Ickathu
Veteran
Cool thanks! I've always wondered that...
Mathematics Thread
- Thread starter IanTheCuber
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Cool thanks! I've always wondered that...
xXxMCCALLxXx
Member
Probability and Statistics should be one of the choices for the poll.
vcuber13
Veteran
can someone explain what the triple equal sign means
edit: ≡
edit: ≡
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tozies24
Veteran
It means congruence in some contexts. In my modern algebra class that I am currently in, congruence occurs when you have two numbers a and b and they differ by a multiple of m. (a-b=mk for some k that is an integer). We write a (congruent or triple equal sign) b mod m. We use this when we are telling time or doing stuff with clocks. When we want to know what time it will be in 25 hours, we will just add 1 hour to the current time and then be done. Our time system is in modulo 12.
if you want more info: http://en.wikipedia.org/wiki/Modular_arithmetic
EDIT: should have quoted last post, oh well
if you want more info: http://en.wikipedia.org/wiki/Modular_arithmetic
EDIT: should have quoted last post, oh well
asportking
Veteran
I used to be a math enthusiast like you, then I took a Calculus in the head.
cubingawsumness
Veteran
yup me too
I'm no math genius, but I'm semi-decent at it. I'm in 8th, and I'm currently in geometry.
In 6th I took what they call "pre-algebra A", but then that summer I took a "challenge test" which allowed me to skip "pre-algebra B", and so in 7th grade I took Algebra.
Next year, I'll take Algebra 2/Trig, then pre-calc or "math analysis", then calc, then maybe stats or something.
In our middle school, we have 3 different "tracks". In 6th grade, there are only two: normal and advanced. I was in normal.
In 7th grade the people who failed in normal in 6th grade get put into "Math 7 skills". Normal people continue in normal unless they pass the "challenge test" like me, which would put them to advanced.
From geometry, there are 3 different courses. I you get D or F (overall grade) then you get put into geometry again (back onto normal track). If you get B or lower, you get into Algebra 2. B or higher, Algebra 2/trig. (so many complications...)
Then after algebra 2 or algebra 2/trig, you can either go to "math analysis" or pre-calc. Apparently math analysis is a fancy word for the first half of pre-calc (slower pace).
Then after math analysis you can go to precalc or if you want you can skip to calc. From precalc you can either go to calc or AP stats. There are 2 calc classes : AB and BC.
Apparently they just go at different paces.
phew...
why do math courses/tracks have to be so complicated...
bump, got buried by Ian's "riddle"
Sa967St
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6!/(4!2!)=6C2=6C4
He travels 6 blocks in total, and he chooses 2 spots where he goes one block west, or you can think of it as he chooses 4 spots where he goes one block north. The solution follows the same pattern as the other questions in that section (I used the same textbook for that course
). Check out question 7 on page 296 for a laugh.
He travels 6 blocks in total, and he chooses 2 spots where he goes one block west, or you can think of it as he chooses 4 spots where he goes one block north. The solution follows the same pattern as the other questions in that section (I used the same textbook for that course
). Check out question 7 on page 296 for a laugh.y(n)=8*sqrt(n)
y(n,x)=(x+1)*sqrt(n)
That was fun.
y(n,x)=(x+1)*sqrt(n)
That was fun.
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asportking
Veteran
Ok, can someone explain to me why the "dx" at the end of integrals are needed? I can integrate easily enough, but I could never quite understand why the "dx" always needed to be at the end, and now that we're starting to mess around with the dx, I figured it would be a good time to understand it better.
samehsameh
Regular
Basically it tells you what your integrating with respect to. If you were given \( y=x \) and told to differentiate it you might say it equals 1 but w.r.t. what? You should write \( dy/dx=1 \) if you differentiated w.r.t. \( x \) This forms one of the simplest differential equation \( dy/dx=1 \) which is equivalent to \( \int dy = \int dx \) so you integrate the LHS w.r.t. \( y \) and the RHS w.r.t. \( x \) and obviously end up with \( y=x+c_1 \). That was easy, now say your differential equation is \( dy/dx=x^2+3x+2 \) this is equivalent to \( \int dy = \int (x^2 +3x + 2)dx \) Evaluating just the LHS you end up with questions like your probably given i.e. \( y=\int (x^2+3x+2) dx \) so, basically the \( dx \) comes from the differential equation of the function you find by evaluating the integral given and could be anything i.e. \( dt \) is another common one if your function is a variable of time.
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samehsameh
Regular
next problem. Find the area between \( y=\frac{1}{x^4+4x^2+3} \) and \( y=0 \). Solutions please, dont just wolfram it.
