#help-10
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ok
so what does that even tell us
c = -a b =-2a
av1 -2av2 - av3 = 0 oooh
but how do we now from this if its linearly dependent
bruh what
its dependent if a b and c
are 0
we dont know if it is
so how can u tell that
it's INdependent if a b c are 0
..
we're not solving for a
we are proving the solution exist
the solution doesn't matter that much
for example let a = 1
then you get one solution
a = 2 is another solution
etc
this is killing my braincells
💀
my prof is solving for c1 c2 c3
to figure out if its depedent or not
you see?
scalar triple product != 0
the general solution is -2t,0,t
-2,0,1 is just one of the solution
again what's important is that a solution exist
ik but i didnt get
where she got that from
like how can you tell if a solution exist
if you dont know the number
thats what i mean
you don't know the value of x and y though
we need to know if the values are all 0 or if 1 of them is not 0
to determing if dependent
or independent
bro
correct?
they are not mutually exclusive
every system have the trivial solution
of all 0
s
we want to find if there are non trivial ones
btw I gtg
alr man u tried ur best
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How do I do this one?
@errant crane Has your question been resolved?
loss in ke = gain in potential energy
whats 2 + 2 ?
22
4
The formula would be K.E= kq1q2/r, take r on one side and put the values
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is the first one a bijection? I am confused . since all of the negative integers would map to -2 which isnt in the codomain
even if you ignore the issue with the -2
it doesnt map to any positive even integers so its not surjective anyway
If it only maps to -2 for negative integers then its also not injective
shouldnt here be a x<0 ?
it says x>=0 so the -2 is basically reserved for (else do this)
which would only be x<0
oh yeah thank you
what about the third one
From what i figured out it maps between 0 and 1 only
@latent walrus I need your guidance
is it also not a bijection?
it doesnt map to any negatives or even 0 so it cant be a bijection
as
nvm
real numbers
but somehow the question is asking
which one of these is not a bijection
the first one isnt even a valid function so it may not count that
makes sense
thank you AZ empty set!
is there a way i could give you some sort of points?
also how do i close thsi channel
.close
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.reopen
How do i calculate and get the 0.12% between this two prices ? 🤔 most easy way
(changed_price-original_price)/original_price
when i do * 100 giving me then 0.1126307320997586
meaning it's exactly 0.112 % right ?
Thx ❤️
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how would i do b?
,rotate
i gotchu. lemme see what I can do for u.
alright thanks!
2 squares?
Krispeh
i got to 4integrate(sec theta tan^2 theta)
no idea how to continue
isnt dx 2secutanu?
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Hey guys can anyone help me please?
-
How do I work out on a calculator? what is 2/3 thirds of [X meaning the total number].
-
how do I work out what is X% of amount.
-
how to I Calculate my currency to another at the BUY rate. and then how do I Calculate the buy / SELL. of the Foreign currency where I to SELL it. If you can give me an Example for each with the icons on a calculator. Thank you very much.
if you want 2/3 of x then do 2/3 multiplied by x
if you want x% then multiply by x/100
for 3 id assume its just the above two but youd have to see the exchange rates and calculate it
cheers how do I do the 2/3 on the calculator
i see
- 2/3 of x will be (2÷3)*x
(2÷3)*x whats the *
the *?
yep
thats multiplication, its the x above the -
so times
yeah
is / ÷ on a calculator on all sums.
it can always be used as a replacement for it yes
/ generally means fraction, but you can consider it ÷
they do the same thing
what do you mean on all sums?
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Trying to write a sci-fi novel, and I'm having trouble with some math. Consider a galaxy 100,000 light-years in diameter. How fast would one need to go in order to traverse the entire length in one week? (This would likely be the max speed of the warp in my book)
,w light year
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Yo can someone help rq
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'm told it's wrong
4. I got an answer and would like my work checked
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6. None of the above
Show your work, and if possible, explain where you are stuck.
@full crown
idk where to begin
aight
which part do u need help in?
