heck i even doubt myself
#help-13
cold briar
dire geode
cedar kilnBOT
cedar kilnBOT
Available help channel!
frigid flare
not sure how to approach this question and the relevance of the horizontal distance given. thanks!
boreal epoch
seems like physics not math
cedar kilnBOT
@frigid flare Has your question been resolved?
cedar kilnBOT
@frigid flare Has your question been resolved?
frigid flare
.close
surreal cave
So I saw a video about Cauchy's Proof for $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$.\
It involves by beginning with an inequality $\sin\theta<\theta<\tan\theta$ for $0<\theta<\frac{\pi}{2}$.\
We then take the squared reciprocal of all parts of the compound inequality. \
Then Cauchy makes, what appears to be, a random substitution: $\theta=\frac{n\pi}{2N+1}$ for $1<n<N$.\
This yields $\displaystyle\cot^2\left(\frac{n\pi}{2N+1}\right)<\frac{(2N+1)^2}{n^2\pi^2}<1+\cot^2\left(\frac{n\pi}{2N+1}\right)$.\
Next, Cauchy multiplies everything by $\frac{\pi^2}{2N+1}$. This was bit more obvious as it cancels nicely with the middle term of the compound inequality.\
So one yields \fbox{$\frac{\pi^2}{2N+1}\cot^2\left(\frac{n\pi}{2N+1}\right)<\frac{1}{n^2}<\frac{\pi^2}{2N+1}+\frac{\pi^2}{2N+1}\cot^2\left(\frac{n\pi}{2N+1}\right)$}.\~\
I get all the other algebra stuff in the video, but this substitution seems to be random :( I know Cauchy had \textit{some} reason but it's going over my head and it probably obvious if anyone could tell me where it comes from that be great :)
wraith daggerBOT
XxMrFancyu2xX
surreal cave
this inequality getting cut off:\
\fbox{$\frac{\pi^2}{2N+1}\cot^2\left(\frac{n\pi}{2N+1}\right)<\frac{1}{n^2}<\frac{\pi^2}{2N+1}+\frac{\pi^2}{2N+1}\cot^2\left(\frac{n\pi}{2N+1}\right)$}
wraith daggerBOT
XxMrFancyu2xX
cedar kilnBOT
@surreal cave Has your question been resolved?
dire geode
**XxMrFancyu2xX**
It's nothing more than just trying a bunch of stuff and seeing if it gets you anywhere
It's not like they teach you the hundreds of things that didn't work
slate lintel
and often it's presented backwards, that factor came up after some manipulation and maybe not all at once
it's like a retcon
cedar kilnBOT
@surreal cave Has your question been resolved?
surreal cave
It's nothing more than just trying a bunch of stuff and seeing if it gets you an...
wow not as elegant as I would've liked 
dire geode
Most of these inequalities are like that
cedar kilnBOT
@surreal cave Has your question been resolved?
thin roost
Hi again
thin roost
This is the question
This is the solution
And I don't understand the green part
cedar kilnBOT
@thin roost Has your question been resolved?
slate lintel
oh I see it
they're asserting that f(6k+2) >= 2f(3k+1)
which is not immediately obvious but it uses that m,n thing again
thin roost
So it's like
f(6k+2) = f(6k)+f(2)+(0,1)
= f(3×2k)+(0,1)
= (2k,2k+1)
2f(3k+1) = 2(f(3k)+(0,1))
= 2k+ (0,2)
=(2k,2k+2)
How'd this helping here??
wraith daggerBOT
_basudev
thin roost
f(6k+2) = f(6k)+f(2)+(0,1)
= f(3×2k)+(0,1)
= (2k,2k+1)
f(3k+1) = f(3k)+(0,1)
= k+ (0,1)
=(k,k+1)
f(3k+1) = f(3k) + (0,1)
= (k,k+1)
adding them will give
(k,k+1) + (k,k+1)
= (2k,2k+1, 2k+2)
cedar kilnBOT
@thin roost Has your question been resolved?
cedar kilnBOT
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river ermine
Could someone check if this proof was done correctly?
cedar kilnBOT
Could someone check if this proof was done correctly?
vague rapids
The process is correct
dusk finch
Prove rigorously that if A ⊆ B, then A ∩ B = A:
(1) Given: A ⊆ B
(2) Let x ∈ A ∩ B
(3) By definition of intersection and (2): x ∈ A
(4) By (2), (3) and definition of subset: A ∩ B ⊆ A
(5) Let x ∈ A
(6) By (1), (5) and definition of subset: x ∈ B
(7) By (5), (6) and definition of intersection: x ∈ A ∩ B
(8) By (5), (7) and definition of subset: A ⊆ A ∩ B
(9) By (4), (8) and extensionality: A = A ∩ B
Is this proof rigorous, formal, and clear enough? What could I make better? And in (2) and (5), should I use "let", "assume", "suppose" or something else?
tropic oxide
yeah sure
dusk finch
And the wording "let" is correct as well? I am kinda confused about where should I use "let" "assume" and "suppose"
tropic oxide
"assume" and "suppose" are for statements you introduce that you aim to contradict
dusk finch
oh, so here I should use "let", right?
tropic oxide
yes your "let"s are fine here
crimson delta
"let bla" is very classic for variable introduction
dusk finch
alright, tysm. I think I understand it now
were you typing something or can I close it?
