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#help-41
amber waspBOT
Available help channel!
stray charm
This is an example in my ordinary differential equations textbook, I don't see why the example of a linearly independent set of vector-valued functions is actually linearly independent.
stray charm
First we're given 3 vectors with two entries, so that's kind of red flag number one for me, we only need two vectors to span the space traditionally, 2, I worked on it by hand:
And I don't have a row of free variables, is there a subtlety I'm missing here?
amber waspBOT
@stray charm Has your question been resolved?
amber waspBOT
@stray charm Has your question been resolved?
amber waspBOT
@stray charm Has your question been resolved?
amber waspBOT
@stray charm Has your question been resolved?
pastel goblet
keen pawn
so I'm trying to prove the followin
amber waspBOT
so I'm trying to prove the followin
bitter bluff
?
grizzled plume
so I'm trying to prove the followin
sometimes i wonder what class you go for
Analytic and Algebraic Topology of Locally Euclidean Metrications of Infinitely Differentiable Riemannian Manifolds dosent cut it
keen pawn
Analytic and Algebraic Topology of Locally Euclidean Metrications of Infinitely ...
It's a joke
grizzled plume
ik
untold zodiac
so I'm trying to prove the followin
The answer is optiom B
grizzled plume
Analytic and Algebraic Topology of Locally Euclidean Metrications of Infinitely ...
this was a joke as well
keen pawn
if $1<gcd(a,b)<a$, then $lcm(a,b) \cdot gcd(a,b)=ab$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
grizzled plume
okay
keen pawn
so I essentially want to prove that $lcm(a,b) = \frac{ab}{gcd(a,b)}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
amber waspBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
grizzled plume
fine
grizzled plume
okay np
untold zodiac
Consider a = product of primes p1, p2, p3........ with powers a1,a2,a3
Similarly consider b as product of p1 p2 p3...... with powers b1 b2 b3..........
Now notice that the hcf will be product of Pi 's and the power of Pi will be min{ai,bi}
Similarly
Lcm will be product of Pi 's and the power of Pi will be max{ai,bi}
Try to continue from here
keen pawn
Can't use prime factorisation
grizzled plume
Consider a = product of primes p1, p2, p3........ with powers a1,a2,a3
Similarl...
bro is thinking
keen pawn
So we have $a|lcm(a,b) \land b|lcm(a,b)$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
so that's hallf the defn
we now want to prove if $\exists d; a \mid d \land b \mid d$ then $d>lcm(a,b)$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
untold zodiac
Yea
Like here's an idea you can continue on
Let hcf be k
So a=km
b=kn
Where (m,n)=1
I don't even understand what u trying to do
grizzled plume
uhhh
let $a=x(gcd(a,b)), b=y(gcd(a,b)$ then $gcd(x,y)=1$. now $lcm(a,b)$ is clearly $xy*gcd(a,b)$ and the conclusion follows.
grizzled plume
...
untold zodiac
Yeah nice
keen pawn
The defn of lcm(a,b) is as follows :- $a\mid lcm(a,b) \land b\mid lcm(a,b)$ and if $e \mid lcm(a,b) \land e\mid lcm(a,b) \implies e \geq lcm(a,b)$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
so I want to prove $e \geq \frac{ab}{gcd(a,b)}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
amber waspBOT
@keen pawn Has your question been resolved?
keen pawn
**Death.cv**
I can't use the fact that every number is $kp$ where p is a prime either
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
Without proving it
keen pawn
Exactly, that's why I'm struggling so much lol
grizzled plume
then i cant do anything with no experience sorry
keen pawn
We first wish to prove that every number can be exprssed in the form $kp$; where $p$ is a prime
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
We wish to prove the following axiomatically :- $lcm(a,b)= \frac{ab}{gcd(a,b)}$.|
\
We use the following definition to do the same:- A number $n$ is said to be the lcm of two numbers $a,b$ if
(i) $a \mid n \land b\mid n$
\
(ii) $e \mid n \land e \mid b \implies e \geq lcm(a,b)$.
\
Proof:- It follows that $a \mid \frac{ab}{gcd(a,b)} \land b \mid \frac{ab}{gcd(a,b)}$ trivially.
\
\
We now wish to prove :- $a \mid e \land b\mid e \implies e \geq \frac{ab}{\gcd(a,b)}$.
\
$e=ak_1; e=bk_2$
\
we thus have $e^2 =ab k_1 k_2$
\
$\implies \frac{e^2}{\gcd(a,b)} = \frac{abk_1k_2}{gcd(a,b)}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
Compile Error! Click the 
reaction for more information.
