31 Jan '04 16:30>
as my trig class dose not nearly allow me to understand half the poasts on this fourm, can someone explain the diffrence between algibraic and transendual irratinal numbers?
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?
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]
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.
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?
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: 🙂
Originally posted by Fiathahelthe first course of a subject in u. s. colleges is usually numbered 101.
[edit]
What does 101 mean?
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
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.
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.
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.