Aight, I guess I will close it, feel free to reopen this and ping me if there is something more I should know about. Thanks again
.close
cedar kilnBOT
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warm vector
how can I do this using pascal’s triangle?
cedar kilnBOT
how can I do this using pascal’s triangle?
warm vector
so i’m thinking of expanding this
tropic oxide
well yeah expand it
do you know how to expand binomials using pascal's triangle in general?
livid hound
after everything simplifies, yes
warm vector
cedar kilnBOT
warm vector
so i expanded the binomial
I don’t know when to stop as it says ‘as far as the term in a^3’
dreamy sleet
It means stop when you see a^3
warm vector
is it this part that I only include?
dreamy sleet
Or it could be the rightmost part, either way, it depends on the next part of the question, where you substitute things in
dreamy sleet
Have you done questions like this before?
The idea is that 1.002^5 = (1 + 0.002)^5 and the terms that have high powers of 0.002 are so small you don’t need them
So you can find it using binomials
dreamy sleet
Expand it
magic solar
if i get an expression of tanx=a, when i solve for x by arctan on both sides, what restriction am i left with
magic solar
is there even any?
magic solar
why is this range not just simply -pi/2 to pi/2 ?
cedar kilnBOT
why is this range not just simply -pi/2 to pi/2 ?
magic solar
i didnt know the inside of the brackets affected the range
dusk finch
It does, would range of sin(pi + 0x) be (-1, 1)? No, it wouldnt since sin(pi + 0x) is just 0 for all x
magic solar
yeah that makes sense
the textbook didnt really give an explanation as to how to find the range
dusk finch
So you will be interested in range of arctan(x) between x=0 and x=1
dusk finch
Well partially, if the range of inside function was e.g. [0, 2] and the inverse trig was arcsin, you would have to keep in mind that domain of arcsin is [-1, 1], and therefore limit [0, 2] to [0, 1]
dreamy sleet
applicable to all composed functions
dusk finch
so if the range is -inf to inf, does it just become -pi/2 to pi/2
What do you mean? And what's your final answer to your original question?
dusk finch
Good, just wanted to make sure you understood it
dusk finch
so if the range is -inf to inf, does it just become -pi/2 to pi/2
And what did you mean here?
magic solar
And what did you mean here?
for example arccos(f(x)) where f(x) has all real numbers as range, would the range of arccos(f(x)) then be -pi/2 to pi/2
dusk finch
yep
magic solar
ah thats helpful
then all odd funcctions within arccos or sin would be just its normal range
and tan i think
dusk finch
then all odd funcctions within arccos or sin would be just its normal range
are you sure? Maybe all odd polynomials with degree > 0, but definitely not all odd functions
magic solar
well all odd functions with all real numbers as its range?
dusk finch
well all odd functions with all real numbers as its range?
yes, this would be true.
cedar kilnBOT
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Available help channel!
austere widget
Hi there, this is a question about complex numbers
I can’t understand why for some scenarios like z^5=cis(-pi/4), you add 2kpi to the angle but for z^4=cis(pi/12) I can’t do the same
dire geode
Hi there, this is a question about complex numbers
I can’t understand why for s...
Depends. What's the question
austere widget
z^4=cis(pi/12)
austere widget
z^4=cis(pi/12)
The pictures are similar to this qn just that I didn’t include 2^(1/8) from the original qn
dire geode
z^4=cis(pi/12)
And what are you supposed to be doing with this equation
austere widget
Solve for the roots of z
If I added 2kpi to the eqn, my k=1,-1 (right pic) is different from the answer (left pic)
I get (25/48)pi and -(23/48)pi when it should be (47/48)pi and -(1/48)pi
wraith daggerBOT
Result:
-1.4791666666667
,calc -23/48
wraith daggerBOT
Result:
-0.47916666666667
Oh arithmetic error
cedar kilnBOT
@austere widget Has your question been resolved?
austere widget
Sorry I’m not quite sure I understand what you mean?
dire geode
In the first one they subtracted pi on the 5th line
But you add/subtracted 2pi on the second image
austere widget
So which one is the correct way?
cedar kilnBOT
@austere widget Has your question been resolved?
dire geode
So which one is the correct way?
What makes you think either are wrong?
austere widget
What makes you think either are wrong?
the left pic is the correct answer given by the school but doesnt use the 2kpi method
whereas the right pic is the answer im getting using the 2kpi method
yet both answers are different, so im not sure what method i should be using
dire geode
Check your answer by raising it to the 4th power and seeing if it equals exp(ipi/12)
austere widget
Check your answer by raising it to the 4th power and seeing if it equals exp(ipi...
z=e^i(25/48)pi
z^4=e^i(25/12)pi
z^4=e^i(2+1/12)pi
Do you mean like this?
dire geode
Yea for all your solutions
austere widget
z=e^i(1/48)pi
z^4=e^i(1/12)pi
z=e^i(25/48)pi
z^4=e^i(25/12)pi
z^4=e^i(2+1/12)pi
z=e^i(-23/48)pi
z^4=e^i(-23/12)pi
z^4=e^i(-2+1/12)pi
z=e^i(-47/48)pi
z^4=e^i(-47/12)pi
z^4=e^i(-4+1/12)pi
Im not exactly sure where i went wrong
It would help tremendously if you were able to point out my mistake because im gonna have to go sleep and cant keep this channel open
dire geode
Im not exactly sure where i went wrong
You didn't. That's what I'm trying to tell you
dire geode
Check your answer by raising it to the 4th power and seeing if it equals exp(ipi...