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keen pawn
We wish to prove the following axiomatically :- $lcm(a,b)= \frac{ab}{gcd(a,b)}$.|
\
We use the following definition to do the same:- A number $n$ is said to be the lcm of two numbers $a,b$ if
(i) $a \mid n \land b\mid n$
\
(ii) $e \mid n \land e \mid b \implies e \geq lcm(a,b)$.
\
Proof:- It follows that $a \mid \frac{ab}{gcd(a,b)} \land b \mid \frac{ab}{gcd(a,b)}$ trivially.
\
\
We now wish to prove :- $a \mid e \land b\mid e \implies e \geq \frac{ab}{\gcd(a,b)}$.
\
$e=ak_1; e=bk_2$
\
we thus have $e^2 =ab k_1 k_2 ; k_1,k_2 \in \N$
\
$\implies \frac{e^2}{\gcd(a,b)} = \frac{abk_1k_2}{gcd(a,b)}$
\
\
We now propose the following lemma :- $e^2\geq e \geq \frac{e^}{\gcd(a,b)}$
From this we can conclude that
\
$e^2 \geq\frac{e^2}{\gcd(a,b)} = \frac{abk_1k_2}{\gcd(a,b)} \geq \frac{ab}{\gcd(a,b)}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
the issue is I have to prove this for $e$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
We wish to prove the following axiomatically :- $lcm(a,b)= \frac{ab}{gcd(a,b)}$.|
\
We use the following definition to do the same:- A number $n$ is said to be the lcm of two numbers $a,b$ if
(i) $a \mid n \land b\mid n$
\
(ii) $e \mid n \land e \mid b \implies e \geq lcm(a,b)$.
\
Proof:- It follows that $a \mid \frac{ab}{gcd(a,b)} \land b \mid \frac{ab}{gcd(a,b)}$ trivially.
\
\
We now wish to prove :- $a \mid e \land b\mid e \implies e \geq \frac{ab}{\gcd(a,b)}$.
\
$e=ak_1; e=bk_2$
\
we thus have $e^2 =ab k_1 k_2 ; k_1,k_2 \in \N$
\
$\implies \frac{e^2}{\gcd(a,b)} = \frac{abk_1k_2}{gcd(a,b)}$
\
\
It's evident that
\
$\frac{e^2}{\gcd(a,b)} \geq \frac{ab}{\gcd(a,b)}$
\
We now wish to show that $ \frac{e}{gcd(a,b)} \geq \frac{ab}{gcd(a,b)}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
We wish to prove the following axiomatically :- $lcm(a,b)= \frac{ab}{gcd(a,b)}$.|
\
We use the following definition to do the same:- A number $n$ is said to be the lcm of two numbers $a,b$ if
(i) $a \mid n \land b\mid n$
\
(ii) $e \mid n \land e \mid b \implies e \geq lcm(a,b)$.
\
Proof:- It follows that $a \mid \frac{ab}{gcd(a,b)} \land b \mid \frac{ab}{gcd(a,b)}$ trivially.
\
\
We now wish to prove :- $a \mid e \land b\mid e \implies e \geq \frac{ab}{\gcd(a,b)}$.
\
$e=ak_1; e=bk_2$
\
we thus have $e^2 =ab k_1 k_2 ; k_1,k_2 \in \N$
\
$\implies \frac{e^2}{\gcd(a,b)} = \frac{abk_1k_2}{gcd(a,b)}$
\
\
It's evident that
\
$\frac{e^2}{\gcd(a,b)} \geq \frac{ab}{\gcd(a,b)}$
\
\
so
\
$e^ 2\geq (ab) \implies e \geq \frac{ab}{e}$
\
But it's evident that $e \geq gcd(a,b)$.
\
so
$e \geq \frac{ab}{e} \geq \frac{ab}{\gcd(a,b)}$
<@&286206848099549185>
ocean widget
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
So We wish to prove the following axiomatically :- $lcm(a,b)= \frac{ab}{gcd(a,b)}$.|
\
We use the following definition to do the same:- A number $n$ is said to be the lcm of two numbers $a,b$ if
(i) $a \mid n \land b\mid n$
\
(ii) $e \mid n \land e \mid b \implies e \geq lcm(a,b)$.
\
Proof:- It follows that $a \mid \frac{ab}{gcd(a,b)} \land b \mid \frac{ab}{gcd(a,b)}$ trivially.
\
\
We now wish to prove :- $a \mid e \land b\mid e \implies e \geq \frac{ab}{\gcd(a,b)}$.
\
$e=ak_1; e=bk_2$
\
we thus have $e^2 =ab k_1 k_2 ; k_1,k_2 \in \N$
\
$\implies \frac{e^2}{\gcd(a,b)} = \frac{abk_1k_2}{gcd(a,b)}$
\
\
It's evident that
\
$\frac{e^2}{\gcd(a,b)} \geq \frac{ab}{\gcd(a,b)}$
\
\
so
\
$e^ 2\geq (ab) \implies e \geq \frac{ab}{e}$
\
But it's evident that $e \geq gcd(a,b)$.
