help-23

[clever ridge]

ooh yea

wait i setup that peiewise wrong i think

my book has it like this instead

do you know why?

[winged pecan]

wtf

[clever ridge]

oh shit

neverminmd

[winged pecan]

must've wrote it wrong

xd

[clever ridge]

thats how they have the derivatiive setup

so this doesnt have an absoulte max does it?? im wondering if it has a local either?

[winged pecan]

no absolute or local max is my belief

[clever ridge]

hwo would i show that wit this?

[winged pecan]

??

[clever ridge]

so to show critical points

you make f'(x) = 0

how would i do thta here?

[winged pecan]

we know f'(x) for both when x > 0 and x < 0 and none of them are 0 since they are 1 and -1

so there are no critical points

and we agreed that f is not differentiable at 0

[clever ridge]

oh so i can write 1 doesnt = 0 and -1 doesnt = 0...

but how will i show that the min is at 0

[winged pecan]

f(0) = 0, we know that.....

[clever ridge]

ok

[winged pecan]

for x > 0, f(x) = x and so f > 0 for all x > 0 and f(x) = -x when x < 0, negative times negative is positive so f > 0 for x < 0 as well

[clever ridge]

yea

so for f(x) = e^x

has no min or max right?

[winged pecan]

nope

[clever ridge]

how would you do f'(x) = 0 for that then?

like e^x = 0?

[winged pecan]

no solutions

[clever ridge]

would it be like $e = ^xsqrt(0)$?

[flat frigate]

wakamole

[clever ridge]

but how do you show that?

[winged pecan]

maybe like ln(e^x) = ln(0) but ln(0) is undefined

[clever ridge]

ok yea ty

i keep forgetting these properties!!!

i feel like im going to fail my teset because of this

[winged pecan]

u took a long time to respond to this one? did u do it by urself? 😅

[clever ridge]

i just did this

[winged pecan]

well that is not a complete proof (insert fitting emoji)

[clever ridge]

y

[winged pecan]

u want to show that (0, 0) is an absolute minimum right?

[clever ridge]

wait before we go there, if we do ln(e^x) = ln(0) we can simplify further.. to x = ln(0) right?

[clever ridge]

yea

[winged pecan]

ig , then x = ln0 which is undefined or like ln 0 is not a value so there is no value for x

[clever ridge]

cool just making sure lol

ok bak to the othter 1!

[winged pecan]

but yeah, u just wrote f(0) = 0, how does that show (0, 0) is an absolute min? Because for it to be an absolute minimum f(0) must be less than or equal to f(x) for all x ∈ R

[clever ridge]

idk actually

ur right

hahaah

i got the derivative then did that

how does this also show theres no max as well?

[winged pecan]

maybe we can just logic our way to that conclusion

f'(x) for x > 0 is 1 and x < 0 is -1 meaning for x > 0, the function is always increasing

for x < 0, we know that as x ---> -∞, f(x) ---> ∞ and because f'(x) = -1 for x < 0, it's just always decreasing until we reach x = 0. So hopefully this logically explains why there is no local or abs max

[clever ridge]

ill just write it

waitr i already did

in the ppiecewise

but hwo did i get the '0' to plug int f(x) ?

like why

oh well

honestly idc at this point

[winged pecan]

lol, i have to go now it was nice talking or helping u

[clever ridge]

Does this look right? Its asking to find the critical numbers of the function

Ohh ok

Bye that’s fine I’ll open a. New one!

.close

[safe radish]

Channel closed

Closed by @clever ridge

Use .reopen if this was a mistake.

[winged pecan]

wait i'll look at this

[clever ridge]

.reopen

[safe radish]

✅

[winged pecan]

at v^2 = 4

v = (+ or -) sqrt(4)

[clever ridge]

shittttttttt

im gna fail seriusly

[winged pecan]

v = (+ or -) 2

[clever ridge]

no if its -2 it would be 20

[winged pecan]

3(-2)^2 - 12

[clever ridge]

no u plug it bak into original equation

oh wait

so maybe the answer is just 2 , -2?

yeah

ok idk what that extra step is for tbh

[winged pecan]

what extra step

[clever ridge]

i plugged them abk into the derivative

i guess its for the y values

v is the x axis

are you good at deriving?

[winged pecan]

differentiate*

[clever ridge]

right haha

Should I do ln and do implicitly here

cause that looks like a nasty quotine rule

[winged pecan]

so this isn't asking u to differentiate, it wants to solve for p?

[clever ridge]

no its same question different

equation

critical numbers

i need to differentate remember/

[winged pecan]

oh is it h(p)?

i thought it was ln(p) at first

[clever ridge]

no h(p)

im saying theres a rule for implicit i think i can add ln to both sides

so i can instead of doing quotient rulw

i just use log properties

isnt that what its for?

[winged pecan]

u want to do implicit differentiation?

[clever ridge]

would it be easier that way

[winged pecan]

I remember when I took calc, there was a trick with ln but honestly I forgot

[clever ridge]

well by looking

at it

qoutient rule seems annoying

g' needs chain rule

right?

o wait nevermind

[winged pecan]

I don't think u need chain rule, quotient rule is good enough

ig u could look on youtube for the implicit differentiation with ln thingy

[clever ridge]

Does this look right?

oops i added this on it

[winged pecan]

I think u have a good grasp of the rules and such so if it is wrong then it means u made a silly mistake. You can check on desmos the graphing calculator to compare if the graphs are the same

[safe radish]

@clever ridge Has your question been resolved?