You should have identified that exp(2pi i n) = 1 in all your solutions so you do get exp(ipi/12)=z^4
cedar kilnBOT
@austere widget Has your question been resolved?
summer valley
cedar kilnBOT
turbid mesa
Yes
cedar kilnBOT
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Available help channel!
marsh pond
are there are a lot of examples in math that lead to ambiguity?
marsh pond
and for this example, who is to blame? the person asking, or the mathematics (notation itself) for getting to this point in the first place
royal loom
whoever posed the question is to blame
the question didn't just arise out of thin air
dire geode
Can you not waste time with meme questions
marsh pond
For example, JavaScript can do some crazy stuff, notation wise. but it was written in 10 days. It's too late to change now.
royal loom
the question asker must've known what order they wanted it to be solved in
marsh pond
it's an honest question
violet flume
you been asking some really open ended questions today runner
royal loom
and anyone who is very worked up over choosing the order, is also at fault
dire geode
And it's an honest response. Who cares who's to blame? It's a meme
marsh pond
ambiguity is a meme?
it's a real thing in math
we see it from time to time, as I explained earlier I found it on my Calculus 1 final exam
violet flume
it seems counter to the spirit of these channels
livid hound
blame people in the past that didn't know better for the origin
blame the people now that still currently make ambiguous stuff
violet flume
but i am not the channel protector
dire geode
There are very few actual ambiguous questions in math education
marsh pond
alright good, I hope not, was just curious how often it comes up
dire geode
You're deliberately wasting time on one because you found out about it through a meme
Not an actual problem from a book
marsh pond
it was a problem from a final exam, which is even more frightening (worth marks)
limpid plume
Almost 99% of the time an ambiguity in math comes from imprecise formulations of a query.
That's the answer you need.
dire geode
it was a problem from a final exam, which is even more frightening (worth marks)
But not your final exam
marsh pond
my final exam
dire geode
Doubt
marsh pond
alright, so at this point we live with it, but if it comes up we blame the teacher
marsh pond
Doubt
why?
I'm telling the truth, the entire class was asking the professor
she had to write it out on the board to clarify what the question meant
due to ambiguity
dire geode
Cuz it's a damn meme
limpid plume
Ah Riemann
marsh pond
doesn't mean it's impossible to see questions like this ocassionally
livid hound
how was the question presented
limpid plume
but yeah avid, these type of questions are usually waste of time for yourself
dire geode
we see it from time to time, as I explained earlier I found it on my Calculus 1 ...
Meme question on a calc final?
marsh pond
multi exponent for trig function without any argument brackets whatsoever
limpid plume
multi exponent for trig function without any argument brackets whatsoever
Yeah there's your answer
livid hound
do you have a pic,
royal loom
multi exponent for trig function without any argument brackets whatsoever
okay these I too actually always hate
marsh pond
do you have a pic,
no the university does not let us take exam questions home
but you'll just have to believe me, entire class had their hand up for the question and she wrote it on the board to clarify
livid hound
can you remember / reproduce/ rewrite the originally ambiguous expression
marsh pond
it was something like sinx^-1^-1
royal loom
if it was something like $e^{x^{2}}$ then you can interpret it as $e^{(x^2)}$ because if what was meant was $(e^{x})^{2}$ then it would be simplest to write it as $e^{2x}$
wraith daggerBOT
austinu
royal loom
does that help with those?
livid hound
$\sin x^{{-1}^{-1}}?$
marsh pond
no
i rewrote it
it was the question of wether it was arcsin or sin
the dreaded -1 exponent for sin
no argument brackets
not clear at all
wraith daggerBOT
ℝamonov
marsh pond
it was something like that, I think
royal loom
that is (one) of the reasons for functions like cscx
to represent 1/sin(x) without writing it fractionally, and without using an exponent that may or may not mean inverse
royal loom
so if something like that happened, it'd be on the question asker to clarify if needed
nothing really to do with the math
it is like assumptions about notation that we make
royal loom
ambiguity that has arisen because the question asker was not clear
limpid plume
**ℝamonov**
wtf
marsh pond
ambiguity that has arisen because the question asker was not clear
isn't that what the meme is about?
the math ambiguity meme
royal loom
yes
marsh pond
that's what i thought
so it is really poor practice to have questions you need clarification on, for the final exam
it should be crystal clear what is being asked for
so you can solve without your own personal interpretation of what you think the teacher meant
limpid plume
yes, definitely, but where are you going with this?
marsh pond
just a general question of how often ambiguity comes up in math
limpid plume
:wg:
marsh pond
for those that have taken math 5-10+ years, maybe once in a blue moon?
livid hound
when the writer is lazy
royal loom
not that often, and if it ever does it isn't the fault of the math
livid hound
or they can't type math
marsh pond
yeah, it's like a syntax error.. programmer's fault
not the language
alright thanks
was just curious
i will close this now
royal loom
that is kind of the goal of math, to be true and logical
marsh pond
yeah, i have made it a rule to always use argument brackets, just to be extra clear
some may say it's overkill, but it's a habit now, I see no harm in doing so, and it's extra clear
dire geode
it was something like that, I think
This is a far departure from your original expression. Next time just lead with your actual expression rather than opening with a meme
livid hound
not overkill in any way
dire geode
marsh pond
This is a far departure from your original expression. Next time just lead with ...