\
so
$e \geq \frac{ab}{e} \geq \frac{ab}{\gcd(a,b)}$
\
Which proves that
$e \geq \frac{ab}{\gcd(a,b)}$
\
This proves $lcm(a,b) = \frac{ab}{\gcd(a,b)}$
\
\
I have to note this proof assumes $lcm(a,b) > gcd(a,b)$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
.close
amber waspBOT
amber waspBOT
Available help channel!
keen pawn
Show that if $e \mid a$ and $e \mid b$ then $e \mid gcd(a,b)$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
amber waspBOT
amber waspBOT
Available help channel!
lapis tapir
Hi I wanted to know if I can get a book recommendation for MAT examination for oxford entrance
normal dove
i have the solution but i dont get it
amber waspBOT
i have the solution but i dont get it
spare belfry
i have the solution but i dont get it
Alr, could you show us the given solution?
normal dove
i get the use of total probability but im not sure about writing those nCr's in this case
amber waspBOT
@normal dove Has your question been resolved?
keen pawn
Let $a,b \in Z$. Prove $\exists a,b \in\Z st gcd(a,b)= \alpha a+ \beta b$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
Proof:- Let $\gcd(a,b) = e$.
\
so $e=\frac{a}{\alpha} ; e = \frac{b}{\beta}$
OK, this doesn't help too much
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
so $\beta a - \alpha b=0$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
we also know that $2e = \frac{\beta a + \alpha b}{\alpha \beta}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
substituting $\beta a = \alpha b$ we find $2\mid \beta a + \alpha b$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
cunning birch
what
indigo cloud
you cant just randomly substitute stuff
indigo cloud
was there a hint given in the question?
cunning birch
instead of showing the gcd is a linear combination, think about showing that a certain linear combination is the gcd
keen pawn
I was actual postulating, given $a,b \in \Z$, any number $1\leq n \leq b$ can can be written as a linear combnation
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
cunning birch
only true if a,b are coprime
indigo cloud
clearly for example if a=2, b=4, then you cant write an odd number with them
try out some examples
see which numbers you can write
keen pawn
I mean we proved this in class using WOP, was hoping there would be another approach
indigo cloud
there is with the extended euclidean algorithm
also there was an approach recently discussed in #elementary-number-theory
and there are probably others
but eea and wop are the classic approaches
keen pawn
I guess I could attempt to prove that $\alpha(a)+ \beta b \mid gcd(a,b)$ and vice versa
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
indigo cloud
well one of those is not always true
keen pawn
yeah, the vice versa is for some \alpha, \beta
indigo cloud
but surely thats what you did in the wop proof
keen pawn
Hmm, yeah
$\gcd(a,b) \mid \alpha a+ \beta b$ is trivially true.
\
We now find the conditions , if any under which $\gcd(a,b) \mid \alpha a + \beta b$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
indigo cloud
you always say trivially. is it trivially true?
keen pawn
Yes
indigo cloud
prove it
keen pawn
$k= \frac{a\alpha + b \beta }{gcd(a,b)} = \frac{\alpha k_1gcd(a,b) + \beta k_2 gcd(a,b)}{gcd(a,b)}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
so $k= \alpha k_1 + \beta k_2$, where $k_1,k_2,\alpha, \beta \in \Z$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
thus $k \in \Z$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
indigo cloud
ok
keen pawn
now let $\frac {\gcd(a,b)}{\alpha a + \beta b}= k$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
so we have $\frac{1}{\alpha k_1 + \beta k_2} = k$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
or $1= \alpha k k_1 + \beta k k_2$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
chilly jackal
**Veni, vidi, perii is $\\R \- \\Q$**
you can use | for st, i think
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
you can use | for st, i think
yeah, I know
chilly jackal
but eea and wop are the classic approaches
whats wop?
keen pawn
Well ordering principal
chilly jackal
oh I havent seen that one
btw i believe this called bezout's identity/theorem or smthin?
indigo cloud
sketch of that proof: consider the set {alpha a + beta b > 0} and take the smallest one
indigo cloud
btw i believe this called bezout's identity/theorem or smthin?