Im pretty sure u just estimate this stuff but idk if thr is a formula for these
how do i find out how many people are between 120-150
its given in the question
240 ^^
no like
for part a
how many had a systolic blood pressure reading 120 and 150, including 120 but not 15
150
u have to find that
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can anyone see what i did wrong?
Um look at second line
The x will cancel top and bottom
So you have 2
what
second like as in when i waa cancelling the 2x?
line*
here is the original question
Your variable should be canceled out on the third line
So it should be 2/1
On the second step of the third line
yeah, you cancelled $\Delta x$ but then its still there and the 2 is missing
AℤØ
should have just been lim(2)=2
ooooooooh
i just cancelled it wrong
thank you
i’m not sure where to start
delta x towards zero ***
Multiply the top and bottom by the conjugate
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What am I misunderstanding about Vectors? Ive watched about an hour if video and tried looking in my text book but I just cant seem to make any sense of what's the right way to do it
The question just wants the magnitude and direction of the resultant
What are the bearings relative to?
I see it as go 5 units on a 100° bearing like a compass and then go 6 units on a 60° bearing from that point
Last picture is horrible, I realised how wrong it is, but the first two I just cant understand whether its right or not
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ok
ok
so
this is the hw
and
specificallty
i wanna understand how to do
9ab
i dont get it
because
Don’t dox urself homie
idk
no but might not be impossible to figure out a location
oh thats fine
Which one?
start with the summation definition of multiplication
wdym
do you go to stanford?
its not stanford.edu
its ulo.stanford.edu
its like a program for highschoolers
for UG math
oh i see
ye
costed way too much for me to fail 😭
i think the midterm will have proofs so
i gotta get good at these
ima latex now.
$AB = BA$ \
$A^n = 0$\
$B^n = 0$\
For some $n \in \mathbb N$
amukh1 | JS,Axler Fanboy
thats slightly incorrect
as a first pass
?
yea ik
and you shouldnt assuem theya re
yeah but why bother tbh
cuz it may be simpler?
amukh1 | JS,Axler Fanboy
😭
so you assume A^n=0 and B^n=0 and AB=BA
yup
sure
what do you mean?
is (AB)^n not B^nA^n
not in general
then what is it
its actually just $ABABABAB\cdot AB$ n times
llspacebarll
right..
aw
we have commutativity
yes
which is 0 . 0
yeah
= 0
so QED
now generalize tho
um
oh yeah thats fine
A^n is 0
0 * B^k
is gonna be 0
and same the other way
A^n * 0
= 0

right
well yeah but you should write a formal proof i guess
i also think its worth noting that (AB)^n=A^nB^n
might not be taken for granted
meaning you could try to prove it as well
using induction is best
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I need to prove whether the set of all fractions p(x)/q(x), where p(x) and q(x) are polynomials with coefficients of complex numbers. (and q(x) is not the zero-polynomial q(x)=0 for all x). Is or isn't a field. I'm stuck deciphering the language of the question and I don't quite understand how to begin writing all the axioms. So far, I have been able to determine that q(x) cannot have a degree equal to 1.
why not?
Because then q(x) is guarenteed to have an x value that causes it to be equal to 0.
so you need to prove the field axioms?
Yes, I do.
that's true for all polynomials
do you know the field axioms
what q can't be is identically zero
yeah
Right! Oh my goodness, I forgot the fundamental theorem of algebra. I forgot that complex x values exist and can make q 0.
wow
I feel like that I might have an idea on how to start.
