12:03 AM
Hmm
Still can't figure out how I'm supposed to do numerical quadratures on antiderivatives of complex functions
Doesn't seem to just be the signed area under the curve starting from a fixed $a$ :P

it gives direction which i thought was bearing and then gives bearing again
really confused

12:19 AM
Question, not sure if I'm doing this right:

@MATHASKER I'm not sure, but I think the direction it's going is 24 degrees past north, but the line from the ship to the cape is 280 degrees past north
Like, the 24 is the path of travel and the 280 is where to face if you want to look at the lighthouse/cape
(I assume the lighthouse and the cape are the same place?)

Given a function $F(x)$ such that $F(x)=\int_0^xf(x)$ for $x\geq0$ and $F(x)=\int_x^0f(x)$ for $x<0$, then does $F^\prime(x)=f(x)$?
($f$ and $F$ here are real in this case)

@LegionMammal978 You'd want $-\int_x^0f(x)\operatorname d\!x$ for $x<0$.

handed in my homework yesterday (or the day before) and I just took at look at it, I completely screwed up.

" Given the vector equation of a line in 2-space, (x,y)=(3,2)+t(2,4), write a scalar equation for the line. (4 marks)
slope=4/2=2
y=m(x-x_1 )+y_1
y=2(x-3)+2
y=3x-9+2
y=3x-7
3x-y-7=0
Therefore, the scalar equation of the line is 3x-y-7=0"

I see.

12:22 AM
Not only did I use the completely wrong approach, but I then did the math wrong.

Now, how to apply this to the complex antiderivative...

You're supposed to make a system of equations right?

Oh @AkivaWeinberger yes i thought the cape was a light house too, let me draw a picture

Wait, I think I did it correctly?
But then my math is off.
Thinking about recalling submission and fixing it.

got upto this then again got confused because if thats the light house, it can't be 205 degrees from the ship @AkivaWeinberger

12:29 AM
360-205 = 155
@MATHASKER Sorry for that math thing, I hadn't read your question. I think that the problem with this question is how it's worded.

i only had some words from the question remembered so i just went from that
why did u do 360-205

oh its fine np @Dodsy

@MATHASKER I just don't know what the question means by a 24 degree course, but I think that if you boil this question down it's basically a "x + y = 6" "x-y = 2" type of question. I wouldn't at all know where to start, because of the language. But try to make a system of equations from the facts you are given.

@MATHASKER Right now, the lighthouse is 280 degrees from the ship. But the ship is still moving, so maybe at some point in the future it'll be 205 degrees from the ship
By the way, a cape is a geographical feature, which I'm assuming the lighthouse is built on.

12:35 AM
oh it did say after 40 minutes

Oh yes.

Like Cape Cod in Massachussets

It is a triangle.

i thought to but I didn't know how to draw the third side

you are finding the position of the ship in cartesian coordinates
well you know the magnitude and direction of the vector.
you know the two angles.
Start by drawing the lines on a cartesian plane.
Triangulation!

12:38 AM
but like wheres the third line, it says that after 280 degrees of bearing from the cape the light house is 205 degrees in bearing of the light house which makes it so wierd

So imagine that the origin is the light house.

oh the origin is the light house....

yes.
wherever you want really.

ohhhh let me do what u said then try to make a triange

yeah give it a shot.
30(40/60) = the distance traveled by the boat at a bearring of 24 degrees.
now you have an angle inside the triangle
and the opposite side
right?
take 280 and subtract 205
that's 75.

12:44 AM
Hey @Ted!

Heya Demonark :)

How's it going?

I think I may be turning the corner on health ... of course, I already postponed my oral surgery tomorrow. How be you?

Oh, you're finally starting to get better? That's good, this stuck for quite some time
Hope you recover well!
And I'm doing alright, thanks!

no still confused about the third side @Dodsy can't make it in a drawing

12:46 AM
@MATHASKER does it say what the boat is travelling 23 degrees to?
relative to.
Well we already know the third side, its has a velocity of 20 and a direction of 23 degrees, in fact let's just say relative to the origin
which means there are 3 points, the origin the boat and the light house
from the origin to the boat is 23 degrees
from the light house to the origin is 280 degrees
from the light house to the boat is 205 degrees

@Dodsy: I hope you don't think the vector $(2,4)$ has slope 3 !!

giving us an angle between the origin and the boat from the light house of 75 degees.
@TedShifrin I KNOW!!!!!
@TedShifrin Luckily I was able to resubmit the assignment.

shakes finger at Dodsy for sloppiness
It's OK, Dodsy. I'm retired now. So I can't be officially too mean :P

12:51 AM
the only picture i could get

@MATHASKER we have a side and an angle! You can solve for it, no problem.