yes
keen pawn
I think this approach isn't feasible
By the division algorithm, we have $gcd(a,b) = q(\alpha a + \beta b)+ r; 0\leq r < \alpha a + \beta b$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
amber waspBOT
amber waspBOT
✅
keen pawn
$gcd(a,b) = q(\alpha a + \beta b)+ r; 0\leq r < \alpha a + \beta b$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
Hmm, I can assume $r=0$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
gcd(a,b)=e. $a= gcd(a,b) k_1, b = gcd(a,b)k_2$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
so $1= q(\alpha k_1 + \beta k_2)$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
$1=\alpha (qk_1)+ \beta (qk_2)$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
Let $qk_1= P; qk_2= U$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
so $1=P\alpha + U \beta$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
This is rather messy
indigo cloud
**Veni, vidi, perii is $\\R \- \\Q$**
it doesnt help that you didnt stop here and think a second what this equality means
but honestly this will lead nowhere probably
keen pawn
**Veni, vidi, perii is $\\R \- \\Q$**
This means that $q$ isn't an integer , or $\alpha k_1 + \beta k_2=1$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
so $\alpha k_1 + \beta k_2=1$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
indigo cloud
ok so great under the assumption that r=0 (which is completely unfounded) you got this
keen pawn
I guess I'll just stick to the WOP proof
Let's consider the set $W={a\alpha + b \beta \mid \alpha ; \beta \in \Z\ \land \alpha a + \beta b >0}$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
WLOG we assume $a<b$ so $a \in W$
grizzled pagodaBOT
Veni, vidi, perii is $\R - \Q$
keen pawn
.close
amber waspBOT
amber waspBOT
Available help channel!
minor fable
Hey, can someone help me solve this integral, with integral by parts.
I got this below, but I am not sure if this is the right way to start
u = e^x^2
du = e^x^2 * 2x dx
v = x4/4
du = x^3 dx
warm glacier
Do this instead
minor fable
$$ u = e^x^2$$
grizzled pagodaBOT
Frank
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
minor fable
Integrate u = xe^x²
Is this another way or how did you come up with this?
I have only learned this formula for integration by parts
warm glacier
I have only learned this formula for integration by parts
Yes this formula only
here
u = x²
v = xe^x²
minor fable
so why is it u = x^2 and not e^x^2
warm glacier
Because u need something which will end up becoming zero
So u can stop using the formula infinite times
minor fable
ahh
warm glacier
Differentiating algebraic stuff will eventually become zero
minor fable
and what about the V, because in the lecture we only picked dv and then the integral of that
so if dv = x^3 dx, then v = x^4/4
warm glacier
so if dv = x^3 dx, then v = x^4/4
Yes so here
U integrate algebraic thing
It will keep on increasing its degree
Which is basically useless
minor fable
and how do you come up with a better placeholder
what is the intuition to choose one
minor fable
ILATE Rule?
No
we only got this integration by parts and partial integration this lecture
warm glacier
I- inverse trigonometric functions
L- logarithms
A- Algebraic
T- trigonometric
E- exponential
warm glacier
we only got this integration by parts and partial integration this lecture
This should be told in by parts only
Not a rule exactly
But a way to solve it easily
warm glacier
I- inverse trigonometric functions
L- logarithms
A- Algebraic
T- trigonometric
E...
For choosing u
This should be followed
minor fable
so it is like an order?
the first thing I see right now is exponential and that is the x^3 right?
minor fable
Wtf
warm glacier
Yes
minor fable
Yes
Alright so I go from left to right and the first thing is algebraic right
x^3
no inverse trig and no log
minor fable
so my dv should be x^3?
warm glacier
Other way
minor fable
how did you come up with xe^x^2
warm glacier
how did you come up with xe^x^2
Because u can't integrate e^x²
U choose xe^x²
So u can put t=x²
And rest is x² which is algebraic
minor fable
I understand everything but I don't get why we choose xe^x^2
minor fable
yes
warm glacier
So u put xe^x²
So u can use substitution for that part of integration
t=x²
So the integral becomes easier
amber waspBOT
amber waspBOT
Available help channel!
split sail
ABC is a triangle AD is a median and E. is the mid point of AD. BE is joined and produced to intersect AC in a point F. Prove that
AF = 1/3 AC
split sail
!status
amber waspBOT
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
inland pulsar
Draw diagram
tender cipher
Think in similar triangles
If there's a line parallel to a side of a triangle which cuts one side in half it must also cut other side in halg
amber waspBOT
@split sail Has your question been resolved?
split sail
What
tender cipher
Say you have a triangle ABX
ABC*
If we have a line DE so that DE parallel to BC
And D is midpoint of AB
What does it tell about E
I think you know about the various congruence criteria for triangles
Side angle side
split sail
E is the midpoint of AC
tender cipher
Yep
So if we have a line parallel to BF beginning at D interesting FC
It will intersect FC at its midpoint
split sail
I think we have to use midpoint theorem twice
hollow ginkgo
Menelaus theorem for triangle ACD and BF
tender cipher
I think we have to use midpoint theorem twice
Yup
You can see two triangles forming
tender cipher
So if we have a line parallel to BF beginning at D interesting FC
With this
Which break the line AC into 3 equal parts
tender cipher
If there's a line parallel to a side of a triangle which cuts one side in half i...