[clever ridge]

i got thtat part right but have no idea how to solve for p

for critical points

[grave lodge]

yo what is that you hash

[winged pecan]

yooooooo

deez rules of freakn engineer

[grave lodge]

lmao

nice to see u here

[winged pecan]

bruh we both got banned from gotham city

[clever ridge]

hwo do you solve for p?

[winged pecan]

hru bruh

[grave lodge]

no i just left

[grave lodge]

since the r.h.s is 0 you can get rid of the denominator

and solve the quadratic

[clever ridge]

r.h.s ?

[grave lodge]

right hand side

[clever ridge]

y does that mean get rid of the denpominator?

[winged pecan]

0 = (2p^2 + 2p) / (p^4 + 4) multiply both sides by p^4 + 4

divide both sides by 2

0 = p^2 + p

[clever ridge]

oh[

[winged pecan]

0 = p(p + 1)

[clever ridge]

ya

i c

unless you can make the denominator a VA right?

cause im trying to find the critical pooints

why divide both sides by 2

[winged pecan]

make it simpler

[clever ridge]

its -p^2 + 2p + 4

not -2p^2 + 2p + 4

[winged pecan]

im looking at the wrong thing

[grave lodge]

the critical points in this case aren't integers

[flat frigate]

TheRuleOfEngineering

[winged pecan]

yeah ^

[clever ridge]

yea fugg

[winged pecan]

oh cool, i got banned for calling one of the mods gay

[grave lodge]

[flat frigate]

TheRuleOfEngineering

[clever ridge]

i got -1 + sqrt(5)

cause denominator is -2 and b is 2

damn its -b

[grave lodge]

you can also flip the signs to make it simpler

[flat frigate]

TheRuleOfEngineering

[clever ridge]

i actually had that and the neg sign just dissappeared halfway through

its ok i got it

[grave lodge]

lol relatable

[clever ridge]

so that is the critical pooints then... 1 + sqrt(5), 1 - aqrt(5)

[grave lodge]

yep

can u send the original question

just to make sure

[clever ridge]

yeah

#36

[scarlet oracle]

Hello, does anyone know how to identify circuits in series, parallel and with?

[clever ridge]

channel is currently occupied

not that i care but i just wana finish this lol

[grave lodge]

ping me in an open channel

[winged pecan]

im probably gonna submit an unban request to gotham city, the stem channel in that server is better than the helping mechanism for this server I think

[grave lodge]

nope this server is better

[clever ridge]

does that look right?

[grave lodge]

yep the critical points check out

[clever ridge]

kk

.close

[safe radish]

Channel closed

Closed by @clever ridge

Use .reopen if this was a mistake.

[clever ridge]

.reopen

[safe radish]

✅

[clever ridge]

they just wana check my algebra at this point 😦

40

ill open a new channel

.close

[safe radish]

Channel closed

Closed by @clever ridge

Use .reopen if this was a mistake.

[clever ridge]

.reopen

[safe radish]

✅

[grave lodge]

u can keep using this one

[clever ridge]

ok

i reworote it as a fraction and going to use chainrule

[grave lodge]

yep

[scarlet oracle]

I sent you a private message

[grave lodge]

just go to #help-45 and ping me

<@&268886789983436800> advertising

[winged pecan]

if it's in series, then the two components are connected only at one terminal and if parallel then both terminals are connected im pretty sure

[clever ridge]

can you please help him in a dif channel, no offense

[grave lodge]

this is almost correct

the denominator would be $3(4-x^2)^{\frac{2}{3}}$

ok cbrt is not a command nice

[clever ridge]

thats what i have

[flat frigate]

TheRuleOfEngineering

[grave lodge]

just gotta add the 3

[clever ridge]

ohhh

damn

ok so the first crit point is 0 because it is undefined at 0

then i can solve for the horizontal slopes = 0

[scarlet oracle]

Yes, but in the image they sent me I cannot identify them.

[grave lodge]

bruh go to an open help channel, ill help you there

[winged pecan]

lmao, this channel has two helpees

[clever ridge]

ok this is confusing

am i wrong about it being undefined at 0 in the denominator?

oh fuq

i meant 2

2 is undefined in the denominator

and neg 2?

or just 2

[grave lodge]

its undefined at 2 and -2 yep

but those arent critical points

i think the only critical point is (0,0)

[clever ridge]

i forgot how the parenthases work in that situation

yea i got that as well

[grave lodge]

you can always graph the equation to verify ur answers

[clever ridge]

yea

[winged pecan]

engineer, do u know randomized testing?

[grave lodge]

yea

[winged pecan]

so u know statistics?

[clever ridge]

is the DNE not critical?

i guess they cant be if they DNE

[grave lodge]

nope

[grave lodge]

yeah

if by randomized testing you mean the software-testing technique with random input generation

[winged pecan]

that seems like something that would be part of software engineering but im saying randomized testing in statistics

[grave lodge]

oh

in that case, not really

im software eng

[winged pecan]

btw, u still talk to oops?

[grave lodge]

oh damn i forgot about oops

[winged pecan]

@clever ridge sorry it's supposed to be ur help channel, im gonna head out since im not helping

lol u talk to jace?

[clever ridge]

Is this looking ok?

[grave lodge]

it would stop at $2x\ln x + x$

[flat frigate]

TheRuleOfEngineering

[grave lodge]

cant add those two up

[clever ridge]

oh really why not

ru sure abt thta

[grave lodge]

2x + x would become 3x

but 2xy + x, would that become 3xy

[clever ridge]

True

[flat frigate]

TheRuleOfEngineering

[grave lodge]

but yeah thats about it

[clever ridge]

Right

which is better equation to have when solviing for x?

not the factored one ya?

