help-27

[umbral creek]

sqrt 3

[mild granite]

thats what i meant to put

[umbral creek]

ohhh hahahaha

what happened

okay so yes tan is sqrt3

but remember

in second quadrant

tan is positive

or negative

[mild granite]

so it would be -sqr3

[umbral creek]

yes

[mild granite]

so i do solve for theta

[umbral creek]

okay for theta

remember they say $sinθ= \frac{\sqrt{3}}{2}$

[woven radish]

duckiescute!

[umbral creek]

do you know what theta will give you sqr3/2

do you know special angles

30,45,60

[mild granite]

umm

not really is that the ones that you add together to get 180

[umbral creek]

oh nono

it’s special angles

that means if you take sin of that special angle

you get a nice value kind of

for example

sin30 = 1/2

[mild granite]

ok

[umbral creek]

cos60 =1/2

cos30 = sqr3/2

etc etc

do you remember

[mild granite]

a little

[umbral creek]

okay

so sin of something

will give you sqr3/2

see this

what sin gives you sqr3/2

[mild granite]

60

[umbral creek]

yes

[mild granite]

so theta =60

[umbral creek]

yea

[mild granite]

like this

[umbral creek]

yes

[mild granite]

ok thank you so much

[umbral creek]

no prob!

[mild granite]

theta was 120

bc i was supposed to subtract 180-60

[umbral creek]

oh right I forgot you were in second quadrant

my bad!

[mild granite]

it fine

[devout snow]

@mild granite Has your question been resolved?

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[summer folio]

hello so i have in issue in solving this question

[summer folio]

i dont understand why the solution wants to multiply 1/x^3

how did it come to the decision to multiply with that

[lunar harbor]

That's the greatest exponent

[summer folio]

yea but y did they use 1 to divide it with

why was it not simply x^3

[lunar harbor]

dividing by $x^3$ is the same as multiplying $1/x^3$ tbh

[woven radish]

Civil Service Pigeon

[devout snow]

@summer folio Has your question been resolved?

[summer folio]

oh

not yet

last question

[summer folio]

how did it become 1-0-0

where did the 0 come from

[granite lichen]

1/x^2 will become superduper small when x gets large. As good as 0!

[summer folio]

ok ok i see

idk y i messed up

but i did

and dont get why

[devout snow]

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[agile narwhal]

x^(2x) vs (2x)^x

[agile narwhal]

which grows faster ?

[stone stump]

they are both of the form x^x * (something)

what is that something

use exponent laws

[agile narwhal]

(x^2)^x

[agile narwhal]

this ^

i found out that x^2x is bigger

[devout snow]

@agile narwhal Has your question been resolved?

[summer folio]

i dont understand how we got from here to here

[devout snow]

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[copper cloud]

what is the answer rounded to the nearest hundredth because on the calendar it’s 5.09433

[winter patrol]

nearest hundredth means the want you to round to 2dp

[copper cloud]

so it’s 5.09 rounded?

[winter patrol]

yeh

[copper cloud]

aight thank u

.close

[devout snow]

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[rapid sedge]

Discrete Math help

[rapid sedge]

I'm trying to prove or disprove this and I'm completely stuck Im not sure wheather to do it by proof by cases or contrapostive

[devout snow]

@rapid sedge Has your question been resolved?

[rapid sedge]

No

.close

[devout snow]

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[chrome junco]

if the one sided limits are different does that mean the limit doesn't exist?

[prime egret]

Yeah

[chrome junco]

here they are different because there's an asymptote, but is it because both limits are different that theres an asymptote or is it just a consequence of there being an asymptote?

.close

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[short gulch]

.open

[cunning sky]

Can anyone help me with this

[devout snow]

@cunning sky Has your question been resolved?

[devout snow]

@cunning sky Has your question been resolved?

[sand dove]

This is called the De Morgan law for sets. You can prove this using the De Morgan law on predicates.
"Consider x.
x not in (AuB) <=> ||not(x in A OR x in B)
Using the De Morgan laws on the predicates "x in A" and "x in B" :
x not in (AuB) <=>|| ||not(x in A) AND not(x in B)
<=> x not in A AND x not in B
We found that the two sets considered have the exact same elements, so they're equal.||"

[devout snow]

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[vivid rover]

Im trying to go from (x^k - y^k) to (x^k+1 - y^k+1) can anybody point me in the right direction?

[vivid rover]

i've tried multiplying by (x+y) but it doesn't give me what i want

[devout snow]

@vivid rover Has your question been resolved?

[winter patrol]

look up difference of two powers

and/or consider geometric series

[devout snow]

@vivid rover Has your question been resolved?

[devout snow]

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[woeful cloud]

question on how to approach this induction of a for loop

@restive river greetings first time using this channel, not sure if you are the person to ping on this

[chrome ocean]

I would very much encourage you to read the channel #❓how-to-get-help and follow appropriate procedure.
Anyway, on the induction, have you written our out your basis?

[woeful cloud]

can i say basis 1 ∈ [1,1]

can i saw that?

for basis since it does start at [1,1]

[chrome ocean]

yeah, the base case is if you sum the first odd number you get one back

then you need to write out your inductive hypothesis and prove it.
What is the inductive hypothesis?

[woeful cloud]

in this part i know i need to change the variable to another character

like n = k

[chrome ocean]

not yet, we need to write a statement that we want to prove

[woeful cloud]

and i explain the previous and current step?

oh

we want to prove pair[1] = pair[0]^2

[chrome ocean]

yep and in layman's language, what is the value of pair[1] and pair[0]**2

[woeful cloud]

btw this is the code modified so it compiles

# Implements: Sum of first n odd numbers is nˆ2
import operator
n = input('Pick a number greater than or equal to 0: ') # upper limit
print ('You entered n = %s. \n' % n)
pair = [1, 1]
# the basis pair
print ('pair is %s \n' % pair)
new_n = int(n)
for i in range (1,new_n+1):
# executes for i = 1, 2, ..., n
    pair[0] = pair[0] + 1
    pair[1] = pair[1] + pair[0]*2 - 1
    print ('pair is %s \n' % pair)
if operator.eq (pair[1],pair[0]**2):
    print ('True: the second member is the square of the first.')
else:
    print ('False: the second member is not the square of the first.')

[chrome ocean]

if you were to write a mathematical equation that the if statement is testing, what would it be?

we want an equation that has the same variable on both sides; pair[1] and pair[2] needs to be translated into their values.

