help-49

[lavish venture]

then using back substitution

you’ll have a third equation in this instance with x3= a number

then you’ll go the second equation and use that number to find x2

and do the same for the first

using both x2 and x3 to find x1

this will eliminate the two variables in the third equation and one variable in the second so that you can use back substitution

[twilit field]

Hmm, haven't learnt back substitution yet

oh well

[lavish venture]

it’s simple

like if you ended up with idk x3=1

and the second equation in reduced form was x2+x3=4

then you just put 1 in for x3

then solve for x2

and use those two values for the first equation

make sense?

[twilit field]

So it's just doing the obvious

yeah

okay

thanks

[lavish venture]

mhm

you’re welcome

[twilit field]

Didn't realise it had a name

.coose

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[twilit field]

This will only have trivial solns , right?

[prime hornet]

what makes you think so?

I'm not gonna say whether you're right or not

[twilit field]

There a theorem that says so

[prime hornet]

can you show the theorem?

[twilit field]

The number of equations should be more than the number of variables

[prime hornet]

for what to happen?

and this system doesn’t even satisfy that

4 eqs, 4 variables

[twilit field]

sorry, using this

[prime hornet]

well, this system of equations does not satisfy the hypothesis

and it looks like you're assuming the inverse is true too, when that's not necessarily true

[twilit field]

so I have to perform ERTs to determine its solutions( if any)?

[prime hornet]

probably

what are ERTs?

elementary row...

what?

[twilit field]

Elementary row transformations

[prime hornet]

I see

yeah, you'll likely need them

this is a pretty easy system to reduce though

[twilit field]

yeah, true.

[modern sapphire]

Well, not really. At the first glance, what can you comment atbout the coefficients of x2 and x4?

[twilit field]

I have to get used to writing these steps fast though

[prime hornet]

mhm, you can do this too

though I suspect wai has more to answer than "does this system only have trivial solutions?"

[twilit field]

I have to find its solns

all solutions

[modern sapphire]

I mean, he has to check if the system is homogeneous before applying that theorem no?

[twilit field]

I haven't done homogeniety yet

that's the last part of this chapter

[prime hornet]

I did say that this system does not even satisfy the hypotheses of the theorem earlier (since it doesn't have more variables than equations)

so the theorem is moot

[twilit field]

I suspect no solutions

,w solve w-x+y+z=0;-w+x+y+z=0;w+x-y+z=0;w+x+y+z=0

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by physicsrocks

[twilit field]

oops

The reduced matrix I got was

[modern sapphire]

well, first off, this is very bad to do when you are solving problems about something you just learnt

And secondly, the equations are wrong

[twilit field]

oh right

[modern sapphire]

You should always calculate the answer yourself before looking at the solution

[twilit field]

\begin{bmatrix}
0 & 0 & 2 & 0 \
0 & 2 & 0 & 2 \
2& 2 & 0 & 2 \
1 & 1& 1& 1
\end{bmatrix}

is what I got

so y=00

[grand pond]

(why am i here )= idk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[modern sapphire]

yeah, doesnt seem wrong

[twilit field]

now I perform further row reductions, I guess?

[modern sapphire]

But the again, why dont you do convert it to RREF

[twilit field]

I could , yeah

[modern sapphire]

why you have to make the matrix so hideously monstrous

[twilit field]

I'll start with $R_1 \leftrightarrow R_4$

[grand pond]

(why am i here )= idk

[modern sapphire]

Always go for the most lucid solutions so others can comprehend them too

[twilit field]

Now $R_1-R_4 \to R_1$

[grand pond]

(why am i here )= idk

[twilit field]

\begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 1 \
1& 1& 0 & 1 \
0& 0& 1& 0
\end{bmatrix}

[grand pond]

(why am i here )= idk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

Now w=0

so all entires in column one can be replaced by 0

[modern sapphire]

Great

Now onwards to rows 2 and 4 (3 is already done)

[twilit field]

hmm

Now firstly this is the same as

\begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 1 & 0 & 1 \
0& 1& 0 & 1 \
0& 0& 1& 0
\end{bmatrix}

[grand pond]

(why am i here )= idk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[modern sapphire]

👀

[twilit field]

as x=0

I mean w

[modern sapphire]

where did the leading 1 go?

[twilit field]

a_11=0, right

[modern sapphire]

from where?

[twilit field]

This

they are all equal to 0

[modern sapphire]

what operation did you do? r1 - r3 + r2?

ok

[twilit field]

so from this we can conclude y=0

so this is

$\begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 1 & 0 & 1 \
0& 1& 0 & 1 \
0& 0& 0& 0
\end{bmatrix}$

[modern sapphire]

There is an even more relevant conclusion to be made here imo

[grand pond]

(why am i here )= idk

[twilit field]

so we have two identical equations

that intersect at infinitely many points

[modern sapphire]

yes. Infinite solutions

[twilit field]

so the solution set is $w=0,x=-z, y=0$, $x \in \R$

[grand pond]

(why am i here )= idk

[modern sapphire]

Yep

,w solve w-x+y-z=0;-w+x+y+z=0;w+x-y+z=0;w+x+y+z=0

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by bacter10fr4g

[modern sapphire]

Now is the time you should be using online calcs

so you can verify your work

[twilit field]

I thought I had already solved it

oops

sorry

hmm, but I failed to RREF it

Thanks a lot for the help though!

[modern sapphire]

np

[twilit field]

Can you help with one more problem ?