Hint
Hint
evaulate \( \int_{-\infty}^{\infty} \! \frac{\mathrm{d} x}{x^4+4x^2+3}. \)
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ben1996123
Banned
Identical. Random example (I think this is what it's used for, not completely sure though):
\( \frac{x}{x}=1 \), but \( \frac{x}{x} \) is not \( \equiv1 \) because \( \frac{0}{0}\neq1 \)
random calculus just for fun:
\( \frac{sin(x)}{2^{xy}}=x+y \)
\( \frac{d}{dx}(\frac{sin(x)}{2^{xy}})=\frac{d}{dx}(x+y) \)
\( \frac{d}{dx}(x+y)=1+\frac{dy}{dx} \)
\( a=\frac{b}{c} \)
\( b=sin(x) \)
\( c=2^{xy} \)
\( \frac{da}{dx}=\frac{c\frac{db}{dx}-b\frac{dc}{dx}}{c^{2}} \)
\( \frac{db}{dx}=cos(x) \)
\( c=2^d \)
\( d=xy \)
\( \frac{dc}{dx}=\frac{dc}{dd}\frac{dd}{dx} \)
\( \frac{dc}{dd}=2^{d}ln(2)=2^{xy}ln(2) \)
\( \frac{dd}{dx}=y\frac{d}{dx}(x)+x\frac{d}{dx}(y)=y+x\frac{dy}{dx} \)
\( \frac{dc}{dx}=2^{xy}ln(2)(y+x\frac{dy}{dx})=2^{xy}yln(2)+2^{xy}xln(2)\frac{dy}{dx} \)
\( c\frac{db}{dx}=2^{xy}cos(x) \)
\( b\frac{dc}{dx}=sin(x)(2^{xy}yln(2)+2^{xy}xln(2)\frac{dy}{dx})=2^{xy}ysin(x)ln(2)+2^{xy}xsin(x)ln(2)\frac{dy}{dx} \)
\( c^{2}=2^{2xy} \)
\( \frac{c\frac{db}{dx}-b\frac{dc}{dx}}{c^{2}}=\frac{2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{xy}xsin(x)ln(2)\frac{dy}{dx}}{2^{2xy}} \)
\( \frac{2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{xy}xsin(x)ln(2)\frac{dy}{dx}}{2^{2xy}}=1+\frac{dy}{dx} \)
\( 2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{xy}xsin(x)ln(2)\frac{dy}{dx}=2^{2xy}+2^{2xy}\frac{dy}{dx} \)
\( 2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{2xy}=\frac{dy}{dx}(2^{2xy}+2^{xy}xsin(x)ln(2)) \)
\( \frac{dy}{dx}=\frac{2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{2xy}}{2^{2xy}+2^{xy}xsin(x)ln(2)} \)
\( \frac{dy}{dx}=\frac{cos(x)-ysin(x)ln(2)-2^{xy}}{2^{xy}+xsin(x)ln(2)} \)
\( \frac{d}{dx}(x+y)=1+\frac{dy}{dx} \)
\( a=\frac{b}{c} \)
\( b=sin(x) \)
\( c=2^{xy} \)
\( \frac{da}{dx}=\frac{c\frac{db}{dx}-b\frac{dc}{dx}}{c^{2}} \)
\( \frac{db}{dx}=cos(x) \)
\( c=2^d \)
\( d=xy \)
\( \frac{dc}{dx}=\frac{dc}{dd}\frac{dd}{dx} \)
\( \frac{dc}{dd}=2^{d}ln(2)=2^{xy}ln(2) \)
\( \frac{dd}{dx}=y\frac{d}{dx}(x)+x\frac{d}{dx}(y)=y+x\frac{dy}{dx} \)
\( \frac{dc}{dx}=2^{xy}ln(2)(y+x\frac{dy}{dx})=2^{xy}yln(2)+2^{xy}xln(2)\frac{dy}{dx} \)
\( c\frac{db}{dx}=2^{xy}cos(x) \)
\( b\frac{dc}{dx}=sin(x)(2^{xy}yln(2)+2^{xy}xln(2)\frac{dy}{dx})=2^{xy}ysin(x)ln(2)+2^{xy}xsin(x)ln(2)\frac{dy}{dx} \)
\( c^{2}=2^{2xy} \)
\( \frac{c\frac{db}{dx}-b\frac{dc}{dx}}{c^{2}}=\frac{2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{xy}xsin(x)ln(2)\frac{dy}{dx}}{2^{2xy}} \)
\( \frac{2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{xy}xsin(x)ln(2)\frac{dy}{dx}}{2^{2xy}}=1+\frac{dy}{dx} \)
\( 2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{xy}xsin(x)ln(2)\frac{dy}{dx}=2^{2xy}+2^{2xy}\frac{dy}{dx} \)
\( 2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{2xy}=\frac{dy}{dx}(2^{2xy}+2^{xy}xsin(x)ln(2)) \)
\( \frac{dy}{dx}=\frac{2^{xy}cos(x)-2^{xy}ysin(x)ln(2)-2^{2xy}}{2^{2xy}+2^{xy}xsin(x)ln(2)} \)
\( \frac{dy}{dx}=\frac{cos(x)-ysin(x)ln(2)-2^{xy}}{2^{xy}+xsin(x)ln(2)} \)
Hopefully that's right.
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Sa967St
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Ickathu
Veteran
This is a lot of information to condense into a forum post, so...
"So what's up with that triple equal sign?" (Stanford)
"So what's up with that triple equal sign?" (Stanford)
samehsameh
Regular
Very nice solution. I thought i preferred the complex analytic solution only because its much shorter than yours but yours can be understood by anyone who's done calculus which i guess is better.
Also what are you using to lay that out, LaTeX?
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samehsameh
Regular
International Mathematical Olympiad?
well at least i'm not being ignored anymore... has anyone here gone to IMO?
aaronb
Veteran
My middle school did the AMC competitions that led to the IMO, but I don't think my high school does, so I pretty much have no chance of going, unless I want to take the AMC outside of school, and set it up myself somehow. But I'm assuming you have a chance of being able to go, so good luck