give it a try
.close
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I was hoping someone could check my work for this
Since $2|xy|\leq (x^2+y^2)$ for all $x,y$ then we should have that
Austin
$$\frac{x^2y}{x^2+y^2}\leq \frac{x^2y}{2|xy|}=\frac{|x|^{2}y}{2|x||y|}=\frac{|x|y}{2|y|}$$
Austin
if $y<0$ then $y=-|y|$ and $y\geq 0$ then $y=|y|$ so either way we have that
Austin
$$\frac{x^2y}{x^2+y^2}\leq \frac{|x|}{2}$$ or $$\frac{x^2y}{x^2+y^2}\leq \frac{-|x|}{2}$$
Austin
which when we then take the limit as (x,y)->(0,0) then these go to 0 so so does our original function
so lim (x,y)->(0,0) f(x,y) =f(0,0) so it is cont at (0,0)
my only issue with this is that when you put |xy| in the denominator, you have issues if x=0 or y=0
what if instead of applying the hint to the denominator, you applied it to the numerator?
how can we rewrite x^2y into a form involving a factor of |xy| though
if we write |xy|x=x^2*y this is false
$|x^2y| = |xy||x|$
Bungo
then we immediately have |f(x,y)|<= |x|/2
taking limit gives us it goes to 0
yep
your way works too but you'd have to have special cases if x=0 or y=0
@astral aurora is this truth?
what the
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Yes@fathom flicker
Please don't occupy multiple help channels.
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isn't it speeding up when the acceleration is positive
yes s is displacement, v is velocity and a is acceleration
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how do I evaluate this?
what's e^(-infinity)?
0?
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confused on how this doesnt count as orthogonal
any vector times an orthogonal vector of the original vector is 0
so taking one of the parallel vectors and finding values that would make the dot product 0 should work
that's not totally true.
you're vector isn't orthogonal to the other parallel line, for example
ok
with the info I currently have aka 2 points and 2 parallel vectors of a plan, how would I find an orthogonal vector
.close
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Ihello, I am learning the limits
To solve this I have to apply Identity lt x-> 0 Sinx/x = 1
I want to make x-> 0 as ax -> 0. can I equate x -> 0 as we do in equations
x-> 0
ax -> a*0
ax -> 0
So i can apply the identity
And what it is called I want to learn more about it
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Solve the equation for :
x^n=n!
!status
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
also did you mean x^n = n! with a lowercase x?
I think the answer is 1 and I want correction
What?
the more i learn about math the more complicated it gets
the answer is not x = 1
ann, how many languages do you know
just curious cuz i feel like i remember you knowing polish and i saw someone mention you knew hindi
perhaps we should move to #discussion if you want to talk about this further?
sure
What's the difference between X and x
X is a capital letter, x is a lowercase letter
How will it make difference in solving this equation
solving X^n = n! (with big X) for x (small x) makes no sense
if the question was genuinely solve:
$$X^n = n!$$
for $x$,
its nonsense because there is no $x$ to solve for
ℝam()n()v
Solve the equation for x :
x^n=n!
yall are missing the invisible x
Sorry I didn't notice
$x*0 + X^n = n!$, duh
how'd you get x=1
accialto
note: "how did you get ___" does NOT mean " ___ is wrong, redo your work without showing it to us"
I didn't do anything I thought of a stupid thing
Never mind that
How to even solve this?
@royal basin
i mean raising both sides to the power of 1/n should do the trick, no?
Hamdy Hisham
$x = (n!)^{1/n}$
Ann
What's next
So I cannot simplify this more?
not significantly no
i smell http://xyproblem.info/ btw.
Asking about your attempted solution rather than your actual problem
The original one was $(x)^{2021-n}=(2021-n)!$
Hamdy Hisham
@royal basin
feels like some kind of 2021 olympiad problem
like it was not just an equation
there had to have been something else written along with it
Wait I'll give you what I've done
i do not understand this, i'm afraid.
don't give me what you've done
give me a clear and understandable problem statement first
Ah I wrote it wrong
It's $y=x^{2021}$ then ...... =2021!
a.$y^{(2019)}$
b.$y^{(2020)}$
c.$y^{(2021)}$
d.$y^{(2022)}$
Hamdy Hisham
@royal basin
this is literally the same as before
oh wait
those things are derivatives are they
Yup
so you're asked which order derivative of x^2021 results in 2021!