$\begin{bmatrix}
1 & -1 & 1 & -2 \
-1 & 1 & 1 & 1 \
-1& 2& 3 & -1 \
1& -1& 2& 1
\end{bmatrix}$

[grand pond]

(why am i here )= idk

[twilit field]

then after a few row operations I got

$\begin{bmatrix}
0 & 0 & 1 & -3 \
0 & 0 & 3 & 2 \
0& 1& 5 & 1 \
1& -1& 2& 1
\end{bmatrix}$

[grand pond]

(why am i here )= idk

[twilit field]

so that means y=3z

and 3y=-2z

which is only possible when they are 0

so the matrix can further be reduced to

$\begin{bmatrix}
0 & 0 & 0& 0 \
0 & 0 & 0 & 0 \
0& 1& 0 & 0\
1& -1& 0& 0
\end{bmatrix}$

[grand pond]

(why am i here )= idk

[twilit field]

oops

This is supposed to be an augmentex matrix

my bad

$\left[\begin{array}{@{}cccc|c@{}}
1& -1&1&-2&1\
-1&1&1&1&-1\
-1&2&3&-1&2\
1&-1&2&1&1\
\end{array}\right]$
\

[grand pond]

(why am i here )= idk

[twilit field]

so on performing a few row operations, I get

$\left[\begin{array}{@{}cccc|c@{}}
0& 0&1&-3&0\
0&0&3&2&0\
0&1&5&0&3\
1&-1&2&1&1\
\end{array}\right]$

[grand pond]

(why am i here )= idk

[twilit field]

wait

so that gives me y=3z

[modern sapphire]

if its still the same ptoblem you are solving, i gottasay this matrix is wrong

[twilit field]

and 2y=-z

[twilit field]

This is the problem

[modern sapphire]

oops, didnt see that one mb

[modern sapphire]

How did it change from this to what you got now?

you went wrong somewhere on one of those

and ffs dude, just go with the RREF 😭

[twilit field]

I think I should

yeah

[modern sapphire]

Goin thru rref steps is just brainless work

you just cant go wrong

[twilit field]

I want to think though

😭

I really struggle with RREF

[modern sapphire]

I mean, you arent 'thinking' here. Just complicating life unnecessarily

I wouldnt mind if you were using some innovative thoughts to do it

[twilit field]

I mean I do think about which rows to combine and in what manner, but point taken

[modern sapphire]

Jus swap rows in that case

swapping is also ert

[twilit field]

I got this as my second step in the ERTs

wait a min please

$\left[\begin{array}{@{}cccc|c@{}}
1& -1&1&-2&1\
0&0&3&2&0\
0&2&3&-1&2\
1&-1&2&1&1\
\end{array}\right]$

[grand pond]

(why am i here )= idk

[prime hornet]

you just need practice

[twilit field]

I'm also panciking rn

idk

[prime hornet]

trust me, it becomes insanely intuitive after you get used to the algo

[prime hornet]

is everything okay?

[modern sapphire]

Take deep breaths, and do stuff one small operation at time. No need to think ahead

[twilit field]

One of my classmates took my solutions to a practice problem set for cross verification yesterday, just realised that a) They may have not done it themselves, b) It may be academic dishonesty for which I can get into trouble though this isn't graded.

[twilit field]

Anyways, from this I get

$\left[\begin{array}{@{}cccc|c@{}}
1& -1&1&-2&1\
0&0&3&2&0\
0&1&5&0&3\
1&-1&2&1&1\
\end{array}\right]$

[grand pond]

(why am i here )= idk

[modern sapphire]

ok, so rref steps say, r4 becomes r4-r1

so you can use this now

[twilit field]

Is this right so far

[modern sapphire]

Hmm, can you show the row op you did for row 3?

where did its -1 go in 1st column, and yet other coeff stayed same?

[twilit field]

I did $R_3+ R_4 \to R_3$

[grand pond]

(why am i here )= idk

[modern sapphire]

oops, my bad. I looked at wrong one

Yep, it looks good

[twilit field]

so after that I do $R_1+R_3 \to R_3$

[grand pond]

(why am i here )= idk

[modern sapphire]

r1 and r4

so you get a 0 on 4th row

[twilit field]

$\left[\begin{array}{@{}cccc|c@{}}
0& 0&1&-3&0\
0&0&3&2&0\
0&1&5&0&3\
1&-1&2&1&1\
\end{array}\right]$

[grand pond]

(why am i here )= idk

[modern sapphire]

ok, you want lower triangular matrix

thats fine too

[twilit field]

My argument now is that

y=3z

and 3y=-3z

the only way this is possible if y=z=0

[modern sapphire]

Uhh, it should be y = 3z and 3y = -2z

but yeah, sure, 0,0

[twilit field]

so this matrix is equivalent to

[modern sapphire]

so you sub back and get your answers

[twilit field]

$\left[\begin{array}{@{}cccc|c@{}}
0& 0&0&0&0\
0&0&0&0&0\
0&1&0&0&3\
1&-1&0&0&1\
\end{array}\right]$

[grand pond]

(why am i here )= idk

[modern sapphire]

And what row transformation is this?

[twilit field]

No row transformation

[modern sapphire]

you cant just replace solution in place pf coefficients

[twilit field]

just using the fact that y=z=0

[modern sapphire]

they arent equivalent

[twilit field]

Why can't I?

[modern sapphire]

solution is value of a variable

coefficient is not

[twilit field]

I know

[modern sapphire]

think of units. Your x is say 5 kg

then the 2x doesnt mean 2 is kg

[twilit field]

Hmm, okay

[modern sapphire]

you cant put 5 in place of 2

[twilit field]

but I now know some solutions

So I can now solve it algebrically, no?

[modern sapphire]

yes, and you put them in the x vector of system Ax = b

not A matrix

the augmented matrix is A|b

[twilit field]

I get that

[modern sapphire]

so why are you putting x in A

two are different

[twilit field]

so I have x+5(y)=3

w-x++2y+z=1

[modern sapphire]

once you put in the values of variable, you should not perform the row operations on it

coz the stuff is now multiplication written out, which makes it a scalar in rows, and not a vector

[twilit field]

Got it.

Thanks so much!

[modern sapphire]

np

glad i could help

and dont stress out too much over this stuff. It is easy, once you are in the right mindset

[twilit field]

This wasn't stressful, the fact that I like to help people caused a panic attack.

[modern sapphire]

so take breaks and relax whenever you feel like you are panicking

[twilit field]

I clarified with my prof, so it's fine now

[twilit field]

Yeah, got it

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mossy dawn]

hi, need to proof this please. Just came across during trig identities practice.

[mossy dawn]

I don't know where to start.

