31 Jan 04
1 edit
Originally posted by fearlessleaderminor spelling errors in your message corrected (as if i'm a good one to do that since i make a huge number of typos).
as my trig class does not nearly allow me to understand half the posts on this forum, can someone explain the difference between algebraic and transcendental irratinal numbers?
ok, don't say you weren't warned!!
an algebraic number is any real irrational number that is a rational power of an integer, ratinal number, or other algebraic number. examples are: 2^0.5, 4^(2/3), 1001^(1/7), 3072^(16/5), and even something like (2^0.5 + 3^(1/4))^(1/7)--at least i'm fairly sure that last one is in this category. (for this discussion, we're considering real roots of the above numbers, no complex ones.) this set has cardinality (number of elements) equivalent to that of the rational numbers, usually designated as "aleph-null".
transcendental numbers are the irrational numbers that do not fit the above definition. in this area, you have all logarithms, most of the trig functions, and irrational powers of numbers. two well-known transcendentals are e (the basis of all logarithms, even if they use a different base) and pi (ratio of circumference to diameter of a circle, used mostly in trigonometry). another would be 2^(2^0.5), a number formed by an irrational power. in addition, almost all numbers created by series sums fall into this category. my favorite number in this class is the decimal represented by a 9 in all (2^n) positions and an 8 otherwise; it looks like: 0.9989888988888889... and some of its offshoots. this set is much larger than the algebraics, having cardinality equivalent to that of the reals, usualy written sa "aleph-one" or "C"
the old definition of an irrational number, "a decimal that doesn't terminate and doesn't ever repeat in blocks", is a simplified way of considering this fascinating set of numbers.
hope that helps!
02 Feb 04
Originally posted by BarefootChessPlayerAn equivalent definition of algebraic numbers is that an algrebaic number is a real solution to P(x) = 0, where P is a polynomial (ie ax^n + bx^n-1 + cx^n-2 .... + yx + z) with integer coefficients. In fact, even if the coefficients in P are arbitrary albegraic numbers, the real solutions will still be algebraic, and all the complex soutions will be of the form A + Bi, where A and B are algebraic.
minor spelling errors in your message corrected (as if i'm a good one to do that since i make a huge number of typos).
ok, don't say you weren't warned!!
an algebraic number is any real irrational number that is a [b]rational power of an integer, ratinal number, or other algebraic number. examples are: 2^0.5, 4^(2/3), 1001^(1/7), ...[text shortened]... blocks", is a simplified way of considering this fascinating set of numbers.
hope that helps![/b]
07 Feb 04
Originally posted by Acolytei think barefood did me a bit more good, but thanks anyway.🙂
An equivalent definition of algebraic numbers is that an algrebaic number is a real solution to P(x) = 0, where P is a polynomial (ie ax^n + bx^n-1 + cx^n-2 .... + yx + z) with integer coefficients. In fact, even if the coefficients in P are arbitrary albegraic numbers, the real solutions will still be algebraic, and all the complex soutions will be of the form A + Bi, where A and B are algebraic.
now that i have a clue, next question: what is a logerithum?
08 Feb 04
2 edits
Originally posted by fearlessleaderyou might think of a logarithm this way.
i think barefood did me a bit more good, but thanks anyway.🙂
now that i have a clue, next question: what is a logerithum?
take any two real positive numbers, a and b, then ask, "to what power do i need to raise a to get b?". for staters, we'll use 2 and 0.25. to what power do we need to raise two to get one fourth? the answer is -2, since 2^(-2) = 1/2^2, which is 1/4. we would write "log[2] 1/4 = -2," where the number in brackets indicates the number, in this case 2, of which we are trying to find the power (this is ordinarily done using subscripts but i don't have that option here); this number is called the base.
how about log[10] 5000? here we see the additive property of logs. if we cube ten, we get 1000. to get 5000, we raise it to the log[10] 5, which is almost .7, so in very rough terms, the answer is 3.7. (the actual value is a transcendental number slightly less than 3.7.)