Yup
this doesn't need any of that equation-solving bullshit
how many times do you need to differentiate x^n to make a constant
yes
I think I'm stupid
and what's the n'th derivative of x^n?
nPn?
But what's that? @royal basin
well maybe you could tell me what nPk is or admit you don't know
I don't know
if you start writing numbers from n and go down by 1 every time,
and you write out n numbers
what'll be the last number you write down?
@weary heart Has your question been resolved?
Aha n!??
no, the last number won't be n!.
It'll be 1 so nPn=n!
yes
geogebra?
Can it integrate stuff?
dunno
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Hey guys
I need help to calculate and understand this
find the endpoints and equations of the two lines
yea, i know, but how do I find it?
Help each time this type of q come up I don't understand it
Oh did not saw that thanks
at (1,2), the line on the left is the first line
that's the right end point
does it contain 1?
I mean does the line contain that end point?
yes?
@soft pond Has your question been resolved?
<@&286206848099549185>
is there any x you can't plug in here?
i dont know
well, do you kind of know how the function looks like?
no i do not
for these kind of questions you only have to think about whether or not there is a mathematical rule you're going against when plugging in certain values
rules like "can't divide by zero"
(a+b)^3
I am a freshman high schooler. I am trying to learn this stuff.
Is there a value for x you can't plug in here? Meaning: If you plug in 5 you will get an answer with h(5)=(5-3)^2=4. Is there an x so when you plug it in, you won't get an h(x) value
I still don't know how to find the answer
Hmm okay. The answer is, the domain of the function is every x in the real numbers.
There is no number we can't plug in here so that h(x) is not defined. Why is this even important?
There are functions where we have fractions and x is in the denominator. So there x couldn't be 0 and therefore the domain would be all x in the reals but without x being 0
Thanks
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HELLO
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⚔️ Guys
Yes to both questions
woah
yo im confuse @sage geode just 1 more question
i know u not economic profesional but i also wonder why p*f(L,K) is also not that rule
Is p a constant?
Yes
Even if you were to apply the product rule to p * f(L, K)
You would end up with the same thing
you expert
yo
yo wait can you just help with 1 more question
with your iq5000
Sure
1min
this is perfect competition and diference is
there is no complex derivation there
but your probably cant use iq5000 for this
but i try wrap my head around this
first example i gave
second picture was first example
THIS FUNCTION IS MONPSONY (SECOND PICTURE)
first picture is without complex derivation
shiiii
so market supply in second picture ( Monopsony) is rising because it increases with wage and labour increases (each another worker hired, monopsony has to pay him more, but also pays every other worker more
that why w(L)??
but in first picture w is constant cos its perfect competition (firm cant affect wage)
so thats why no complex derivation in perfect competition
am i right?! 😮
I don't know econ 
yo but
like does it make sense
from iq5000 perspective
its just lines and graph and math like
i gave like econ premise
but in math like does it connect
i make econ bold
just 2 rules
I suppose it connects
woah
im enlightedn now
i used to look at those function and think damn
and thought i gotta memorise
in my brains
but now i knwo the truth
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Try and use Newton-Raphson Method to compute all the zeros of f1(x) = 0.
What would the steps be to compute all the zeros?
have you studied the newton-raphson method yet?
what is f_1
also that yea
Yes..^^
I know how to do it
I have also implemented it in python
but what is f1(x)
I'd agree with that
definetly has one
you can even solve the eq explicitly without the use of any special methods
I did
which ig means you have a way to doublecheck yourself
I got x=0
But, exercise is to use the method
So
But so would the idea be to find two values one negative and one positive
f(x)
and then you know by IVT that inbetween this there must be a value for f(x)=0
oh sure then
You find a negative f(x) and a positive f(x) and then you know by the intermediate value theorem that inbetween these two there must exists x such that f(x)=0
Because the assignments says "compute all the zeros of"
but as you said,theres only one
Yes, but how would I know that using the newtons raphson?