the original post was via Reddit
marsh pond
ah it's a famous example, maybe it triggers some ppl for seeing so often, lol
my example is ambiguous too I believe
that's why I was asking, but good to know it doesn't come up too much
livid hound
an expression itself could be unambiguous
but the ambiguity arises from the existence of that in the context of the question
cedar kilnBOT
dire geode
Read
https://xyproblem.info/
@marsh pond read this
cedar kilnBOT
Available help channel!
opal pebble
im currently doing a mathamatical induction and im on step 3 and im stuck at (k+1)! - 1 + (k+1)(k+1)! = (k+2)! - 1
opal pebble
and im stuck on how to prove that
austere widget
Are you able to send over your working?
tropic oxide
consider: 1 * (k+1)! + (k+1) * (k+1)! = ?
opal pebble
There's not much besides that besides proving that base 1 gives equal terms
opal pebble
consider: 1 * (k+1)! + (k+1) * (k+1)! = ?
could that be... (1+(k+1)) x (k+1)! ?
what would i do next than if i have (k+2)(k+1)!-1=(k+2)!-1
tropic oxide
could that be... (1+(k+1)) x (k+1)! ?
don't use the letter x as a multiplication symbol.
also that's ?= not just =. don't conflate goals with assertions.
can you prove that (k+2) * (k+1)! = (k+2)! ?
opal pebble
To be honest, I don't know much about multiplying terms with factorials
So I really don't have a clue on how to prove (k+2) * (k+1)! = (k+2)! ?
tropic oxide
To be honest, I don't know much about multiplying terms with factorials
do you know what a factorial is?
tropic oxide
ok tell me what n! means
tropic oxide
i never said n wasn't an integer...
n! is the product of all integers from n down to 1.
(k+1)! is the product of all integers from k+1 down to 1.
(k+2)! is the product of all integers from k+2 down to 1.
by what do those last two differ
opal pebble
theres an additional factor of (k+2)?
tropic oxide
yes exactly
granite eagle
(3+7x)^2 = 0; so someone asked this and i told them to divide both sides by (3+7x) but isnt it already known that its equal to 0....
granite eagle
are we evalutaing 0/0 here?
granite eagle
oh like 3+7x = 0
tropic oxide
yes
cedar kilnBOT
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Available help channel!
magic solar
cedar kilnBOT
rotund jasper
but where do they get cosy=+root1-sin^2y ?
cos^2 y + sin^2 y = 1
cos y = sqrt ( 1 - sin^2 y )
cedar kilnBOT
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Available help channel!
keen sluice
I have answers and want to be corrected because my answer is different when i checked the answer key
drifting marlin
Send them and we'll take a look
tropic oxide
ok, show us your answer and the key.
keen sluice
my answer on 1 is 2x² - 2x + 7 = 0 a=2 b= -2 c=7
fresh quarry
my answer on 1 is 2x² - 2x + 7 = 0 a=2 b= -2 c=7
Uh don't forget the signs
While converting to standard form
livid hound
conversion itself was fine, identifying the values was not
livid hound
why did you put -7 for your c
keen sluice
oops
livid hound
you mean on Q1?
livid hound
notice that the terms in the equation have a common factor
thus the equation can be simplified
livid hound
why did you change your answer
keen sluice
on?
keen sluice
6 at first
livid hound
ok, so why did you change from 6 to 2 then 3
keen sluice
okay im confused my answer should be 3 because the 6 is devisible by 3
livid hound
no
keen sluice
right?
livid hound
what are the factors of 6
keen sluice
1 2 3 6
livid hound
what are the factors of 2
keen sluice
1 2
livid hound
and what's the greatest factor present in both lists
keen sluice
2
livid hound
yes
keen sluice
and 1
keen sluice
ok sorry
livid hound
so divide both sides by 2 to simplify the equation
keen sluice
what side
livid hound
of the equation
livid hound
the equation you have is
$$2x^2 + 6x - 6= 0$$
all terms have a common factor of 2, thus the equation can be simplified by dividing both sides of this equation by 2
wraith daggerBOT
ℝamonov
livid hound
yes, 0 included, same operations to both sides of the equation
livid hound
yes, the issue was you didn't simplify
keen sluice
i can simplify by dividing it
livid hound
yes, as i've mentioned
keen sluice
(2x^2 + 6x - 6) 0
÷ 2. ÷2
is this right?
okay ty i solve the q2 now
my ans on q4
get it now
thanks @livid hound
its just the gcf thing and divide
.close
cedar kilnBOT
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Available help channel!
raw drift
is it me or is the last factorize incorrect? https://i.imgur.com/ASMzaVg.png
livid hound
what makes you think its wrong
raw drift
-4 (3m - 2) should be -12m + 8 no?
livid hound
yeh, it is...
so what's the problem
raw drift
the end result is m pow 2 - 8m +12 im not seeing the step between and how
livid hound
did you expand (m+2)^2?
livid hound
yes
cedar kilnBOT
cedar kilnBOT
Available help channel!
vagrant hamlet
hello
vagrant hamlet
i have a feeling my answer for a) ii) is wrong
i am very confused on how to find the n term
valid yacht
looks good to me..
vagrant hamlet
really?
