help-19

[amber schooner]

,w lim x to -inf arctan(x)

[clever fjord]

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Requested by kniefy

[amber schooner]

and clearly arctan(0) = 0

[shy grove]

I see

so it evaluates to 0 instead of π

[amber schooner]

what does?

[shy grove]

wait nvm

isnt the inequality false here

[amber schooner]

both the a and b limits are pi/2

so you add them and get pi

[amber schooner]

what inequality

[shy grove]

so this is false for f(x) = f'(arctan(x))

[amber schooner]

well for f(x) = 1/(1 + x^2) yea you actually have equality

[shy grove]

but you did claim this inequality is sometimes true, right?

[amber schooner]

don't say inequality btw

inequality refers to <, <= etc

[shy grove]

oh

right

so its a nequality as per \neq

[amber schooner]

but yes they are not always equal

so long as the left side is convergent

so both the integrals exist

[shy grove]

then what I'm asking is if there exists f(x) such that the \neq I wrote holds, but where both LHS and RHS evaluate to determinate forms

[narrow crypt]

X is odd. It’s from the rhs

[amber schooner]

no if both the integrals on the left converge then the cauchy principal value will be equal

[shy grove]

oh well

but I guess I get it now

[amber schooner]

this is interpreted to mean does equality hold

[shy grove]

for all intents and purposes I can write improper integrals from -∞ to ∞ as lim(a→∞) from -a to a

[shy grove]

I see. My wording was terrible lol

[odd edge]

@shy grove Has your question been resolved?

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[odd edge]

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[weary orchid]

Can somebody please provide its solutions with proper steps ...

[weary orchid]

<@&286206848099549185> help me

[narrow crypt]

holy notation

[solar peak]

what have u tried

it simplifies kind of nicely

[mystic saffron]

L

L'hopital thez

[solar peak]

??

[weary orchid]

nothing i cant think how to approach t

it*

[warped badger]

My thoughts exactly

[mystic saffron]

I meant to type l'hopital

[solar peak]

ok first write that arctan in terms of arcsin so u can remove the sin term and do the same for the arcsin in the other bracket

[warped badger]

No way they put sin(tan(arcsin(x)))

[weary orchid]

i was thinking about it but i was not sure

[solar peak]

first do that and tell me what u get

[weary orchid]

thanks

[weary orchid]

ok

[narrow crypt]

[ \arctan \left(\frac{x}{1} \right) = \arcsin \left(\frac{x}{\sqrt{1+x^2}}\right)]

[clever fjord]

k

[narrow crypt]

${\tan x \sim x}$ when ${x \to 0}$?

[clever fjord]

k

[solar peak]

u cant use this

u have to expand it upto 2 terms

1 term wont do

[narrow crypt]

i see

[narrow crypt]

ye doesnt work :/

[solar peak]

yeah

[solar peak]

lol

not even 2 terms

3 terms

thats crazy

this is a nice question

@weary orchid are u here?

[weary orchid]

it is a pyq of jee advanced

[weary orchid]

yes sir

[solar peak]

ohh makes sense

[solar peak]

did u do what i said

or are u having some problem with that?

[weary orchid]

i am trying to solve i will send you the steps

[solar peak]

ok sure

[solar peak]

do u know which year by any chance

[weary orchid]

ah it is not written in pdf

[solar peak]

i dont think it is a pyq tho

[weary orchid]

but there is hint availaible i am sending it

[solar peak]

yeah

but to use the expansion ull first have to simplify the trignometric expressions

[solar peak]

like this

now i gtg now so wait for someone else to help u or else u can ping @ helpers

[weary orchid]

ok thanks for the help

[odd edge]

@weary orchid Has your question been resolved?

[weary orchid]

yes

[subtle raft]

You can .close if you are done

[weary orchid]

.close

[odd edge]

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[versed knoll]

Can someone explain why to find the number of divisions, we do n+2 choose 2 instead of n-1 choose k -1, where k is the number of divisions

[late dust]

Can you elaborate

Is it example 22?

[versed knoll]

Yes

Refer to the explanation under the problem please

[late dust]

I'm not sure what you're referring to with "divisions"

[versed knoll]

Like if I have a sum

Let’s say 3

How many ways can I add up to 3

I could have 1+1+1

Or 2+1+0

Or 3+0+0

Right

That’s what I mean

[late dust]

Well, yeah, so two divisions

[versed knoll]

No 3

[late dust]

Right, I'd call that 3 parts, not divisions, but sure

[versed knoll]

Oh sure

[late dust]

So ok, k = 3

So it's something choose 2

[versed knoll]

Right but in my problem they did something + 2 right for k

That’s what I didn’t get

[late dust]

Let me double check something

Right so imagine you have three boxes into which to distribute 5 items

[versed knoll]

Oh ok yeah

[late dust]

(and boxes can be left empty)

[versed knoll]

K

So this is just boxes and balls

[late dust]

I'm guessing you're talking about stars and bars but learned another name for it

[versed knoll]

Yeah

[late dust]

Anyway, you can equivalently lay out the 5 items in a row, and place two separators somewhere

Separators can both be in the same place, and can be between two items or at the beginning or the end

How many ways to do that?

