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?

- Joined
- 03 Oct '03
- Moves
- 671

my head- Joined
- 22 Jul '03
- Moves
- 20133

central usa31 Jan '04 19:491 edit

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).*Originally posted by fearlessleader***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!- Joined
- 04 Jul '02
- Moves
- 3790

Loughborough02 Feb '04 00:00

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 BarefootChessPlayer***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).**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.

ok, don't say you weren't warned!!

an*algebraic*number is any real irrational number that is a [b]rational power

hope that helps![/b]- Joined
- 03 Oct '03
- Moves
- 671

my head07 Feb '04 21:04

i think barefood did me a bit more good, but thanks anyway.ðŸ™‚*Originally posted by Acolyte***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?- Joined
- 28 Sep '01
- Moves
- 40665

Your Kingside- Joined
- 22 Jul '03
- Moves
- 20133

central usa08 Feb '04 16:402 edits

you might think of a logarithm this way.*Originally posted by fearlessleader***i think barefood did me a bit more good, but thanks anyway.ðŸ™‚**

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**the only difference is that, to get any other base, you take reciprocal of log[*e*(roughly 2.718281828459...), regardless of their actual base.*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: ðŸ™‚- Joined
- 03 Oct '03
- Moves
- 671

my head10 Feb '04 12:35

i sure am!ðŸ˜€*Originally posted by BarefootChessPlayer***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: ðŸ™‚

i'll read it though a few dozen times until i have a clue.- Joined
- 21 May '03
- Moves
- 9766

in your fridge- Joined
- 25 Oct '02
- Moves
- 20443

Out of my mind- Joined
- 22 Jul '03
- Moves
- 20133

central usa11 Feb '04 03:473 edits

the first course of a subject in u. s. colleges is usually numbered 101.*Originally posted by Fiathahel*

[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.- Joined
- 03 Oct '03
- Moves
- 671

my head- Joined
- 22 Jul '03
- Moves
- 20133

central usa12 Feb '04 23:044 edits

the transcendental number*Originally posted by fearlessleader***explain this concept of***e**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*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*.- Joined
- 21 May '03
- Moves
- 9766

in your fridge13 Feb '04 10:43

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.*Originally posted by BarefootChessPlayer***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.

- Joined
- 22 Jul '03
- Moves
- 20133

central usa13 Feb '04 15:381 edit

i agree, except i think the last expression should start with n=0, rather than 1.*Originally posted by Fiathahel***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.- Joined
- 03 Oct '03
- Moves
- 671

my head23 Feb '04 16:31

i understand (the first parts at least)*Originally posted by BarefootChessPlayer***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?