Hi all!

hi @BAYMAX.

hi@TedShifrin

@TedShifrin I was lucky to have been looking over my work when I found the error.

12:52 AM
we have the 30 km/ hr but what other angle do we have, maybe subtract 90 -24 ?

Well, @Dodsy, I'll save picking on you for another offense, then :)

@MATHASKER Subtract 205 from 280 to get 75 degrees.

Hello chaps

Hi @TimTheEnchanter.

@Dodsy why subtract 205 from 280 ?

12:54 AM
Because the differences between the angles.
Imagine the starting point of the boat being the origin
the barring of the light house is 280 degrees.
draw a line.

Hi@TimTheEnchanter

then draw the second bearing of 205.

What are these 'bearings' you speak of?

ball bearings ?

ah some physics problem I presume.
Perhaps vectors.

12:55 AM
See the question.

no, trig

Yeah, I'm not sure what exactly a 'bearing' is
which angle are we talking about?

@TimTheEnchanter: Nautical people seem to be contrary, so they measure angles clockwise (rather than counterclockwise) from north.

I think a bearing would be the old school method non cartesian coordination system of finding the angle of something

12:57 AM
Sorry.

I'm thinking about writing the putnam this year, but my mathematical maturity isn't too great, if I do poorly this year will I be judged badly for it?

of course not

@Dodsy wasn't the origin the light house?

does your school have a get-together to practice for it, working old problems and such?

A continuous function on a connected set that is never zero is either positive or negative ?

12:59 AM
@MATHASKER it's really whatever you choose.
@MATHASKER But just know you have an angle and a side.

yes

oh hol up let me try doing it one more time and ill send the picture

OK, so yes.

but how can prove it?

1:00 AM
@TedShifrin They have biweekly meetings where they discuss problems, and have lectures.

it seems ok in 1D case

Start going to those and work on problems for a few hours every week. It also takes a certain skill at presenting solutions to get credit.

So I could write it every year and if I do good then good but if I do poorly then it's not that big of a deal?

but in general how we can show that!

But it's a real-valued function, @BAYMAX, so the image of the function is a connected set in $\Bbb R$.

1:02 AM
Ok

Do you know what all connected sets in $\Bbb R$ are?

connected sets in $\mathbb{R}$ are like intervals ?
as we can connect any two pints in an interval

@BAYMAX: Somewhere you should have seen a proof of that, yes. The only connected subsets of $\Bbb R$ are intervals.

1:04 AM
no cape

ohhhh

ok,next step?
@TedShifrin

you know how it says after it passes cape at a bearing of 280

yeah
so here's how I would imagine it.
put the light house in the fourth quadrant of a cartesian coordinate system

Then you're done, @BAYMAX. Suppose you had a positive number and a negative number both in the image.

1:06 AM
put the starting position of the boat at the origin
draw a like at 280 degrees from the light house to the origin.
put the line that the boat goes from the origin at 24 degrees from the origin
write "20" above that line
that's the magnitude.

then I can take (negative number - 1 , positive number + 1) as an interval @TedShifrin

draw a line from wherever you chose your light house to be to the finishing line where the boat is, now you can see that the bottom angle being made is 280-205.
=75

No, no. You know the connected set contains [negative number,positive number]. So?

oh sorry, this is still all messed up. hahaah.

how did u connect the line to the fourth quadrant?

1:11 AM
and it is an interval meaning it must contain zero somewhere in the interval [-ve , +ve],but function cannot take zero values as given ,@TedShifrin

So, therefore, must have all positive or all negative values!

imagine isntead that the boat is coming towards the origin. and it works.

ha ha ... that is cool , thank you @TedShifrin

Heres the thing. Let's label the sides.
x, y, and z.

oh ok ill try one more time, if i can't ill give up thanks for helping @Dodsy, 2moro Ill get a really good look at the question and see if that helps

1:14 AM
imgur.com/a/eZksd vector x plus vector y equal vector z.
I am probably confusing you more. I'd just plug in the numbers and write down the answer given that you know the opposite and you need the adjacent.
tan(theta) = o/a
@TedShifrin hey the rest of it was right though, right?
@TedShifrin that question I was doing.

I didn't look, @Dodsy :P

@TedShifrin That's alright, no worries.
@TedShifrin When did you realize you wanted to be a mathematician?

Do a good job with vector calculus and linear algebra. They're very important to understand.
Oh, at a pretty young age, I guess.