[versed knoll]

Wait why only 2 separators

[late dust]

Because that makes three sets

[versed knoll]

Oh I see yeah

[late dust]

That's why I didn't like the term "divisions"

[versed knoll]

So I guess 4 choose 2

Cause you have 4 spaces between the numbers?

[late dust]

Yes but both separators can be in the same place

I'm trying to find a nice way to explain how that makes it different

[versed knoll]

So it’s like 4 choose 2 plus 4

[late dust]

Ok so if the separators had to be in different places, and between two items, you'd just get 4 choose 2, because 4 spots between items and 2 separators

Here however, separators can be in the same spot, and can be at either end of the item row

[versed knoll]

Right

[late dust]

So instead of trying to place the separators, think about the items and the separators together as a row of objects

Now you choose two of these objects to be the separators

[versed knoll]

So it’s like 7 choose 2

[late dust]

How many objects? 5 + 2

So 7, choose 2

[versed knoll]

Right

Ok tysm

[late dust]

So to go back to the original, you have three boxes, that's k=3

And n=5 for example

So you get n+k-1 choose k-1

That's the general formula

[versed knoll]

Oh ok

This is for any divisions right

Where order matters

[late dust]

Yeah

I recommend a read through this: https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

Some parts might be a bit complicated but overall it's a good explanation with proofs

[versed knoll]

Ok I will

Tysm

[odd edge]

@versed knoll Has your question been resolved?

[odd edge]

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[crystal charm]

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[somber river]

can't even begin

tried to factorize into eh... x+1 form but doesn't work

[weary orchid]

is this a sphere equation

[somber river]

you figure that out

this the complete question

No diagram attached

[weary orchid]

ok i am trying to solve it

can you tell me the answer

[somber river]

ehhh

you already got?

[weary orchid]

no for my answer suerity

[somber river]

Given is 0

[weary orchid]

ok

[mystic saffron]

hint: x²+3x+2=0

[somber river]

what...... where did ya get that

[mystic saffron]

basically

i would group like

everything

[somber river]

Okay......

[weary orchid]

you can use legendre three square theorem also i got the answer

[somber river]

☠️

[mystic saffron]

x²+y²+z²+3(x+y+z)+5= 0 can be taken into x²+3x+2+y²+z²+3y+3z+3=0

and x²+3x+2 can be factorised

same for y and z

should take u somewhere

[somber river]

ehhh

Sowwy I cant be as far sighted as ya

lie let's say I facotrise it to x-a and x-b

so what

Y^2 won't have +2 will it and same for z^2

basically I can't repeat this for that

[mystic saffron]

you can:)

[somber river]

Hmm..... Okie I shall try

[mystic saffron]

try

[somber river]

x+2)(x+1) +3(y+z+1) = 0 hmm.....

[mystic saffron]

no

[somber river]

eh...... what t do now

hello?

wth

<@&286206848099549185>

[abstract epoch]

maybe try converting into (x+1.5)^2, (y+1.5)^2 and (z+1.5)^2

[crystal charm]

I was thinking of rewriting the expression as something like $\sum_{cyc}^{}(x+\frac{3}{2})^2=\frac{7}{4}$

[crystal charm]

yeah basically this

[abstract epoch]

its not x+3 tho, right?

[clever fjord]

឵឵MxRgD

[abstract epoch]

yeah that seems right

[crystal charm]

I forgot to include the fraction yeah

[abstract epoch]

but what do we do after this?

[abstract epoch]

hit and trial might work

but im not sure about it

oh wait multiplying on both sides by 4, then applying mod (remainder) 8 on both sides gives no real solutions

because a^2 can have the values =0,4,1(mod 8) and no combination of the sum of 3 squares gives 7(mod 8)

[odd edge]

@somber river Has your question been resolved?

[odd edge]

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[humble basalt]

I have currently proven that the left formula is less than or equal to 2 and the right is greater than or equal to the square root of two. And the condition for the equality of the inequality is that the n numbers are equal, and both sides of the inequality are equal to the square root of two.

[bitter folio]

头晕

[humble basalt]

Forgive me, I don't have an English version of this question for the time being.

[humble basalt]

遇到中国人了

啊

[bitter folio]

新加坡

[humble basalt]

哦那没事了

会中文就好😃

[bitter folio]

我代数挺烂

只会几何

[humble basalt]

那很厉害了

这题应该也是偏难的

[bitter folio]

我的不等式只能 muirhead 或用笨方法

[bitter folio]

看起来是

[humble basalt]

听说有东南亚奥林匹克的难度（）

[bitter folio]

[humble basalt]

蹲人

[bitter folio]

在这里

用了titu 不等式可以吧

[bitter folio]

你上面那两个是哪怎么推出来的

[humble basalt]

右式的缩放我已经完成了

[humble basalt]

rhd 用 cauchy

[bitter folio]

看起来我无法帮你

不好意思

[humble basalt]

没关系哒哈哈

每个人都有自己擅长的领域嘛

[bitter folio]

希望你能得到帮助

我去搞几何了 bye

[humble basalt]

好的呐

[odd edge]

@humble basalt Has your question been resolved?

