@placid trail Has your question been resolved?
#help-27
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wild vector
i'm not sure if this is exactly the way the person derived this but here's how i found the function
you basically want a function g that fulfils following conditions:
g(0) = 0 (obviously, putting in 0 should actually give you 0)
g(1)=1
g'(0) = 0 (slope at 0 should also be 0, this would make your function "smooth" around that point)
g'(1) = 1 (same here)
g''(0) = 0 (slope of the derivative should be 0 - you can imagine this making it even smoother at this point)
g''(1)=1 (^)
now you've got six conditions, therefore six unknowns. you're basically asking yourself: find the function g(x)=ax^5+bx^4+cx^3+dx^2+ex+f
that satisfies above conditions
try solving it and see the magic happen :)
devout snowBOT
@placid trail Has your question been resolved?
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restive river
Show the hypergeometric probability distribution for an experiment with:
n = 15, r=4, a=7
restive river
i understand how to do this using the the p(x) formula
but how many trials will there be
7 right?
including 0 of course
devout snowBOT
@restive river Has your question been resolved?
devout snowBOT
@restive river Has your question been resolved?
restive river
<@&286206848099549185>
devout snowBOT
@restive river Has your question been resolved?
supple knot
Show the hypergeometric probability distribution for an experiment with:
n = 15...
What do these parameters even mean
And show what exactly
faint zinc
Moreover, how would one "show" a distribution? Via the formula for the PDF?
A graph?
A table of cumulative probabilities?
devout snowBOT
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echo ember
devout snowBOT
granite arch
f(x) = x^(1/3)?
winter torrent
negative bases for noninteger exponents are really weird
echo ember
f(x) = x^(1/3)?
yes
granite arch
granite arch
granite arch
whats wrong with mine
No idea
does your calculator go into the negative x (it should)
granite arch
that's weird
granite arch
okay
devout snowBOT
granite arch
yw
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Available help channel!
wanton granite
can u help me with these two limits?
devout snowBOT
can u help me with these two limits?
finite briar
can u help me with these two limits?
!status
devout snowBOT
can u help me with these two limits?
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
show work
wanton granite
ok give me one sec
finite briar
👍
finite briar
lemme see
Instead of l'hopping directly, I think the substitution e^(1/x) = u can help
also make sure to change what the limit approaches to when u do so
wanton granite
I try with this and I let u know, give me 1 sec again
finite briar
👍
wanton granite
*than not then
wanton granite
aaah sorry
wanton granite
i said e^(1/x) = t
I do it again
finite briar
I do it again
and?
wanton granite
i'm doing the limit in this moment
finite briar
hmm
devout snowBOT
@wanton granite Has your question been resolved?
finite briar
looks correct to me
wanton granite
looks correct to me
and if it's 0- how can I do?
in this case i don't know where to start
crisp niche
in this case i don't know where to start
$\text{note that:}\\\lim_{x \to 0^{-}}e^{\frac{1}{x}}=e^{-\infty}=0$
woven radishBOT
Joanna Angel
devout snowBOT
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Available help channel!
old crane
Hi, i have this relation defined as:
$$ R = {((a,b), (c,d)) \in \mathbb{N}^2 \times \mathbb{N}^2 | ad = bc}$$
Now im trying to prove the transisitive property, which in essence is pretty easy:
$adcf=bcde$ and then remove c and d. But this can happen only if $c,d \neq 0$. how can i take care of these two cases where c=0 and d = 0?
devout snowBOT
Hi, i have this relation defined as:
$$ R = \{((a,b), (c,d)) \in \mathbb{N}^2 \t...
woven radishBOT
sup sup
graceful cosmos
You should end up with two equations, not one
Actually I guess you multiplied them together, huh
old crane
$$ad=bc \land cf=de \Rightarrow af=eb$$ multiply side by side
$$ adcf = bcde$$
woven radishBOT
sup sup
old crane
even if you factor out c in first eq, and replace it at the second equation, you still end up with 2 cases where b,d!=0
now to prove transistive i should also prove these 0 cases, but i have no idea how
graceful cosmos
Are we sure N includes 0?
old crane
ah sorry, initial problem says integers
shouldve written Z
wait
nevermind
damn i didnt think of that
says positive integers
graceful cosmos
Indeed, I think this is no longer transitive if we allow 0s here
old crane
yeah well im lucky to have proven only pozitive ints
thanks
gotta observe better next time
thanks again!
.close
sacred roost
hey, here im just supposed to find n such that x^6n = e, where e is x^28, x^56, ... and so on ryt
devout snowBOT
hey, here im just supposed to find n such that x^6n = e, where e is x^28, x^56, ...
sacred roost
so just n = 14
pseudo basin
could phrase it that way sure
sacred roost
thanks
rocky mural
if anyone could help
supple knot
if anyone could help
! occupied
devout snowBOT
if anyone could help
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
@sacred roost Has your question been resolved?
devout snowBOT
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Available help channel!
restive river
Hi
restive river
pls help!!
yall r smart so ye
TYSM!!!
- Tom and Jerry started walking at constant speeds from the same spot in opposite directions along a straight path. After they had walked 45 min, they were 5.25 km apart. Tom speed is 3 km/h. What was Jerry's speed?