[grave lodge]

you just gotta make sure whichever equation you use, you consider all possible cases for x where the function=0

you could use either equation

[clever ridge]

hmm

im trying to do both but i cant seem to get the same answer

in the factored one you can do x = 0

but i think that's it

in the non factored one i get ln(x) = -1/2

im not sure what properites i could use though for ln

i am rly bad at that

[winged pecan]

I don't think that LHS division was done properly, and in addition dividing both sides by 2x here isn't very helpful

[clever ridge]

i need to use ln properties here

it is done right

[winged pecan]

like (2 + lnx) / 2x ≠ ln x

[clever ridge]

?

thats not what i have written

wait

it does though

[winged pecan]

(a + b) / c = a/c + b/c

[clever ridge]

i am getting rid of 2x on the left side

and moving it to the right side

brb im taking a shower

[winged pecan]

u have -x / 2x which already simplifies to -1/2 if im not mistaken

[grave lodge]

this is kinda why i prefer using factorized equations when the other side is 0

actually hold on just a minute

[flat frigate]

TheRuleOfEngineering

[winged pecan]

idk, im pretty sure the algebra is wrong but i saw on the desmos, that the answer is the same, so idk what's going on

[grave lodge]

i think the algebra is fine

@clever ridge at the end u got ln(x) = -1/2
both the factorized and non factorized equations give you the same result

[flat frigate]

TheRuleOfEngineering

[safe radish]

@clever ridge Has your question been resolved?

[safe radish]

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[clever ridge]

yea but cant be 0 cause that would be undefined in the original equation

[shadow glade]

.close

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[tender kernel]

Hey there 👋 I need help with Calc 2

I understand that this is an integration by parts problem, but I am unsure on how to do this.

[light shoal]

but you got the right answer, how did you find it

[tender kernel]

there is an answer key posted, but I need to show how to get to my awnser

[light shoal]

ah i see

[tender kernel]

but I'm not too sure how to get to that

[light shoal]

do you know how to do integration by parts in general?

[tender kernel]

In general; I know the concept is kinda splitting it into two?

and then going downish

[light shoal]

yeah it should be something like $\int u,dv = uv - \int v,du$

[flat frigate]

Bungo

[light shoal]

if you're familiar with that form

[tender kernel]

I think I've seen it yeah

errrrr

so I'm assuming its like

U = t
u' = 1

[light shoal]

generally idea is that you let u be part of your integrand, and dv the rest of the integrand
and you want to choose u to be something that gets simpler (ideally just becomes a constant) when you differentiate it

so yea u = t is a good choice

[tender kernel]

v = e^4t
v' = e^4t/4

[light shoal]

except you're integrating

dv = e^(4t)

v = (1/4)e^(4t)

right idea but wrong notation

[tender kernel]

we're only integrating the last part right?

[light shoal]

so now plug all that into the earlier formula

yep the uv part doesn't get integrated

just the v du part

[tender kernel]

okkk

soooo

that means

( t * e^(4t) ) - Integration of (1/4)e^(4t) * 1)

[safe radish]

@tender kernel Has your question been resolved?

[safe radish]

@tender kernel Has your question been resolved?

[safe radish]

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[flat frigate]

jesko

jesko

[gusty trench]

sqrt is the same thing as raising to the 1/2 power

[rigid valley]

uh huh

[gusty trench]

so you can use the properties that (xy)^z = x^z * y^z, and the property that (x^y)^z = x^(yz)

[rigid valley]

wait im processing

[odd arch]

pls do

while @gusty trench is explaing in language

[gusty trench]

(81a^7b^12c^9)^(1/2) is the same as if you distributed the 1/2 power to the various factors

[rigid valley]

so b^6 and c^4?

[rigid valley]

uhhhhh

[gusty trench]

so it's equivalent to (81)^(1/2) * (a^7)^(1/2) * (b^12)^(1/2) * (c^9)^(1/2)

[flat frigate]

jesko

[rigid valley]

oh wha tthe heck

[gusty trench]

and a power to a power is the same as if you multiplied the powers

the 81^(1/2) is just sqrt(81) = 9

(a^7)^(1/2) = a^(7/2), etc

but the idea is to try to bring out as many whole copies of a as you can

a^(7/2) = a^3 * a^(1/2)

[flat frigate]

jesko

[gusty trench]

which is what @odd arch is doing but bringing out an even power under the square root

[subtle yew]

$\sqrt{a^7}=\sqrt{a^6\cdot a}=\sqrt{a^6}\cdot\sqrt{a}$

[flat frigate]

Vѳrtєx-

jesko

[rigid valley]

oh wait i see it kind of

[gusty trench]

the idea is to write the stuff under the radical as a bunch of perfect squares

[rigid valley]

uh huh

[gusty trench]

distribute the square root, simplify taking the square root of a perfect square, and just leave what's left under the radical

[rigid valley]

so you just divide the exponent inside by 2 because it's square root right

[subtle yew]

yeah start with $=\sqrt{81}\cdot\sqrt{a^7}\cdot\sqrt{b^{12}}\cdot\sqrt{c^9}$

[subtle yew]

which exponent

[rigid valley]

like all fo them basically i guess

[gusty trench]

so for example, sqrt(x^5y^4) is the same as sqrt((x^2y^2)^2 * x), which is the same as sqrt((x^2y^2)^2) * sqrt(x), which is the same as x^2y^2 * sqrt(x)

[flat frigate]

Vѳrtєx-

[subtle yew]

yea except for the odd numbers like 9 and 7

[rigid valley]

i see

[subtle yew]

then split it up to make it even (like you did with a^7)

[rigid valley]

uh huh

i get it

thank you goats.

should i use khan academy or soemhting else

cuase im kind of iffy on this thing

[flat frigate]

jesko

[subtle yew]

yea or other practice websites prolly as well

[rigid valley]

eughhhh okay

thank you

[rigid valley]

any recs :3

[odd arch]

prof. Leonard ig

on youtube

[rigid valley]

ok thank you goat.