vagrant hamlet
if it is okay, then part iii)
how to dooo
because of part 3
i am very confused on how to solve it
or that i need to use
the
valid yacht
solve for n. n can only be a natural number. if n is not a natural number, then the 4 digit number cannot be a part of the sequence
vagrant hamlet
let me try
okay
its not part of the sequence
because n is not a natural number
valid yacht
,w solve{2n^2 + n - 1174 = 0}
wraith daggerBOT
vagrant hamlet
yap
valid yacht
vagrant hamlet
thank u
valid yacht
welcome
vagrant hamlet
.close
white urchin
I need some practice problems of Locus and Stright lines in coordinate geometry
hoary chasm
What level
white urchin
grade 11
hoary chasm
Try jee lol
white urchin
ok
white urchin
This kind?
i dont have normal form in my syllabus
hoary chasm
Not gaokao...it's not written in chinese
vagrant elbow
I really don't understand the obsession over overly hard coordinate geometry
cerulean narwhal
Hong Kong Certificate of education exmaination
hoary chasm
Then mhtcet
cerulean narwhal
math and Additional math
hoary chasm
I need some practice problems of Locus and Stright lines in coordinate geometry
Try mhtcet pyq
To start
With
cerulean narwhal
If you wanna do damn hard questions in Calculus try HK advanced level
white urchin
ykw
cerulean narwhal
espcially 1980s
white urchin
ima just do COP
hoary chasm
If you wanna do damn hard questions in Calculus try HK advanced level
We got imo for that
Olympiads
cerulean narwhal
I'm just a normal DSE candidate who wanna get A in math IMO is too hard for me
hoary chasm
i mean RD sharma
Ig mhtcet is that level
Mcqs are fun...Rd got some theoretical ones
white urchin
Ig mhtcet is that level
Mcqs are fun...Rd got some theoretical ones
ye
so what book
deepthi publications?
white urchin
ok
hoary chasm
Search on YouTube best books for mhtcet pyq
hoary chasm
Idk about srces where u get paid stuffs for free
Ig telegram
Or Maybe browse a bit more
white urchin
kk
white urchin
ik
rotund jasper
cedar kilnBOT
limpid plume
If you don't get answers here, feel free to ask on #probability-statistics. Just don't spam helper pings there, or here, for that matter.
rotund jasper
If you don't get answers here, feel free to ask on <#326138783181438976>. Just d...
done that
sry for ping
cedar kilnBOT
@rotund jasper Has your question been resolved?
cedar kilnBOT
@rotund jasper Has your question been resolved?
topaz tartan
Heyo, what is the statement in equation form. I've spent hours trying to decipher it and i just couldn't get a single solution for one being in her teens. I let j be janes age and b be Betty's age, and x be the difference of ages (i.e. b-j)
topaz tartan
exactly my reaction dw
tropic oxide
theres like 3 different points in time being mentioned here like. eh?
topaz tartan
nah more than that
but im pre sure it can be boiled down to those 3 variables
because obviously difference of age will remain constant over any period of time
tropic oxide
yeah true
x := b - j
Time | Betty | Jane |
-----+----------+----------+
Pres | b | j |
-----+----------+----------+
Pt 1 | 2b+x | 2b | [when Jane is twice as old as Betty is at present]
-----+----------+----------+
Pt 2 | b+3x/2 | b+x/2 | [when Jane is half as old as Betty will be at Pt 1]
-----+----------+----------+
Pt 3 | b+5x/2-1 | b+3x/2-1 | [when Jane is 1 year younger than Betty will be at Pt 2]
-----+----------+----------+
Pt 4 | j | j-x | [when Betty was as old as Jane is now]
-----+----------+----------+
then the problem reads:
Betty's age at Pt 3 is 3 times Jane's age at Pt 4.
young flower
Chat gpt it
tropic oxide
Chat gpt it
NO.
chatgpt cannot do math.
and posting chatGPT output is bannable here.
and it definitely WILL get confused at such labyrinthine problems here.
dusk finch
Chat gpt it
I actually tried it. "So, currently, Jane is 2 years old, and Betty is 3 years old."
(As you can see it's solution is incorrect)
tropic oxide
@topaz tartan ok so basically the problem references 5 distinct points in time lmfao
topaz tartan
I actually tried it. "So, currently, Jane is 2 years old, and Betty is 3 years o...
damn teenages really be young now a days
tropic oxide
yeah so like
tropic oxide
so what we end up with is $b + \frac{5x}{2} - 1 = 3(j-x)$...
wraith daggerBOT
Ann
topaz tartan
alright alright
also pre sure u take an extra x on RHS to make sure its Jane= Jane
topaz tartan
LHS refers to Janes age at the final point in time
and RHS refers to Betty's age at final point in time
unless im dumb
cuz like
LHS
RHS
tropic oxide
```
x := b - j
Time | Betty | Jane |
-----+----------+----------+
Pres |...
i wrote down which point in time is introduced by which clause
topaz tartan
would 19 be teenage?
proven thistle
Yes
topaz tartan
i mean, ig it sorta could be, but maybe not
ig ill just go with it
tropic oxide
anyway it looks like this equation transforms into:
2b + 5x - 2 = 6j - 6x
2b - 6j + 11x = 2
13b - 17j = 2
assuming i did not fuck up any of my arithmetic
,w 13b - 17j = 2 integer solutions
wraith daggerBOT
topaz tartan
yeah i checked with wolfram ahaha
tropic oxide
hm.
topaz tartan
it gave 19 for n=1
tropic oxide
25 and 19?