Could anyone help? 😊
devout snowBOT
@restive river Has your question been resolved?
restive river
No
guys please help
Nvm :(
45 = 0.75 and 5.25/0.75 = 7
7-3 = 4km/h right
But is it sure if u get the 5.25/ 0.75 and like use it in diff numbers the same type of question, and minus with the tom’s speed, will it work with the diff type of question
fast stirrup
devout snowBOT
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fleet steppe
When $2700 is deposited, a bank offers two types of savings accounts:
Account A: After depositing $2700 the account earns 2.9% simple interest p.a.
Account B: After depositing $1650 the account earns 2.8% compound interest p.a.
After how many years will Account B have a greater balance than Account A?
I've got 1650(1.028)^t>2700+78.3t, but have no idea how the get the answer of t=50.5
soft nest
When $2700 is deposited, a bank offers two types of savings accounts:
Account A...
How did you set up the right side?
And you have the formula for compound vs simple mixed up
fleet steppe
How did you set up the right side?
Ah, right, 2700 is deposited into both
But then is the compound interest only earnt on the amount deposited above 1650 or on the 2700?
This question is confusing me
fleet steppe
And you have the formula for compound vs simple mixed up
Do I?
I'm pretty sure that compound is P(1+r/n)^tn, and that simple is P(1+rt)
The compound has r=0.028 and n=1, and simple has r=0.029
So it should be Acc B: P(1.028)^t and Acc A: P(1+0.029t)
soft nest
Do I?
I'm pretty sure that compound is P(1+r/n)^tn, and that simple is P(1+rt)
T...
Oh no I was mixing up simple and compound interest for compound and continuously compounded interest
fleet steppe
Ah okay
soft nest
So either way this is a logarithms question
fleet steppe
Wait, so for acc b the amount is 2700(1.028)^t?
2700(1.028)^t>2700(1+0.029t)
1.028^t>1+0.029t
...?
soft nest
Wait, so for acc b the amount is 2700(1.028)^t?
Is it not p=1650 ?
soft nest
What is the original problem?
fleet steppe
When $2700 is deposited, a bank offers two types of savings accounts:
Account A: After depositing $2700 the account earns 2.9% simple interest p.a.
Account B: After depositing $1650 the account earns 2.8% compound interest p.a.
After how many years will Account B have a greater balance than Account A?
soft nest
As it was presented to you?
fleet steppe
Yes
soft nest
And those are the formulas you have from class?
fleet steppe
I was helping a younger sibling answer the question
And then I got stumped too
But I assume they would’ve learnt it?
fleet steppe
At this point I don’t even know what the question is saying
Is P for acc B 1650 or 2700?
soft nest
I keep getting negative answers
fleet steppe
I tried plotting it on desmos, got nothing near 50.5
soft nest
I tried plotting it on desmos, got nothing near 50.5
Lol me too, -23.4
fleet steppe
Desmos didn’t pull through for me so I figured I’d ask here
soft nest
Desmos didn’t pull through for me so I figured I’d ask here
Ask your sibling if they can send a picture of the question because I think they transcribed it wrong
soft nest
That’s weird
fleet steppe
I know, right?
soft nest
Are we supposed to do something else with the difference between 2700 and 1650?
soft nest
Okay so 50.5 is a correct answer, now we need to figure out how that happened
fleet steppe
Okay so 50.5 is a correct answer, now we need to figure out how that happened
Yes
I'm not getting anything anywhere like that, though
soft nest
I'm not getting anything anywhere like that, though
I put it into excel and got it to work, idk how they got the 50.5 though
It’s 6654.15 in A and 6654.18 in B after 50.5 years
fleet steppe
Oh okay
soft nest
But that’s solving (2700/1650)=(1.028^t)/(1+0.029t) for t
fleet steppe
Can a year 10 student even solve that for t?
soft nest
Considering symbolab can’t solve it, I wouldn’t expect them to
soft nest
Is this for a pre Calc class?
fleet steppe
We have a diff schooling system
We don't have pre calc classes
Though we do start calc in year 11
soft nest
Tell your sibling to tell their teacher you solved it numerically and the solution is actually 50.493658
So the boom is wrong anyway ✨
forest mural
G'night!, I'm trying to show this limit is equal to 0. I see how the x^n factor dominates the equations and it goes to 0 but I can't show formally how given α>0 (when α≤0 it's inmediate) the limits is in fact 0
devout snowBOT
G'night!, I'm trying to show this limit is equal to 0. I see how the x^n factor ...
forest mural
elfin cargo
forest mural
So I first tried to use L'Hôpital's rule but that works only in domains that are intervals and this sequences of functions (actually the limit) is not such case , I also see the sequence is bounded from above by n^α but I don't see if that could help in any way.