pleae note that 1 is the zero power of any other positive number so log[a] 1 = 0 for all positive real a.
here is an intresting side light: all logarithms are based on the transcendental number e (roughly 2.718281828459...), regardless of their actual base. the only difference is that, to get any other base, you take reciprocal of log[e] of the base (a above) and multiply by log[e] of the number b above. to simplify notation, the operation "log to base e" is written "ln" or "ln" by most authors but i learned a different way. using this, we could write log[2] 8 as (ln 8)/ln 2. it will still be 3.
thoroughly confused yet: 🙂
10 Feb 04
Originally posted by BarefootChessPlayeri sure am!😀
you might think of a logarithm this way.
take any two real positive numbers, a and b, then ask, "to what power do i need to raise a to get b?". for staters, we'll use 2 and 0.25. to what power do we need to raise two to get one fourth? the answer is -2, since 2^(-2) = 1/2^2, which is 1/4. we would write "log[2] 1/ ...[text shortened]... g this, we could write log[2] 8 as (ln 8)/ln 2. it will still be 3.
thoroughly confused yet: 🙂
i'll read it though a few dozen times until i have a clue.
11 Feb 04
3 edits
Originally posted by Fiathahelthe first course of a subject in u. s. colleges is usually numbered 101.
[edit]
What does 101 mean?
some colleges start their courses at 1, but most universities i've seen have their beginning courses at 101, and anything below that is "remedial" material.
12 Feb 04
4 edits
Originally posted by fearlessleaderthe transcendental number e is most simply derived as the sum of the reciprocals of the factorials of integers starting with zero (remember, 0! = 1! =1). in sum notation (as modified for the keyboard): sum(n=0 to infinity) (1/n!).
explain this concept of e
this will produce about 2.718281828459....
there are two other ways of reaching it: the limit as x goes to zero of (1+x)^(1/x) or, as x goes to infinity, of (1 + 1/x)^x, a case of 1^infinity.
the function e^x is called the exponential function and is written often as exp x or exp(x), and is the only nonlinear function which is its own derivative.
the inverse of exp x is ln x--the number you to which you must raise e to get x.
13 Feb 04
Originally posted by BarefootChessPlayerlim{x->inf} (1+1/x)^x is the best (most common) way to define e, and has the advantage that lim{x->inf} (1+t/x)^x = e^t.
the transcendental number e is most simply derived as the sum of the reciprocals of the factorials of integers starting with zero (remember, 0! = 1! =1). in sum notation (as modified for the keyboard): sum(n=0 to infinity) (1/n!).
this will produce about 2.718281828459....
there are two other ways of reaching it: the limit as x go ...[text shortened]...
the inverse of exp x is ln x--the number you to which you must raise e to get x.
also sum{n=1 to inf} t^n/(n!) = e^t
with which it is easy to prove it is its own derivative.
13 Feb 04
1 edit
Originally posted by Fiathaheli agree, except i think the last expression should start with n=0, rather than 1.
lim{x->inf} (1+1/x)^x is the best (most common) way to define e, and has the advantage that lim{x->inf} (1+t/x)^x = e^t.
also sum{n=1 to inf} t^n/(n!) = e^t
with which it is easy to prove it is its own derivative.
the frist way i learned it was lim(x->0) (1+x)^(1/x), and subsitutting y=1/x gives the corresponding formula.
it wsa somewhat after that that i learned the factorial formulae for e^x, sin x, cos x, tan x, etc., and that e^(2*pi*i) = 1.
23 Feb 04
Originally posted by BarefootChessPlayeri understand (the first parts at least)
the transcendental number e is most simply derived as the sum of the reciprocals of the factorials of integers starting with zero (remember, 0! = 1! =1). in sum notation (as modified for the keyboard): sum(n=0 to infinity) (1/n!).
this will produce about 2.718281828459....
there are two other ways of reaching it: the limit as x go ...[text shortened]...
the inverse of exp x is ln x--the number you to which you must raise e to get x.
new line: how dose one find the system max of a system of non-linear inequalities?