!close

oops

[odd arch]

the prefix is .

[rigid valley]

thank you again

.close

[odd arch]

.

[safe radish]

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[tepid walrus]

don't spam smh my head

<@&268886789983436800> minor amount of spam in a couple channels

[gray flume]

Dead

[safe radish]

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[fluid prairie]

just to be sure, is this 10 square units?

[subtle yew]

yuh

[fluid prairie]

thanks

.close

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[safe radish]

Please don't occupy multiple help channels.

@mental moon Has your question been resolved?

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@mental moon Has your question been resolved?

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[jolly cipher]

this is from a friend of mine. Just trying to get him some help 🙏

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@jolly cipher Has your question been resolved?

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@jolly cipher Has your question been resolved?

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[noble flicker]

how do i find the residue of 3/(z-i ) at i

[noble flicker]

i tried doing this : 3/i(-1+z/i)

3/-i(1-z/i)

then 3i/(1-z/i)

in order to use this

[low vector]

have you had the residue theorem?

[noble flicker]

yes

[toxic stratus]

the residue of a laurent series is the coefficient of the -1th term

[noble flicker]

yes i know that

but why what i'm doing is wrong

[toxic stratus]

well the laurent expansion of $\df 3 {z - i}$ around $i$ is $\df 3 {z - i}$

[noble flicker]

but how

[toxic stratus]

do you know what a laurent series looks like?

[noble flicker]

this doesn't give me that

[noble flicker]

series with negative powers of z

and residue is a_-1

[toxic stratus]

no

the expansion is around i

so its powers of (z - i)

,, \f 3 {z - i} = \sum_{k = -\infty}^\infty a_k (z - i)^k = \f 3 {z - i}

[toxic stratus]

3 is the coefficient of the -1th power

[noble flicker]

ok i get it

thank you

can i also ask about somethin similar

i don't understand this reasoning

[toxic stratus]

do you know what it means for g to have a simple pole at z = 1

[noble flicker]

it depends what you mean

i think i understand this

[toxic stratus]

it depends

[noble flicker]

that there is a lauren expansion at z=1?

[toxic stratus]

they have a typo there

not just that there is a laurent expansion

,, g(z) = \2r{\underbrace {\f {a_{-1}} {z - 1}}{\text{principal part}}} + \2b{\underbrace {\sum{k = 0}^\infty a_k (z - 1)^k}_{\text{analytic part}}}

[noble flicker]

yes

i agree

[toxic stratus]

the fact that g has a simple pole at z = 1 means that the principal part is order 1

so the negative powers only go up to -1

[noble flicker]

how can you be sure for that

[toxic stratus]

thats the definition of simple pole

[noble flicker]

so for a double pole the negative powers go up to -2

[toxic stratus]

yes

also a_{-1} has to be nonzero here

anyway

you just sub this into the expression for f

,align f(z) & = \f 2 {\2g{z - 1}} \parens [\bigg] {\2b{\f {a_{-1}} {z - 1} + \sum_{k = 0}^\infty a_k (z - 1)^k}} + \f 1 {(z - 1)^2} \
& = \f {2a_{-1}} {(z - 1)^2} + \f {2a_0} {z - 1} + \sum_{l = 0}^\infty 2a_{\2r{l - 1}} (z - 1)^l + \2o{\f 1 {(z - 1)^2}} \
& = \f {2a_{-1} + \2o1} {(z - 1)^2} + \f {2a_0} {z - 1} + \sum_{l = 0}^\infty 2a_{\2r{l - 1}} (z - 1)^l

[noble flicker]

ok

but how does this give the answer

[toxic stratus]

the 1/(z - 1) shifts all the powers down by 1

[noble flicker]

and at the second row should k start from 1?

[toxic stratus]

ive shifted everything already

[noble flicker]

ok

[toxic stratus]

the part in green causes all the powers of the blue bit to shift down by 1

and the part in red is the shift

[noble flicker]

ok

[toxic stratus]

okay so

you now need to analyse what happens in the various cases presented to you

[noble flicker]

it is a simple pole

we see this

[toxic stratus]

the orange part gets combined into the fraction with (z - 1)^2 denom

[noble flicker]

yes

[toxic stratus]

its not a simple pole if the (z - 1)^-2 term remains

so the first option is false

so long as this term is nonzero, you have a double pole

this is false because you can see the residue of f is 2a_0, which is not necessarily 0

and that means this is also false

since $\Res_f(1) = 2a_0 \ne 2\Res_g(1) = 2a_{-1}$

[toxic stratus]

finally this is true because

[noble flicker]

the first option said what you already said before

[toxic stratus]

because when $\Res_g(1) = a_{-1} \ne -\f12$, you have a double pole

[noble flicker]

.

[toxic stratus]

oh it says not

well

when $\Res_g(1) = a_{-1} = -\f12$, you don't have a double pole

[toxic stratus]

this term will vanish

[noble flicker]

ok

[noble flicker]

is this false because a0 can be anything?