Or is that just not possible
Or not using that
but like without solving the equation
not possible really
f(x)=0
its just a lot easier to do algebraicaly
Ye, I did solve f(x)=0
i just don't feel like thats what the assignment is telling me to do
thats true
Try and use Newton-Raphson solver to compute all the zeros
is that a given solution?
But that does not furfill that tho
It's the question
2) Try and use Newton-Raphson solver to compute all the zeros of f(x)=0
Anyway, perhaps it's fine what i've done
thanks for the help
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Howdy
!show
Show your work, and if possible, explain where you are stuck.
I dont know where to start
I know they are parallel lines.
I know the formula (x-x)^2 + (y-y)^2 Root all this stuff
Is the shortest distance between two points
Its also not the Green Line (Difference in Y axis)
yeah you need the perpendicular line
the shortest distance would be the perpendicular
there is also a formula for it
which always works
Alright?
I dont think ive done that formula before
and chat gpt isnt helping, usless as always 0/10 i rate out of the 5 times i used it
try |C1–C2|/√(A^2 + B^2 )
what would A and B be?
both equations can be written as ax + by + c = 0 and ax + by + c1 = 0
Also yes dont use chatgpt for maths questions, its gonna blurt out garbage and apologise when you tell it that its wrong every time
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hello
what's confusing you?
!status
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
Because I’m division when u divide anything by a fraction
U make the think ur dividing by its own reciprocal
And the signs becomes multiplication
uhm

can u explain simpler pls
if i divide fraction the second fraction go upside down and it become multiply
cooks told me that before
i remember now
$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} * \frac{d}{c}$
TooManyCooks
@edgy cedar Has your question been resolved?
Hi
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!help
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for this I got 8y+6 for the width but im not sure what to do next to find the area
do i just do (8y+6)*2y
@edgy cargo Has your question been resolved?
ty
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For any integer n>2, Show that there are at least two elements in U(n) that satisfy x^2=1.
U(n) is the set of all least residues modulo n which are relatively prime to n
1 is element of U as 1 is always relatively prime to n
and -1?
Different context
that’s one of my confusion too, if think they are taking solutions modulo m i guess
definitely -1 not in U(n), -1 cannot be a least residue aka remainder when divided by m.
<@&286206848099549185>
Question 20.
@runic void Has your question been resolved?
what number in U(n) is congruent to -1 mod n?
also what's m
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Alpha is contained in irrational n.group. just how?
it doesn't matter that alpha is irrational
c'est vrai pour tout alpha réel
ça matière pas qu'il est irrationnel
pour le premier pas au moins
Can you speak English?
What I don't get is how `Alpha^(n+1)+Alpha^n = Alp^(n) * (Alp+1)
This ^
alpha +1 = alpha²
$\alpha^{n+1} + \alpha^n = \alpha^n \cdot \alpha + \alpha^n \cdot 1$
Ann
si tu te souviens du polynome qui donne le nombre d'or
x²-x-1
x² = x+1
vrai pour alpha
Yes, of course.
Ann
agree or disagree?
agree.
Thanks but without this info, it is impossible?
i mean...
sure?
like alpha was INTRODUCED as one of the roots of that equation
so it's a strange question to ask
Well, this will take me to another topic rather math that if you want to know.
which is I'm going to learn math using French in Uni while I'm not that cool with the language.
Anyway, thank you.
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hello guys! I think that the correct answer from the book to the challenge question 20 was totally unclear for me (didn't even visualize or get the first requirement that scalars should be positive)
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anyone can explain me the sum of sequences formula ? the 1-q^n+1 / 1-q ?
on geometrical sequences
A Lonely Bean
Then $qS = q + q^2 + q^3 + \dots + q^{n+1}$, right?