This is the crux
Proving that seems tricky
But the idea is definitely congruence
tender cipher
Because of the Side angle side congruence criteria, we can talk about the other side based on just two sides and an angle
Isn't there also angle side angle
Oh yea
Based on that
You can immediately see that if there's a parallel line, we get a smaller inner triangle congruent to the larger outer triangle
split sail
I think what you're talking about is similarilty
tender cipher
If a parallel line is there with it intersecting one side on midpoint
Pretty sure ASA is congruence
split sail
i mean yes, but congruent triangles are identical
tender cipher
Oh right
Yea but
We have an exact ratio here
Yea yea ur right
But u get why the fact is true right
split sail
i have got it
tender cipher
So now you just need a segment starting at D parallel to BF
amber waspBOT
amber waspBOT
Available help channel!
worthy aurora
Hello I need help
amber waspBOT
Hello I need help
worthy aurora
I will send some photos in a second
It is about inequalities
I was solving a double inequality
and this is what x is a member of
if I did everything correctly
I am supposed to combine those intervals together
I don't want the answer outright
I want to know how to think
Me and a fellow course-goer tried to describe it as "the largest value in one interval is restricted by the smaller larger value in the other" and "the smallest value in one interval is restricted by the larger smaller value in the other."
But that doesn't seem to be correct
rocky vapor
u could try draw a number line?
worthy aurora
that's an idea
wait no that's not an idea
how would i do that for a double inequality
worthy aurora
Me and a fellow course-goer tried to describe it as "the largest value in one in...
Using this I got x is a member of (-9, -1) U (1, 7]
rocky vapor
how would i do that for a double inequality
Just draw 2 lines and then find the common area (intersection) between them
smthing like this (hope i am correct)
the 1st shows (-9,-1) U (1 , inf)
2nd shows (-inf , -3] U (-1 , 7)
and just find the common region between the 2
worthy aurora
and the common region is (9, -3] U (1, 7]
rocky vapor
ok apparently its 7] , so that circle should be coloured
rocky vapor
and yeah ur answer seems good , according to number line
worthy aurora
thank you for sharing
my original answer was wrong
I got -1) instead of -3]
still trying to understand what was wrong with my logic
whatever this method works
when I have a double in equality
is it always the upside down between the two intervals for the individual inequalities?
rocky vapor
is it always the upside down between the two intervals for the individual inequa...
yeah should be , cuz x needs to satisfy both the inequalitys
and "upside down U" (intersection) basically gives us the common between the 2 regions
rocky vapor
yes
worthy aurora
ok
rocky vapor
"U" is union
"upside U" is intersection
worthy aurora
thanks I knew about union because it is the same in my language
thanks for helping me
I am happy now
!!!
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gilded wedge
This is a mid term assignment that I'm stuck on. I'm stuck on most questions but I need help with this
fierce tiger
This is a mid term assignment that I'm stuck on. I'm stuck on most questions but...
the first thing that came to mind is the inscribed angle theorem
the measure of an inscribed angle is 1/2 the measure of the central angle
gilded wedge
the measure of an inscribed angle is 1/2 the measure of the central angle
I'm still not understanding....
fierce tiger
I'm still not understanding....
do you know what the theorem is?
gilded wedge
I thought of the cyclic quadilateral at first but since not all four vertices are touching the circumference, I gave up on it
fierce tiger
what happens if we draw in line OB?
we get 2 isosceles triangles
and then you can use angle measurements to solve
gilded wedge
So will the angle of 140 be 70 on each triangle or will it stay as 140
fierce tiger
yep 70 on each
gilded wedge
tysm
fierce tiger
(180 - 70)/2 =55 then 55x2 =110
gilded wedge
but how do I explain??
fierce tiger
the same process we just did
draw in line OB, splitting angle AOC into two 70 degree angles
then use isosceles triangles to solve
fierce tiger
OA and OD are both radii so triangle OAD must be an isosceles
angle ODA = angle OAD
for the first one
angle OAD = angle CAD
i’ve gotta go sorry someone else can help you with the rest
gilded wedge
i’ve gotta go sorry someone else can help you with the rest
Tyy anywaysss
I would appreciate assistance to anyone who is free
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tough mica
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tough mica
How to write the matrix M(f).?
Tal que = such that
The standard basis vectors
And we get a system
Right?
cloud scarab
Well, in case of standard basis vectors it's easy
You just write the output in columns
cloud scarab
No, I meant (c)
In c you just write outputs in columns
In (b) you write coefficients of outputs in rows
cloud scarab
What about d)
In (d) you should write inputs and outputs in rows, then row reduce, take the right 3 columns as a separate matrix and transpose it
tough mica
Wow
cloud scarab
That's the general method
But in case of (d) we can note that the inputs are slightly scaled standard basis vectors
tough mica
,w rref {{2,0,0,4,2,2},{0,4,0,1,1,1},{0,0,3,0,0,-1}}
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tough mica
But in case of (d) we can note that the inputs are slightly scaled standard basi...