[robust forge]

Use cosa - cosb = 2sin(a+b/2) sin(b-a/2)

[shy ice]

or maybe make x into 2A (A = x/2) and same for x+2a

cuz the stem wants u to use that identity

idk

[versed fractal]

what is x and alpha? just any numbers?

[mossy dawn]

so @shy ice I tried ur way but the answer is supposed to be in terms of x.

[robust forge]

Variables

[mossy dawn]

@robust forge we are supposed to use the trig identity given

[robust forge]

Atleast ur allowed to open cos(x+2a)

[mossy dawn]

ok

Oh I am still confused

when I open up the brackets, then when do I involve the identity given in the question. Either way I am not getting answer.

[versed fractal]

Are you only allowed to use the double angle for cos?

if so then I have no idea

[mossy dawn]

yes

but what is your idea? can you sort of like do a few steps.

[versed fractal]

if you allow using the other addition formulae:

expand cos(x+2a) in the left

[mossy dawn]

yes, yes this is assumed knowledge

[versed fractal]

with this you can expand cos(x+2a) to get cos(x)cos(2a) - sin(x)sin(2a)

now expand cos(2a) and sin(2a) in this

when you do that and write verything out

it all cancels, then there is a common factor of 2sin(a) in the top which cancels with the denominator

then you just get sin(x+a)

[mossy dawn]

OH I see!! Do I expand cos 2a into 1-2sin^2A? or is it just double angle cos as you were saying?

[versed fractal]

expand to 1-2sin^2 A as given in the question

the reason being is that this will cancel with the cosx on the top left

and then you can simplify everything easily afterwards

the question likely gave that cos2a = 1-sin^2a to save you the trouble of doing this, and then needing to sit and try and rearrange until things cancel

when I expanded i got $$\cos(x) - [\cos(x) - 2\cos(x)\sin^2(\alpha) - \sin(2\alpha)\cos(x)]$$

[grand pond]

turtl

[versed fractal]

so the two cos(x)'s cancel

[mossy dawn]

yup I got that too

after cancelling the cos out, do you expand the whole thing

[versed fractal]

yea you have to expand sin(2a) after then aswell

when you do that, you should be able to cancel with the denominator

[mossy dawn]

expand sin2a through identity?

[versed fractal]

yep

[mossy dawn]

so the fraction I got was 2sin^2a -1/1-2cos^2a

is that right?

[versed fractal]

after expanding sin(2a)?

do you mean $$\frac{2\sin^2(\alpha) - 1}{1 - 2\cos^2(\alpha)}$$

[grand pond]

turtl

[mossy dawn]

yes

[versed fractal]

how did you get 1-2cos^2 a in the denominator? shouldnt it be 2sin a

[mossy dawn]

I expanded 2sina using identity given

so can't we write 2sinA as 1-2cos^2a

[versed fractal]

do you mean sin(2A)?

[mossy dawn]

yes

[versed fractal]

bottom should not be changed, it is just 2 * sinA

[mossy dawn]

ok

yh scrap that

[versed fractal]

i dont think this is right

identity is only for cos, not sin

sin2a = 2sin(a)cos(a)

[mossy dawn]

yes. sorry.

so now the fraction is 2cos^2A / 2sinAcosA

[versed fractal]

bottom should be 2 sinA

bottom should not change as it is 2 sinA, not sin(2A)

[mossy dawn]

oh alright 2cos^2A/2sinA

[versed fractal]

i got something different for the top,

[versed fractal]

working from here, the cos cancels, and after expanding sin(2a), you get $$2\cos(x)\sin^2(\alpha) - 2\sin(a)\cos(a)\cos(x)$$

[grand pond]

turtl

[versed fractal]

this is what I got for the top, and then it is over 2sin(a) on the bottom

give me a second, i can type out the whole tihng

\begin{equation*}
\begin{split}
\frac{\cos(x) - \cos(x+2\alpha)}{2\sin(\alpha)} &= \frac{\cos(x) - [\cos(x)\cos(2\alpha) - \sin(2\alpha)\sin(x)]}{2\sin(\alpha)}\
&= \frac{\cos(x) - [\cos(x) - 2\cos(x)\sin^2(
\alpha)- \sin(2\alpha)\sin(x)]}{2\sin(\alpha)}\
&= \frac{2\cos(x)\sin^2(\alpha) + 2\sin(\alpha)\cos(\alpha)\sin(x)}{2\sin(\alpha)}\
&= \cos(x)\sin(\alpha) - \cos(\alpha)\sin(x)
\end{split}
\end{equation*}

[grand pond]

turtl

[versed fractal]

sorry took so long i made a typo in my work actually

[mossy dawn]

oh! and then it fits in the final identity of sin(x + a)

[versed fractal]

yep

[mossy dawn]

CrAzY! Absolutely bizarre! you are truly a maths olympiad. I honestly have never struggled that much with an identity. Thank you soooo much!!!! I understand completely!

Really made it easy for me.

Thankyou for the hardwork

[versed fractal]

usually these identities just need practice, but then you will get the idea

in my experience if you dont see something to do, expand everything out and then write everything in sin,cos if there are tan or others involved

sometimes you can get stuck but just dont stress

[mossy dawn]

definitely good advice

the stress built here as helpers kept coming and going away, confused

so yes absolutely

Once again, a huge thanks

.close

[midnight plank]

Channel closed

Closed by @mossy dawn

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hazy flint]

as a partial fraction doesnt it become 1/V + 1/(1-V)

so lnV + ln(1-V)

[grim vector]

It does

But you can simplify it further

[hazy flint]

how

[grim vector]

Ln(a) - ln(b) = ln(a/b)

[hazy flint]

ok so?

how can lnV + ln(1-V) become ln(V) - ln(1-V)

surely the solution is wrong?

this is the question btw

what to do? @grim vector

oh wait im so stupid

int 1/(1-V) is -ln(1-v)

oops

.close

[midnight plank]

Channel closed

Closed by @hazy flint

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[supple inlet]

is there a neater way of doing this without expanding everything

[slender walrus]

binomail theorem

[supple inlet]

do you mean expanding everything with binomial theorem?