[woeful cloud]

n = n^2

i believe

[chrome ocean]

😢 i feel like you'll struggle to prove that no matter the technique,
if you look at the for loop, what is the value of pair[0] at the n'th iteration of the loop

and what is the value of pair[1] at the n'th iteration of the loop

[woeful cloud]

or actually pair[n] = pair[n^2]

i do agree, i am struggling i think i am use to doing induction with actual math problems

[chrome ocean]

do you want to reconsider? have you tried running the program and printing pair[0], pair[1]
It prints it out if you run the program.

[woeful cloud]

but when you toss in code, i have to think a little more creatively

[chrome ocean]

and the first sentence of problem 1 gives you a massive hint on what your inductive statement should be

[woeful cloud]

i modified the problem and added the following line

print (pair[0], pair[1]) = 1 1

before the for loop

[chrome ocean]

yeah, it prints two ones, and what will it print in the first iteration of the for loop? what is the mathematical formula it is following to get the next value for pair[0]

again, the first sentence of problem 1 has a massive hint

[woeful cloud]

it will print the first pair

which is [1, 1]

oh

wait sum of the first n numbers

so if i select 5

1+3+5

so 3 numbers?

[chrome ocean]

yeah, so it'll be the sum of the first n odd numbers

which is the left side of your inductive statement

[woven radish]

WombatInCombat

[woeful cloud]

ah i had to get a online latex editor

to see what this is

[chrome ocean]

lmao,

now what is on the other side of the sum of the first n odd numbers?

what are you trying to prove it's equal to

[woeful cloud]

n^2

[chrome ocean]

yep

,, \sum_{1}^{n}{2n-1} = n^2

[woven radish]

WombatInCombat

[chrome ocean]

nice, now we can do induction on our hypothesis!
What we want to do is assume it is true for some hypothetical value of n and then using that, prove that it is true for the value n+1

[woeful cloud]

perfect!

do we change the variable in the induction

like change to k+1

were there is an n

(k+1) sum 1

then 2(k+1)-1 = (k+1)^2

[chrome ocean]

uh, you can? but it doesn't do anything useful, because n=k in order for your proof to work

you might as well just use stick with n

[woeful cloud]

so if i stick with n

just add 1

were there is an n?

[chrome ocean]

yes, except in one specific location

[woeful cloud]

oh

the right

side

only the left

leave right alone

[chrome ocean]

,, \sum_{1}^{n+1} {2n-1} = (n+1)^{2}

[woven radish]

WombatInCombat

[chrome ocean]

you don't want to change the inside of the summation; oh maybe that's why they change to k? for clarity?

[woeful cloud]

can we change it to k

for consistence?

[chrome ocean]

might as well

[woeful cloud]

i replaced the n above with a k

[chrome ocean]

anyway, from there, you just manipulate the left side and it eventually looks like the right

or you can manipulate the right and make it look like the left

the second might be easier

if you FOIL and then convert to a summation using the inductive hypothesis that you are assuming to be true

[woeful cloud]

nice so this is the equation to use and from here build more on top

this is amazing

i greatly appreciate this

do you code in python?

[chrome ocean]

yes,

[woeful cloud]

i was having such difficulty converting a for loop to math

not sure if its just practice or is there actual material on converting code to math?

[chrome ocean]

if you are done, you should close the channel.
yeah, I mean you just need to write more for loops.
eventually you'll learn to make a lot of functions based on loops because loops are one of the main tools to call the same function with different parameters.

[woeful cloud]

yes we are good sorry for the delay and of course i will continue to practice, do i type .close to close?

[chrome ocean]

mostly people convert math into code, which is a bit easier than dissecting other people's code, but yeah practice and maybe get a book? online one's are free if you use tor and z-library, but you didn't hear that from me

[dot] close, read the #❓how-to-get-help

[woeful cloud]

will do, thank you so much again, see you next time 🙂

bye for now

.close

[devout snow]

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[devout snow]

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[autumn grove]

How is this incorrect?

[wanton pumice]

u forgot to apply the chain rule

[grand flare]

A = 5??

[autumn grove]

Shoot you guys are right. I applied chain rule, got 5x(8-x^2)^3/2

So I put A = 5 and apparently it was incorrect

[wanton pumice]

its still incorrect

whats the derivative of the inside function?

[autumn grove]

3x

2x

wait a min

-2x

so is it -5?

[wanton pumice]

yeah

[autumn grove]

To find the second derivative, would i need to chain rule again

First derivative was -5x(8-x^2)^(1/2)

For second, I got

-5(8-x^2)^(3/2) + -3x(8-x^2)^(1/2)

[devout snow]

@autumn grove Has your question been resolved?

[wanton pumice]

yes product and chain rule

everything u see a compostion of functions eg f(g(x)), u need to apply the chain rule when differntiating

[autumn grove]

where would i product rule this?

[wanton pumice]

first derivative is -5x(8 -x^2)^(1/2)

its a product of two functions

so u need to apply the product rule

and when computing the derivative of the latter function, u need to apply the chain rule as it is a composite function

[autumn grove]

but they are both x variables?

So what would this be? u = -5x and v = -x^2?

[devout snow]

@autumn grove Has your question been resolved?

[devout snow]

@autumn grove Has your question been resolved?

[devout snow]

@autumn grove Has your question been resolved?

[devout snow]

@autumn grove Has your question been resolved?

[devout snow]

@autumn grove Has your question been resolved?

[wanton pumice]

f'(x)= -5x(8-x^2)^(1/2)
let u =-5x, v =(8-x^2)^(1/2)
so f'(x)=uv
differentiate both sides to get f''(x) and apply product rule to uv, and since u,v are functions of x, u need to use chain rule on them too

[devout snow]

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[final vessel]

I understand how to prove that the inner product must be positive for scalar multiples of v

I'm not sure how that condition would prove that all vectors in v would have a positive inner product

Maybe I want to look at vectors orthogonal to v but I don't think I'm allowed to do that

I know that if there was <u,u>=-<v,v> such that u and v are orthogonal and u != -v then <u+v,u+v> = 0 which can't be possible

[devout snow]

@final vessel Has your question been resolved?

[devout snow]

@final vessel Has your question been resolved?

[final vessel]

<@&286206848099549185>

[final vessel]

rip

[devout snow]

@final vessel Has your question been resolved?