If anyone has an idea or maybe throw a lemma that could share some light in this limit, I'll be here!
warm bolt
If anyone has an idea or maybe throw a lemma that could share some light in this...
i’m looking at it
forest mural
Should I resend this in the analysis channel?
warm bolt
sure , what math are you in?
warm bolt
just what class yeah
so i’m thinking this :
m^x approaches infinity if m>1 and approaches 0 if 0<m<1
forest mural
I think I didn't write the limit clearly
warm bolt
okay just let me know
forest mural
devout snowBOT
@forest mural Has your question been resolved?
warm bolt
i don’t got it
forest mural
mmmm, it's ok
supple knot
Are you supposed to use eps N proof?
forest mural
I tried working through that way and got stuck trying to find a bound to n^α given ε>0
But if you got an idea, let me know
<@&286206848099549185>
forest mural
Yeah, as I said earlier I see that clearly but haven't managed to formally show it
supple knot
Exponential grows faster than polynomial
You weren't proven this in class?
devout snowBOT
@forest mural Has your question been resolved?
devout snowBOT
devout snowBOT
Available help channel!
worthy girder
How would I go about proving something like this:
devout snowBOT
How would I go about proving something like this:
worthy girder
$\sum_{i=1}^nx_i^2 \leq (\sum_{i=1}^n|x_i|)^2$
woven radishBOT
Bob Goldham
worthy girder
yes that
basically a variation on the Cauchy-Schwarz inequation I think? But assuming I can't just say "according to the cauchy-schwarz inequation, qed"?
low holly
well you can note that $\sum_{i=1}^nx_i^2 = \sum_{i=1}^n|x_i|^2$, yes?
woven radishBOT
Tushar
low holly
then note that $(\sum_{i=1}^n|x_i|)^2$ contains each term on the RHS
woven radishBOT
Tushar
worthy girder
intuitively, every term in the left sum is also in the square of the right sum, but there could be more terms on the right side, so since all terms are positive RHS >= LHS
is that statement sufficient as a proof?
low holly
I mean you can prove it by induction on n, if you want
you can use Cauchy-Schwarz if you're allowed to use it
worthy girder
since this is effectively asking to prove the cauchy-schwarz inequation, I'm thinking taht's not allowed... can't do circular proofs
now for the somewhat less intuitive second part
to show that
$||x||_1 \leq \sqrt{n}||x||_2$
woven radishBOT
Bob Goldham
woven radishBOT
Bob Goldham
worthy girder
at which point I'm stuck... why would this always be true? for any x in R^n
devout snowBOT
@worthy girder Has your question been resolved?
low holly
might be worth trying induction then
supple knot
basically a variation on the Cauchy-Schwarz inequation I think? But assuming I c...
Why can't you just mimic the proof of Cauchy Schwarz
crisp niche
How would I go about proving something like this:
first at all:
$\text{Note that:}\\\sum_{k=1}^{n}a_{k}^{2}\sum_{k=1}^{n}b_{k}^{2}-\left( \sum_{k=1}^{n}a_{k}b_{k} \right)^{2}=\\=\sum_{k,j=1}^{n}a_{k}^{2}b_{j}^{2}-\sum_{k,j=1}^{n}a_{k}b_{k}a_{j}b_{j}=\\=\frac{1}{2}\sum_{k,j=1}^{n}\left( a_{k}b_{j}-b_{k}a_{j} \right)^{2}\ge0\text{ }\text{ thus:}\\\left( \sum_{k=1}^{n}a_{k}b_{k} \right)^{2}\le\sum_{k=1}^{n}a_{k}^{2}\sum_{k=1}^{n}b_{k}^{2}$
woven radishBOT
Joanna Angel
crisp niche
How would I go about proving something like this:
hence, consider the consequences.
devout snowBOT
@worthy girder Has your question been resolved?
lost laurel
the reminder on dividing 5^99 by 11
devout snowBOT
the reminder on dividing 5^99 by 11
lost laurel
how would I do this using the binomial theorem
stone stump
why do you want to use that
lost laurel
That's in my syllabus, modular arithmatic, which I think is an alternative isn't
lost laurel
I don't know modular arithmatic so I'd really rather use the binomial theorm for now
I tried expressing it as (5^n-1)^m
but I'm unable to do that
the closest I got is (5^3-2)^33
stone stump
well then good news because now you have a chance to learn modular arithmetic
lost laurel
have any resource suggestions for the same in that case?
stone stump
I dont see any particularly nice way to use the binomial theorem here
compute the first few results of 5^k mod 11
lost laurel
how would I do that? are there any good resources to learn modular arithmatic from scratch?
stone stump
what is 5^2 mod 11
stone stump
rephrased, what is the remainder of 5^2 when divided by 11
lost laurel
4
stone stump
check again
lost laurel
,3
stone stump
ok
so, now the nice thing about remainders
lets say we have two numbers a and b. and they have the remainders n and m
then a*b has the same remainder has n*m
so, for example, 25 and 5 have remainders 3 and 5 respectively
then 25*5 has the same remainder as 3*5. which has remainder 4
lost laurel
huh, that;s really neat
I'll try using that here, then, thanks!
so I was thinking I could rewrite it as (625)^33, 625mod11 is 9, so I would essentially have to find 9^33mod11, right?
lost laurel
did you actually calculate 625? cause thats not necessary
yes
stone stump
(sure its not a big number but still)
we want to find 5^4 mod 11
we previously saw that 5^2 mod 11 is 3
and that 5^3=5^2*5=3*5=4 mod 11
so now 5^4 = 5^3*5 = 4*5=20=9 mod 11
by replacing the 5^3 with 4
that is the strength of this modulo stuff. you never need to deal with big numbers
lost laurel
so the answer is 9? (that's 9 mod 11, right?)
stone stump
9 mod 11 is 9
or what do you mean
5^4 mod 11 is 9. like you previously had
can you tell me what 5^5 mod 11 is?
lost laurel
that would be 27mod 11, right?
lost laurel
then a\*b has the same remainder has n\*m
5^5 has the same reminder as 45 when divided by 11?