[toxic stratus]

well 2a_0 and 2a_-1 are just not equal in general

[toxic stratus]

see here

[noble flicker]

how do you know that 2a_-1 is the residue of g

[toxic stratus]

its 2 times the residue of g

kind of literally by definition

[noble flicker]

but why specificaly not -1/2

[toxic stratus]

again, its all to do with this term

,, 2a_{-1} + 1 \ne 0 \Iff a_{-1} \ne -\f12

[toxic stratus]

the existence of the (z - 1)^-2 term causes a double pole

[noble flicker]

ok i get it

thank you

[toxic stratus]

👍

[safe radish]

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[knotty jungle]

when they say like 'given any (4×4)-matrix A and the (4x4)-matrix M1' and they want u to check out which given statements r correct whats like the easiest matrix to pick for A?

[knotty jungle]

if you're going to multiply A*M1 for example

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[finite yacht]

It depends on the statement

[quick crater]

huh i drilled trough like 10 questions in a channel once

[knotty jungle]

Which of the matrices below has
as its first column the first column of matrix A,
as the second column the double of the fourth column of matrix A,
as the third column the third column of matrix A,
as the fourth column the second column of matrix A?

what would be the best option to use for A

[finite yacht]

You don't have to choose a matrix A here

This is about the properties of the product of matrices in general

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[finite yacht]

If you multiply two matrices A and B then each column of AB equals A times the corresponding column of B

And A times a vector b is a linear combination of the columns of A, with the numbers of b corresponding to the coefficients

For example, consider the product
$$
\begin{pmatrix}
a_1 & b_1 \\
a_2 & b_2
\end{pmatrix}
\begin{pmatrix}
2 & 0 & 1\\
1 & 1 & -1
\end{pmatrix}
$$

[flat frigate]

d

[finite yacht]

The result is a matrix with two rows and three columns

The first column is 2 times (a_1 a_2) plus 1 times (b_1 b_2)

Because you have the vector (2 1)

And the second and third are similar

[finite yacht]

So here you have to use this idea but for 4x4 matrices

[knotty jungle]

yeah but you still used a matrix

what if i want to use numbers

[topaz tree]

when you use a matrix with numbers you will prove it for that particular matrix only

instead use a matrix with variables

[knotty jungle]

ok

thx

.close

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[cyan vale]

any progress or stuck

[dense sand]

stuck

i dont know how to apply the identity

[safe radish]

Show your work, and if possible, explain where you are stuck.

[dense sand]

sorry abt the mess
idk what I'm doing

[safe radish]

@dense sand Has your question been resolved?

[wooden dirge]

Use the definition

What are (n over r) and (n over r - 1) ?

(By definition)

[dense sand]

mhm?

[wooden dirge]

Have you seen what this symbol mean ?

[dense sand]

nCr

ohh hold on
so it would be something similar to this

[wooden dirge]

Oh no, there is much simpler

In my opinion

What is the definition of nCr ?

Actually, when I say "there is much simpler", now that I read your screen, I realize that it has nothing to do with your exercise. Your screen prove a property based on what you want to show, so you cannot use it to solve your exercise

(And no, the solution of your exercise has nothing to do with your screen. It is simply writing definitions)

So, I insist, what is the definition of nCr ?

[dense sand]

choosing r things out of n?

[wooden dirge]

Yes, that is the intuition/motivation

And the mathematical definition ?

(Or the formula to compute it, if you prefer)

[dense sand]

n! / r!(n-r)!

?

ohh the problem is, using n! / r!(n-r)! is part of another question

i know how to prove it using n! / r!(n-r)! but idk how to prove it with that identity

[wooden dirge]

Oops, I removed the wrong answer

[wooden dirge]

But yes

[wooden dirge]

I don't understand

Oooh, okey

Now I understand

[wooden dirge]

Thus, you will use this Newton's binomial formula, yes

First, compute
(1 + x)^(n+1) with it
Then, compute
(1 + x)^n with it
After, multiply this last result by (1 + x) and compare

[dense sand]

so i have to use this?

OHH OH I GOT IT NOW

[wooden dirge]

Yes

[dense sand]

THIS HELPED ALOT

[wooden dirge]

xD Perfect xD

[dense sand]

i had smth written in my book during my lecture but reading it now i didnt know what it said 💀

so i had to redo the whole proof

oh thank goodness oh my gosh

[wooden dirge]

No problem xD

[dense sand]

THANK YOU SO SO MUCH

LIFESAVER

ABSOLUTE LEGEND ✨

[wooden dirge]

Oh, I did not expect your teacher to be so dangerous xD

[dense sand]

i have a test tmrw and i needed to know how to do this 🥲

thank you so much for helping 🫶

[wooden dirge]

You are welcome ^^

[dense sand]

Take care ❤️

.close

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[vestal bridge]

Couldnt find help

[vestal bridge]

algebra2

If for all $x ||Tx|| = ||x||$then T is unitary

[flat frigate]

homer

[cunning pasture]

what's this?

[vestal bridge]

Unitary opreator

I want to show $TT^*=T^*T=I$

[flat frigate]

homer

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@vestal bridge Has your question been resolved?

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@vestal bridge Has your question been resolved?

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[empty pumice]

Can anyone solve this by easiest methods

[daring cedar]

!nosols

[safe radish]

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.

[daring cedar]

which number do you need help with

[safe radish]

@empty pumice Has your question been resolved?