[devout snow]

@final vessel Has your question been resolved?

[devout snow]

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[ionic rover]

Hey for my homework I have to prove that two sets don't intersect with each other. I know I have to somehow prove the two sets are a disjoint sets but I don't know where to start

[violet wind]

that means the same thing

disjoint $\iff$ don't intersect

[woven radish]

monikanicity

[violet wind]

oh ok

nvm

[elfin hill]

Suppose that there is a common element and find a contradiction

[violet wind]

ye that

or directly if an element is in one set then it can't be in the other

[ionic rover]

Ok

[ionic rover]

Ok so make sure that each set have different elements in them

[violet wind]

yea

[ionic rover]

Is that it?

[violet wind]

yea

sets intersect if they share elements

[ionic rover]

Ohhhh

[violet wind]

so show that if an element is in one set it can't be in the other

[ionic rover]

Thank you

Also would a set minus just translate from A - B to A interest not (A Intersect B)

[violet wind]

yes

A intersect not B is simpler

[ionic rover]

Ok thanks

[violet wind]

$A \cap B^C$

[woven radish]

monikanicity

[violet wind]

idk if there's a better way to do complement symbols lol

$\overline{B}$

[woven radish]

monikanicity

[violet wind]

nice

[ionic rover]

Ok

.close

[devout snow]

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[devout snow]

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[restive river]

Hey can someone help me

[devout snow]

Please don't occupy multiple help channels.

[restive river]

This is what I need help with

[long kettle]

Why did you open another channel

[restive river]

I did not mean to can you help me

[long kettle]

Close this channel since you already have one open

[restive river]

I thought you were smart can you help me

[long kettle]

You have two channels open. You don't need two channels. Close this one.

[restive river]

How do I close it

[long kettle]

".close"

[restive river]

I did it

Can u help me now

[long kettle]

I can't. The questions are beyond my scope

But I'm sure someone here can if you wait

[restive river]

So you are dumb

[long kettle]

You can't either, that's why you're here ¯\_(ツ)_/¯

[restive river]

Because this is my math quiz

You dumb

[long kettle]

Quiz?

[restive river]

Yes kid

[long kettle]

We don't help on quizzes, tests, or exams

[restive river]

Why you being such such pussy about it

Help me dumb ass

I have to study

I thought it was big bairn

[long kettle]

If you don't want to be timed out/banned, I'd suggest more polite language

[restive river]

Thought I discord moderator I don’t need your help

I’m just going to get someone’s help ugly fat disgusting bitch

.close

[devout snow]

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[random temple]

💀

calm down buddy

[restive river]

Shut up

[long kettle]

<@&268886789983436800>

[cyan obsidian]

@restive river gets a 24h mute to contemplate how not to be rude.

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[sonic sedge]

close out the old one then bot

[long kettle]

Close it yourself

[sonic sedge]

it has to prompt me again

[long kettle]

No

You can type ".close"

[main gull]

No it doesn't

[sonic sedge]

I closed it

[midnight dirge]

lol

[supple knot]

.close

[devout snow]

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[devout snow]

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[chrome furnace]

Where did I I mess up yo

[supple knot]

Where did the 12x go

[devout snow]

@chrome furnace Has your question been resolved?

[chrome furnace]

I don’t think I even got one

[supple knot]

It's in the numerator in j)

[chrome furnace]

I think I didn’t do something

Missed of messed up a step

[devout snow]

@chrome furnace Has your question been resolved?

[narrow badger]

Yeah. The 12x has gone missing

[devout snow]

@chrome furnace Has your question been resolved?

[devout snow]

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[devout snow]

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[restive river]

How do I change sinx/x to piecewise function? Please help

[stone stump]

do you mean $f(x) = \begin{cases} \frac{\sin x}{x} & x\neq 0 \ 1 & \text{ else } \end{cases}$ ?

[woven radish]

Denascite

[restive river]

No f(x) = 1, |x| <1

And

F(x)=0 when | x| >1

[stone stump]

how does that have anything to do with sinx/x

[restive river]

Yeah

I have to convert sinx/x to piecewise

So I could solve fourier integral

[stone stump]

oh you are talking about the fourier transform

[restive river]

This is from fourier integrals topic

[stone stump]

because clearly this is obvious from your earlier message

you could use that the fourier transform is more or less self inverse

and calculate the fourier transform of $f(x) = \begin{cases} 1 & |x|<1 \ 0 & |x|\geq 1 \end{cases}$ first

[woven radish]

Denascite

[restive river]

Wait they have given me

Intergal of sinx/x (dx) from 0 to inf =pi/2

I need to use fourier integral represention

But for that I need to know the piecewise function

Im not sure how they got that piecewise function

From sinx/x

[stone stump]

I really dont get what you mean with the piecewise function

the fourier transform of $f(x) = \begin{cases} 1 & |x|<1 \ 0 & |x|\geq 1 \end{cases}$ is (up to scaling) sin(x)/x. so therefore we can reason backwards that the fourier transform of sin(x)/x is (up to scaling) again that piecewise function

[woven radish]

Denascite

[restive river]

I don't get how during exam I can say that sinx/x is that piecewise function?
Like do I have to memorise it

Or is there any way to figure out that sinx/x will result in that piecewise

Like in my question they just said

Find following expression using fourier integral represention of a function

Integral from 0 to inf of sinx/x dx = pi/2

Idk how to go about it
But if they had given me the piecewise function I would have calculated a0,an etc

[devout snow]

@restive river Has your question been resolved?

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.reopen

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✅

[restive river]

<@&286206848099549185>

Find following expression using fourier integral represention of a function

Integral from 0 to inf of sinx/x dx = pi/2

How can I solve this when they haven't given me the piecewise function?

[devout snow]

@restive river Has your question been resolved?

[humble perch]

complex analysis?

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[restive river]

Advanced Differential equations

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[old snow]

hello can anyone check that my derivatives are correct and i set it up correct on my D-Test

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[fiery sun]

i got a question

[fiery sun]

how do i solve |3x-12|<5.1

my alg hw has a bunch of innequalities but whenever i try to solve them like a reg alg problem like usual it marks it wrong

i got x<17.1/3

[deep cove]

Split cases.

Either case 1, $3x - 12 \ge 0$ and $3x - 12 < 5.1$.