[vapid valve]

maybe just the ( )^4 and ()^6

and then see what will sum to the constant

[supple inlet]

hmm alright

i was wondering if theres a way to do it just by perms and combs or something simiolar

[summer ravine]

I have done it my.

it was easy

[vapid valve]

yes I found a way

it will please you

[midnight plank]

@supple inlet Has your question been resolved?

[vapid valve]

to get a term you make a binary choise in each bracket

from the left side you can chose:
x or -2
from right and:
2/x or 1

we notice that the number of brackets with x must be equal to the number of 2/x

lets analize when we have chosen k number of x

edit: forgot about x*2/x =2 you have to add it to the value

and you know what to do

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[twilit field]

So I'm trying to find the Rank of this matrix

[twilit field]

I first simplified it to

I've closed the previous one

I used to post all my questions in one channel

Please do

They can decide what to do

$\begin{bmatrix}
1 & 2& -1 & 0 \
0 &a & 1-a & a^2+1 \
0 & a & 0 & 2a^2\
\end{bmatrix}$

which can further be simplified to

$\begin{bmatrix}
0 & 1 & -2 & -1 & -2 &-1&1 \
0 &0 & 0 & 3 & 9 &3&3 \
0 & -2 & 4 & 3& 7 & 1 &0\
0 &3&-6&1&6&4&1
\end{bmatrix}$

oops

$\begin{bmatrix}
1 & 2& -1 & 0 \
0 & a & 0 & 2a^2\
0 &a& 1-a & a^2+1 \

\end{bmatrix}$

This can further be simplified to

[dreamy lichen]

Where did the a from (2,2) go?

[twilit field]

oops

[grand pond]

(why am i here )= idk

(why am i here )= idk

[twilit field]

$\begin{bmatrix}
1 & 2& -1 & 0 \
0 & 1 & 0 & 2a\
0 &a& 1-a & a^2+1 \

\end{bmatrix}$

[grand pond]

(why am i here )= idk

[twilit field]

Hmm

This is nearly in REF but not quite

Maybe another row transfomation

$aR_1 -2 R_3 \to R_3$

[grand pond]

(why am i here )= idk

[dreamy lichen]

R1?

[twilit field]

yeah

R_1

I'm trying to eliminate that a in 1-a

[dreamy lichen]

That will also move something here though

[twilit field]

True

yeah

[sudden yacht]

Use second row instead

[dreamy lichen]

always use the row above it

[twilit field]

then $aR_2 -R_3 \to R_3$ is probably a better idea

[dreamy lichen]

-R_3 I suppose

[grand pond]

(why am i here )= idk

[dreamy lichen]

yes

definitely

[twilit field]

$\begin{bmatrix}
1 & 2& -1 & 0 \
0 & 1 & 0 & 2a\
0 &0& 1-a & a^2-1 \ \

\end{bmatrix}$

[grand pond]

(why am i here )= idk

[twilit field]

ooh

the rank is 3

right

thanks a lot!

[sudden yacht]

What happens if a=1?

[dreamy lichen]

usualy

and also a = 0

[twilit field]

Limiting case

same for a=0

it's the limiting case

[dreamy lichen]

Tbf I don't know what that is, but sounds right probably

[twilit field]

I'm saying a=0 and a=1 are points at which it doesn't make sense to talk about the matrix's rank

But the limit of the rank is 3

[sudden yacht]

No

The rank is 2 if a=0 or a=1

[twilit field]

right

Makes sense

[sudden yacht]

I've never heard the word limit in this context

[twilit field]

I was just guessing for when a=1 and a=0 tbh

sorry

Understood though

Thanks!

[midnight plank]

@twilit field Has your question been resolved?

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[elfin fox]

Is this first step for finding the Laplce Transform of this function correct? If so, how should the result be interpreted. I wonder this because the left integral for example will give a different result to the normal L{t}.

[midnight plank]

@elfin fox Has your question been resolved?

[midnight plank]

@elfin fox Has your question been resolved?

[elfin fox]

<@&286206848099549185>

[carmine sigil]

No need for the first limit imo

And no reason you can't just do an improper integral for the second.

Though the limit in the second case is the actual definition.

[midnight plank]

@elfin fox Has your question been resolved?

[summer terrace]

what did you find, and what is the correct answer?

seems like youre away

anyway, i did the integrals and i ended up with the following

$F(s)=\frac 1{s^2} + (1-\frac 1 s) , \frac{e^{-4s}}s$

[grand pond]

Emily

[summer terrace]

compare with what you found and with the correct answer, ping me if needed

[midnight plank]

@elfin fox Has your question been resolved?

[elfin fox]

Thank you

.close

[midnight plank]

Channel closed

Closed by @elfin fox

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lament fox]

does there exist any function $f$ such that $f(f(x)) = -x$ other than $f(x)=ix$ for $x\in\mathbf{R}$?

[grand pond]

The د

[midnight plank]

@lament fox Has your question been resolved?

[exotic pelican]

f(x) = x+i for x in R
f(x) = -x+i for x in C\R

any more restrictions on f? Theres many ways one could construct it using complex numbers

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hasty vortex]

How do you get from S_k = k(k+1) /2 to S_(k+1) = S_k + (k+1) by adding k+1?

[hasty vortex]

Nvm

.close

[midnight plank]

Channel closed

Closed by @hasty vortex

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[idle karma]

im wondering here this function f(x) im struggling to understand intuitively what it means

im struggling to understand how a "a probability density function can take on values greater than one" and so i think im misunderstanding exactly what a probability density is, i think of these two as distinct measures and i dont know what is meant when they are "together"

[small jasper]

im struggling to understand how a "a probability density function can take on values greater than one"

The only requirements are that the density is always non-negative and that the area under the entire probability density curve is 1. The probability density represents the density of the continuous random variable, which gives the relative likelihood of a specific outcome occurring. You can think of it as the individual densities not mattering, but rather how they compare with other densities.

and so i think im misunderstanding exactly what a probability density is, i think of these two as distinct measures and i dont know what is meant when they are "together"

see above, the total probability (which should be 1) is given by the total area under the probability density curve

[idle karma]

thank you i understand the premise here now

[small jasper]

[idle karma]

.close

[small jasper]

!done

[midnight plank]

If you are done with this channel, please mark your problem as solved by typing .close

Channel closed

Closed by @idle karma

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tight jewel]

Let's assume we have a puzzle pure white in colour with no patterns.

