Post all difficulties with Math here, and hopefully someone will be able to help you.
I have a good question. I need some help with some basic math operations but when dealing with radians. For example, I have to graph a sine wave that's in the general form. Easy, until you have to generate 5 new points. I understand WHAT I have to do, but not how the book comes up with the answers. I am really looking for a shortcut on how to do it because it took me like .. 30 minutes to come up with the answer because I suck at pi/pi reductions.
I have a good question. I need some help with some basic math operations but when dealing with radians. For example, I have to graph a sine wave that's in the general form. Easy, until you have to generate 5 new points. I understand WHAT I have to do, but not how the book comes up with the answers. I am really looking for a shortcut on how to do it because it took me like .. 30 minutes to come up with the answer because I suck at pi/pi reductions.
memorize
sin(0)=0
sin(2π)=0
so in the middle as well,
sin(π)=0
then further split those (sine goes up then down),
sin(π/2)=1
sin(3π/2)=-1
then you get into roots (you can use the pythagorean thm here).
is this what you're talking about?
sin(0)=0
sin(2π)=0
so in the middle as well,
sin(π)=0
then further split those (sine goes up then down),
sin(π/2)=1
sin(3π/2)=-1
then you get into roots (you can use the pythagorean thm here).
is this what you're talking about?
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Can TTF do pi?
calculate or display?
Display
Display
π ftw!
π equations coming soon. I'm at work now.
I don't think I need help with anything right now, but we just started doing derivaties and stuff. just finished up with limits. I'm digging calc.
derivatives, rather.
CALCULUS IS LIMITS
EPSILONS
DELTAS
``I SHATTERED THEIR ALGEBRA WITH CALCULUS'' -- blackalicious
EPSILONS
DELTAS
``I SHATTERED THEIR ALGEBRA WITH CALCULUS'' -- blackalicious
Ok
Say I want to get
(π/2) / 3
Is there any easier way to get the radian equivalent without algebraically manipulating the equation?
Note - I do know this is essentially (180/2) / 3. However, the radian equivalent of 30 is an ass shit.
Say I want to get
(π/2) / 3
Is there any easier way to get the radian equivalent without algebraically manipulating the equation?
Note - I do know this is essentially (180/2) / 3. However, the radian equivalent of 30 is an ass shit.
(π/2) / 3
= (π/2)(1/3)
= (π/6)
= (π/2)(1/3)
= (π/6)
Well yea that was an easy example.
I just realized there is no way around it.
I just realized there is no way around it.
probably not, but i wasn't quite sure what you meant.
Whoa ... Old school.
One question:
In statistics, how do I determine whether to use a two-tailed or one-tailed test when attempting to find a T statistic or T distribution?
One question:
In statistics, how do I determine whether to use a two-tailed or one-tailed test when attempting to find a T statistic or T distribution?
iirc it is intuitive... like whether what you are testing is a standard of error (e.g. deviance from a standard in manufacturing) or something that is tolerant in one direction (all i can think of is black-scholes option pricing.. probably not useful).
http://www.mathsrevision.net/alevel/pages.php?page=64
has some pics
http://www.mathsrevision.net/alevel/pages.php?page=64
has some pics
I dunno.
In our stats class, it tells us what test to use. But I'm really looking more for theory on HOW to determine the proper test, one tailed or two.
In our stats class, it tells us what test to use. But I'm really looking more for theory on HOW to determine the proper test, one tailed or two.
it is pretty intuitive... like absolute deviance vs. in one direction
So it really depends on the type of study.
Gotcha. Oh well ..
Poker results?
Gotcha. Oh well ..
Poker results?
guess i forgot to hit the click to post button. anyways i was having some difficulty understanding this one probability problem. to sum up, with N mobile users in a cell, each user has probability P to transmit. Each mobile user transmits independently of the others. A: what is the probability no one transmits? B: what is the probability that only 1 user transmits? C: what is the probability 2 or more users transmits causing a collision?
I figured out the answers are (or i think are)
A: (1-P)^N
B: probability mass function of the binomial distribution
C: 1-(answer from B)
I figured out the answers are (or i think are)
A: (1-P)^N
B: probability mass function of the binomial distribution
C: 1-(answer from B)
so I've been doing some binomial and negative binomial experiments. if I understand them correctly, a binomial experiment can tell you the probability of x successes after n trials, and a negative binomial experiment can tell you the probability of x successes on the nth trial. but what if one of the outcomes of the experiment is not independent, i.e., one success (say, death) prevents any further trials from taking place. it's easy to figure out the probability of success (death) on any given trial, say the tenth trial, but what if I want to know the cumulative probability of death in any of the first ten trials?
I know I can just do a binomial experiment, define success as living, and subtract that result from 1, but is there another type of experiment that can give me the cumulative probability if a success precludes any further trials?
I may have fucked up terminology but hopefully this makes sense.
I know I can just do a binomial experiment, define success as living, and subtract that result from 1, but is there another type of experiment that can give me the cumulative probability if a success precludes any further trials?
I may have fucked up terminology but hopefully this makes sense.
this also might just be a moot question because there is an easy way around such a situation.