topaz tartan
which just seems suspicious yk
tropic oxide
let's see how well this works
we get x = 6
b = 25, j = 19, x = 6
Time | Betty | Jane |
-----+-------+------+
Pres | 25 | 19 |
-----+-------+------+
Pt 1 | 56 | 50 | [when Jane is twice as old as Betty is at present]
-----+-------+------+
Pt 2 | 34 | 28 | [when Jane is half as old as Betty will be at Pt 1]
-----+-------+------+
Pt 3 | 39 | 33 | [when Jane is 1 year younger than Betty will be at Pt 2]
-----+-------+------+
Pt 4 | 19 | 13 | [when Betty was as old as Jane is now]
-----+-------+------+
ok yeah works out actually
topaz tartan
hmmm okok
imma try it with the words and see if it works out
25*2=50
50+6=56
28->34->33=LHS
19->13->39 which is 6 different
yeah ig it does work huh
tytyyyy
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cedar kilnBOT
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thin hedge
does anyone know how to graph natural logs on a ti nspire cx 2
tropic oxide
show exact input & output
slate lintel
yeah i mean
if you graphed y = 7 you'd get a straight line as well
you're graphing f(x) = a constant
cedar kilnBOT
@thin hedge Has your question been resolved?
cedar kilnBOT
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slim finch
if the radius of the ball increases by 3 times, its surface area will increase by
rare vault
if the radius of the ball increases by 3 times, its surface area will increase b...
do you know the formula for radius of a ball and surface area of a ball?
well rather do you know the relationship between radius and surface area of a ball*?
slim finch
4 pi r2
rare vault
okay
slim finch
well rather do you know the relationship between radius and surface area of a ba...
didnt understand you
rare vault
$4\pi r^2$
wraith daggerBOT
slim finch
yes
slate lintel
mixa have you tried this problem at all or did you just immediately post it here?
slim finch
i tried them before
rare vault
is this something mixa should know
(you should do 3r instead of r and see how much larger it it)
slim finch
and now im going trough the problems i didnt understand or i didnt know
rare vault
$4\pi (3r)^2$ has 3 times the radius right?
wraith daggerBOT
slim finch
so
let me write it
4pi(3)^2 x 3
i remove the 4pi
get 3^2
thats 9?
and 9 by 3
27?
rare vault
4pi(3)^2 x 3
in this one
rare vault
its 4pi x (3)^2 x 3
I don't know what this expression is from
rare vault
$4\pi 3^2 \cdot 3$
wraith daggerBOT
slim finch
yes
rare vault
the original formula is 4 pi r^2 right?
so if we want the radius to be 3 times bigger, we have to use (3r)
instead of r
rare vault
why do you care about the ball area?
slate lintel
it might be helpful to think about two balls, one of which has a radius of a and the other of which has a radius of b, which is three times a
slate lintel
no? where'd that 3 come from
slim finch
and ball2=4pi3^2 *2
slate lintel
first ball has a radius of a, what's its surface area?
surreal cave
If a ball has radius r, scaling the radius by 3 will result in a new radius of 3r compute the new surface area and compared it to the original :)
slim finch
<:what:561568026210730004>
me rn
wraith daggerBOT
Hayley
slim finch
yes
slate lintel
ok and the second ball? the one with the radius of b?
slate lintel
great okay so now we know that b is three times as big as a
can you write that as an equation?
slim finch
idk how
have a problem with turning textual questions to equations
i know what you want i just dont know how to put it on paper
slate lintel
okay, this one is pretty straightforward: b = 3a
slate lintel
so now we can find the surface area of the second ball in terms of a by substituting that into our equation
slate lintel
<@&268886789983436800>
slate lintel
start with $4\pi b^2$ and substitute $b = (3a)$
wraith daggerBOT
Hayley
slate lintel
take the equation on the left, and everywhere you see b replace it with (3a)
slate lintel
where'd the ^2 go?
wraith daggerBOT
Mixa
slim finch
like so?
slate lintel
yeah
slim finch
now i replace a w 3=radius?
slate lintel
now we just divide the new surface area by the old one
slate lintel
$\frac{4\pi(3a)^2}{4\pi a^2}$
wraith daggerBOT
Hayley
wraith daggerBOT
Hayley
slate lintel
in particular the "distributivity" rule will be useful
slim finch
this is how ive did it
so the difference in the size of the ball when we increased its radius is 9
slate lintel
yep, when it gets 3 times wider it gets 9 times as much surface area
slim finch
okay hayley thank you very much on helping me today it ment a lot i look forward in expanding my knowledge in math and yes i was asking a for a lot of help here today but i needed to clear some walls i had in these problems and i was doing them all day today since 10am +1 gmt time and now its 12:15 am next day xd
@slate lintel
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viral fox
Hey math legends, sorry me again, the big brain math lover.
let say we have x1+x2+x3+x4+x5+x6=12 where x3=4, and x5>2 . And finding how many solutions does this equation have using combination with repetition is appreciated.
So I removed x3 from the equation, as it's essentially a const number, and as x5>2, I assumed it must be at-least 3, so subtract the result by 3 as well. I came up with x1+x2+x4+x5+x6=5 (12 => 12-(4+3)) . Then I used combination with repetition formula, as C(n+k-1,k-1) where n=5, and k=5, and came up as C(9,4), but another solution tells it's C(5,4)
solid juniper
Hey math legends, sorry me again, the big brain math lover.
let say we have `x1+...
and as x5>2, I assumed it must be at-least 3, so subtract the result by 3 as well.
does that mean you didn't count cases where x_5 > 3?
oh
can it not even be >3
viral fox
> and as x5>2, I assumed it must be at-least 3, so subtract the result by 3 as w...