[odd edge]

@humble basalt Has your question been resolved?

[vagrant oxide]

问题是什么

？

[odd edge]

@humble basalt Has your question been resolved?

[odd edge]

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[gusty cave]

can someone help me here nobody replied to me in other cannel so closed it and came here

[mystic saffron]

don't open multiple

[gusty cave]

I closed it

[mystic saffron]

alr for this

[mystic saffron]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[marble laurel]

Please don't close and reopen the same question in different channels without waiting for help

[gusty cave]

oo

sorry

[marble laurel]

Nw

Now, the problem tells you that you have a present day price

And the annual rate of depreciation and the time are given

Can you identify the present value, the time, and the rate of depreciation?

[gusty cave]

yes..
the rate of depreciation is 16% per annum, Present Value is 9260 rupees and Time is 3 years

[marble laurel]

Time is 3 years to the past or to the future?

[gusty cave]

past

[marble laurel]

Alright so how would you write it? t = ???

[gusty cave]

t= -3 maybe

[marble laurel]

Correct

Now do you remember the formula?

[gusty cave]

maybe it is v/(1-2/100) the whole square

[marble laurel]

Hmm...what is v?

[gusty cave]

v is present value

[marble laurel]

Okay

So is it like this?

$v \left(1-\frac{2}{100}\right)^2$

[clever fjord]

VulcanOne

[gusty cave]

yes

[marble laurel]

Okay you got the form, but you didn't place the given info properly

[gusty cave]

yeah

so instead of v its should be 9261 and 2 should be 16 maybe and raised to whole sq is 3

[marble laurel]

Right

But is it 3?

Or -3?

[gusty cave]

-3 but like

isn't it like u cant have power negative

[marble laurel]

You can

[gusty cave]

oo

my bad so it is -3

[marble laurel]

$a^{-3} = \frac{1}{a^3}$

[clever fjord]

VulcanOne

[marble laurel]

Remember

[gusty cave]

we did not learn this yet

[marble laurel]

Huh

Alright

Keep it in mind

[gusty cave]

ok

[marble laurel]

So can you find the value 3 years in the past?

[gusty cave]

yeah now I got it

thanks so much

for helping me out

[marble laurel]

Nw :)

[gusty cave]

I wish there were teachers like you

shall I close it now?

[marble laurel]

Sure

[gusty cave]

thx bye

.close

[odd edge]

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[odd edge]

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[dull tundra]

Can anyone message me and help me with my geometry its literally the basics

[twin inlet]

just ask ur question

[dull tundra]

Ok

[twin inlet]

here

[dull tundra]

My question is

How do I figure out the planes on a question like this

I'm on question c

@twin inlet

[merry kestrel]

Think like a prism placed on the paper

triangular prism

[dull tundra]

I would think 8 ? Would that be right

[marble laurel]

It asks how many sides does the triangle prism have

[merry kestrel]

yeah count

If I gave you an actual triangular prism in hand

How many sides does it have?

Let’s start, there is one that is the base

[dull tundra]

So the answer is 8

@merry kestrel @merry kestrel @marble laurel

[marble laurel]

Hmm

How do you reach 8?

This is the prism right?

[dull tundra]

Yea

[marble laurel]

Can you point out in a remix of the image the 8 sides?

Note:
the solid line means the line is in the front
the dotted line means the line is behind the shapes

[odd edge]

@dull tundra Has your question been resolved?

[odd edge]

@dull tundra Has your question been resolved?

[odd edge]

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[mortal trench]

Can some1 explain (C,alpha) summation to me

[mortal trench]

So I understand C1 that makes intuitive sense

Just the average of partial sums

But like I don't understand 2 and up

The wikipedia article is unbelievably dense about this

It's not just me being dumb I swear

<@&286206848099549185>

<@&286206848099549185>

[manic sleet]

!15m

[odd edge]

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.

[manic sleet]

oh wait nvm I cannot read time

[quasi sparrow]

would be helpful if you posted the wiki link

[tidal matrix]

the link

the header is (C, alpha) summation

from this link:

Note that [the (C,2) sum] would not be the Cesaro sum of the Cesaro sum but rather the limit of the Cesaro transform of the Cesaro transform of the sequence s_1, s_2, ⋯
but this kinda remains unhelpful

[dull tundra]

@mortal trench calculate the y value of the x intercept them reverse then should be the answer

[tidal matrix]

For the Cesàro sum of a series?

[tidal matrix]

Hardy, Divergent Series seems helpful! here is pp. 96:

so it seems like you would want to take the sum of the sum, alpha times, and take a limit of that

[graceful viper]

https://encyclopediaofmath.org/wiki/Cesàro_summation_methods

There is a way to view it using binomial coefficients

[mortal trench]

What?