@TedShifrin Vectors are fun!
It took me longer than I'm willing to admit.

I'm a big fan of vectors. (Since you're new, I'll tell you that I have a bunch of lectures on YouTube on linear algebra and multivariable calculus [done with proofs] if you're interested at some point.)

1:20 AM
That'd be great!

Glad to have you in chat. There's room for people at all levels, and it's great to get a chance to help other people out, too.

I am not starting school until september, so I'll have all summer to study math and do fun things. Right now I need to finish chemistry to be accepted. It's literally the worst thing in the world.
Thanks for the warm welcome!
I find that teaching/helping keeps you solid in your foundations.
well, to become enrolled - not accepted.

Oh, absolutely. One reason I aced the GRE my senior year of college was that I'd spent 10 hours a week or more helping people with math at all levels.
Oh, were you less motivated in high school?

That's awesome!
Oh boy, could I tell you one heck of a long story!

Let us consider $S = \{(x,y) \in \mathbb{R}^{2}, -1 \leq x \leq 1 , -1 \leq y \leq 1 \}$

1:23 AM
But I won't, I'll just tell you that I'm 22 and am just starting university this september (in Canada)

I like squares, go ahead @BAYMAX

and $T = S - (0,0)$

That's cool, @Dodsy. You're more motivated by far now, which will lead to greater results.

I had to work very hard to even get this conditional acceptance. My high school average was abysmal.

LOL @TimTheEnchanter.

1:25 AM
$f$ be a continuous function from $T$ to $\mathbb{R}$

Well, @Dodsy. You have a great opportunity and you will work to make sure you don't squander it. :)

from Ted's answer previously I got that image of $f$ must be connected!

Thanks, @TedShifrin . I really appreciate your words of encouragment.

why can the image of f be compact ?

It won't be in general, @BAYMAX.
@Dodsy: I am being sincere in encouraging you. I taught for something like 40 years, and part of being a good teacher is to be a good cheer-leader.

1:27 AM
a continuous function pulls back open sets to open sets!
@TedShifrin

And a continuous function maps compact sets to compact sets, but $T$ isn't compact!

as there is an open puncture at 0 ?

Right, so you removed a limit point.

@TedShifrin Haha! Well, I can't tell you how much I appreciate it. Words of encouragement can often be few and far between, and I could use the motivation at this moment.

I am thinking now compact sets and limit point?

1:30 AM
Well, I'm glad to help, @Dodsy. Stay determined!

a topological space X is said to be limit point compact or weakly countably compact if every infinite subset of X has a limit point in X

@BAYMAX: In order for a set to be compact, it must contain all of its limit points.
We're in $\Bbb R^2$, @BAYMAX. A set is compact iff it's both closed and bounded.

yes yes

@TedShifrin I guess I was asking about the Putnam Comp a bit early- but I was set on writing it in 3rd or 4th year to save any embarrassment, it's great to know that if you do very poorly you will not be judged harshly for it.

@Dodsy: Work hard and don't waste time being embarrassed.

1:32 AM
Those are good words of advice!

Can any such continuous function $f$ can be extended to a continous from $S$ to $\mathbb{R}$ ?

Great question, @BAYMAX. What do you think?
Is it true for a function from $(0,1]$? Can you necessarily extend it to a continuous function on $[0,1]$?

Speaking of working hard, I better get back to chemistry- I have 168 hours of work due in 7 days.

Yikes. Good luck, @Dodsy, and I look forward to mathing with you.

we are in $\mathbb{R}^{2}$ ?
@ted

1:35 AM
But I was asking you a simpler question.

sorry I have to go , I feel sad , I will ping you after I read something about extending concepts , can yu please give me some reference,i will get back to you!
@TedShifrin

Just think in terms of pictures, @BAYMAX. Graphs of functions (of one variable).
Talk later.

actually extending functions is a concept ?
@TedShifrin
may be it will be violating some concepts of a functtion like some properties ?

It is. But this is not so hard. If you have a continuous function $f$ on $(0,1]$, must $\lim\limits_{x\to 0+} f(x)$ exist?

no no let me try! , will get back to you!

1:39 AM
See ya :)

bye @TedShifrin , have a good day

Hi @s.harp .. oh and I didn't even see DogAteMy enter.
Balarka, aren't you supposed to still be un-sleeping?

Awake a little early.

Uh huh.

I'll have to decide whether I'd take another nap or use this un-un-slept time to temporarily fix my sleep cycle.