lost laurel
1
lost laurel
the reminder when 5^5 is divided by 11 is 1, and when 5 is dvided by 11 is 5, so 5mod11 which is 5?
lost laurel
25 mod 11 which is 3
stone stump
5^8 ?
lost laurel
15mod 11, which is 4
stone stump
do you notice something?
lost laurel
5^n mod 11 is periodic
tame pumice
you can use binomial here tho
by writing it as 5^4 * 5^95 works i think
stone stump
and then what
lost laurel
so the first time we get 4 is when n=3, so we can get an AP 99=3+(n-1)4, which gives me n=24
worthy girder
thats just some other way using mod
dividing with remainder is using mod
lost laurel
which would give me 5^99mod11=4
stone stump
oh wait you want to write 3125=3124+1
tame pumice
this is not the binomial theorem
(3124+1)^11 on expanding
stone stump
I suppose that works
tame pumice
yea
stone stump
not easy to see
tame pumice
well 5^5 is 1 mod 11 so yea
stone stump
which is much easier to see than knowing that 3124 is divisible by 11
tame pumice
well true it's hard to see
but that's probably the method they need if they had to use binomial
lost laurel
so the first time we get 4 is when n=3, so we can get an AP 99=3+(n-1)4, which g...
would the answer be 4 according to this series?
stone stump
oh I didnt actually check your result. no 5^99 mod 11 is not 4
lost laurel
why would using the AP series to determine its mod be wrong
stone stump
I am not quite sure what you even mean by this. 5^n is 1 when n is divisible by 4
so 5^96 is 1
and then from there go through the pattern
lost laurel
I mean 5^3mod11 is 4 as is 5^7mod11, and thus as is 5^11mod11
so any term satisfying 5^(3+4n) mod 11 is 4
stone stump
5^7 mod 11 is not 4
stone stump
wait wait sry. 5^n is 1 if n is divisible by 5
so 5^95 is 1
and then from there continue the pattern. at least thats how I like to think about it
lost laurel
I see, thanks . Thanks for all the help, its kind of late here, so I'll continue this problem tomorrow, thanks for the introduction to modular arithmatic @stone stump !
proven agate
this was really interesting
thank you @stone stump
stone stump
youre welcome
I want to note, there are of course smarter ways to do this
instead of just bruteforce calculating some stuff and then finding the pattern
but for now, this should be quite enough
opal falcon
1.05 min. is 1 minute and 5 seconds?
devout snowBOT
1.05 min. is 1 minute and 5 seconds?
devout snowBOT
@opal falcon Has your question been resolved?
opal falcon
?
devout snowBOT
@opal falcon Has your question been resolved?
fervent needle
devout snowBOT
fervent needle
I'm a bit confused:
If we take the base case of n = 1 the set of A = { 0 }
So therefore how can B be a proper subset of A, surely they'll be equal?
spice raft
$\subseteq$ doesn't mean proper subset
woven radishBOT
Daniel
fervent needle
oh shit im being silly
But even then, surely the cardinality wont always be = to the size of A?
fervent needle
But it's confined to k < n?
so surely
oh wait no it's just a coincidence that the base case is valid
Sorry, I think I'm confusing myselef
and you
spice raft
like that the base case fulfils $|{B\subseteq A_n|0\in B}|=|A_n|$?
woven radishBOT
Daniel
fervent needle
yes
spice raft
oh yeah that's just a coincidence
fervent needle
ah ok
spice raft
law of small numbers
fervent needle
So what is it saying, The cardinality of { B is a subset of A where 0 is a element of B } is 2^(n-1)
spice raft
yes
fervent needle
But surely the subset could just be { 0 } every time
spice raft
yeah that's one of them
but it's counting all of the subsets that have 0 in them
so e.g. with $A_2$ you have ${0}$ and ${0,1}$
woven radishBOT
Daniel
fervent needle
But what 'symbol' indicates that it's all subsets?
spice raft
that's the set builder notation
fervent needle
Yea, but isn't that just A subset not all?
spice raft
${B\subseteq A_n | 0\in B}$ should be read as "the set of all subsets of A such that 0 is in the subset"
woven radishBOT
Daniel
fervent needle
Ah ok
spice raft
like how we can write ${n\in\mathbb{N}|n<5}$ to mean "the set of all natural numbers that are less than 5"
woven radishBOT
Daniel
fervent needle
Ah that makes alot of sense
Would that be equivalent to the P(A) | 0 is an element of B?
or is that not valid notation
woven radishBOT
Daniel
spice raft
but you have the right idea
spice raft
that's honestly a better way to write it imo
fervent needle
That's the way I've seen
Ig they just wanted to be different
So now the proving part
fervent needle
Why it's 2^(n-1)?
spice raft
yes
fervent needle
Well I know that the power set has 2^n elements
And ig you've removed the pathway where you consider adding/removing 0
So you only have 2^(n-1) choices?
fervent needle
that's not really inductive though
yea that's what I was thinking
How would you explain it?
spice raft
my thought was that once we introduce a new element to $A_{n}$, we get twice the amount of subsets which have $0$, because for each subset of $A_n$ with a zero in it, we can now add another element to the subset
woven radishBOT
Daniel
spice raft
so each valid subset of $A_n$ gives rise to two valid subsets of $A_{n+1}$
woven radishBOT
Daniel
fervent needle
my thought was that once we introduce a new element to $A_{n}$, we get twice the...
