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Unfortunately, some people do composition on the left, and some do it on the right. In your example, you're doing composition from the left -- i.e. (1 2) is done first, then (1 3):
Applying (1 2):
1 -> 2
2 -> 1
(3 -> 3)
Then applying (1 3):
1 -> 2 -> 2 i.e. 1 -> 2
2 -> 1 -> 3 i.e. 2 -> 3
3 -> 3 -> 1 i.e. 3 -> 1
which, of course, is the same as (1 2 3).
I myself prefer composition on the right, although I can understand why many don't.
Check the letter positions.
Yes -- I even give the next two digits in my 'solution' (see the bolded letters).
Try sharpening razors on denim jeans before throwing them out.
And don't let a woman use them to shave her legs.
For the first question, how would you solve that system of equations if A + B and A - B were numbers, rather than matrices?
For the second, can you combine all of those terms into a single matrix?
For the last question, do you know how to perform Gaussian elimination?
Math is a subject that is totally based on formulas there are thousands of formulas used in maths, it is not possible to remember all of those formulas you can take help from some apps like Math formulas and Physics formulas.
I'd agree with you at the high-school level, yes; most of the maths I was taught back then was more or less formulaic, and entirely context-driven. But as a subject in and of itself -- and, perhaps even as an art, some might say -- it is built up with axioms, definitions, theorems and proofs. It is certainly not totally based on formulae, and I'm glad that it isn't, because that would sap the fun right out of it for me.
Well, i was just considering only in context of shape.
Shape wise they might look similar.
anyway.. thanks guys for having a very nice discussuion
Well, as shapes, they are homeomorphic. We define a homeomorphism as follows.
If f satisfies these conditions, then we say that A and B are homeomorphic.
Intuitively, think about being able to 'continuously deform' one shape into the other.
Whoops only ok for x>0. Back to the drawing board.
Bob
How about f(t) = constant?
Have you been given any other information? What is H?
The Weierstrass function comes to mind, but I think that has a countable number of strict extrema. You might be able to show this via a proof by contradiction (i.e. assuming that there exists an uncountable number of strict extrema).
Of course, the Weierstrass function has no such extrema if they are required to be smooth. I'll have to think about this one.
I do not get it. What is the period of e^x?
If one extends the definition of exp to the complex numbers, then it is 2πi-periodic, making it very useful for parametrising contours in the complex plane.
Because Alpha is saying that those polynomials are coprime -- but that does not necessarily mean that the evaluations of those polynomials at some number are also coprime.
Hi;
What is the answer to this question?
For how many integers
satisfying is it true that the fraction cannot be simplified?Thanks,
SolarDevil
If that fraction cannot be simplified, then what can you say about the numbers (2n + 7) and (3n - 4)?
What must gcd(3n - 4, 2n + 7) be, in that case? And what does this tell you about the possible values of n? (Hint: Use the Euclidean algorithm.)
What does the unique factorisation theorem tell you?
The book is wrong.
Can I just assume that the det=1 because it's just a re-arrangment of the identity matrix? Or does the mix up of order affect the determinent? thanks.
The determinant is indeed 1, but you must be careful. Here are some rules about how row (or column) operations affect the determinant of a matrix.
Let the matrix B be obtained from the matrix A by applying the row operation e. Then:
.These results hold true for the corresponding column operations.
For your matrix, the identity matrix has had two row operations: P(1,4) and P(2,3). So the determinant of your new matrix is (-1)*(-1)*1 = 1.
Have you encountered partial fractions before?
This is clearly true as you claimed, since
How else can you write x^x?
Hi guys, wouldn't it be better if there was a 20 instead of a 19 there? You can then get up to 47.
How would you make 22 and 24?
For the first question, it might help you to consider the product f(n-1)f(n), and then the product f(n-2)f(n-1)f(n), f(n-3)f(n-2)f(n-1)f(n), etc... you should then get a neat formula for f(2)f(3)...f(n-1)f(n) which can be plugged into your inequality and solved for n.
You seem to have dropped the differential operator (d/dx) in lines 2 and 3, but otherwise, that looks good to me.