1:44 AM

Maybe I'll just try to un-nap
See ya

How are people

2:13 AM
[Integral symmetries] To be investigated:
Consider a function $f(x)$ which obeys some relation $g(f(x))=h(x)$ and is uniquely determined by $n$ points (where $n$ can be from a finite or infinite set $S$). Now consider the indefinite integral $$F(x)=\int f(x)dx$$. An interesting problem is to determine whether given a known $h(x)$ and $f,g$, the number (or better, the set of points) that uniquely determines $F$ (modulo constant of integration) exists
Footnote: The above is actually inspired from the chemistry paper I am currently reading, where in order to test how density of a solvent affects the calculated energetics of a reaction, the authors pick two numbers. They found the overall result does not exhibit dependence to said densities
Overgeneralisation of this observation mixed with the munkres stuff that we went through roughly a year ago (in particular how there are countable number of continous infinitely differentiable functions because they are restrained by the number of rationals) then expand it into the problem of given a function where its domain is restricted to some set (may be a disjointed union), whether it consists of points that restrict its behaviour enough to ensure uniqueness in that region
This generalised problem then mixed with the integral symmetry investigation and become the problem shown in the beginning of this post
===
As an illustration, let's consider one of the simplest relation of $f$, that is, $f(x)=f(x-a), a\in \Bbb{C}$. For this we had the following nice symmetry:

$$F(x)=\int f(x)dx=\int f(x-a) dx$$

2:37 AM

Im completely lost:
Useâ€‹ Newton's method to approximate the indicated root of the given equation accurate to five decimal places. Begin by sketching a graph.
The root of $x - 6 + 6cosx=0$ that lies on the interval â€‹[4â€‹, 6â€‹]

Hey everyone!

@Daminark Taxicab metric

@WillNjundong pick a starting point in that interval, then apply newton's method until you get two subsequent iterate to agree to 5 d.p.

Kek @Akiva

2:45 AM
Actually, a better caption would just be "???"
or Alan Davies making a confused face

also, a brief inspection of that region of the graph suggest the root is going to be an attractor for all values in [4,6] thus you should always converge regardless of the initial conditions

Dunno who that is, but why not?

3:33 AM
hi @TedShifrin @Daminark @AkivaWeinberger

4:17 AM
Joke: This is a magic square

1 2
3 4

But it is actually not if you realise what is really going on

So what's the joke?

Consider f(x)=5x mod 5. Then you get 5, 10 ,15 ,20. In mod 5, their magic sum are indeed equal ( = 0) but no numbers are unique in that grid, hence it is not a magic square
In fact, 2x2 magic squares don't exist
The joke is that via mod 5, I label what is essentially 0 into 4 different numbers, thus for those unaware, they will think a 2x2 magic square exists despite when they tried to add the magic sum, they cannot find any
And now, consider if you have these two interpretations in superposition, then you can say that this is a magic square because a magic sum exist (interpret it in mod n), and all numbers are unique (interpret it in terms of the labels which runs from 1 to n^2 on what are basically the exact same numbers)
Now for something more sane, there are actually no 2x2 magic n-cubes, because once you solve the system of linear equations, you find the only answer that can satisfy it is having all entries equal

4:34 AM
good remove ;-)

It does not make sense anyway, thus I should remove it

I think keeping "engaged" is important.

4:46 AM
@Secret thank you, that page helped me get it right

2 hours later…
6:39 AM
Lol well you're not here

6:57 AM
Hi pal @Danu
How's it going?

2 hours later…
8:44 AM
Hi guys, anyone here that could help me with a permutation problem?

2 hours later…
10:35 AM

10:45 AM
hi
can someone explain me this plz

11:23 AM
@Fawad $x^2+y^2-c=0$ is a circle with radius $\sqrt c$, which only makes sense when $c\ge0$. $~x^2+y^2+c=0$ is a circle of radius $\sqrt{-c}$, which only makes sense when $c\le0$.

@AkivaWeinberger when centre of circle(say g,f) point is given and radius of circle(say r),then while calculating "C" of equation of circle, $r=\sqrt(f^2+g^2-c)$
then value of $c$ is plus or minus $c$ or just plus $+c$ ??

The circles $x^2+y^2-c=0$ and $x^2+y^2+c=0$ are both centered at the origin
To center it on $(f,g)$ (and with radius $r$), you want $(x-f)^2+(y-g)^2-r^2=0$

@AkivaWeinberger i mean while solving numericals

Write down a random number between $0$ and $1$. Write down another one and add it to the previous one. If the total exceeds $1$, stop. Otherwise keep adding such random numbers until the total exceeds $1$ and stop.