Assuming we build a subsection of puzzle and keep building onto it by trying pieces, would the complexity of this be n! where n represent the number of pieces?

[torpid garnet]

It seems so, yes

Brute force seems O(n!)

[tight jewel]

Ok sweet I had the right idea

Is there a way to optimize this assuming no patterns on the board and we can't tell the difference between pieces* to help us solve

[midnight plank]

@tight jewel Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[steel sandal]

At what value of n does using Stirling's approximation in equations involving factorials start to yield correct results when solving for n, where the resulting n is a rounded integer?

[visual tiger]

so you're asking when does the difference between n! and stirling approx get smaller than 1?

the thing is, n! - stirling approximation is of the order of constant*(n-1)!

so not completely sure you'll ever get 100% correct results with that approximation

[midnight plank]

@steel sandal Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tardy anchor]

does anyone have an wolfram pro account, could someone get me a graph of an expression?

[runic hamlet]

well it would be much faster if you just posted the expression so if someone had a pro account they could immediately graph it

have you tried desmos or something aswell?

[slender walrus]

don't need pro to get graphs

[runic hamlet]

maybe computation time or something

[tardy anchor]

inverse Laplace transform (Divide[12,s(40)Power[s,2]+19180s+10000(41)])

[tardy anchor]

it only gives you a small preview

[tardy anchor]

doesn't work for some reason

[runic hamlet]

the graphs look like bullshit

[midnight plank]

@tardy anchor Has your question been resolved?

[unreal fjord]

btw, if you are college students and your university subscribe Mathematica, you probably have WolframAlpha Pro (not really but comparable) for you to use within Mathematica

here you go. the right one

[midnight plank]

@tardy anchor Has your question been resolved?

Channel closed

Closed by @tardy anchor

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[unreal cedar]

is the solution as simple as 4x2x6! = 5,760 ? or do i have to account somehow for the times when A is taken already in the first 4 letters?

[unreal cedar]

right

[pearl hull]

End with what?

[unreal cedar]

A or T

[pearl hull]

Bruh

@unreal cedar Your solution is incorrect, this question isn’t as simple as you thought. Hint: Determine the last letter in the first place

[unreal cedar]

ah...

im still confused

[unreal cedar]

if we do the last letter first, it would be 2 x ?? x 6! ??

[pearl hull]

wdym, elaborate

Work what? You haven’t provide your solution yet

[unreal cedar]

im not sure how i would show that in an equation?

[pearl hull]

Have you tried my hint?

Taiwanese

[unreal cedar]

i dont know how to show the splitting of probailities

[pearl hull]

ok stop

[pearl hull]

Alr, do you know the concept of my method?

Pick the last one n pick the first one

[unreal cedar]

for last letter being a we would write the number of arrangements in an equation as 1 x 4 x 6!

for hte last letter being t we would write the number of arrangements in an equation as 1 x 5 x 6!

i think

[pearl hull]

But the last letter can only be either A or T

[unreal cedar]

errr, thers no i in this set of letters

ill try again i think i sort of see what your method is showing

with the picture

[pearl hull]

Sure, are you in a hurry rn?

[unreal cedar]

not in a hurry

[pearl hull]

If you don’t mind, I can wait till you finish the question

Bro stop….this is a help channel, not a place for chitchat

[unreal cedar]

when the last letter ends in A the formula is 1 x 3 x 6!

when the last letter ends in T the formula is 1 x 4 x 6!

is the combination of both together the amount of rearrangements? (1 x 3 x 6!) + (1 x 4 x 6!) = solution

[pearl hull]

I mean, if you’re looking for something to chill with you. You should consider moving to #discussion

[pearl hull]

That’s all, correct

Which one are you referring to?

I don’t find any problem with their approach

[unreal cedar]

yeah, it logically looks right to me now compared to my first attempt

[unreal cedar]

ty for this

[pearl hull]

np

[unreal cedar]

.close

[midnight plank]

Channel closed

Closed by @unreal cedar

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wide marsh]

why is my answer double the correct one

[wide marsh]

question is to find the arclength of the cardioid

[robust forge]

Where did that 2 come from while taking the integral of cos

[wide marsh]

which step?

when i took the half angle identity?

bruh nvm i literally had the right answer

my textbook asked for half the arclength aka from 0 to pi only lol

so my answer being double is correct

.close

[midnight plank]

Channel closed

Closed by @wide marsh

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[twilit field]

Part c

[twilit field]

So here's my proof

Consider $$ 1 \leq i \leq m; \sum_{j=1}^{n} a_{ij}x_j =k_i; k \in \R$$
Now let's assume that $$x_j=0_v$$ where $$0_v$$ is the $0$ vector.
From this we get any $$\sum_{j=1}^{n}=0$$.
which implies $$j_i=0$$.
so the system is homogeneous

[grand pond]

(why am i here )= idk

[rose trout]

From what I understand, x_j would be one of the coordinates of the solution vector, so not a vector itself. I'd just write it as x_j = 0.
You also have an empty sum and I guess you mean k_i = 0 at the bottom?

[twilit field]

Yes

That should be $x_{ij}$

[grand pond]

(why am i here )= idk

[twilit field]

My bad

[rose trout]

But the same coordinates come up in the linear system, so why would they be double indexed?

[twilit field]

As this is a system of LEs

ooh

sorry

<@&286206848099549185>

[rose trout]

The sum is still empty

[twilit field]

I don't follow

[rose trout]

And it should be k_i = 0, there isn't a j_i

[twilit field]

is my proof wrong?

[rose trout]

This is meaningless

[twilit field]

oops

Consider $$ 1 \leq i \leq m; \sum{j=1}^{n} a{ij}xj =k_i; k \in \R$$
Now let's assume that $$x_j=0_v$$ where $$0_v$$ is the $0$ vector.
From this we get any $$\sum_{j=1}^{n} a_{ij} \cdot 0=0$$.
which implies $$j_i=0$$.