OR case 2, $3x - 12 < 0$ and $-(3x - 12) < 5.1$

[woven radish]

jimmy1234

[patent marsh]

You can also recognize that $|3x - 12| < 5.1 \implies -5.1 < 3x - 12 < 5.1$. Then you can solve for x by adding 12 to everything and then dividing everything by 3

[woven radish]

MellowDramaLlama

[patent marsh]

it's the same as the split cases but more "concise"

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[restive river]

I have to show by using the def of limit that

Lim ((n^2-1)/(2n^2+3))= 1/2

Here's what I tried
For e>0
Absolute of ((n^2-1)/(2n^2+3)) -(1/2) < e

Absolute of (-5/(4n^2+12)) = (5/(4n^2+12))< 1/n^2 < 1/n

Now Absolute of ((n^2-1)/(2n^2+3)) -(1/2) < e if 1/n <e

Now by archiemedian property 1/k <e

Then if n>k
1/n < 1/k < e

Thus Absolute of ((n^2-1)/(2n^2+3)) -(1/2) < 1/n < 1/k < e

And therefore the limit is equal to 1/2

[restive river]

Could someone please confirm if I'm on the right track

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

[devout snow]

@restive river Has your question been resolved?

[torpid condor]

you starded from the result no ?

"Absolute of ((n^2-1)/(2n^2+3)) -(1/2) < e"

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[unreal lynx]

how do I figure out what constant this approaches. I know it is 2 but how do I know that other than just adding up the first handful? 1 + 1/2 + 1/4 +1/8 + 1/16 + ...

[prime egret]

Geometric series

[spare hawk]

so lets define a partial sum, S= 1+ 1/2+... 1/2^n
Basically if n goes to infinity we have the original sum now we can do some fancy math with S
imagine the difference S-(1/2)S and what this would be. you can there after manipulate that expression to get your answer

[unreal lynx]

what level of math is this? calc 2?

[prime egret]

I’m from uk so idk about that system but like when you’re 15-16 probably

[unreal lynx]

is it in calculus?

[prime egret]

I guess in a way

Not really probably

[unreal lynx]

you lost me at partial sum

[spare hawk]

okay so imagine a sum that is like the n first terms of your infinite sum
example: if you have the sum T=1+2+3+4... then we can define S=1+2+...+n. now we have n->inf then S=T

[unreal lynx]

I follow

[spare hawk]

then if S=1 +1/2 + ... 1/(2^n)
we would have (1/2)S=1/2+1/4+...1/(2^n+1)
so S-(1/2)S=1+1/ (2^n+1)

also S-(1/2)S=S(1-1/2)=S/2

[unreal lynx]

ahh, gotcha

I think I get it

[spare hawk]

just don't forget all the lim n-> inf
I skipped them here since they take alot of time to write and are cumbersome on computer but for S to equal your sum you have to include it

[unreal lynx]

thank you @spare hawk

[spare hawk]

no problem! take care mate

[unreal lynx]

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[high nebula]

I have a theorem that states "Let $f: I \to \mathbb{R}$ $n$ times differentiable at the point $a \in I$. The function $r: J \to \mathbb{R}$, defined in the interval $J = {h \in \mathbb{R}; a+h \in I}$ by the equality $\\f(a+h) = f(a) + f'(a)\cdot h + \frac{f"(a)}{2}\cdot h^2 + \hdots + \frac{f^{(n)}(a)}{n!} \cdot h^n + r(h)\\$ satisfies $\lim_{h \to 0} \frac{r(h)}{h^n} = 0$. Reciprocally, if $p(h)$ is a polynomial of degree $\leq n$ such that $r(h) + f(a+h) - p(h)$ satisfies $\lim_{h \to 0} \frac{r(h)}{h^n} = 0$ then $p(h)$ is the Taylor polynomial of degree $n$ of $f$ at point $a$, that is, $\\p(h) = \sum_{i=0}^n \frac{f^{(i)}(a)}{i!} \cdot h^i$ " $\\$
So, with that theorem, how can i show that when $n$ approaches infinity, $r(h) = 0$ for any $h$?

[woven radish]

Amarinya

[high nebula]

you can think of r(h) as an error function that either adds or subtracts to make sure the taylor series is equal to the function at a+h

so naturally, as you add infinitely many terms, the error function will always be 0, but i don't know how to prove that

[devout snow]

@high nebula Has your question been resolved?

[patent tartan]

the error term does not always converge to zero for any input, only for "nice" functions

[high nebula]

sorry, i forgot to say that there exists a $K > 0$ such that for any $x \in I$ and $n \in \mathbb{N}, |f^{(n)}(x)| \leq K$

[woven radish]

Amarinya

[patent tartan]

you might be able to use the explicit lagrange or cauchy formula for the remainder term

then bound the derivative term and prove lim h^k / k! = 0

[high nebula]

oh there's here a theorem of taylor formula with lagrange remainder

but there's a little problem with it

"$Let f:[a,b] \to \mathbb{R}$ be $n$ times differentiable in the open interval $(a,b)$, with $f^{(n-1)}$ continuous in $[a,b]$. There exists $c \in (a,b)$ such that $\\f(b) = f(a) + f'(a)(b-a) + \hdots + \frac{f^{(n-1)}(a)}{(n-1)!}(b-a)^{n-1} + \frac{f^{(n)}(c)}{n!}(b-a)^n$ " $\\$My problem with it is that 1) i can't find a $c$ in between $a$ and $b$ that makes the derivative equal to 0, and 2) $b$ has to be explicitly greater than $a$.

[woven radish]

Amarinya

[patent tartan]

why would you need to find c if you have the < K condition for any x?

it will also hold for c, whatever it might be

[high nebula]

but if i change f^(n)(c) with K, it'll be f(b) <= the new value now, not equal

so taking the limit would still make it <=

[wild creek]

you should maybe try to prove that the taylor series converges

[wild creek]

thx to this

[high nebula]

well, if i replace all f terms with K, i'll have the taylor series for the e^x function, but i can't use that because to replace the taylor series to the e^x function i need to use what i'm trying to prove

[wild creek]

you have taylor series (but with absolute values)

smaller than some quantity

thats finite

[high nebula]

yeah, that's true

but the infinite sum might be infinite

[wild creek]

its not

think of the sujm

[high nebula]

i know it's not, but i can't realy prove it

[wild creek]

for any n

the sum is smaller than the same quantity

if you take the limit

[high nebula]

i just have to take the limit as n approaches infinity

[wild creek]

are you familiar with series and proofs of convergence

[high nebula]

yes i am

[wild creek]