A Lonely Bean
@fickle flame If you have any questions about what I am writing/saying, just ask
yes, until now i didn't study that, so i'm a bit confiosing
You will understand the proof once I am finished, I hope, but for now I shall wait for the helpee's response
for what u said yes
but then saying that my sequence start at u1 = 1
does it change something to the formula
Once you know the formula for 1 + q + q^2 + ... + q^n, you can just multiple it by u1 and you will get the formula for any geometric sequence
But let's first be done with simplifying S
Alright, so, you may notice that these two sums (S and qS) share a lot of terms in common
yes
In fact, every term is shared except for 1 and q^{n+1}
What would happen if we were to subtract S from qS?
just q ?
yes
No no q gets cancelled out as well
It's present in this sum
And that sum too
I mentioned this fact because all of those shared terms will cancel out when we subtract the sums
how tho i m interested
So q, q^2, q^3 and so on up until q^n disappears when we take qS - S
\begin{align*}
S &= 1 + q + q^2 + q^3 + \dots + q^n \
qS &= \phantom{1 + } q + q^2 + q^3 + \dots + q^n + q^{n+1} \
S - qS &= 1 + \cancel q - \cancel q + \cancel{q^2} - \cancel{q^2} + \cancel{q^3} - \cancel{q^3} + \dots + \cancel {q^n} - \cancel {q^n} - q^{n+1}
\end{align*}
Ugh wait
A Lonely Bean
okay i see and how does it give the 1-q^n+1/1-q formula tho
So the last equation is just the same as S(1 - q) = 1 - q^{n+1}
And you just divide both sides by (1 - q)
1 - 2 is -1
knowing first term is =1
So 1 - 2^64, when divided by -1, becomes 1/(-1) - 2^64/(-1)
Which is -1 + 2^64
Or just 2^64 - 1
Because the formula gives us $\frac{1-2^{64}}{1-2}$, everything up until this makes sense, right?
A Lonely Bean
the why we do this no but ig it s just applying formula
like i could get there too by just following the formula
but then the after for results i don t get it
do u think i should understand the formula?
like the why
Well I have shown you the proof of the formula already
can you explain why there is 1 in the beginning of S
That's just how I defined S to be
so wait instead of putting the 1 at the beginning u just put it in qS ?
to explain it s q^next one always?
i m sorry i lack a lot in some stuffs in maths i m trying to figure out
so i can understand the exercice better
and why it gives 2^64 -1
No I defined S to be 1 + q + q^2 + ... + q^n because any geometric sum can be evaluated by knowing the formula for 1 + q + q^2 + ... + q^n
Let's say we have some geometric sequence with the first term and the common ratio given as $u_1$ and $q$ respectively
A Lonely Bean
The $k$th term of the sequence can be calculated using $u_k = u_1q^{k-1}$, right?
A Lonely Bean
Ah, so you start with 0
it s another system that might be the reason as well
Alright one moment I'll fix that
in my today exercice it starts from u1
but i would like to understand the difference
of what it does to the formula
oh
The end result is the same no matter what you call your variables
and no matter from when does it start?
the sequence
no matter like first term is ?
Yeah, as long as you have done everything else correctly, there should be no problem
So u0 or u1, which one shall we start with?
u1
Let's say we are given some geometric sequence starting with $u_0$ and the common ratio as $q$. The $k$th term of the sequence can be aclculated using $u_k = u_0q^k$, right?
A Lonely Bean
yes
So if we were to add up some amount of first terms in the sequence, we would have [ u_0 + u_1 + u_2 + u_3 + \dots + u_n ] For some natural number $n$. According to the formula for the $k$th term, we are able to rewrite that as [ u_0 + u_0q + u_0q^2 + u_0q^3 + \dots + u_0q^n ] Can you see anything that can be factored out?
A Lonely Bean
u0?
Yes, when we factor it out, we get $u_0(1 + q + q^2 + q^3 + \dots + q^n)$
A Lonely Bean
And that sum in the parenthesis is some geometric sequence starting with 1 as its first term
This is why you can calculate any geometric sum by knowing how to calculate geometric sums whose first term is 1
Hence I defined S to be 1 + q + ... + q^n
oh my god