists, or you have no claim to its existence.
"That which can be asserted without proof, can be dismissed without proof."
given a function f(k) defined on the reals and a complex constant z0, what is the maximum of following function?
You first need to phrase your question in a way that it is possible to answer. Since z_0f(k) is of a complex value, there is no concept of maximum or minimum. In fact, you can't even say one complex number is greater than another. For example, which is greater, 1 or i? You can talk about which modulus is greater (in which case, |1| = |i| = 1), and you can talk about when z_0f(k) achieves its maximum modulus. Is this what you want?
Dummit and Foote have a great text for Abstract Algebra. Durin is easier to follow, but also no where near as in depth as Dummit and Foote. Many people recommend Munkres for topology, but I haven't worked out of it. Calculus I, II, and III are covered by Stewart, this is what I learned from and it was fine. It also covers some analytic geometry (as all multivariable texts do), but not in depth.
Your solution appears to be correct.
A function is increasing if f'(x) > 0. It is decreasing if f'(x) < 0. In order to find the intervals where f'(x) is increasing, we first look for all the points where:
(i) f'(x) = 0
OR
(ii) f'(x) has a discontinuity
Note that it is very possible for f'(x) to exist and have a discontinuity, but certainly any place where f'(x) does not exist will also be a discontinuity. Once you've found these points, they will be the intervals where f'(x) is either greater than or less than 0. Now all you need to do is test a point in each one of these.
Concavity is exactly the same except we're now using f''(x).
Thanks for the answers - so it seems if one tried to define some "new maths" involving ln(-1) it would turn out that sqrt(-1) could be defined using the "new" value ln(-1).
That is precisely correct. In fact, in complex analysis we can define a square root if and only if we can define a log.
Where l(z) is a branch of the log. Intuitively
But there are problems with just saying "z^1/2".
What I mean to say is I dont have the patitence to read wikipedia(lol!)
Then why do you expect to have the patience to read whatever it is we write? This site is mostly for answering questions, not replacing your textbook.
MathGuise was Anthony R. Brown, again harassing the forums.
No L'Hopital does not apply. Ive thought about it
Huh? The most you can say is "I don't see how to apply it".
I have a deductive argument proving that there is only one solution (and finding it). Unfortunately it's about 6 pages or so (but I think it can be shortened quite a bit). It was actually quite fun, Sudoku like.
That is the only solution.
For[a2 = 0, a2 <= 6, a2++,
For[a3 = 0, a3 <= 2, a3++,
For[a5 = 0, a5 <= 6, a5++,
For[b2 = 0, b2 <= 6 - a2, b2++,
For[b3 = 0, b3 <= 2 - a3, b3++,
For[b5 = 0, b5 <= 6 - a5, b5++,
For[c2 = 0, c2 <= 6 - a2 - b2, c2++,
For[c3 = 0, c3 <= 2 - a3 - b3, c3++,
For[c5 = 0, c5 <= 6 - a5 - b5, c5++,
d2 = 6 - a2 - b2 - c2;
d3 = 2 - a3 - b3 - c3;
d5 = 6 - a5 - b5 - c5;
If[2^a2*3^a3*5^a5 + 2^b2*3^b3*5^b5 +
2^c2*3^c3*5^c5 + 79*2^d2*3^d3*5^d5 == 711,
Print[{a2, a3, a5, b2, b3, b5, c2, c3, c5}]];
];
];
];
];
];
];
];
];
];L'Hopital is the way to go. Do some algebraic manipulation till you get it into this form:
And note that the limit of the right factor is zero. Now L'Hopital, and then a little bit more algebraic manipulation and you're done.
The limit actually turns out to be about
Lashko, just use the fact that:
Just because the limit of f(x) as x approaches c exists does not mean that f(c) is equal to this limit.
The answer will always be 9. Can someone prove how?
1111-1111 = 0
A corrollary should follow from a preceeding theorem, so that's the first place to look. What book/page number is this?
I'm guessing that L_n is an nth order differential equation? That determinant is almost exactly the same as the Wronskian, I would first calculate that out. You should be able to come up with a pretty simple closed form for it.
Word: Definition
1. Recondite: Abstruse
2. Abstruse: Enigmatic
3. Enigmatic: Cabalistic
4. Cabalistic: Arcane
5. Arcane: Esoteric
6. Esoteric: Hermetic
7. Hermetic: Recondite
(This all started when I saw the definition of recondite including the word "abstruse")
There was a game like this, I believe it was called "Black box". The only difference is that it would tell you how many you got right (i.e. in the right place as well), and how many numbers you guessed but were in the wrong place.
Unfortunately soroban, the question was to prove that it converges to 0, not just to show convergence.
Huh, weird, but I guess it's ok. I've never heard of derivatives talk about in this way.
Just write the entire thing out. Fix an i and j, and in terms of a_ij, what is A(t) - A(gamma)? And then what is this when it is divided by t - gamma?
What is X?
It looks to me like you're trying to define the derivative of a n to m dimensional function, but this doesn't make sense because you say that I is an interval. So maybe you are attempting to define the derivative of a path?
Either way, please remember that there is no such thing as a derivative of a matrix. A derivative of a function can be represented by a matrix, but this is still a derivative of a function.