?

yeah ok

[high nebula]

i've seen a bit of absolute convergence

[wild creek]

so we have the absolute convergence of the taylor series here

the infinite sum of absolute values

is smaller than Ke^|h|

[high nebula]

yeah, so you're replacing all the |f| terms with K, right?

yeah but see, here's the problem

[wild creek]

so it proves the convergence

[high nebula]

you can't just replace it with e^|h| because you're using what you're trying to prove to prove it

[wild creek]

why is that so?

its litteraly the definition of exp

if you dont want to you can prove that the exponential series converges

its really easy

[high nebula]

because you're using the fact that $\sum_{n=0}^\infty \frac{x^n}{n!} = e^x$ and that formula is a consequence of what we're proving

[woven radish]

Amarinya

[wild creek]

well it depends on how you define exponential i guess

just prove that this series converge then

ratio test for exemple works

[high nebula]

the |xn+1/xn| test?

[wild creek]

yeah

you see one way to define exponential is just to say that the series defining it converges for any real number

and just write the result e^x

[high nebula]

this book defines e as the limit (1+1/n)^n

[wild creek]

yeah but how does the book defines power of real numbers then?

[high nebula]

so that's why i said

[high nebula]

what do you mean?

[wild creek]

because afaik the common def comes from exponential

like a^b = exp(b ln(a))

[high nebula]

it doesn't define exp() i think

[wild creek]

e^sqrt(2)

is not very well defined if you just want the value

but w/e

you can still prove the series converge

if you dont even want to talk about e

just prove the specific series you have here converges

[high nebula]

so

i can just do it like this?

hold on i'm writing it

[wild creek]

i know the rickroll url by heart and still got rickrolled

[high nebula]

LMAO

$f(b) = \frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^n + \sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(b-a)^i\\$
$f(b) = \lim_{n \to \infty}\left(\frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^n + \sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(b-a)^i\right)\\$
$f(b) = \lim_{n \to \infty}\left(\frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^n\right) + \lim_{n \to \infty} \left(\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(b-a)^i\right)\\$
$f(b) = 0 + \sum_{i=0}^\infty \frac{f^{(i)}(a)}{i!}(b-a)^i\\$
$f(b) = \sum_{i=0}^\infty \frac{f^{(i)}(a)}{i!}(b-a)^i\\$

[woven radish]

Amarinya

[wild creek]

yeah this works too

[high nebula]

did i do something wrong?

[wild creek]

nope but you used the other theorem

nothing wrong with it

[high nebula]

ahh alrighty

there's just one small little issue though

the lagrange theorem assumes b > a, so if i want b < a, i don't know how to fix that

[high nebula]

because the interval is [a,b], so if i want b < a, then i need to replace it with f(a), which i do not want

[wild creek]

the theorem you are using comes from the integral remainder version of taylor

[high nebula]

yeah, it says the theorem is the taylor formula with lagrange remainder

[wild creek]

and then applying the mean value theorem to it

which is where the "c" comes from

[high nebula]

funnily enough it used the rolle's theorem to prove that

[wild creek]

to prove what?

[high nebula]

the taylor formula with lagrange remainder

[arctic field]

taylors theorem with lagranges remainder is typically proved with MVT or rolle's

[wild creek]

because MVT is just rolle but you work less

MVT proof uses rolle's

[high nebula]

yeah, i had to use rolle to prove mvt

[arctic field]

i think rolle's is applied for the most common proof ive seen for taylor's

[high nebula]

i think it's pretty cool

[arctic field]

the expression comes out of nowhere tho

somebody just pulled it out of their ass

[high nebula]

rolle must've drawn the graphs of functions a lot and noticed a pattern

[wild creek]

rolle makes intuitive sense i think

[high nebula]

yeah and rolle didn't know that much, because he only proved for polynomials functions. tbh we didn't even have calculus at his time

but uhh, what should i do in case i have b < a?

but still need it to be f(b) instead of f(a)

[wild creek]

you take the integral form and do the proof almost the same as you just did

[high nebula]

i haven't reached integration yet

it's the next chapter

i could maybe consider the interval [-a, -b], but i don't know if f(-b) = f(b)

[wild creek]

how did you prove the lagrange remainder version then?

[high nebula]

you used integration to prove it?

[wild creek]

the way my teacher taught me yes

and its the most useful version of taylor's

i'm not against it but i'm wondering

[high nebula]

Let $ϕ: [a,b] \to \mathbb{R}$ defined by $\\ϕ(x) = f(b) - f(x) - f'(x)(b-x) - \hdots - \frac{f^{(n-1)}(x)}{(n-1)!}(b-x)^{n-1} - \frac{K}{n!}(b-x)^n$, $\\$where the constant $K$ is choosen in a way that $ϕ(a) = 0$. Then $ϕ$ is continuous in $[a,b]$, differentiable in $(a,b)$, with $ϕ(a) = ϕ(b) = 0$. Taking the derivative, you can see that it telescopically cancels most of the terms and what remains is $\\ϕ'(x) = \frac{K - f^{(n)}(x)}{(n-1)!}(b-x)^{n-1}\\$By Rolle's Theorem, there exists $c \in (a,b)$ such that $ϕ'(c) = 0$. This means that $K = f^{(n)}(c)$. The theorem is obtained by doing $x=a$ in the definition of $ϕ$ and remembering that $ϕ(a) = 0$.

[woven radish]

Amarinya

[wild creek]

oooh yeah thats a bit weird

its not very natural

like why this formula?

[high nebula]

yeah it's pretty unintuitive

[wild creek]

https://brilliant.org/wiki/taylors-theorem-with-lagrange-remainder/

i found something if you wanna check it out

it derives the formula from well known theorems

you can click outside the pop up

or esc

[high nebula]

oh wow

tbh it's weird for me that you can prove it with integrals lmao

[wild creek]

but the integrals actually give you a "reason"

for the terms in the formula

like this formula only comes from FTC

and fubini

[high nebula]

yeah but still XD

[wild creek]

so for your specific problem

i would just do it my way

i'm not very good at finding something else if i already have something that suits my neeeds

[wild creek]

or you could prove this theorem but with I=[b,a]

like do the same proof

but with small details changing

[high nebula]

yeah the thing is that in the theorem, the upper bound is the value that goes in the function at the left side of the equality

[wild creek]

if you just replace [a,b] by [b,a] where in the proff do you need to change something?