Oh that makes sense
fervent needle
**Daniel**
Ah ok
spice raft
do you think you're able to take it from here?
fervent needle
I'm still struggling to see how we can prove this inductively, I've got the base case I think
Excuse my bad hand writing
spice raft
yes
that's good
now if it holds for $n$, then we know that there are $2^{n-1}$ valid subsets for $A_n$. Consider any valid subset $S$. Then both $S$ and $S\cup{n}$ are valid subsets for $A_{n+1}$
woven radishBOT
Daniel
fervent needle
Yes ok
spice raft
you also have to show that there aren't any additional subsets apart from these
but i think you're able to do that
fervent needle
but i think you're able to do that
Uhh
Well it's all the subsets that don't have n + the one that does
spice raft
yeah
so if it has $n$, then we can remove the $n$ and map it back to a valid subset of $A_n$, and if it doesn't have $n$ then we can just map it directly back to $A_n$
woven radishBOT
Daniel
spice raft
so all valid subsets of $A_{n+1}$ are of the form $S$ or $S\cup{n}$ where $S$ is a valid subset of $A_n$
woven radishBOT
Daniel
fervent needle
Ok that makes sense
I used p cause k is already being used
I just don't see how we'd prove the 2^p part
spice raft
I just don't see how we'd prove the 2^p part
tbh you will probably have to use words
idk maybe there is a cleverer way
you can directly prove it without induction pretty easily because there are $2^n$ subsets and to pick a subset without $0$ there are $2^{n-1}$ possibilities
woven radishBOT
Daniel
spice raft
it wouldn't cause it isn't inductive
fervent needle
yea but to get the 2^(p-1) + 2^p bit we are looking for
spice raft
i'd write
Consider a set $S\subseteq A_{n+1}$ that includes $0$. There are two cases. $S\subseteq A_n$. Since $A_n\subseteq A_{n+1}$, there are as many of these as there are subsets of $A_n$ that include $0$. Hence this case contributes $2^{n-1}$ subsets. If $S\not\subseteq A_n$, then $n\in S$. The set $S\setminus {n}\subseteq A_n$ and if $0\in S$, $0\in S\setminus{n}$. There are $2^{n-1}$ subsets of $A_n$ that include $0$, so this case also contributes $2^{n-1}$ subsets. In total there are $2^{n-1}+2^{n-1}=2^n$ subsets of $A_{n+1}$ that include $0$.
woven radishBOT
Daniel
spice raft
maybe this is too wordy
fervent needle
No that makes intuitive sense
I just dont see how you'd prove it by induction, your way makes it much easier
I might have to ask my professor
spice raft
that is inductive
spice raft
**Daniel**
this is inductive
fervent needle
this is inductive
wait is it
spice raft
i didn't explicitly say it but i implicitly assume there are $2^{n-1}$ valid subsets of $A_n$
woven radishBOT
Daniel
spice raft
this is the inductive hypothesis
fervent needle
ohhhh ok
God this is tricky
Thanks so much tho, I'll write that down and consult my professor about it
That actually makes sense
spice raft
you could also just say there is a bijection between subsets of $A_{n+1}$ that include zero and $n$ and subsets of $A_n$ that include zero, and that there is a bijection between subsets of $A_{n+1]$ that include zero and don't include $n$ and subsets of $A_n$ that include zero
woven radishBOT
Daniel
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spice raft
since every subset of $A_{n+1}$ that includes zero either includes $n$ or doesn't, there are $2^{n-1}+2^{n-1}$ such subsets
woven radishBOT
Daniel
spice raft
(assuming $A_n$ has $2^{n-1}$ valid subsets)
woven radishBOT
Daniel
fervent needle
hmm ok
spice raft
that's basically the same but faster
idk i think it's kinda dumb to prove this by induction
fervent needle
There was a similiar queston in a previous past paper, I'm assuming there's a method to this madness
So there must be some format to this that we haven't been taught
spice raft
yeah you can solve that in the exact same way as what i just described
fervent needle
exactly
spice raft
it's basically the same problem lol
fervent needle
yea
so tehre must be some framework
it cant be a coincidence
anyway, thanks so much - I really appreciate your help
fervent needle
yea for sure
I'll inform you if I hear anything from my professor, is it ok if I show them your work
spice raft
yeah absolutely
weak coyote
y=2x+3/(x-3)
devout snowBOT
y=2x+3/(x-3)
weak coyote
Q-3 to Q
So it will be one -one
I checked it through x1=x2
Is there any other method?
weak coyote
what does this mean
Domain and range
weak coyote
Yes
pseudo basin
and what other method do you want to know besides the definition of one to one
weak coyote
Any method which can be helpful or speedy
If any exists
Okay so nevermind, next task is to check whether onto/into?
So here Our Range is Q and for onto we want all elements image into Q/3
So as we can see for x=2 will not have any image
So it is not onto..right?