[mortal trench]

Ok this is actually somewhat digestible, thanks

[tidal matrix]

np :) i also didnt find it yet, but apparently this book contains a (C,2) sum of 1 + 2 - 3 + 4...

so if you can find that, maybe it is a good concrete example

• dj's method seems easiest practically speaking

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[ember shell]

My screenshot is loading immensely slowly

[ember shell]

hey why do i need to factor the -1?

Yippee

So basically midway through solving the problem I got a bit confused

So I took a look at the answer key because I knew that b and a had to be reversed to cross out

So the way they did it was by multiplying the bottom by negative 1

But like wouldn't that just be changing the equation to get the outcome you want?

How is that even possible

[forest sky]

b - a = -1(a - b)

it's factoring

[ember shell]

Uhmmmm

Lemme think about that

Omg!

Wait that makes sense kinda?

So u remove the negative 1

So then it all becomes negative reasonable

Cuz ur just taking it out

Okay thank uu

.close

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[patent sun]

what i need to do exactliy

[odd edge]

Please don't occupy multiple help channels.

[patent sun]

ok

[wanton bison]

Is y a real or natural number?

[patent sun]

both ?

[rough birch]

try taking a logarithm on both sides, and using exponent properties of logarithms

[patent sun]

ok thx

<3

.close

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[twin vigil]

Question 8 how to continue from here onwards

[odd edge]

@twin vigil Has your question been resolved?

[faint knot]

I found a BS way you couldve solved this

this is cos(70 - x) = 2 sin x cos 40
now just assume that cos(70 - x) = cos 40, and notice that sin 30 = 1/2

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[gloomy shell]

can someone help me

[gloomy shell]

for trigonometry

[unkempt lichen]

please send your question

[unkempt lichen]

your job is to find a?

[gloomy shell]

im not sure about the answers,

yeah

[unkempt lichen]

what have you learnt and tried?

[gloomy shell]

i got 16.7cm as my answer

[unkempt lichen]

ok, let's check it

[gloomy shell]

but the answer key says 11cm

[gloomy shell]

alr

[unkempt lichen]

does your answer key have workings?

[gloomy shell]

nope

it only says "a=11.15cm"

[unkempt lichen]

how did you get 16.7cm?

because if it's the same method as mine, i don't see what's wrong here personally

[gloomy shell]

15sin48/sin42 = 16.7

the sine rule

a/sin A = b/sin B

[unkempt lichen]

ok that was a little overcomplicated, you could have just done 15 tan 48

[gloomy shell]

yeah

i figured

[unkempt lichen]

but i got the same as you that way

[gloomy shell]

alr

so the answer key "11.15cm" is incorrect?

[unkempt lichen]

sure you looking at the answer to the correct question?

[gloomy shell]

yeah

[unkempt lichen]

then i can't explain the discrepancy

i agree with your answer though

[gloomy shell]

okay thank you so much

[unkempt lichen]

sure

!done

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[gloomy shell]

.close

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[unkempt lichen]

<@&268886789983436800>

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[snow socket]

.close

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[left mist]

how come there are 4 values although sin(thetre) = +1/2 only has 2 values i thought?

[nimble blaze]

firstly theta, not thetre
and how do you only have sin(theta) = 1/2

[left mist]

oh wait nvm

i figured it out

my b

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[tulip cradle]

In the graph you see a half circle with radius one and a curve f(x) = sqrt(x/2). I don't understand why they subtract the second integral and not add it.

[tulip cradle]

It expresses the area of the grey zone btw

[wooden python]

subtracting the second integral is equivalent to adding the integral of sqrt(x/2)-sqrt(1-(x-1)^2)

[tulip cradle]

Oh okay

because then f(x) lies above the circle?

[wooden python]

yes

[tulip cradle]

Alright

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[unkempt lichen]

you're not stuck, right?

[lone elbow]

hmm

it would be useful to determine all the pairs in R

[lone elbow]

for reflexive 4 pairs,
d R d, b R b, c R c, f R f

for symmetric 1 pair,
c R b

for transitive 1 pair
d R d

for reflexive and symmetric, 5 pairs
d R d, b R b, c R c, f R f, c R b

for symmetric and transitive, 4 pairs
d R d, b R b, c R c, c R b

for equicalence relation, 5 pairs
d R d, f R f, b R b, c R c, c R b

[odd edge]

@lone elbow Has your question been resolved?

[rapid hound]

probably helps if it’s in english

[odd edge]

@lone elbow Has your question been resolved?

[void yew]

ok you've got a)right
b)right
c)right
d)right
e)right
f)right

Imagine I used Roman numerals instead of letters

Haha

@lone elbow

[lone elbow]

i appreciate it

@void yew

.solved

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[rich wasp]

$$ \iint_R dx dy $$
$$ R = [0, a] \cross [0, b] $$

[clever fjord]

fluX

[rich wasp]

idk which expression the integral would be equal to

hold on im sending u a photo

acc nvm

it's just ab

right

[manic sleet]

[rich wasp]

ty

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[rich wasp]

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[keen bison]

Guys, I have an exam tmr and I am cooked. 1st year uni math. calc and algebra. there are about 30 questions. I would really appreciate if someone could explain and help me get through them.