[wild creek]

ok but just do the same with the lower bound

where is it used that b>a?

[high nebula]

but [5, 0] doesn't make sense, does it?

[wild creek]

indeed but it has nothing to do with our case

a has the role of point of reference

b is the value at whichc you wanna apply taylor's

now either b<a or b>a

[high nebula]

#help-27 message

after the Let

[wild creek]

its not used

its supposed

like i can suppose w=3 at the beginning

but it would have nothing to do with the proof

Let $ϕ: [b,a] \to \mathbb{R}$ defined by $\ϕ(x) = f(b) - f(x) - f'(x)(b-x) - \hdots - \frac{f^{(n-1)}(x)}{(n-1)!}(b-x)^{n-1} - \frac{K}{n!}(b-x)^n$, $\$where the constant $K$ is choosen in a way that $ϕ(a) = 0$. Then $ϕ$ is continuous in $[a,b]$, differentiable in $(a,b)$, with $ϕ(a) = ϕ(b) = 0$. Taking the derivative, you can see that it telescopically cancels most of the terms and what remains is $\ϕ'(x) = \frac{K - f^{(n)}(x)}{(n-1)!}(b-x)^{n-1}\$By Rolle's Theorem, there exists $c \in (a,b)$ such that $ϕ'(c) = 0$. This means that $K = f^{(n)}(c)$. The theorem is obtained by doing $x=a$ in the definition of $ϕ$ and remembering that $ϕ(a) = 0$.

[high nebula]

but if the theorem uses [a,b] to make it work, you can't assume it'll work if you do [3, 2]

[wild creek]

you are not understanding what i mean

i'm not saying b>a and [b,a]

i'm saying

if b<a

it works aswell

but with [b,a]

instead of [a,b]

Let $ϕ: [b,a] \to \mathbb{R}$ defined by $\ϕ(x) = f(b) - f(x) - f'(x)(b-x) - \hdots - \frac{f^{(n-1)}(x)}{(n-1)!}(b-x)^{n-1} - \frac{K}{n!}(b-x)^n$, $$where the constant $K$ is chosen in a way that $ϕ(a) = 0$. Then $ϕ$ is continuous in $[b,a]$, differentiable in $(b,a)$, with $ϕ(a) = ϕ(b) = 0$. Taking the derivative, you can see that it telescopically cancels most of the terms and what remains is $\ϕ'(x) = \frac{K - f^{(n)}(x)}{(n-1)!}(b-x)^{n-1}$By Rolle's Theorem, there exists $c \in (b,a)$ such that $ϕ'(c) = 0$. This means that $K = f^{(n)}(c)$. The theorem is obtained by doing $x=a$ in the definition of $ϕ$ and remembering that $ϕ(a) = 0$.

[woven radish]