What about it
tough mica
What do I do with this result btw
cloud scarab
Transpose this and you're done
grizzled pagodaBOT
cloud scarab
What about it
Well if e1, e2, e3 are standard basis vectors, then (d) says that
f(2 e1) = (4,2,2)
f(4 e2) = (1,1,1)
f(3 e3) = (0,0,-1)
Due to linearity this is equivalent to
f(e1) = (2,1,1)
f(e2) = (1/4,1/4,1/4)
f(e3) = (0,0,-1/3)
Now we're in a situation similar to (c) and we just write the outputs in columns
tough mica
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tough mica
We know f(1,0,0)=(1,2,-1)
f(0,1,0)=(3,1,2)
f(0,0,1)=(-2,0,-2)
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cloud scarab
Do you know how to multiply a vector by a matrix?
tough mica
right
I forgot linear transformation can be expressed as matrix multiplication f(x) =Ax
,w {{1,2,-1},{3,1,2},{-2,0,-2}} * {{1},{2},{1}}
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tough mica
a + c = 0 ==> a = -c
b - c = 0 ==> b = c
(a,b,c)=(-c,c,c)
ker(f)=<(-1,1,1)>
,w rref {{1,2,-1},{3,1,2},{-2,0,-2}}^T
the image contains two vectors
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tough mica
Im(f)=<(1,2,-1),(3,1,2)>
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edgy cave
I just have a slight question on how to input this into my online homework portal I think I have successfully done the taylor polynomial but I do not understand why the error bound formula is incorrect
edgy cave
The answer is not correct on this version because of a previous mistake but I am just scared I will not give them the answer that they want
they want it to three decimal points and 10^-1
I just want to make sure first of all that I have correctly determined the maximum possible size of the error
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light ore
Is there a way to make the bot do the calculation for me when I say to substitute for x?
Like ex
x³+5x²-x, and I say that x=1,
coral wigeon
,w x^3+5x^2-x, x = 1
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meager gale
$\int \frac{3}{xlnx} ,dx$
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Sho
shadow stump
try it and find out
obsidian helm
Here
obsidian helm
$1000 into an account with 3.5 annual interest. How much if compounded semi-annually, quarterly, monthly, weekly, continuously.
Algebra 2 Honors
pliant copper
Algebra 2 Honors
you learned a formula for this in class
find your compound interest formula
pliant copper
wheres the 3 coming from
obsidian helm
The answer is around 1030. I got around 1035
pliant copper
,calc 1000 * (1+0.03/2)^2
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Result:
1030.225
seems to be correct
obsidian helm
^2?
pliant copper
its ^nt, t = 1 (year) and n is number of times per year (which is 2 for semiannually)
2 * 1 = 2
pliant copper
what
notably this formula works for the first 4 parts, but you need to switch formulas for continuously compounding interest
obsidian helm
F(x)=-3^x+1 -2. Identify 2 points on the line. The first one if 0,-5 but how is the second one -1,-3
pliant copper
do you mean
obsidian helm
(1,-11)? Yes
pliant copper
$f(x) =-3^{x+1}-2$ or $f(x) =(-3)^{x+1}-2$
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obsidian helm
Right one with no parenthesis
pliant copper
the right one with no parenthesis is the left one
obsidian helm
Oh sorry
obsidian helm
Where does -1 come from?
pliant copper
(1,-11)? Yes
this is also a point on the line
pliant copper
Where does -1 come from?
you said the second point is (-1, -3)
therefore x = -1 to plug in, and you get -3 out
so its a valid point on the line
pliant copper
I don't know what "original" means, but for a line every input had an output
so both your new one and the previous two are all points on the line
there are infinite points on the line
@obsidian helm
obsidian helm
The formula said use 0,1 to find the first set and 1,b for the second
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fresh ocean
at maximum height, the clip will have 0 velocity
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fresh ocean
it leaves your hand at t=0 ...