[echo relic]

Can someone please help

[keen bison]

please dm me if you are able to hop on a call with me

[glass vault]

!occupied

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[soft terrace]

does this help?

[glass vault]

bro i feel bad for chetti getting his channel robbed lol

[echo relic]

?

[glass vault]

eddy open ur own help channel

[soft terrace]

oop, good point

[echo relic]

How

[soft terrace]

!occupied

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[echo relic]

!open channel

Open channel

@mod open channel

!occupied

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[soft terrace]

@echo relic just repost your question in one of the unoccupied help channels 🙂

[echo relic]

Thanks

[indigo dirge]

@keen bison if you have a specific problem that you need help with, post it here! Don’t ask to ask :)

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@keen bison Has your question been resolved?

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[mystic saffron]

i got 6 but its wrong

[odd edge]

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

[mystic saffron]

i got help from this server

and um

[marble laurel]

Well could you post how you got 6?

[mystic saffron]

ok

[mystic nova]

Let me see

[mystic saffron]

ye lol

[marble laurel]

Wait what?

Why is Pythagoras in that form?

[mystic saffron]

weren't u the one who helped

[zinc glacier]

it follows from similar triangles

[mystic saffron]

i might have messed up

[zinc glacier]

where does the first formula come from?

[mystic nova]

It's not Pythagoras

[mystic nova]

I used area to come up with this

[marble laurel]

Ah

Alright

[mystic nova]

He's not allowed to use trig tho

[zinc glacier]

you familiar with angle bisector theorem?

[mystic saffron]

ya

ray that divides an angle into two equal angles

[mystic nova]

Using area then [ABC]=[ABQ]+[AQC]=>AB.AC=sin(45).AQ.AB+sin(45)AQ.AC

[mystic saffron]

mhm

[mystic nova]

That's how I got the first equation in this

[mystic saffron]

oh

we did that

[mystic saffron]

we got BC is 12

[mystic nova]

What's the answer

Do you have it?

[mystic saffron]

we got 6

[mystic saffron]

its wrng

[mystic nova]

I mean the correct one

Do you have one?

[mystic saffron]

no

[mystic nova]

I think it's 24

[mystic saffron]

thats too high

ngl

[mystic nova]

Hmm

True

[mystic saffron]

idk lol

so what do we do

try trig

[mystic saffron]

i tried 5 but its wrng too

[mystic nova]

Try 24

I mean everything i did lead to 24 lol

[mystic saffron]

ihave an idea

nvm

[mystic nova]

Could you try 24 for me

[mystic saffron]

ima do 24

alr its loading

dang'

its correct

but how

tysm

im done

ty @mystic nova

bye

.clos

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[last junco]

If i have to find the growth speed for a certain point in a function, do i just differentiate the function and insert my x?

[amber schooner]

if by growth speed you mean the derivative then yes

[last junco]

Yes, thank you

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[frozen musk]

tried following this

I got y=0 and y=3 and I drew those lines

then I use sign analysis and found positive for y<0 and y>3

and negative from 0<y<3

idk what to do now

[odd edge]

@frozen musk Has your question been resolved?

[wanton bison]

You can draw a few vectors of the form (x',y') = (1, y(y-3))

Can do a table

[frozen musk]

hmm

okay let me ask this first

y'(t)=f(y) which means for my problem f(y)=y^2 -3y

so f'(y) = 2y-3?

[wanton bison]

No

You dont need f' wrt y

[frozen musk]

do we need to find stability?

[wanton bison]

Well it says that in the task

[frozen musk]

I meant for the sketching aspect

sorry this topic is literally brand new to me

jumped head first into this problem before trying to learn the theory

[wanton bison]

Oh then no

[frozen musk]

okay

[frozen musk]

okay ig let me try this

is this what you meant?

[wanton bison]

Yes

[frozen musk]

okay cool

[wanton bison]

Well x' is always 1

Oh mb i misread

So for y=1 you would have the vectors (1,-2)

And as you noticed has negative orientation for 0<y<3

[frozen musk]

okay bet yeah

makes sense

for y=2 (2,-2)?

[wanton bison]

x'=1 always

[frozen musk]

so (1,-2)

?

[wanton bison]

You plot (x',y') not (y,y')

[frozen musk]

oh

[wanton bison]

Yes

[frozen musk]

right

okay cool

so just plot all these and draw lines connecting them?

[wanton bison]

Not lines

Like this (old example of my classes)

[frozen musk]

oh

this is from my book

idk how they got the lines

honestly I prob needa watch some videos on this before I can continue doing more problems

[wanton bison]

They drew the vectors see the small gray lines

They draw example functions that fit the direction field

Like for example in y=0 you just have the direction vectors (1,0) that implies a constant function like y=0 representing a solution

[frozen musk]

oh omg

[frozen musk]

yeah

I get it now

okay cool yeah

I should be able to finish the problem then. and I'll try to figure out the stability and part b. if I can't I'll just come back in here

tysm

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[mystic saffron]

parts?