[grand pond]

(why am i here )= idk

[rose trout]

I mean it's messed up at the top, and it should be $k_i \in \R$.
And you still have $j_i = 0$ written

[grand pond]

Azyrashacorki

[rose trout]

But apart from that I think it's fine

[twilit field]

Awesome, thanks!

[rose trout]

Oh and yeah what I mentioned at the beginning. You shoudl be comparing a particular coordinate to 0, not the zero vector

[twilit field]

👍

thanks

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[last slate]

any thoughts?

[merry pewter]

no, braindead

[merry pewter]

is there any more information?

[last slate]

well leeme give u the options i js dk how to translate it back to radius and height

[merry pewter]

oh i see, do you know the formula for the surface area?

[last slate]

2,3,4,5,6,7 for radius 2,3,4,5,6,7 for height

[last slate]

isn't it like

2pirh+2pir^2?

[merry pewter]

yeah

[last slate]

how do you convert back tho

[merry pewter]

well we want to solve 2pirh + 2pir^2 = 219.91 right?

so first divide by 2pi

[last slate]

ah yeah

to eliminate both pis?

[merry pewter]

yeah

[last slate]

ah interesting

leeme try

got 35.01

but would u get

rid of

squared

[merry pewter]

so probably rh + r^2 = 35

[last slate]

right?

[merry pewter]

you can try, but what would that give you?

[last slate]

cause u r trying to find r and h alone

unless im mistaken

5.9

maybe set height

=0?

[merry pewter]

well you can't find them without using the information that r and h are one of 2,3,4,5,6,7

[last slate]

IDK IF THATS ALLOWED

TO FIND r?

[merry pewter]

well the issue is that any value of h, will give you a valid value of r

so you need to check all of the values of h in the set 2,3,4,5,6,7 to see which will give you an r that is in 2,3,4,5,6,7

[last slate]

to which formula

original or simplified one

[merry pewter]

the simplified one of course, though both are identical

we just simplified so that we could work with integers

[last slate]

?

[merry pewter]

you know that r,h are integers

so they have their product and sum is also an integer

35 is pretty close to whatever you get when you divide by 2pi

[last slate]

yee

ok lets try

ok so basically

h=2

if im not wrong ofc

and assuming thats true we shall plug in other values

[merry pewter]

yeah

[last slate]

oh but wait

[merry pewter]

or actually checking r would be easier i think

[last slate]

5

[merry pewter]

yeah

[last slate]

cause i plugged em in lol

js realized that

so h=2 and r=5

tysm

.close

[midnight plank]

Channel closed

Closed by @chrome forge

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[waxen silo]

This is a physics question but it should be simple enough to ask here, how would I go with proceeding to do this

[waxen silo]

I know you got to use mc delta(T) +mL but the time is what's throwing me off

idk what you should do with that or how to implement it into it

[flat spire]

hi

[waxen silo]

yo

[gray totem]

Pt = mcdT + mL

[waxen silo]

I see now

[gray totem]

power = Energy / time, yes?

so power * time = energy

total energy provided to the water is mcdT + mL

[waxen silo]

then solve for m from the mL part

[gray totem]

what im wondering is why is no initial temperature given

[waxen silo]

it's either you assume it starts from 0 or you just keep it at 100

not sure

[gray totem]

if it stays at 100 then the mcdT part can be ignored

cuz dT = 0 then

so the question really isnt that clear

[waxen silo]

:/

[gray totem]

from my limited english comprehension skill

[waxen silo]

wait ig that works, because we are given the power which is the time it take to go from whatever q in some time t

[gray totem]

its just pt = mL cuz the button is pressed to "continue boiling", ie its alrd at 100

[waxen silo]

yea ig so

but why would they provide the specific heat then?

wait, theres the 2nd part, probably for that

[gray totem]

yeah they provided c for that then

[waxen silo]

I see alr

thanks for the help good :))

.close

[midnight plank]

Channel closed

Closed by @waxen silo

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[twilit field]

So I started with teh matrix

$\left[\begin{array}{@{}ccccc|c@{}}
1& 2&-1&2&1&0\
1&2&2&0&1&0\
2&4&-2&3&1&0\
\end{array}\right]$

[grand pond]

(why am i here )= idk

[twilit field]

Then

$\left[\begin{array}{@{}ccccc|c@{}}
1& 2&-1&2&1&0\
1&2&2&0&1&0\
0&0&-6&3&-1&0\
\end{array}\right]$

[grand pond]

(why am i here )= idk

[twilit field]

Followed by

$\left[\begin{array}{@{}ccccc|c@{}}
1& 2&-1&2&1&0\
0&0&4&-2&-1&0\
0&0&-6&3&-1&0\
\end{array}\right]$

[grand pond]

(why am i here )= idk

[twilit field]

Which is then followed by

$\left[\begin{array}{@{}ccccc|c@{}}
1& 2&-1&2&1&0\
0&0&4&-2&0&0\
0&0&0&0&-1&0\
\end{array}\right]$

now that gives me $x_5=0$

[grand pond]

(why am i here )= idk

[twilit field]

So I have

[grand pond]

(why am i here )= idk

[twilit field]

$x_1+2x_2-x_3+2x_4+x_5=0 ; \
4x_3-2x_4=0$

[grand pond]

(why am i here )= idk

[polar mortar]

looks good so far

[twilit field]

But how do I achieve what they ask of me

[polar mortar]

well... maybe not, i don't get x5 = 0

have they defined what a 'basic solution' is?

[twilit field]

The gaussian algorithm systemically produces solutions to any homogenous system called basic solutions, one for each parameter.

[polar mortar]

so they want you to reduce the matrix completely before converting back to equations

[twilit field]

I mean here I just divide row 2 by 4, and multiply the last row by -1

[polar mortar]

you haven't done anything to the first row though

[twilit field]

It's already in REF isn't it?

[polar mortar]

and also, i think you messed up your addition somewhere.

I don't get x5 = 0

[polar mortar]

sure, have you learned RREF?

[twilit field]

Yes

do I reduce this to RREF?