[safe radish]

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[lean otter]

Hello

[lean otter]

I think this is a really basic question but I got confused about it

I'm doing this problem right now "A driver sets out on a journey. For the first half of the distance, she drives at the
leisurely pace of 30 mi/h; during the second half she drives 60 mi/h. What is her
average speed on this trip?"
So while looking at the steps at solving problems the books says that I need to multypy the denominator and numerator by 60... But I don't understand why that number and not another one

!help

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lean otter]

<@&286206848099549185>

[gusty trench]

!15mins

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[cunning pasture]

to remove the fraction ig?

lcm

L.C.M will be the lowest number so that the fraction goes extinct

[lean otter]

IS that possible in this case because the fractions are being added

I thought canceling is only possible when you're multiplying{

[cunning pasture]

Idu

[lean otter]

What is idu?

[cunning pasture]

i don't understand your question

[lean otter]

Oh I got it!

I understand

the problem

!close

.close

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[elfin folio]

If I'm guessing correctly, are polynomial functions derived from various behaviors, and can the input values be applied to see how the function behaves? And factoring them somehow, due to equivalence, would let you see various other parts of that function when graphed?

[elfin folio]

?

<@&286206848099549185>

[daring oyster]

Yes

and Understanding Polynomial Functions:

When graphed, the polynomial function shows a curve that can be analyzed for various characteristics like intercepts, turning points, and end behavior.

[elfin folio]

How are the values relations to the X, Y, or Z axis's determined?

[daring oyster]

The roots (found by factoring) are the points where the graph intersects the
𝑥
x-axis.

[elfin folio]

sqrt?

[daring oyster]

No the solutions

[cunning pasture]

https://tenor.com/view/growing-plant-tree-roots-gif-10041802

[cunning pasture]

the value of x so that y =0

comes at the root of the y-axis

which is y=0

[daring oyster]

That's a interesting way to say it

Basically the roots are the solutions you get after factorisation

[elfin folio]

I don't understand how any of those are related, as I was just asking how they got x or y in the first place.

[daring oyster]

They get x by factorisation

[elfin folio]

Are each X value considered as a different axis?

[daring oyster]

Each x value is plotted on a graph on the x-axis

[elfin folio]

Because are the X values considered different types of data or points of interst?

So what bout the C value?

would that be the Y axis?

[daring oyster]

Constant?

[elfin folio]

No, C as a variable.

Or are constants what increase the Y value?

[daring oyster]

Where is C here?

[elfin folio]

I made it up

[daring oyster]

No

[elfin folio]

If you had a variable named C would that increase the Y value?

[daring oyster]

C instead of x you mean?

[cunning pasture]

depends

[daring oyster]

If you meant C instead of y, it would become a linear equation with different points on different axis

[cunning pasture]

ofc

[elfin folio]

Meta Calculations

[elfin folio]

I'm going to ask this again.

What do the variables represent, and how are they related to the X and Y axis's? Do you specefically need to name it Y for it to exist on both axis's?

Just using a function alone you wouldn't get the Y axis if you were only adding to the X axis.

[daring oyster]

You equations you sent above are polynomial functions

[elfin folio]

And?

WAIT

NO

WHAT????????????????????????????????? THEY WERE DIFFERENT?

[daring oyster]

This one I mean

Yes

[elfin folio]

Ok.

Meta Calculations

[elfin folio]

3rd adjustment:

How do polynomials and quadratics differ?

[daring oyster]

A quadratic function is a polynomial function

[cunning pasture]

the highest degree is 2

[daring oyster]

But not all polynomial functions are a quadratic function

[cunning pasture]

yeah

[elfin folio]

Ok then.

So, as I am still asking, where exactly did the Y axis appear?

[daring oyster]

Y is the range or co-domain of the function

That is, the set of all possible solutions of the given equation

If you need in simple terms,

[elfin folio]

Calculations

[daring oyster]

You basically find the all the values here and you'll get a value for y after inputting them

That gives the graph along with x axis

[cunning pasture]

yea

he'll now come tomorrow.

[daring oyster]

Lmao

He didn't even close it

[elfin folio]

I'm still thinking about what you said, and trying to think of a model that would explain all this.

[daring oyster]

Alright

[cunning pasture]

he did it twice yesterday and day before it. but nvm.

[daring oyster]

You're secretly lurking everywhere I c

[cunning pasture]

let's help him

What?

[daring oyster]

Tell your problem clearly

[elfin folio]

is this the y axis

the x without ^2

[cunning pasture]

y-axis is the line that is perpendicular to x-axis

the whole line

[daring oyster]

[elfin folio]

meolve how are your answers so off topic i swear

like its hard not to mention it

like how earlier i was asking for the answer to how they got the x and y axis and it looked like you answered by using the knowledge i was already asking about to try to explain it

sigh

i dont get it

so thats just normal to you?

[cunning pasture]

what?

[elfin folio]

the y axis will simply magically appear

and graph what?

[cunning pasture]

wdym 'y-axis'?

[elfin folio]

the numbers already move linearly up or down along the X axis

[daring oyster]

You get you basically get (x-5)(x+4) that's the solution

[elfin folio]

what the hell would the y axis graph if x was positive or negative direction

o h

[elfin folio]

so they are 2 separate answers and the 2 axis's are both of them graphed?

and both of them represent the positive and negative directions?

[daring oyster]

Yes

[elfin folio]

BRO

[daring oyster]

They are just points

[cunning pasture]

,w graph (x-5)(x+4)

[flat frigate]

Results provided by WolframAlpha

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

Requested by anshulyolo121

[daring oyster]

Yup this is it

[elfin folio]

Meta Calculations

4th adjustment

[daring oyster]

A parabola that connects the 2 points give the graph

[cunning pasture]

2 roots

uhm

[elfin folio]

i dont get it how do you NOT see that?