[mystic saffron]

im pretty sure its doable by parts

like u = lnx and v = arctg(x)

?

hm not really

arctan(x)/x is nightmare

oh nvm i figured it out

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[frosty mango]

No l’hopital allowed or tripple angle trigonometric identity

Any hint?

[subtle raft]

Huh

[bitter lodge]

i would ||multiply top and bottom by x + sin(x)|| then consider ||appropriate small angle approximations||

[frosty mango]

Oh, approximations looks like the way, i can use taylor but then i dont need this, i can simply replace sinx and solve

Thx

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[bitter lodge]

oh yeah haha i was over complicating

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[tame folio]

I need help with how to solve these, which are called pythagorean and quotient identities?

[tame folio]

Help-

My precalculus quiz s tommorow and I still cannot solve a single equation

[lilac galleon]

its just basic trigno

[tame folio]

Yeah but it's kinda confusing

[lilac galleon]

see tan theta = sin theta / cos theta , so we are multiplying cos theta to tan theta , in which the cos theta of tan theta will get cancelled

[tame folio]

Which number is that?

[lilac galleon]

$tan theta = \frac{sin theta}{cos theta}$

[clever fjord]

Nobita18

[lilac galleon]

$tan (theta) = \frac{sin (theta)}{cos (theta)}$

[clever fjord]

Nobita18

[lilac galleon]

theta is θ as you can see on your board

[lilac galleon]

now we are multiplying cos (theta) to tan (theta) , but we can write tan (theta) like this so then the cos (theta) which we multiplied and the one in denominator will get cancelled

@tame folio

[tame folio]

tan(theta) and cos(theta) gets cancelled if we multiply them?

[lilac galleon]

the cos(theta) will get cancelled in it and we get Sin(theta)

[crystal charm]

you can put an \ before the word theta to get the symbol. Like this, $$tan (\theta) = \frac{sin (\theta)}{cos (\theta)}$$

[clever fjord]

឵឵MxRgD

[tame folio]

How come? I thought if we multiply the same identities then we cancel them

[lilac galleon]

$tan (theta) . cos(theta) = \frac{Sin(theta)}{cos(theta)} . cos(theta) = Sin(theta)$

[clever fjord]

Nobita18

[lilac galleon]

alright thanks!

[tame folio]

Soo we rewrte them?

[lilac galleon]

tan(theta) can be written like this

[unkempt lichen]

quick question to OP: do you understand why this works?

[tame folio]

Nah, I was listening to my teacher explaining this stuff and the only thing I get was it had a pattern similar to center of a circle and parabola equation

Soo, NAH

[unkempt lichen]

would you like a visual explanation?

[tame folio]

Yeah, cuz I've tried to learn on this group but never really get anything😭

[unkempt lichen]

okay. let us work on this arbitrary right triangle.

[tame folio]

can we use this?

I think it's called a super hexagon from what I remembered with our lesson

[mystic nova]

It's the first time I see smth like this

[tame folio]

It's how we base results from multiplying identities

[crystal charm]

I think it's similiar to the cast diagram

or maybe not, i'm not sure

[unkempt lichen]

do you get this sequence?

[tame folio]

Kinda, I get the pattern.

Here's an equation my teachered showed us

[unkempt lichen]

alright, then you may proceed. sorry for interrupting

[crystal charm]

shouldn't it be just sinx rather than 1

[tame folio]

The main thing I get about this topic is that if we multiply identities the result will come out whatever matches

[tame folio]

Idk man

I'm also confused how there was a 1 there

[tame folio]

I think we basing it on the reciprocal identities

[crystal charm]

well you have a sin^2(x) on top which is canceled by the sinx. So that becomes sinx

[tame folio]

Can you anotate the image I sent? cuz I swear i never get things explained if its only in text form

[crystal charm]

sure give me a sec

[tame folio]

Soo pretty much we just cancel out the identities and whatever is left is the final result?

[lilac galleon]

only if we can cancel them , if they cannot be cancelled then the remaining is your answer

[tame folio]

But how did we not cancel out the first sin(x)?

[lilac galleon]

because there are 2 Sinx ( we have sin^2 x ) so we did cancel 1 of them and then the remaining is 1 Sinx

$\frac{sin x . sin x . cos x}{cos x . sin x} = sin x$

[clever fjord]

Nobita18

[tame folio]

I forgot there was 3 sin(x);-;

[unkempt lichen]

the square is above the i and not the n? new notation just dropped

[crystal charm]

it should be above the n, I just realised 😭

[tame folio]

Also, every step is base on the reciprocal, pythagorean, quotient identities

[tame folio]

I also realized that is my friends wrong answer...

Y'all still alive?

[lilac galleon]

hey!

[tame folio]

Yay...

How about this problem?