[polar mortar]

That's the way I learned it, but not totally necessary. You can get there through back substitution from the REF
(It's the same thing just written out differently)

[twilit field]

Pretty sure this is right

the coefficent of a_{55} may be wrong

but other than that, yeah

I'll have my lunch now, con tinue this later

thanks for the help

[midnight plank]

@twilit field Has your question been resolved?

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

I don't understand why this is wrong though

[jagged saffron]

Pro tip: if you want anyone ever to even think about trying to see if your row ops are correct, you should really write down which you've done at each step

[twilit field]

I have, haven't I?

I think I got it

Thanks!

[twilit field]

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fiery plover]

Prove F_n(F_(n+1)+F_n)=(F_(n+1))^2+1 where F_n is the nth fibbanoci number

[fiery plover]

dk how to start i suck with reccurence relations

[hollow oyster]

consider using the properties of the fibbonaci numbers

[delicate helm]

Can't prove that sadly, since it's false
1 1 2 3 5 8 ...
1 (2 + 1) = 2² + 1

[tiny reef]

Some sign is messed up, likely

[fiery plover]

n is odd*

not F_n but n

[tiny reef]

Maybe it helps to try put in F_{n+1} = F_n + F_{n-1}.

[delicate helm]

Whatever your proof is, you will have to use this fact somehow

[delicate helm]

that +1 is slightly concerning

[inland oriole]

i think because it's odd you can better sub in 2n+1 for n

so n can be any integer

[delicate helm]

Because this formula will never give you +1 unless it's actually F_{1} or something

[fiery plover]

nvm found this on wiki lmao

.close

[midnight plank]

Channel closed

Closed by @fiery plover

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[twilit field]

So consider a system of two equations in say 3 variables

a_1x+a_2y+a_2z-a=0

b_1x+b_2y+b_3z-b=0

when equated, this gives us a plane

sorry, a line

For we get $(a_1-b_1)x+(a_2-b_2)y+(a_3-b_3)z-(a+b)=0

wait

Isn't this a plane ?

[dreamy lichen]

Each of them gives plane

[strong lava]

It doesn't capture the full constraints
You have a line as an intersection of two planes

[twilit field]

I know

but when I equate them I get a new plane

[dreamy lichen]

what do you even mean by "equate them"

[strong lava]

If, once removing redundant equations, you have n variables and m equations, your solution set is of dimension n-m

[dreamy lichen]

Usually, this system will give you intersection of 2 planes, which is a line

[twilit field]

I mean equate the two equations I have

[twilit field]

I know that

[dreamy lichen]

equate 2 equations?

[twilit field]

I'm trying to prove the theorm using induction

ykw, I think a matrix may help here

An augmented matrix instead

[worthy wing]

This is a bit weird

[twilit field]

Let me start with m variables , an m-1 equations

[dreamy lichen]

Btw, have you learnt about general = particular + homogenous?

[twilit field]

idts

[dreamy lichen]

hmm okay then

[twilit field]

$\left \begin{array}{@{}cccccccccc|c@{}}
a_{11} & a_{12} & \cdots & a_{1m} & & b_1 \
a_{21} & a_{22} & \cdots & a_{2m} & & b_2 \
\vdots & \vdots & \ddots & \vdots & & \vdots \
a_{(m-1)1} & a_{(m-1)2} & \cdots & a_{(m-1)m} & b_{m-1}
\end{array} \right$

[grand pond]

(why am i here )= idk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

okay

wth

$\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1m} & | & b_1 \
a_{21} & a_{22} & \cdots & a_{2m} & | & b_2 \
\vdots & \vdots & \ddots & \vdots & | & \vdots \
a_{(m-1)1} & a_{(m-1)2} & \cdots & a_{(m-1)m} & | & b_{m-1}
\end{bmatrix}$

[grand pond]

(why am i here )= idk

[twilit field]

I'll just work with this, I guess

so I first assume all elements of the form $a_{ij}; i=j$ aren't 0 for simplicity

[grand pond]

(why am i here )= idk

[twilit field]

I divide the ith row by $a_ij; i=$

[grand pond]

(why am i here )= idk

[twilit field]

To thus obatin

$\begin{bmatrix}
1 & a_{12} & \cdots & a_{1m} & | & \frac{b_1}{a_{11}} \
a_{21} & 1 & \cdots & a_{2m} & | & \frac{b_2}{a_{22}} \
\vdots & \vdots & \ddots & \vdots & | & \vdots \
a_{(m-1)1} & a_{(m-1)2} & \cdots &1 & | & \frac{ b_{m-1}}{a_{m-1,m}}
\end{bmatrix}$

[grand pond]

(why am i here )= idk

[twilit field]

Now I bring it to it's RREF

from that I'll get at most m-1 free constants, and atleast one parameter

giving me infitely many solns

[dreamy lichen]

hmm

you made few assumptions

"for simplicity"

[twilit field]

Yeah, I know.

[dreamy lichen]

Try this

your book already has some theorems that can be very useful

[twilit field]

I mean my argument is essentially the same right

I've just proven it for the extreme case

I can now add this can similarly be done for all other cases too

[dreamy lichen]

how do you know that your case is the extreme one?

[twilit field]

m vars, m-1 equations

[dreamy lichen]

Why not do m equations, n vars where n > m

that would be a more general argument

[twilit field]

Okay, sure.

[dreamy lichen]

and you can argue that in the (R)REF, there can be at most m leading variables

meaning n-m non-leading ones

and n-m parameters

and since n>m, n-m > 0

[twilit field]

So I'm essentially proving this theorm that they expect me to use

[dreamy lichen]

This theorem relies also on rank

but it wouldnt be too hard to reach the conclusion from that theorem

[twilit field]

$\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} & | & b_1 \
a_{21} & a_{22} & \cdots & a_{2n} & | & b_2 \
\vdots & \vdots & \ddots & \vdots & | & \vdots \
a_{m1} & a_{m2} & \cdots & a_{mn} & | & b_m
\end{bmatrix}$

[dreamy lichen]

if you want to avoid it though, you can still do this

[grand pond]

(why am i here )= idk

[twilit field]

here I assume n>m

[dreamy lichen]

Why do you even need to show the matrix?