Don't know, original equation has 6 variables, which x3 is const, so I removed it. Then I got 5 variables(k=5). Also about x5>2, I just subtracted result by 3, and assumed x5 like otehr variables, as it could take 0 or any value
x5>2 means x5 must be at-least 3, or I'm really stupid. Sorry
solid juniper
yes but we could have x_5 = 4, x_3 = 4, and x_1 = x_2 = x_4 = x_6 = 1
but you still counted higher than what they say the answer is
so that's not the only issue
nova glacier
Where is the double counting though?
viral fox
I just assumed x5 = y5+3 , where y5 is new x5.
nova glacier
and x_5 = 4, x_3 = 4, and x_1 = x_2 = x_4 = x_6 = 1 corresponds to y_5=1,x_3=3 x_1,2,4,6=1 right?
Looks fine to me
solid juniper
||with x_5 = 3, you get one solution from setting one of x_1, x_2, x_4, x_6 to 2 and the rest to 1, for 4 total
with x_5 = 4, there's just 1 solution||
viral fox
Let me explain it better, so here we have x1+x2+x3+x4+x5+x6=12 , where x3=4 and x5>2
First I replaced x3 to 4(as it's a const), and x5 to y5+3 (as x5 must be at-least 3 then), so I get x1+x2+4+x4+y5+3+x6=12 which becomes x1+x2+x4+y5+x6=12-(4+3) => x1+x2+x4+y5+x6=5
Now k=5 (number of variables), and n=5, so plugin to C(n+k-1,k-1) => C(5+5-1,5-1) => C(9,4)
nova glacier
Oh, @solid juniper you are saying that all the vars are at least 1?
This explains it
viral fox
No, only given x5, and x3 must be non-zero, as x1 could be 0
nova glacier
But 5 choose 4 = 5... There are much more than 5 solutions if 0 is possible
solid juniper
idk i feel like they should all have to be at least 1 if the answer is 5 choose 4
got an exact statement of the problem?
viral fox
So my solution works? the reason I could not accept provided C(5,4) is because 0 are possible for x1, x2,.. vars, so it's obvious there are more than 5 solutions. The C(5,4) which derived from C(n-1,k-1) works if all variables must be at-least 1
nova glacier
And also I think it should be C(n+k-1,k) isn't it?
viral fox
And also I think it should be C(n+k-1,k) isn't it?
No, it's C(n+k-1,k-1) the star/bar problem. had this issues recently actually
nova glacier
you need to remove 4 more if all vars >1 right?
because you have 4 more vars that arent x5 and x3
so the sum is 1 now right?
and you have y_1,...,y_5 all can be 0 now
5 vars sum to 1? or do I have a mistake somewhere?
viral fox
you need to remove 4 more if all vars >1 right?
If all vars must be over 1, then x5>2 could be replaced by y5+2 (not 3) I guess, besides x3 should be removed anyway, so there will be 5 variables, to become 8.. hmm, a little confusing!
nova glacier
Why y5+2? x5 > 2 means it's at least 3 still
viral fox
5 vars sum to 1? or do I have a mistake somewhere?
No, 5 vars to sum of 5, but 0 could be possible
nova glacier
I'm saying let's do the same trick if x_i >= 1, replace with y_i +1 where y_i>=0
viral fox
Why y5+2? x5 > 2 means it's at least 3 still
Yes, that 3 would be covered by the general rule after equation is simplified I guess. If we substitute x5>2 with y5+3, and keep the at-least 1 for all variable after equation is simplified, then this means x5 must be at-least 4
nova glacier
And also I think it should be C(n+k-1,k) isn't it?
oh I got confused with n and k
if k was the sum... it was true
I thought k is the sum
Nevermind then
=]
viral fox
I'm saying let's do the same trick if x_i >= 1, replace with y_i +1 where y_i>=...
Alright, so x1+x2+x3+x4+x5+x6=12 , where x3=4 and x5>2, and xi>=1
First I replaced x3 to 4(as it's a const), then substitute all other xi as yi so: x1 => y1+1 , x2 => y2+1 , x4 => y4+1 , x5 => y5+3 and x6 => y6+1 so we get
y1+1+y2+1+4+y4+1+y5+3+y6+1=12 => y1+y2+y4+y5+y6=12 - (1+1+4+1+3+1) => y1+y2+y4+y5+y6=1 => n=1 , k=5 => C(n+k-1,k-1) => C(5,4)
Edit: fixed(thank Cain bro), I was stupid
nova glacier
Nope
Again x5 should be replaced with y5+3
otherwise it can get the value 2
when y5=0
Or you should make 2 steps for x5
viral fox
Dang, yes you are right, yes, must be replaced with y5+3, so final result will be C(5,4) then, aweosme bro
nova glacier
Yes, that's a possible explanation for this answer C(5,4)
But, who knows what they really meant... Don't bother yourself too much with ill posed questions..
viral fox
But, who knows what they really meant... Don't bother yourself too much with ill...
Yeah, actually if something like this is dropped in the exam, then I'll explain all possible moves probably
nova glacier
Yep
viral fox
Thank you @nova glacier bro, and @solid juniper very much! Appreciate your time. Math bless us all.