[lilac galleon]

(a-b)/b can be written like (a/b) - b/b

in your case a = 1 , b = cos^2 theta

[crystal charm]

you can rewrite the numerator as sin^2(theta)

[lilac galleon]

1-cos^2 theta = sin^2 theta

[lilac galleon]

see to the right side

[tame folio]

Ohhh

I don't get it.

[lilac galleon]

there's an identity

$sin^2 (\theta) + cos^2(\theta) = 1$

[clever fjord]

Nobita18

[lilac galleon]

so if you shift cos^2 theta on the right side , you get 1-cos^2 theta

[lilac galleon]

your question has 1-cos^2 theta right but we have its value which is sin^2 theta

[tame folio]

Continue

Slr, I was doing something

[lilac galleon]

you do know whats ( Sin theta / Cos theta ) right

[tame folio]

identities?

[lilac galleon]

$tan (\theta) = \frac{sin (\theta)}{cos (\theta)}$

[clever fjord]

Nobita18

[lilac galleon]

so $tan^2 (\theta) = \frac{sin^2 (\theta)}{cos^2 (\theta)}$

[clever fjord]

Nobita18

[lilac galleon]

and when you solve the 1st step of your problem you get the right side , but we know that its value is tan^2 theta

so your answer is tan^2 theta

[tame folio]

The final or nah?

[lilac galleon]

the final answer

did you understand?

[tame folio]

Not entirely, more of a visual learner;-;

[lilac galleon]

$(1-cos^2(\theta)) = sin^2(\theta)$

[unkempt lichen]

think you missed a bracket somewhere

[clever fjord]

Nobita18

[lilac galleon]

$\frac{(1-cos^2(\theta))}{cos^2(\theta)} = \frac{sin^2(\theta)}{cos^2(\theta)} = tan^2 (\theta)$

[clever fjord]

Nobita18

[lilac galleon]

@tame folio

[tame folio]

Soo we cancel out the 2 cos^2?

[lilac galleon]

elaborate

[tame folio]

I'm tryna make sense of this using superhexagon hold on

[crystal charm]

superhexagon?

ngl, I'm thinking of the game when you say superhexagon lol

[lilac galleon]

#help-19 message prolly this

[tame folio]

This

[crystal charm]

Yeah I have no idea what it is

looks confusing

[lilac galleon]

yep

[crystal charm]

is it meant to represent the trig reciprocals?

[tame folio]

Yeah

[lilac galleon]

yep in an easy way , but it aint a easy way

[crystal charm]

Hmm ic

An easier way to remember which one is which is to look at the third letter. So for example.
cot(x) would be 1/tan(x)

sec(x) is 1/cos(x)

cosec(x) is 1/sin(x)

it was how I was taught, but tbh when you do a lot of trig manipulation. You just already have them memorised

[unkempt lichen]

that's an ingenious memorization method

[tame folio]

Soo cooked with what I'm doing

[lilac galleon]

you are just saying this in an indirect way

[lilac galleon]

also the numerator cosine doesnt get cancelled

[tame folio]

I fr thought🥀

[lilac galleon]

ok

[tame folio]

Can you explain in a visual way like annotation?

[odd edge]

@tame folio Has your question been resolved?

[crisp egret]

help
help
Find the coefficient of x^10 in the expansion of (1 + x^2)^20

[ashen yew]

Can I ask if yk the formula for the coefficient of x^r in expansion of (1+x)^n

[crisp egret]

ig its the general expansion for binomial theorem

[ashen yew]

Yeah it’s n choose r

[crystal charm]

btw this is occupied still

[unkempt lichen]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[crystal charm]

and they already have there channel

[unkempt lichen]

also

!1c

[odd edge]

Please stick to your channel.

[ashen yew]

I’m new here what does this mean

Am I helping someone who’s already got someone helping them or smth

[crystal charm]

basically you can get your info of what occurs here #❓how-to-get-help. They already have an help channel opened from #help-36

currently this one is occupied though by someone else

[ashen yew]

Okay I think I get it now

[odd edge]

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[hearty nest]

The question states that f(x) is an increasing function, so why is f'(x) ≥ 0?

[potent nexus]

What would be wrong with that

[hearty nest]

f'(x) should be greater than 0, right?

and not equal

coz its states increasing and not non decreasing

[potent nexus]

I would tend to agree but I think >= 0 is also correct

[bronze citrus]

If it equals to 0 doesnt mean its decreasing

Its constant

[hearty nest]

yh

its constant

but

nvm

[summer wave]

Depend on your terminology but usually constant is increasing also

[bronze citrus]

If the question states the function is stictly increasing then > 0 but if its just increasing >= 0

[hearty nest]

decreasing too?

[summer wave]

Ye

[hearty nest]

acc to the ques?

okok

thanks

.close

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[hearty nest]

f'(x)>0, how to find domain

[wooden python]

domain...?

or do you want to simply solve the inequality f'(x)>0

[hearty nest]

umm

yh

[wooden python]

progress?

[hearty nest]

is domain not the right word??