[daring oyster]

The y value will be the product of the roots

[elfin folio]

how dont you see what im seeing?

also i just realised something

smart people DO NOT weave past experiences to solve current experiences without altering the data because this clearly is used very differently

[daring oyster]

Not necessarily

[elfin folio]

cuz i tried that and that killed me so many times

unless idk they pull a colin galen but nvm

[daring oyster]

Could you phrase it better?

[daring oyster]

There is no data being altered here

[elfin folio]

you cant think oh i saw this somewhere else this is probably whats happening because 99% of the time in math it seems they will literally show you the same thing but different in such a way that if you take the same interpretation as something was prior you can literally get stuck for hours

literally all the knowledge i learnt in math feels like its useless for problem solving

because any time i try to weave the current problem with past problems something dumb happens like they used the same thing but change its semantic context so it means something entirely different

[daring oyster]

You just gotta learn more and more

[elfin folio]

ah yes solve for X by isolating X (unknown value first algebra)

later: OOPS x is now multiple input values and can be whatever you want and you must solve for the quadratic instead

[daring oyster]

I think it's mostly because people use their intuitive sense in problem solving subconsciously and when they don't get it they think like this

[daring oyster]

Yes it can be anything you want, the graph only works if (x-5)(x+4) = 0

[elfin folio]

and then people say stuff like this

and then i end up thinking my teachers are insane

because nobody seems to think it was a good idea to explain why thats the case

so theres nothing to build it on

though recently ive been trying to just focus on interpreting and reinterpreting everything and building models of how everything works and then asking if they are correct to narrow it down

though sorry for personal dialogue its just that bugs me alot

[daring oyster]

Nah it's alright

[elfin folio]

ty

[daring oyster]

Goodluck

[elfin folio]

ty!

welp

[daring oyster]

You're welcome

[elfin folio]

you just gotta

Meta-Flow

[daring oyster]

[elfin folio]

.close

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[cunning pasture]

https://tenor.com/view/good-luck-peace-peace-out-vanish-disappear-gif-17484686

[safe radish]

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[storm umbra]

hi

[storm umbra]

will changing the concentration of a reaction affect how much the substance is formed?

[pseudo scroll]

depends

[storm umbra]

how so?

[pseudo scroll]

generally speaking it affects the rate at how fast a reaction proceeds

[storm umbra]

i understand that

but shouldnt it also affect the final product

[pseudo scroll]

however if you have an equillibrium reaction then it changes in accordance with Le Chatelier's principle

[pseudo scroll]

I mean more reactancts = more products sure

[storm umbra]

so if it's at equilibrium it won't change?

the final product

[pseudo scroll]

no it will

but it more has to do with the moles of gaseuous reactants than the concentration explicitly

but since these terms are related, it will affect

[storm umbra]

shit

no

wait

Question is "Outline why changes in only temp, non-limiting reagent reactant concentration, surface area, and catalyst will not result in line D"

I kinda see why bc the mass of MgO shouldnt be different

but again by changing the concentration it opens up a possibility?

perhaps an error in the question

[pseudo scroll]

what's the solid line

[storm umbra]

uhh

it's the original line

[pseudo scroll]

And what's the y axis

[storm umbra]

Y is mass(g)

[pseudo scroll]

This seems kind of arbitrary to me

Maybe you'll have better luck in the chemistry server

[storm umbra]

ight thanks for helping out

have a nice one

.close

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[unkempt orbit]

I am trying to calculate this through but I have problems with reading the notation and Im not sure why there is a minus at athe very front.

I defined 3 sets:
A={1,2,3,4} B={3,4,5,6} C={1,6,7,8}

I looked at the first term of the equation and did the union of all 3 sets which cardinality equals 6. Due to the minus sign it is -6.
How do I continue from here?

[safe radish]

@unkempt orbit Has your question been resolved?

[safe radish]

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[unkempt orbit]

.reopen

[safe radish]

✅

[unkempt orbit]

<@&286206848099549185>

[safe radish]

@unkempt orbit Has your question been resolved?

[lost vector]

isnt this the Inclusion Exclusion Principle?

[flat frigate]

GigaChad

LaTeX source sent via direct message.
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See the LaTeX manual or LaTeX Companion for explanation.
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[raven gull]

TBH, yes.

Does look like it

[lost vector]

like what

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[lost timber]

Hello, I would like to know if there's a known / defined name for the operation of pancaking values between two limits, so unlike the clamp where we have the values made constant before and after the limits, here the middle is collapsed to a constant (0)
And can be usually expressed as
y = x - clamp(x, a, b)

[lost timber]

I usually call it the "Fold"

[empty gyro]

squeeze theorem?

[lost timber]

And I could represent it in desmos like this https://www.desmos.com/calculator/wnh7kgahvr

Or in a 3D environment, like this

[lost timber]

Lemme websearch it

No, unfortunately

Since I see this operation pop up several times in some algorithms

I wondered if it was named, like the clamp

Basically if a value is between the limits

it will be 0

So a grid like this

Becomes flattned between those two poistions like this

now, what happens if I make a circular hole

If I don't collapse the mesh, just use the collapsed positions for texture, I get a line between the points.

I was just wondering if there was a known name

[safe radish]

@lost timber Has your question been resolved?