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,rotate
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i’m in the bathroom
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quick ridge
😹😹
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quick ridge
nah gotta stay doing math
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knief
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quick ridge
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in the numerator
take it out of the limit
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quick ridge
$2\lim_{x \to -1} \frac{\frac{1+x}{x}}{x+1}$
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knief
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knief
quick ridge
👍🏻
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quick ridge
flabbergasted
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**knief**
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quick ridge
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bruh
verbal lark
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quick ridge
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and it’s x - c in denominator
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quick ridge
not x - g(c)
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quick ridge
try reading
quick ridge
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yes
verbal lark
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quick ridge
yuh
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quick ridge
1/x + 1 = 1/x + x/x = (1+x)/x
^
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quick ridge
which cancels with the x+1
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quick ridge
it’s just multiplied by the limit
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quick ridge
which
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quick ridge
bruh
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quick ridge
your teacher is a moron
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quick ridge
3 - 1.5 ≠ 4.5
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quick ridge
that’s where you messed up
then you did the slope as 3 though
in the next line
when it was 1.5
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quick ridge
it was inconsistent either way
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quick ridge
you said the slope was 4.5 then used 3 in the next line
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quick ridge
the correct slope is 1.5
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quick ridge
then
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quick ridge
you’re welcome
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spark iris
if a and b are consecutive even numbers c=b/a-a/b give the value of c in terms of a and explain that this will always be even/even
spark iris
put values of a and b in this form then you will get the ans
so c=4x-4/x^2-2x
i was kinda confused because when i put a as 2x and b as 2x+2 it simplifes to 2x+1/x^2+x and 2x+1 if a is even would be odd
thorn monolith
If this is what you got after solving the take things common and make the equations in terms of (x-2) which is a.
thorn monolith
so c=4x-4/x^2-2x
here is this in terms of a (2a/b.a) since b-2=a so b=a+2. so its {2a/(a+2).a}
spark iris
i was kinda confused because when i put a as 2x and b as 2x+2 it simplifes to 2x...
@thorn monolith ?
thorn monolith
i was kinda confused because when i put a as 2x and b as 2x+2 it simplifes to 2x...
I dont know but maybe then simplfication maybe wrong or something.
spark iris
I dont know but maybe then simplfication maybe wrong or something.
i used a calculator to double check and it got the same result
thorn monolith
I just simplified and i got this 2(2x+1) whole divided by x(2x+2)
in terms of a it's a/2
I think I am right. and done it correctly
spark iris
hmm ill try to redo it and see
i end up getting 8n+4/4n2+4n
and it simplfies to my answer before
spark iris
I just simplified and i got this 2(2x+1) whole divided by x(2x+2)
can i see your working?
thorn monolith
sending photo from laptop would be difficult but I ill tell you the steps
we have a=2x and b=2x+2 and we need to find b/a-a/b
this after CD
after opening squares I got this which then simplifies to my result
thorn monolith
8x+4
take 4 common dude
divide it then that 2 from denominter will divide 4 by 2 and we get
2(2x+1) whole divided by x(2x+2)
spark iris
but then its not fully simplfied cause 4x+2/2x^2+2x?
thorn monolith
How 4x+2 when you take 4 as common its 2x+1
spark iris
take common 2
thorn monolith
take 4 common
this is what I got at final and when simplifying it to a i get c=a/2
spark iris
take 4 common
expandd
thorn monolith
4
thorn monolith
its' just same only thing is You have multiplied it and I have taken it common
spark iris
its' just same only thing is You have multiplied it and I have taken it common
ye its the same but then why is 2x+1/x^2+x a possibility but isnt even/even
thorn monolith
ye its the same but then why is 2x+1/x^2+x a possibility but isnt even/even
Have you simplified this in terms of a?
thorn monolith
I got a/2
directly
But still doesnt makes sence cause since a=2x. so a/2 is x. Which can be odd.
thorn monolith
ops the ans is just there
instead of puting a=2x or something we directly get c= 4(a+1) whole divided by a(a+2)
and 4 multiply by any number is even and since a is even so a+2 that is b is also even so product of 2 even numbers is even. so yeah c is even/even.
wasn't it easy? we went on putting it as a=2x. but directly put b=a+2 and we get the answer
spark iris
and 4 multiply by any number is even and since a is even so a+2 that is b is als...
i understand that there are answers for like b=a+2 my question is because a can be 2x and b can be 2x+2 but when we simplify it it becomes odd
thorn monolith
i understand that there are answers for like b=a+2 my question is because a can ...
yeah it becomes odd now I dont know why but maybe we are missing something there. But anyway we got the answer
spark iris
Yeah ig,
spark iris
yeah it becomes odd now I dont know why but maybe we are missing something there...
i just got one more question if ur free if not i can wait for somebody else
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pseudo sail
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pseudo sail
Can someone explain this to me why this statement is false
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pseudo sail
Why |omega| is 45!/(6!*39!)
It says we will bet on sequence of numbers. Doesnt that make order important
stoic ivy
Can someone explain this to me why this statement is false
Laplace experiment is defined as?