<@&286206848099549185>
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signal robin
We have $u_{n+1}=u_n+\frac{1}{n^2}$ and $v_{n+1}=\frac{\pi^2}{3}-u_{n+1}$
How to prove that the limit of $v_n-u_n=0$?
devout snowBOT
We have $u_{n+1}=u_n+\frac{1}{n^2}$ and $v_{n+1}=\frac{\pi^2}{3}-u_{n+1}$
How to...
woven radishBOT
Free Palestine 🇵🇸
pulsar sand
I'm not sure why you are asking this considering you basically asked the same question the other day.
signal robin
I'm not sure why you are asking this considering you basically asked the same qu...
i think this is easier
pulsar sand
Well, what is v_n - u_n
pulsar sand
And what is that difference
Surely you can subtract two things without making me do it for you
pulsar sand
That doesn't really matter you can just write the recursion in terms of u_n and v_n instead
pulsar sand
If you're going to take the limit as n tends to infinity, it doesn't matter if you calculate v_n+1 - u_n+1, or v_n -u_n, both are going to be the same under the limit
signal robin
If you're going to take the limit as n tends to infinity, it doesn't matter if y...
yh but
this works if u_n and v_n converge
here we dont know if it converges
we want to calculate the difference of both limits
to
OH YH
I GOT AN IDEA
pulsar sand
Shifting n by a finite number of terms does not matter either way.
signal robin
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orchid wasp
devout snowBOT
wooden wraith
Multiplying by 1.03 represents a 3% increase
orchid wasp
Multiplying by 1.03 represents a 3% increase
3 percent is 0.03 though
wooden wraith
The 1 is the 100% of the money you already had, and the .03 is the extra 3%
if you just multiplied by 0.03, you would be left with only 3% of what you started with
but you want to keep what you started with and increase it by 3%
orchid wasp
What about =2 shouldn’t it be *2?
woven radishBOT
tatpoj
wooden wraith
This equation says, Starting with 1 dollar, how many times do we have to multiply it by 1.03 until it doubles to 2
but the 1 in front can disappear, since multiplying by 1 doesn't do anything
orchid wasp
So do I always start with 1 if it doesn’t state anything?
wooden wraith
Might as well, yeah. If you started with 100 dollars it would look like $100(1.03)^t = 200$
woven radishBOT
tatpoj
wooden wraith
But dividing both sides by 100 would get you right back to (1.03)^t = 2
So really it doesn't matter how much money you start with
orchid wasp
So really it doesn't matter how much money you start with
Would those exponential formulas work with these too? Because originally I thought u were suppose to use those
Like a=(1+r)^t
wooden wraith
Like a=(1+r)^t
usually this is expressed as A=P(1+r)^t
and yeah, that is the same formula you're using here
the 1+r is how the 3% became 1.03
In this formula, P is the starting amount, like $1, and A is the ending amount, like $2
or $100 and $200, or whatever
orchid wasp
Yeah I see it now
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glad kraken
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glad kraken
How would you get this equation to an explicit formula?
orchid sierra
How would you get this equation to an explicit formula?
Recall the formula, see what you need
This is a Geometric Sequence
What does the explicit form of a geometric sequence require
glad kraken
The ratio and the a1
glad kraken
Using geometric formula or the recursive formula?
orchid sierra
Using the recursive form you were given
glad kraken
I’d rewrite it but a1 is already there
orchid sierra
Calculate a2
sand dove
Geometric sequence is an = constant*an-1
glad kraken
Geometric sequence??
I got a bit confused too since my paper’s title is Arithmetic
sand dove
Here we have an = an-1 * an-1
an-1 is not a constant, so this doesn't define a geometric sequence
orchid sierra
I got a bit confused too since my paper’s title is Arithmetic
Read the instructions for this question.
sand dove
However :
orchid sierra
an-1 is not a constant, so this doesn't define a geometric sequence
Ooh...you are right
glad kraken
Read the instructions for this question.
“Write the first five terms of the sequence”
orchid sierra
*“Write the first five terms of the sequence”*
So...it isn't asking you for an explicit formula then
Like you said it was
You assumed you needed one
glad kraken
So...it isn't asking you for an explicit formula then
I was asking for it because wouldn’t I need to turn it into an explicit format?
I’m going off what my professor said, my apologies
orchid sierra
I was asking for it because wouldn’t I need to turn it into an explicit format?
Not if it just wants terms in the sequence you have a recursive definition for
Which include a given term
glad kraken
I see, I’m super sorry about that
ripe plinth
Is this in use?
orchid sierra
I see, I’m super sorry about that
It's ok, I made a mistake too
sand dove
Is this in use?
Yes
sand dove
Is this in use?
#help-30 is free
orchid sierra
Use a1 to calculate a2;
Use a2 to calculate a3;
etc.
glad kraken
Would I plug in 2 for n then?
orchid sierra
Would I plug in 2 for n then?
Yeah
glad kraken
Why is it at the bottom? Like (n-1)
orchid sierra
I’m going off what my professor said, my apologies
You need to go off what the question asks, not necessarily what you think you remember is important
orchid sierra
Why is it at the bottom? Like (n-1)
That's called the index
It's the nth, or n-1 term in the sequence
You have the terms listed in the order they are generated
glad kraken
Would it affect anything? What does it do to the problem?
orchid sierra
$a_1 = 2\
a_2 = (a_{2-1})^2 = (a_1)^2 = 2^2 = 4\
a_3 = (a_{3-1})^2 = (a_2)^2 = 4^2\
etc.$
devout snowBOT
woven radishBOT
Melvin Eugene Punymier
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Available help channel!
delicate arch
How come combinations are used when solving the paths on a grid problem, instead of permutations? Here is an example:
devout snowBOT
How come combinations are used when solving the paths on a grid problem, instead...
delicate arch
From what I know, the order in permutations matter, so paths that use the same "directions" would be counted as different paths. But for combinatiosn, the order of the "directions" wouldn't matter, it would count as the same path?