[safe radish]

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[steep magnet]

I know theres a trick to solve this really easily because the highest coefficient of the numerator is 2 greater than the highest coefficient of the denominator right? So isnt it automatically equal to 1 or zero or something? Or am I mixing this up with something else

Its been a while since I did these so I vaguely remember that trick but I might be mixing it up with something else...

[jaunty spire]

multiply both the numerator and denominator with 1/x

[steep magnet]

Ok I can do that but what about the trick? Am I confusing it with something else

[jaunty spire]

don't really know what trick exactly you're talking about. If the degree of the numerator is higher it should just go to infty or -infty.

[steep magnet]

Ah maybe thats it

[jaunty spire]

you can prove this using this argument basically

[steep magnet]

Yep I got it!

Thanks!

❤️

.close

[safe radish]

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Use .reopen if this was a mistake.

[steep magnet]

.reopen

[safe radish]

✅

[steep magnet]

Oke I think i just got lucky with that one cause I tried some others and got them all wrong lol

[steep magnet]

Whats the next step after that

[jaunty spire]

in the example you provided the denominator would go to 1 so you're left with the numerator goint to -infty

if the degree of the denominator is k you multply both with $x^{-k}$

[flat frigate]

belabutter

[jaunty spire]

that way the denominator will always go to 1 if x goes to infty

[steep magnet]

Hmmm I dont really follow sry

The denominator is 1+1/x

[jaunty spire]

yeah

[steep magnet]

What do you mean it would go to 1 sry

[jaunty spire]

what happens when x goes to infinity?

1+1/x goes to 1

[steep magnet]

Thats what were trying to find out no?

Im a little lost haha

[jaunty spire]

correction: it will converge to some constant, not necessarly 1

i meant for the denominator

remember the limit laws

lim f/g = lim f / lim g but only if lim g is not equal to 0 and g(x) = 0 for finitely many x

1/x^k goes to 0 if x goes to infinity, agreed?

[steep magnet]

I think so haha

[jaunty spire]

so here is an example 1/(x^2 + x^3 +x^4)

multiply both with x^(-4)

[steep magnet]

x^(-4)/(x^-2 + x^-1 + 1) ?

[jaunty spire]

yes

numerator goes to zero

denominator goes to 1

so you have 0/1 for the limit

which is 0

[steep magnet]

Ok i get you so far

[jaunty spire]

now let's say you have (-15x^4)/(x^2 + x^3 +x^4)

what would the limit be?

[steep magnet]

-infinity ?

[jaunty spire]

no

multiply both with x^(-4) again

[steep magnet]

0

[jaunty spire]

no

[steep magnet]

1 ?

[jaunty spire]

no

[steep magnet]

Hmmm

(1)/(x^-2 + x^-1 + 1)

So limit DNE?

[jaunty spire]

why 1 in the numerator?

you're close

[jaunty spire]

it does exist

[steep magnet]

ah -15 lol

[jaunty spire]

yes

[steep magnet]

-15/(x^-2 + x^-1 + 1)

[jaunty spire]

yes

[steep magnet]

Idk whats next tho

[jaunty spire]

take the limit

[jaunty spire]

this

[steep magnet]

So x^-2 and x^-1 both become 0, no?

[jaunty spire]

yes

[steep magnet]

So its just 15?

[jaunty spire]

-15

[steep magnet]

ah ye haha

[jaunty spire]

but yes

[steep magnet]

Ok got it now

I think

[jaunty spire]

so what happens when for (-15x^4 -x^8)/(x^2 + x^3 +x^4)?

can you tell at a glance?

[steep magnet]

DNE?

[jaunty spire]

no

[steep magnet]

But the entire denominator goes to 0 no?

[jaunty spire]

you multiply both den and num with x^(-k) where k is the degree of the denominator

by doing this it will never be zero

that's the point so we can take the limit

[steep magnet]

So what happens sry

[jaunty spire]

you multiply both num and den with x^(-k). What's k?

[steep magnet]

-4

[jaunty spire]

yes

or no it's just 4

but anyway

[steep magnet]

(-15 -x^-4)/(x^-2 + x^-1 + 1)

[jaunty spire]

not quite

[steep magnet]

(-15 -x^4)/(x^-2 + x^-1 + 1)

[jaunty spire]

yes

[steep magnet]

(-15 -x^4)/(0 + 0 + 1)

-15 -x^4

I see now

[jaunty spire]

maybe i should clarify: By multiplying both den and num with x^(-k) it's the same as multiplying with 1, you don't change anything. That's why this works

[jaunty spire]

yes

and what happens to -x^4 as x goes to infinity?

[steep magnet]

Im not sure sry

[jaunty spire]

come on, that's easy

maybe plug in some values

or look at the graph

[steep magnet]

ah yeah

it goes to negative infinity

[jaunty spire]

yess

[steep magnet]

Okkkk getcha now

I will try to solve some examples hopefully I got it now haha

[jaunty spire]

you can do it bro

[steep magnet]

Hmm im already confused about this

(-x^2 + x - 4)/(4x + 4)

[jaunty spire]

note that x goes to negative infinity

[steep magnet]

OHHHHH

I see now

Hmmm what about this one

(-x^3 - x^2 + 8x + 4)/(-x^2 - x + 5)

[jaunty spire]

for the sign you need to pay attention to: does x go to + or - infinity? Is the degree of the numerator odd or even? What's the factor in front of x^r (where r is the degree of num).

[steep magnet]

What does the degree of the numerator being odd or even change?

[jaunty spire]

x^3 for instance

goes to -infinity if x goes to negative infinity