I've got another one. here's the function:
here's the inverse:
I'm having trouble understanding the domain of the inverse. the website says the correct answer is:
but why would -1, for example, be excluded from the domain?
f(x) = sqrt(3 - e^2x)
here's the inverse:
f^-1(x) = ln(-x^2 + 3) / 2
I'm having trouble understanding the domain of the inverse. the website says the correct answer is:
[0, sqrt(3)]
but why would -1, for example, be excluded from the domain?
no idea.
Well, the domain of f^-1 is the range of f, which is [0, sqrt(3)] but I guess this explanation is kinda backwards...
Wasn't there any restriction given to the values of x?
If you plot f^-1 you can see that the domain does include negative numbers: http://fooplot.com/index.php?&type0=0& … amp;ymax=3
Wasn't there any restriction given to the values of x?
If you plot f^-1 you can see that the domain does include negative numbers: http://fooplot.com/index.php?&type0=0& … amp;ymax=3
i don't think that's actually the inverse, then. i think the inverse requires a domain restriction itself to match the co-domain of the original function.
sorry, i don't care enough atm to think about this very much.
sorry, i don't care enough atm to think about this very much.
I also gave this as the inverse and the website said I was correct:
but I got that because I was overzealous with e and changed e^2x <= 3 to 2xe <= 3, which is not right. so maybe the website is messed up and I still don't even have the correct inverse. I'm pretty sure the one I gave is right though.
f^-1(x) = -1/2 (x^2 - 3)/e
but I got that because I was overzealous with e and changed e^2x <= 3 to 2xe <= 3, which is not right. so maybe the website is messed up and I still don't even have the correct inverse. I'm pretty sure the one I gave is right though.
> sorry, i don't care enough atm to think about this very much.
what a dickhead--terribly lazy.
what a dickhead--terribly lazy.
haha, it's ok. I haven't invested a lot of time in thinking this through either. or at least not since the website told me I had the correct answer. I've got some other math homework to do tomorrow so maybe I'll come back to this then.
here's what should be an easy one: find the limit in terms of a as h approaches 0.
wolfram alpha says it's -3/a^2, but how? I'm bad at algebra.
[ 3/(a+h) - 3/a ] / h
wolfram alpha says it's -3/a^2, but how? I'm bad at algebra.
I mean I can do shortcuts and find that f(a) = 3/a = 3a^-1, and f'(a) = -3a^-2 = -3/a^2. but I'm curious how to accomplish this algebraicly.
Check out L'Hopital's rule.
I guess this wasn't too helpful a suggestion?
andre helped me with that last one in #bsd. it was ridiculously easy and betrayed my rustiness with algebra. anyway, I just used L'Hopital's rule for another problem! limit of
as x approaches 2. it made that one very easy. also, check out this ridiculous function I had to differentiate: http://prntscr.com/3cdqh
that took me like an hour.
sin(x-2) / (x^2+2x-8)
as x approaches 2. it made that one very easy. also, check out this ridiculous function I had to differentiate: http://prntscr.com/3cdqh
that took me like an hour.
x^2+2x-8 maybe?
I just found the derivative of the numerator and denominator. so it was
and then taking the limit of that as x->2 was 1/6.
cos(x-2) / 2x+2
and then taking the limit of that as x->2 was 1/6.
this one was fun too.
I meant the denominator is x^2+2x-8, right? Otherwise it would be 0/16=0 when x->2, no need for any calculation.
oh woops. fixed.
this is from my take-home exam in linear algebra. given the matrices:
and determinants: det(A) = 3, det(B) = 4.
find determinant(C).
switching rows 2 and 3 in B makes det(B) = -4. writing out the determinants:
I then wrote this as an augmented matrix with (af-cd), (bd-ae), (ce-bf) as the unknowns:
the reduced row echelon form of this yields:
so (ce-bf) is a free parameter, say s, and:
can I then conclude that the determinant, x, of C is equal to -2? I am unsure.
a b c
A = 1 1 1
d e f
a b c
B = d e f
1 2 3
a b c
C = 4 6 8
d e fand determinants: det(A) = 3, det(B) = 4.
find determinant(C).
switching rows 2 and 3 in B makes det(B) = -4. writing out the determinants:
det(A) = (af-cd)+(bd-ae)+(ce-bf) = 3 det(B) = 2(af-cd)+3(bd-ae)+(ce-bf) = -4 det(C) = 6(af-cd)+8(bd-ae)+4(ce-bf) = x
I then wrote this as an augmented matrix with (af-cd), (bd-ae), (ce-bf) as the unknowns:
1 1 1 | 3 2 3 1 | -4 6 8 4 | x
the reduced row echelon form of this yields:
1 0 2 | 13 0 1 -1 | -10 0 0 0 | x+2
so (ce-bf) is a free parameter, say s, and:
af-cd = 13-2s bd-ae = -10+s ce-bf = s
can I then conclude that the determinant, x, of C is equal to -2? I am unsure.
welp, exam turned in. hope I was right.
Man, I should get on this linear algebra thing.
I kind of hate it, but I mostly blame my professor. lecture consists of lots of proofs and boring examples of algorithmic procedures to find results of which the significance is never explained.
calculus sure is fun though.
calculus sure is fun though.
once the deadline for the exam passes I'm going to post this on math.stackexchange.
got it right. nice.