I'm trying to logic it out but I just don't understand why combinations would be used instead if the order doesn't matter for this question
sand dove
You are looking for the number of possible sequences of 3 'Down's and 3 'Right's
Since we are not labelling them as Down1, Down2, etc...
The order in which the "downs" themselves appear does not matter
So as you put it, when the order between downs doesn't matter (same goes for the rights), we use combinations
You could look at the problem another way:
We have 6 arrows, and we have to put 3 of them in 'down' position, the rest of them will be on 'right' position
So we have to choose 3 down positions out of the 6 arrows
hexed ocean
can someone please help me with question b?
devout snowBOT
can someone please help me with question b?
hexed ocean
I'm not sure how you're supposed to find what P(x) is based on the info they give you
lost laurel
what is $\frac{(x-1)(x+3)}{x+3}$?
woven radishBOT
Why am. I here
hexed ocean
what is $\frac{(x-1)(x+3)}{x+3}$?
(x-1)
lost laurel
no, x+3 is a factor of p(x)
hexed ocean
right
lost laurel
so the reminder is?
hexed ocean
(x-1)(2x+5)
lost laurel
ok, so you have p(x)=Q(x)(x-1)(x+3)+(2x+5)(x-1)(x+3), right? use that
scenic light
Anyone can resolve this system?
lost laurel
$\frac{P(x)}{(x-1)(x-3)}= Q(x) +(2x+5)$, right?
woven radishBOT
Why am. I here
lost laurel
Anyone can resolve this system?
!occupied
devout snowBOT
Anyone can resolve this system?
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
hexed ocean
**Why am. I here**
yes
lost laurel
do you know the reminder theorm ?
ok, I've got to go now, sorry. Can't help anymore for now
split gate
what happens when log(x-4)^2
devout snowBOT
what happens when log(x-4)^2
vast wing
log(x-4)^2 is the same thing as 2log(x-4), if that helps
vast wing
it's a property of log
split gate
wouldnt it have to be log((x-4)^2)
winter patrol
but idk what happens when log(x-4) * log(x-4)
not much you could do with stuff like that
vast wing
wouldnt it have to be log((x-4)^2)
oh yeah, my apologies
winter patrol
notation is a bit ambiguous
split gate
is there a rule for when you have to use a parenthesis that includes everything wihtin a log?
cause ive seen it not happen a few times, my teacher says log 10
vast wing
i think that log(x-4) * log(x-4) would be (log(x-4))^2
winter patrol
ideally arguments of functions should be in parentheses
split gate
i think that log(x-4) * log(x-4) would be (log(x-4))^2
(log(x-4))^2 = log(x-4)^2 the way im looking at it
winter patrol
where there is potential for ambiguity, more parentheses should be used
devout snowBOT
winter patrol
base of the log when only log is written depends on context
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Available help channel!
orchid tulip
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orchid tulip
How do I approach this problem? the average squared distance keeps tripping me up. My logic was to find the center of mass (xbar, y,bar, zbar) which is (0,0,h/2) doing a triple integral of f(x,y,z) over dV divided by the volume of the cylinder. I then found the distance from the center of mass to (x0,y0,z0). But how do I take into account the average distance squared? Is my approach flawed?
devout snowBOT
@orchid tulip Has your question been resolved?
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@orchid tulip Has your question been resolved?
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@orchid tulip Has your question been resolved?
frail igloo
can anyone explain what this means for a 2x2 matrix?
devout snowBOT
can anyone explain what this means for a 2x2 matrix?
frail igloo
can anyone explain what this means for a 2x2 matrix?
so the eigen values here are x=2,2
and one eigenvector is [1, 2]
this fact is supposed to help us find the other eigenvector (i think) but i'm not sure if that's true.
does this just mean in this special case the second eigenvector can be taken as [1,0]?
devout snowBOT
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Available help channel!
fervent needle
This is my induction proof for the question above, I don’t know if my previous steps are valid or how to progress on the induction step
devout snowBOT
This is my induction proof for the question above, I don’t know if my previous s...
fervent needle
I went over this previously with someone else and we managed to get an induction proof but it was quite wordy, so I'm having another attempt using just mathematical notation
anyone?
<@&286206848099549185>
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@fervent needle Has your question been resolved?
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@fervent needle Has your question been resolved?
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.close
last egret
devout snowBOT
last egret
can anyone help with 14
lunar oar
if sinx is negative
and tanx is negative
would cosx be positive or negative?
given than $\tan{x}=\frac{\sin{x}}{\cos{x}}$
woven radishBOT
The Great D
lunar oar
@last egret
lunar oar
yes
last egret
but it says tan is less then 0
last egret
it would be negative
i see
so the answer is negative?
it will be in quadrant 3 then
last egret
oh i think i get it
lunar oar
which quadrants are sin negative in
lunar oar
in this case yes
lunar oar
or sin can be positive and cos negative
last egret
sun is negative
lunar oar
tan is negative in quadrants 2 and 4
last egret
i see
last egret
cos must be pod
lunar oar
we must be inquadrant 4
last egret
pos*
lunar oar
yes exactly
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last egret
.close
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Available help channel!
forest shuttle
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forest shuttle
<@&286206848099549185>
sage burrow
!status
devout snowBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
sage burrow
start such examples with calculating the first few elements (3,4,5,...) to see how the sequence behaves and to so if there is a pattern.