Benjamin

Let $ϕ: [b,a] \to \mathbb{R}$ defined by $\ϕ(x) = f(b) - f(x) - f'(x)(b-x) - \hdots - \frac{f^{(n-1)}(x)}{(n-1)!}(b-x)^{n-1} - \frac{K}{n!}(b-x)^n$, $\$where the constant $K$ is chosen in a way that $ϕ(a) = 0$. Then $ϕ$ is continuous in $[b,a]$, differentiable in $(b,a)$, with $ϕ(a) = ϕ(b) = 0$. Taking the derivative, you can see that it telescopically cancels most of the terms and what remains is $\ϕ'(x) = \frac{K - f^{(n)}(x)}{(n-1)!}(b-x)^{n-1}\$By Rolle's Theorem, there exists $c \in (b,a)$ such that $ϕ'(c) = 0$. This means that $K = f^{(n)}(c)$. The theorem is obtained by doing $x=a$ in the definition of $ϕ$ and remembering that $ϕ(a) = 0$.
```Compilation error:```! Undefined control sequence.
<recently read> \\xcf
                  
l.57 Let $ϕ: [b,a] \to \mathbb{R}$ defined by $\\xcf
                                                  \x95(x) = f(b) - f(x) - f'(x)...
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```

[wild creek]

[high nebula]

but ϕ is just a random function

[restive river]

Yes

[wild creek]

a bit late

[wild creek]

why is it a problem?

[fair parrot]

Please help me

[wild creek]

#❓how-to-get-help

[fair parrot]

I’m gonna lose my server if I don’t get a question right

Ok

[high nebula]

because ϕ is just used to find K. i looked at the proof and tried swapping a and b but it doesn't make sense because then ϕ(b) is not 0, it's f(a) - f(b), which would be 0 if i hadn't swapped a with b

so swapping a and b while maintaining [a,b] would be like this

[wild creek]

dont swap a and b inside phi

[high nebula]

huh?

[wild creek]

taylor's formula doesnt care wether b>a or b<a its the same

[high nebula]

yes but i haven't proven that

[wild creek]

yeah but as you said

phi is a random

function

so you can chose it to be what you want

so keep phi the same

[high nebula]

but then i'm gonna keep with f(b) = f(a) + ... with b>a. if i want to make it work with b<a, i need to make it f(b) = f(a) + ... with [b,a] instead of [a,b]

[wild creek]

you wrote two times the same thing

but yeah

keep it the same

the proof is exactly the same if b<a just each time theres written "a,b" you replace it with "b,a"

[high nebula]

yeah i just did it in my head and you're right..

this question was harder than it was supposed to be

i think that clears everything

[wild creek]

anyway this proof is just bad teaching imho

[high nebula]

why is that?

[wild creek]

it gives no insight

[high nebula]

yeah. i'm kinda left to figure it out on my own

[wild creek]

it a proof by someone who is already very familiar

with taylor series

and for people who are also already familiar

[high nebula]

this book has some questions that are like that

[wild creek]

if you are not it doesnt make sense

[high nebula]

https://tenor.com/view/anime-nichijou-sigh-gif-12179345

it's troublesome

but thank you for your time

[wild creek]

try to read through the link i sent

its self explanatory

[high nebula]

i did. and it was pretty confusing to me

[wild creek]

you dont know about fubini?

[high nebula]

not really

[wild creek]

or is it something else

[high nebula]

i don't know what fubini is

[wild creek]

its "you can swap in which order you integrate when you have nested integrals"

in this case its easy to understand because everythin is bounded and finite

[high nebula]

ah, that's pretty useful

[wild creek]

but its not true all the time

[high nebula]

how come?

[wild creek]

when infinite things come

weird things happen

[high nebula]

example?

[wild creek]

https://math.jhu.edu/~jmb/note/nofub.pdf

notice theyre integrals over R

[high nebula]

oh wow

thanks for all of this information

[wild creek]

keep up the work

analysis is the best its worth the struggle

[high nebula]

after i'm done with real analysis i'm gonna study something i find a bit more practical like linear algebra

[wild creek]

linear algebra is the maximum amount of algebra i can accept

its still very good

though

i'm just bad at abstract algebra

[high nebula]

my teacher told me that the most useful chapter is the last one, which has sequences and series of functions, with simple and uniform convergence, properties of uniform convergence, power series, trigonometric functions and taylor series

[wild creek]

yeah

its very useful

[high nebula]

i thought this was taylor series already. guess it's called taylor formula

[wild creek]

taylor series doesnt always converge

but taylors formula is always right

[high nebula]

ahh i see

[wild creek]

if you have enough differentiations available

[high nebula]

i really gotta learn matrices and stuff

[high nebula]

C infinite?

[wild creek]

not necessarely

i was just saying you need the differentiation up to n to get the formula with n-1 terms

[high nebula]

ahh i see

but thanks a lot ❤️

hope you have a great day

.close

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[rotund gull]

I know it wants a formula, but i really have no idea where to go with this

i solved the previosu question, but i typed in the formula that i used and it was wrong...

.78 was the percentage of the material and 1.3 was the entire material, but here i understand that it wants a formula that doesn't require the beginning amount....

nvm i found it

.close

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[digital phoenix]

If I have a wheel with 7 possibilities on it, -6,-2,-1,0,1,2,3 and have a sample of 1000 spins, if I want to determine the cumulative odds of ending with 4 or less after 12 spins, do I take the average needed to end with 4 - the sample average divided by the sample std dev, use that as the z score to determine the probability?

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[digital phoenix]

If I have a wheel with 7 possibilities on it, -6,-2,-1,0,1,2,3 and have a sample of 1000 spins, if I want to determine the cumulative odds of ending with 4 or less after 12 spins, do I take the average needed to end with 4 - the sample average divided by the sample std dev, use that as the z score to determine the probability?

[devout snow]

@digital phoenix Has your question been resolved?

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@digital phoenix Has your question been resolved?

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[restive river]

whats the first step to solving this

try rearranging

what does that mean sorry

[proud perch]

There is nothing to solve for

Lol

[restive river]

simplify i assume

this is the original question sorry

bru

[proud perch]

Factor out the 2

[restive river]

i make =0 and multiply by reciprocal?

[proud perch]

No

[restive river]

idk how to factor out 2

[proud perch]

2A-2B=2(A-B)

Like that

[restive river]

how do i do that here

replace 2 with 1 and put 2 outside the parenthises?

[proud perch]

Yea

[restive river]

oh ok ty

[devout snow]

@restive river Has your question been resolved?

[restive river]

idk what to do

[winter patrol]

show what you have atm

[devout snow]

@restive river Has your question been resolved?

[restive river]

i have the same thing

[restive river]

https://media.discordapp.net/attachments/903486480075354182/1028547368343322634/E388AFB8-5AA4-4E27-8D6D-222CA85F41E2.jpg?width=810&height=1080

[winter patrol]

combine numerator (of big fraction) into a sinlgle fraction

[restive river]

how can i if the denominators arent the same

[winter patrol]

do something to make them the same

the same way you'd simplify something like
$$\frac12 + \frac13$$

[woven radish]

ℝamonov

[restive river]

ok

[arctic rapids]

You gotta make the denominators the same!

[restive river]

idk how sorry

[arctic rapids]

Well in the example ramonov posted

[restive river]

i would multiply both to get 6 denominator

and that would change numerator as well

[arctic rapids]

Exactly!

[restive river]

i dont get how to do this with +h in one side tho

[arctic rapids]

Well in the example, u multiply the 1st fraction by 3/3 and the second one by 2/2 right?

It’s the exact same principle for your problem

[restive river]

ok i multiply each fraction by the other denominator

?

[arctic rapids]

Yea that’s all there is to it

[restive river]

ok thank u for explaining

[arctic rapids]

No problem