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cedar kilnBOT
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sleek mantle
When finding the area under a curve how do I know when it's under the x axis or above
carmine bronze
When finding the area under a curve how do I know when it's under the x axis or ...
Graph it or you can solve for f(x) = 0 to see if it intersects the x-axis and, if so, determine which side of f(x) = 0 is negative.
cedar kilnBOT
@sleek mantle Has your question been resolved?
crimson sedge
When finding the area under a curve how do I know when it's under the x axis or ...
f(0)
cedar kilnBOT
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jovial forum
Idk how to proceed and I'm starting to think I might have made a mistake
jovial forum
according to the master's theorem this should give T(n) = Theta(n^(log_2(3))
cedar kilnBOT
@jovial forum Has your question been resolved?
sharp flare
hey, how should i go about doing this? not really sure how the x range comes into play here and can only seem to get 1 answer (which i know is wrong)
open nacelle
whats the answer you got?
cedar kilnBOT
@sharp flare Has your question been resolved?
open nacelle
right
you have to produce other solutions from this first one
lets imagine you have sine of some angle
say like sin(pi/2)
and then you rotate the angle all the way around the circle back to its original point
open nacelle
say like sin(pi/2)
the sin of this NEW angle will be the same as this right?
sharp flare
yes
open nacelle
so what youve done, essentially, is add 2pi to this angle
this holds for any angle
if you add or subtract any multiple of 2pi to an angle, its sine and cosine will be the same
because all youre really doing is just rotating it around a full circle
sharp flare
okay
sharp flare
yes?
open nacelle
so whats x + 2pi
sharp flare
wouldnt this have infinite solutions
rotund jasper
uhm sry to disturb ur convo but isn't sin (x + pi) = -sinx
open nacelle
no
open nacelle
wouldnt this have infinite solutions
yes, but this is where the -1 <= x <= 10 comes in
rotund jasper
oh mb
open nacelle
uhm sry to disturb ur convo but isn't sin (x + pi) = -sinx
yes
this is right
once u edited
lolll
rotund jasper
yep
sharp flare
yes, but this is where the -1 <= x <= 10 comes in
how exactly do we use this
open nacelle
it just helps narrow down the range of what x can be
cause as you correctly pointed out, x + 2*pi*n is a solution for sin(x) = -0.7 for any integer n
but some of those integers n would cause x + 2*pi*n to be out of the range theyre asking for
sharp flare
okay
open nacelle
so you just need to find n values that remain in the range
and theres one step after that that we can talk about too
sharp flare
but wouldnt you still have infinite solutions like you cant check all of them to see if their in the range or not
open nacelle
well
x = -0.77539
x + 2*pi = 5.50778
if you add another 2*pi its gonna be greater than 10
so
its not in the range
sharp flare
ok
open nacelle
and if you subtract 2*pi from -0.77539 its gonna be less than -1
so nothing else of the form x + 2*pi*n works
sharp flare
so its the 3rd one then
cedar kilnBOT
sharp flare
k
cedar kilnBOT
✅
open nacelle
do you know the identity sin(x) = sin(pi - x)
sharp flare
it seems familiar
open nacelle
ok good
basically the intuition is
if you have an angle
and then you flip it horizontally across the y-axis
the y value doesnt change
so the sin doesnt change
sharp flare
ok
open nacelle
what youve done is just do pi - x
so sin(x) = sin(pi - x) for any x
does that give you another solution in the range -1 <= x <= 10?
sharp flare
yes
open nacelle
perf
sharp flare
ok thanks
open nacelle
so then you should have your answer
sharp flare
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regal minnow
whats 1+64323423634526346
cedar kilnBOT
whats 1+64323423634526346
slate lintel
regal minnow
help
anyone know anything abt rocekt sicence that can teach me something about it
<@&286206848099549185>
cedar kilnBOT
@regal minnow Has your question been resolved?
dire geode
anyone know anything abt rocekt sicence that can teach me something about it
Stop abusing help channels
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cedar kilnBOT
regal minnow
sorry
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heady mountain
Hi Im having trouble approaching this problem
cedar kilnBOT
Hi Im having trouble approaching this problem
heady mountain
so t = -pi/3 if you just plug the points into the original equation
so would it just be -pi/3 for each?
so ex. x(t) = cos(t-pi/3)
nvm this is wrong
dull oxide
You have t correct
heady mountain
You have t correct
ok but why is the substitution (t-pi/3) incorrect here
heady mountain
so
x'=-sin(t)
y'=cos(t)
z'=1
tropic oxide
those are x', y' and z', not x, y and z
dull oxide
The variable names are wrong, but the calculations are correct
heady mountain
yea sorry im lazy
slate lintel
at least say x=> -sin(t) then
dull oxide
We spent more time correcting you than it would take to write ' three times
dull oxide
Do you remember the formula for a 3D parametric line given a point on the line and it's "slope"?
dull oxide
Do you remember the formula for a 3D parametric line given a point on the line a...
To get the point on the line, and to get the direction, yes
heady mountain
Do you remember the formula for a 3D parametric line given a point on the line a...
x(t) = x'(t) + x(0) for example?
or am i missing smt
dull oxide
I kinda see what you are going for, but it's not quite right
$\vec{r}(t)=\vec{r}_0+\vec{v}\cdot t$
wraith daggerBOT
bonimy