express x3 in terms of x1 and x2, express x4 in terms of x1 and x2, express x5 in terms of x1 and x2, i am sure you will see a pattern.
then proof by induction this pattern/formula where you express x_n in terms of x1 and x2.
then use this formula to calc for which element this formula gets a 0.
forest shuttle
1842/1843 * 19 for x3
1841/1842 * 97 for x4
1840/1841 * (1842/97) for x5
i accidentally flipped the fractions
sage burrow
1842/1843 * 19 for x3
1841/1842 * 97 for x4
1840/1841 * (1842/97) for x5
are you sure? x3=x1-1/x2 = (x1x2-1)/x2=(19x97-1)/97
do the same without using the numbers
forest shuttle
do the same without using the numbers
how
sage burrow
x3=(x1x2-1)/x2, now use this for x4 = x2-1/x3 -> ....
forest shuttle
idk how to use induction to prove this but basically xn will be something in the terms of x1 and x2 multiplied by (x1x2-x2-n+3)/ something else
unless i did something wrong
sage burrow
do it step by step. how does x4 look like?
forest shuttle
i did do it wrong
its just what i said without the -x2
so id just write it as a(x1x2-n+3)/b
sage burrow
i get a x4=x2(x1x2-2)/(x1x2-1)
sage burrow
now do this for x5.
forest shuttle
(x1x2-1)/x2 multiplied by (x1x2-3)/(x1x2-2)
for xn it would be (x1x2-n+2)/(x1x2-n+3) multiplied by something else
and since we know x1 and x2
we can set x1x2-n+2 = 0 and solve
97*19 + 2
ive looked up to solution for the thing now that ive solved it and their way is basically the same but much easier to understand
.close
serene rampart
what did I do wrong?
devout snowBOT
what did I do wrong?
devout snowBOT
@serene rampart Has your question been resolved?
sonic arrow
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sonic arrow
For 3x3 matrixes
Im unsure where they second line of working comes from?
A^2 + AB = I
A + B = A^-1
B = A^-1 - A
I think images are broken, thatswhat the image is
How is A^2 + AB = i the same as
A+B = A^-1
.close
shut haven
Find the sum of the following arithmetic series.
1+1 ¼+1 ½+ …+9¾
I did tn = t1 + d(n-1) and did tn = 4/4 + 1/4(39/4 - 4/4) (i expanded the improper fraction) and got 39 (the nth term)
then sn = n(t1 + tn) / 2, which was 39(1+ 93/4) / 2
i got 195 3/4 from that (39(1 + 9 3/4) / 2 into 39(10 3/4) / 2 into 390 3/4 / 2 into 195 3/4) (the sum of all terms)
but the answer key said S36 = 387/2 or 193 1/2
devout snowBOT
Find the sum of the following arithmetic series.
1+1 ¼+1 ½+ ...
hard dagger
do you know how many numbers are there
shut haven
do you know how many numbers are there
well i did tn = t1 + d(n-1) right? so that gives me the nth (idk i think i mixed tn and n not sure)
hard dagger
its correct but i think there is easier way (?) @-@
or your teacher whats you to do like that
shut haven
its correct but i think there is easier way (?) @-@
well im in gr 11 so we didnt learn anything after that (well i was missing for a week and learning this all now so...)
hard dagger
im grade 7 ;
shut haven
im grade 7 ;
damn huh???
woven radishBOT
Question mark
shut haven
you can change the form like this right
yeah i did
hard dagger
so
woven radishBOT
Question mark
shut haven
**Question mark**
ah alr
hard dagger
and then
shut haven
cancel out the denominator right
shut haven
what is 4+5+6..+39
each and every one individually or tn = t1 + d(n-1)???
hard dagger
did you learned a formula for arithmetic sequence?
shut haven
did you learned a formula for arithmetic sequence?
yeah tn = t1 + d(n-1)
hard dagger
nono, you know $1+2+3+...+n=\frac{n(n+1)}{2}$ ?
woven radishBOT
Question mark
hard dagger
do you know this @-@
shut haven
**Question mark**
thats for series right??
hard dagger
type "yes"
shut haven
yex
hard dagger
ha?
shut haven
yes lmao
shut haven
yeah
hard dagger
so what is 1+2+3+...+39
first
$\frac{4+5+6+7+......+39}{4}=\frac{(1+2+3+4+5+6+7+......+39)-(1+2+3)}{4}$
woven radishBOT
Question mark
shut haven
39 is n right?? so 39 ( 39 + 4) / 2?????
hard dagger
39 is n right?? so 39 ( 39 + 4) / 2?????
why is it +4 lol
+1 lahh
oh, shark whats to teach you
shut haven
bro nvm
hard dagger
er, wre you there
shut haven
.close
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restive river
Hey guys I have my semester 1 tests for math in like 2 days
restive river
I was wondering if I could get help here for functions and proofs
devout snowBOT
Available help channel!
restive river
I seriously don’t understand it and the terminology he uses in general 😢
restive river
There’s not really any fixed one I have in mind but I just get like overwhelmed with it
Like let me show you a question from revision village
restive river
Too hard dude
radiant anvil
already overwhelmed
restive river
already overwhelmed
Yes
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@restive river Has your question been resolved?
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@restive river Has your question been resolved?