Ping if you get stuck again or want your answer checked

[restive river]

oh ok tysm

[restive river]

Whats the asnwer

[devout snow]

@restive river Has your question been resolved?

[restive river]

.close

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[jovial geode]

hi

[jovial geode]

im doing

[restive river]

hi

[jovial geode]

complteting the square

and im stuck on the question

[restive river]

show me

[jovial geode]

: $x^2 + px + 4 = (x+q)^2 -5$

[woven radish]

Poseidon59

[jovial geode]

Wherre i have to find the value for p and q

[restive river]

ok

[jovial geode]

can you help me?

[restive river]

i dont want to tell u the wrong thing to do

[jovial geode]

wdym

[restive river]

idk what if im wrong and i waste ur time

[jovial geode]

i mean you can try

i dont think there will be another helper here

for a long time so

[restive river]

why dont u foil x+q

[jovial geode]

x^2 + 2qx + q^2

[restive river]

idk

[jovial geode]

its ok

<@&286206848099549185>

[restive river]

i got it

[jovial geode]

oooo

[restive river]

do u want me to send u asnwer

[jovial geode]

• steps but yes

[restive river]

this is my answer what do u think

[jovial geode]

yea that would work but the questions needs a constant value

[restive river]

wow

[jovial geode]

yea

[winter patrol]

don't divide both sides by x
instead equate coefficients and constants and solve a system of equations

[jovial geode]

go on....

[winter patrol]

try following those instructions

[jovial geode]

equaste which co efficients tho

[winter patrol]

of your variable x

[jovial geode]

im not sure how to do tha

[winter patrol]

identify the coefficients of x on each side of the equation

set them equal to each other

[jovial geode]

so on the left, the co efficient of x^2 would be 1, the co efficient of px would be p?

[winter patrol]

yeh

[jovial geode]

okok

then on the right

1 for x^2, 2q for 2qx

[winter patrol]

yes

[jovial geode]

ok cool now what

[winter patrol]

identify the coefficients of x on each side of the equation
set them equal to each other

[jovial geode]

idk what that wants me to do

[winter patrol]

what part of set them equal to each other don't you undstand

[jovial geode]

like how do you make p = 2q

[winter patrol]

that's all i wanted you to do for that step

you want the coefficients to be equal
and that's the equation you set up to represent that

you also want the constants to be the same for the expressions to be equivalent

instead equate ... and constants

[jovial geode]

so is this what you want me to do or

[winter patrol]

yes...

[jovial geode]

what about the rest

1 = 1?

[winter patrol]

well 1 = 1, you don't need to worry about the x^2 term

[jovial geode]

okok

so we just have p = 2q?

[winter patrol]

you also want to identify and equate the constant terms on each side

[jovial geode]

4 = -5 .....

that seems a bit wrong

[winter patrol]

the constant in this context is the part that doesn't contain x

[jovial geode]

ye

[winter patrol]

the constant in
x^2 + 2qx + q^2 - 5
isn't just -5

[jovial geode]

oh

what is the other constant then

[winter patrol]

is the part that doesn't contain x

which part(s) of that don't have any x in it

[jovial geode]

q^2

= 1

[winter patrol]

no

stop skipping ahead and doing the wrong thing

[jovial geode]

sorry

[winter patrol]

which part(s) of the right side don't have any x in it

[jovial geode]

: $x^2 + px + 4 = (x+q)^2 -5$

[woven radish]

Poseidon59

[jovial geode]

just the q^2

[winter patrol]

...

which parts of the whole expression on the right side of the equation after the expansion of (x+q)^2 - 5
don't have any x in them

[jovial geode]

-5 and q^2

[winter patrol]

after expanding, you would have had
$$x^2 + 2qx + q^2 - 5$$
which part(s) of that don't have any $x$ in them

yes

[woven radish]

ℝamonov

[jovial geode]

there

[winter patrol]

hence q^2 - 5 would be the constant of that

[jovial geode]

ah ha

[winter patrol]

and like before, set those constants equal to each other
4 = q^2 - 5

[jovial geode]

this is gains,

okok now what do i do

[winter patrol]

you now have a system of equations
$$\begin{cases} p = 2q \
4 = q^2 - 5 \end{cases}$$
solve that system to get your solution(s)

[woven radish]

ℝamonov

[jovial geode]

i lowk forgot how to do these

[winter patrol]

well q^2 - 5 = 4
is a relatively simple equation with 1 variable
try solving that first

[jovial geode]

okay

q = 3

[winter patrol]

not quite

[jovial geode]

why

you add 5 to both sides

[winter patrol]

q = 3 isn't the only soluluition

[jovial geode]

what would the other sol. be

[winter patrol]

have a think about it for a minute

[jovial geode]

ehmm

does it have to do with like

the square rooting bit?

[winter patrol]

yes

[jovial geode]

would it also be -3?

[winter patrol]

yes, q=-3 would also be a solution

[jovial geode]

eyyyy

okay last bit

for the next question, which is x^2 + 4x + p = (x+2)^2 -9,

[winter patrol]

sub that value back into the first equation to determine the respective values of p

[jovial geode]

would i take the same approach?

[winter patrol]

you're not done yet

[jovial geode]

ok wait

p could = 6 or -6

@winter patrol

[winter patrol]

not quite

[jovial geode]

how

[winter patrol]

you'd need to properly indicate your two solutions

[jovial geode]

wdym properly indicate

[winter patrol]

p=6 when q=3
these are separate solutions
p=-6 when q=-3

[jovial geode]

right

tysm

[jovial geode]

.

[devout snow]

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[sterile merlin]

(3x + 2) (x + 1) = 0
so, 3x + 2 = 0 / (x + 1) [which is invalid due to it being 0]

is this true?

[brazen star]

that is not true

why is it divided by 3x + 2
:/

[sonic smelt]

(3x + 2)(x + 1) = 0 <=> 3x + 2 = 0 or x + 1 = 0

[sterile merlin]

my bad it was a typo

read it again

apologies

[brazen star]

well no, in that case, x + 1 is not equal to 0

so what you're doing is taking 2 cases

where: (3x + 2) is 0, and then (x + 1) is 0
when (3x + 2) is 0, (x + 1) is not 0 and vice versa

which will result in two answers that works on the original equation, but the answer wont make both (3x + 2) and (x + 1) both be equal to 0 at the same time

[sterile merlin]

how do i find out which expression is equal to 0

[brazen star]

both of them should be equal to 0, but not using the same x value

for example

(x + 1)(x - 2) = 0

when x = -1, (x + 1) is 0 but (x - 2) is not

when x = 2, (x - 2) is 0 but (x + 1) is not

[sterile merlin]

i see

ty

[devout snow]

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[prisma kestrel]

yo

[prisma kestrel]

i need help

<@&286206848099549185>

[young fossil]

eww degrees....

so the area of a circle is 2 times pi times radius^2 right?

so if we pretend for a moment that this is a whole circle, then we would have a circle with area of 2 times pi times 36 cm^2.... or pi times 72cm^2

but, this is just a slice of circle, its equivalent to 48/360 of the circle we had before....
so we take out original number and times by the fraction.
Lets reduce our fraction first
48/360 = 2/15

so,
pi times 72cm^2 times 2/15

(144pi/15)cm^2

=(48pi/15)cm^2

[devout snow]

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[zenith prism]

hello I have a problem with this question

[zenith prism]

i tried searching for gcf of the given

and no luck

[balmy quest]

compare with $(a+b)^2=a^2+2ab+b^2$

[woven radish]

秋水

[zenith prism]

i can try that

answer is (6y+5z)^2

but how do i find the area?

is it like this? (6+5)(6+5)?

answer is like this (12y^2 + 10z^2)?

i am confused

[balmy quest]

I don't know, I think the area is (6y+5z)^2, the side is (6y+5z)

[zenith prism]

but the formula said A=s^2

so i tried multiplying both polynomials

the answer is supposed to be (6y+5z) without squared

i dont know how it got the answer

[balmy quest]

A = s^2 = (6y+5z)^2 , so s = (6y+5z)

[zenith prism]

can you pls explain it?

[zenith prism]

i ended up here

[balmy quest]

I think the answer should be (6y+5z), or (-6y-5z)

[zenith prism]

why?

I dont understand

[zenith prism]

it isnt like this?

[balmy quest]

the area = (6y+5z)^2, right?

[zenith prism]

no that is my answer to the trinomial

after i answered the trinomial, i will answer the area