[twilit field]

Wdym

[dreamy lichen]

you can just say "Let A be m x n matrix, where n > m"

[dreamy lichen]

then do this argument

and you will immidiately get that it has at least 1 parameter

[twilit field]

I just feel this will help me understand what I'm working with since I learnt most of this only yesterday.

[dreamy lichen]

Hmm, it might be better to use particular examples for that

bringing general matrix to RREF is not particularly easy

[twilit field]

I'm not bringing it to RREF

this is just for my reference

[dreamy lichen]

okay then

[dreamy lichen]

Anyway, you will probably still want to avoid doing this dividing

it only makes you put a new and unnecessary constraint, that a's on diagonal have to be non-zero

[twilit field]

now the elements of the ith row is divided by $a_{ii}$ where none of the $a_{ii}$ are non-zero$ from i=1 to i =m

[grand pond]

(why am i here )= idk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

I think it's fine for now

I'll deal with the other case later

$\begin{bmatrix}
1 & a{12} & \cdots & a{1n} & | & b_1 \
a{21} & a1 & \cdots & a{2n} & | & b2 \
\vdots & \vdots & \ddots & \vdots & | & \vdots \
a{m1} & a{m2} & 1(a_{mm)}&\cdots & a{mn} & | & b_m
\end{bmatrix}$

[grand pond]

(why am i here )= idk

[twilit field]

Now on reducing this to RREF, I get the solutions to m of the n variabes

which leaves me with n-m parameters

and thus infinitely many solns

[dreamy lichen]

Assuming all a's on diagonal are 0...

[twilit field]

wdym

The other case?

[dreamy lichen]

Why not just reduce this to RREF

you can avoid the condition completely

[twilit field]

Even on reducing that to the RREF the same argument applies

[dreamy lichen]

yeah, so why make it a seperate case when the exact same argument works?

It feels so random

It's like if I made case for when riemann hypothesis is true and when it's false

when it's true, reduce it to RREF

when it's false, reduce it to RREF as well

[twilit field]

Oh yeah

xD

Thanks!

[dreamy lichen]

np

[twilit field]

.rank

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[main remnant]

Where are you stuck? This is just solving 123+x=456 4 times

[flat veldt]

Simplify the two matrices multiplying each other

[midnight plank]

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

[frigid turret]

no. 4

[flat veldt]

Oh ok

Then you should show your steps

[frigid turret]

@flat veldt okwait

.close

[midnight plank]

Channel closed

Closed by @frigid turret

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timber arrow]

can someone help explain how to interpret this data

and start me off

[midnight plank]

@timber arrow Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sly nova]

my thing closed

[midnight plank]

Please don't occupy multiple help channels.

[last slate]

which was?

[sly nova]

i had one for help

which i need

but

it closed

[fallen aurora]

https://discord.com/channels/268882317391429632/908078124405755914

.close

[midnight plank]

Channel closed

Closed by @fallen aurora

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timber arrow]

i just learnt about asymptotes

so bit unsure

can anyone guide

[inland oriole]

do you know how to calculate the asymptotes?

[delicate sage]

Do you think it will have an asymptote ?

[inland oriole]

@timber arrow

[timber arrow]

i think so

it cant have any vertical ones right

cuz 2^x is always positive

[inland oriole]

correct

yes

[timber arrow]

so u cant ever make it -1?

ok nice

[inland oriole]

but do you know how to calculate the asymptotes?

[timber arrow]

so

[inland oriole]

the horizontal ones

[timber arrow]

when x is infinity

or - infinity

[inland oriole]

yes

[timber arrow]

so 2 to the infinity is

[inland oriole]

infinity

[timber arrow]

infinity + 1

[inland oriole]

yeah

[timber arrow]

yeah so infinity

[inland oriole]

and if we have infinity in the denominator

what will the function approach?

[timber arrow]

oh then 1/infinity is like close to 0

[inland oriole]

yeah

[timber arrow]

ohhh nice

[inland oriole]

so if x becomes really big it will approach 0

[timber arrow]

so as x becomes large u get y = 0

but it doesnt ever touch 0 right

[inland oriole]

and what happens if x approaches -infinity?

[inland oriole]

yes

[timber arrow]

2 to the -infinity

hm

indicies

so 2/infinity

[inland oriole]

yes

[timber arrow]

then oh wait so y becomes infity

cuz it goes up

to the numerator

[inland oriole]

nono

[timber arrow]

2/2/infinity

[inland oriole]

2/infinity

what's that

just 0 right?

[timber arrow]

waiut but

isnt it 2 to the - infinity

[inland oriole]

so we have $\frac{3}{0+1}$

[grand pond]

pixel

[inland oriole]

because 2^-infinity is 0

[timber arrow]

ohhhhhhhhhhhhh

[inland oriole]

so in the negative direction it will have an asymptote at what y?

[timber arrow]

wait why is 2^-infinity 0

i can j sketch 2^x right

oh yeah

[inland oriole]

yes

[timber arrow]

for negative x it just becomes close to 0 ok yeah

so 3/0+1 = 3

[inland oriole]

yeah

[timber arrow]

ahh so horizontal asymptotes at y=0 and y=3

[inland oriole]

yes

[timber arrow]

hence D

[inland oriole]

yes

[timber arrow]

cuz intercept is at 1/2

[inland oriole]

very good

[timber arrow]

thanks

[inland oriole]

ur welcome

close the channel with .close btw

if you're done

[timber arrow]

oh yh cheers

.close

[midnight plank]

Channel closed

Closed by @timber arrow

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[heavy lynx]

could someone please explain how you get u(x,y) here

[distant heron]

You want to put $x +iy$ in exponential form. How do you do it in general? You find the module and the argument

[grand pond]

red

[distant heron]

What's the module here? and what's the argument?

[heavy lynx]

i used x+iy = e^(i theta)

to find the v(x,y) is just the uhh

arctan y/x

[distant heron]

Yeah but you seem to forget that e^(i theta) has a module of 1

[heavy lynx]

ohh wait

um sorry bare with me one second im a bit slow with this

[distant heron]

Don't worry