[wooden python]

i would say it is not

[hearty nest]

no progress

[wooden python]

maybe if you showed the entire problem exactly as was originally stated to you then we could look for a better word

ok, in that case

[hearty nest]

okok

wait

[wooden python]

make a sign table

for x going from 0 to 2pi

take note 1-tan(x) will change signs when passing through infinity as well as through 0

(sin(x)+cos(x))^2 is always positive and so serves only to exclude points where it's 0 (bc it's in the denominator and you can't divide by 0)

[wooden python]

ok so you just want to pick which of these intervals it is increasing in

[hearty nest]

hmm

[wooden python]

your f'(x) looks wrong

[hearty nest]

oh sorry there's a typo

[wooden python]

this is why we have !xy

[hearty nest]

you fixed it earlier

its this in denominator

[hearty nest]

i am trying this

[wooden python]

ok so then 1+(sin(x)+cos(x))^2 is always positive and even never 0

so can be entirely ignored

also cos(x)(1-tan(x)) is better rewritten back into cos(x)-sin(x)

[hearty nest]

ok

tryng

[wooden python]

now you only need to find which of these intervals cos(x)-sin(x) is positive in

which is considerably easier than your original question

[hearty nest]

i got it!!

thanksss

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[timid brook]

$\frac{(4^3)^{1/3}}{(4^{0.5})^{\frac{4^2}{4\sqrt{4}}}} ;-; \frac{\sqrt{4^2}}{(\sqrt{4})^2}$

[clever fjord]

Phantom

[strange aspen]

!status

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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[timid brook]

4

[golden sonnet]

u wanna simplify this?

[odd viper]

Ig 0

[unkempt lichen]

hang on, you put 4

show your work

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@timid brook Has your question been resolved?

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[mystic nova]

Same radius, M1M2=r\sqrt{2}

[odd edge]

@upper pebble Has your question been resolved?

[mystic nova]

I mean as long as M1M2 is r\sqrt{2} it can always make a square

Why you have to made the diagram using coordinates btw

Using geogebra geometry

It's much easier

No it's not, you don't have to use coord in geogebra geometry

It's different from geogebra calculator btw

There are a lots videos about it, most of geo main use it

It has many features I can't describe all here

Okay,... It's hard for me since the diagram doesn't label all the point

And it's 12am here so I can't open my computer or get a paper

You could ping other helpers for help

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@upper pebble Has your question been resolved?

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[dawn roost]

I thought the answer is D, I’m abit confused differentiation on a graph. I have drawn f’ and f” but it seems wrong

[amber schooner]

if the tangent line is above the curve then the function is concave down

[unkempt lichen]

f' should be ||always decreasing|| in the domain, no? that should tell you something about f''

[dawn roost]

So f’ will be like this right

[amber schooner]

no

f’ is decreasing

[amber schooner]

see here

[dawn roost]

I don’t really know how to draw f’

[amber schooner]

it doesn’t really matter

you don’t need to

[amber schooner]

^

if you don’t understand what it means then just say so

/how to apply it here

[unkempt lichen]

another way maybe you can think about it is to imagine tangents at different points along f in the domain

as you move along from a to b, what do you notice about the slopes of those tangents?

[dawn roost]

So if f’ concave down, then f’’ increasing, like that right

[amber schooner]

you mean decreasing

[amber schooner]

but again our graph here is of f

not of f’

[dawn roost]

f is decreasing -> f’ concave down , f’ <0. Bc f’ concave down them f” becomes increasing -> f” >0 right 🥹🥹🥹

[amber schooner]

no?

f decreasing means f’ < 0. it does not mean f’’’ < 0

[dawn roost]

Yeah, I understand the part f decreasing , f’ <0, f’ will have concave down. And bc it’s concave down f” will increase and > 0

[amber schooner]

no

where is f’ being concave down coming from?

f’ being concave down means it’s second derivative is < 0

which is f’’’

[dawn roost]

I can’t really visualize it

[amber schooner]

if f is concave down and decreasing then f’ < 0 and f’’ < 0

do you agree

[dawn roost]

Why does f’’ <0

[unkempt lichen]

wanna try coming back to this?

maybe it'd help

[amber schooner]

this is a consequence/definition of being concave down

[dawn roost]

It goes down

[unkempt lichen]

sure, but as you move along from a to b and draw tangents at different points

do the slopes of those tangents change?

if so, are they steeper? shallower?

[dawn roost]

They steeper

[unkempt lichen]

they grow steeper downwards, right?

[dawn roost]

Yep

[unkempt lichen]

so that means, f' has a negative rate, aka f' is < 0 and constantly growing in the negative direction

[unkempt lichen]

so f'' is the slope of the graph of f'

[dawn roost]

Yes

[unkempt lichen]

if f' constantly grows in the negative direction, then f'' must be...?

[dawn roost]

Oh, it will also be negative

[unkempt lichen]

there we go

[amber schooner]

(compare this picture to the one i just sent)

[dawn roost]

Okay okay i got it now