I’m unfamiliar with that term
Also I’ve never seen intro probability taught like this I feel like it is just going to lead to confusion to introduce the concept of probability spaces without even invoking the idea of a measure
Just seems very unnecessary
Just a bit of a jab at your teacher’s pedagogy
stoic ivy
It says we will bet on sequence of numbers. Doesnt that make order important
So in the case of a lottery I believe that the order of numbers drawn isn’t relevant
I think you only need to worry about the values
So you would use the combination not permutation so 45 C 6
Or what you have written there
That’s the total number of possible combinations of numbers that could be drawn hence the sample space (ambient space)
Sample space is the term that I see in most textbooks
But I suppose maybe ambient space is used in other countries
In fact I’m pretty sure that ambient space is referring to something vastly more general so sample space ought to be preferred
pseudo sail
I think you only need to worry about the values
Ohhhh now i seee
pseudo sail
Sample space is the term that I see in most textbooks
Yeahh it also confused me too
pseudo sail
Laplace experiment is defined as?
Is it false or true
stoic ivy
I don’t know what a laplace experiment is
So I can’t help you unless you define it for me
pseudo sail
Fair experiment isnt it
pseudo sail
Experiment with finite number of result and all of them have same probability
I think something like this written on my lecture notes
stoic ivy
The power set of {1,2,3} is {{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3},{}}
Does that make sense first off
pseudo sail
Yeahhh
pseudo sail
So all the elements of this have same probability?
pseudo sail
Also i forgot to ask
pseudo sail
Does empty element in the end also have probability?
stoic ivy
We would need every event to have a likelihood of 1/8
But that’s not the case
So it can’t be laplace experiment
pseudo sail
Does empty element in the end also have probability?
Like in this case 1/8 ?
pseudo sail
So it can’t be laplace experiment
I see
stoic ivy
Probability is defined on all the subsets
The empty Is also a subset
So it must have a probability equal to the rest to be a Laplace experiment
At least I assume
I mean maybe Laplace experiments specifically exclude the empty and full sets
Idk
pseudo sail
Thats bit confusing isnt it i thought empty elements will always have 0 probability
stoic ivy
Again I’ve never heard that term
stoic ivy
That’s a good idea
Best of luck
This seems like an awful intro to probability 😦
Just too much confusing stuff for a high schooler
Even for a first year undergraduate
pseudo sail
Im not high schooler
stoic ivy
You don’t need to know about probability spaces
Oh you haven’t updated your tag
Haha sorry
pseudo sail
First year undergraduate
stoic ivy
It still says pre undergraduate
pseudo sail
Ohh didnt notice that. I will change that later
stoic ivy
Either way I think for an intro course this is not needed
But it’s minor
Best of luck
pseudo sail
Thanks again
stoic ivy
Np
pseudo sail
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split sail
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split sail
Which option shall be correct?
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patent raptor
First year undergraduate
Ohhh krass
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royal parcel
im having trouble understanding why the remark is true, could anyone help?
indigo cloud
all powers g^k are elements with (g^k, m)=1
and the first phi(m) of those powers are all different
royal parcel
how do we know that (g^k,m)=1?
indigo cloud
but you also know that there are precisely phi(m) many numbers a with (a,m)=1
royal parcel
or how do we know that (g,m)=1, i guess this will imply (g^k,m)=1
indigo cloud
g has an order mod m
thats only possible if its invertible
which is equivalent to (g,m)=1
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ionic bison
this is half math half chem but i figured i’d ask
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this is half math half chem but i figured i’d ask
ionic bison
I have a problem like this: An example would be:
The element magnesium consists of three stable isotopes
Given the following masses and abundances, calculate the mass of the third isotope.
Mg-25 mass = 24.9858 abundance = 10.0%
Mg-26 mass = 25.9826 abundance = 11.0%
Mg-?? mass = ?? abundance = ??
but with no given element
just 3 isotopes, 2 with abundance and mass, one with no info
i can find abundance but i also need to find mass of isotope 3 AND the most likely element is this even possible?
white sundial
Only if you are also given the molar mass of Mg. That way you can solve for the mass of your missing third isotope.
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flint ember
Hi, how can we approach this graph theory problem? I am trying a proof by contradiction but I feel like I'm not seeing the bigger picture for how to proceed
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abstract canopy
usually hall's matching theorem is proven with induction on the number of vertices
im a bit confused by this context, but try taking some number of vertices in X and adding one to them and see what happens
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flint ember
im a bit confused by this context, but try taking some number of vertices in X a...
what do you mean by "adding one"?
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flint ember
.reopen
abstract canopy
what do you mean by "adding one"?
so say you have a matching for some set of n vertices, then if you add another vertex show that you still have a matching
timid pulsar
Let G be a cyclic group of order n, do we have G admits exactly d cyclic subgroups with d being the number of divisors of n ?
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Let G be a cyclic group of order n, do we have G admits exactly d cyclic subgrou...
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wraith violet
I’m unsure of how I went from the first step to the second step
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pure heron
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Chill my man 👨