Basically, I dont understand the transition between the blue bracket to the purple bracket. I don't understand what kind of manipulation happened there to go from Blue -> Purple in a single operation.
#help-0
brittle sun
lone heartBOT
@brittle sun Has your question been resolved?
brittle sun
<@&286206848099549185> I'd really appreciate any sort of help
sonic kiln
It's just a factorization by (k+1)²
Let me try to give slightly more steps
$\frac{k^2(k+1)^2}{4}=(k+1)^2\cdot\frac{k^2}{4}$
ocean sealBOT
Blaxapate
sonic kiln
$(k+1)^3=(k+1)^2\cdot(k+1)$
ocean sealBOT
Blaxapate
sonic kiln
$\frac{k^2(k+1)^2}{4}+(k+1)^3=(k+1)^2\cdot\left[\frac{k^2}{4}+(k+1)\right]$
ocean sealBOT
Blaxapate
lone heartBOT
@brittle sun Has your question been resolved?
brittle sun
@sonic kiln Oh. And the brackets only serve as a visual cue?
brittle sun
Oh... I even googled what square brackets are for
sonic kiln
It's just that your teacher didn't want to write parenthesis, because that could look confusing. Some people don't like nested parenthesis
But yeah it's exactly the same as $$\frac{k^2(k+1)^2}{4}+(k+1)^3=(k+1)^2\cdot\left(\frac{k^2}{4}+(k+1)\right)$$
ocean sealBOT
Blaxapate
brittle sun
I see. So basically they extracted (k+1)^2 as a common derivative, and then multiplied whatever remained. Do I understand?
Sorry if the terminology is bad, English isnt my native tongue. @sonic kiln
sonic kiln
yeah we would say common "factor", derivative means something else. But that's the idea, yes.
What's your mothertongue?
brittle sun
Ah, I'm sorry
Its pretty foreign and uncommon, doubt you've heard about it haha
Anyway, I understand this now. Many thanks to you
sonic kiln
No problem
brittle sun
.close
dark pelican
lone heartBOT
dark pelican
Does some want to help me with my geometric proofs
ripe rain
I believe it’s called vertical angles theorem
When 2 lines intersect opposite pairs of angles are congruent
Hence 1=2=3=4=5
And see that 5&6 are supplementary
I believe you can write much better proof
alpine sable
Does some want to help me with my geometric proofs
Focus on angles 3 and 4
dark pelican
Yeah I got the answer
alpine sable
If you find the relation, you can easily find your answer (prove the given question*)
👍
dark pelican
I got another math problem I need help with
lone heartBOT
Available help channel!
dark pelican
lone heartBOT
dark pelican
I’m stuck and need help
rapid lava
i never got the answer 💀
wary stream
Is this a test?
desert coral
considering. they didnt reply quick. it might be possible that they may be giving a test
dark pelican
No
desert coral
sure?
desert coral
alright
method:
try to equate
2x = (5-y)
and
(4x-2) = 6y [bisection]
find the value of x in terms of y in one equation
and substitute that value in the second equation'
which will result in you being able to find the value of one of the variables. and thus you can find the value of second variable
wary stream
Sounds pretty sketchy if it says show all work for credit
dark pelican
Cause that’s how strict my teacher is
desert coral
Sounds pretty sketchy if it says show all work for credit
does full credit refer more to ... uh... assignments?
dark pelican
You helped me last weak
wary stream
I don't suggest this because people would still cheat and copy your answers
desert coral
I don't suggest this because people would still cheat and copy your answers
ohk. my bad
dark pelican
At midnight
desert coral
At midnight
?
wary stream
Probably referring to me
dark pelican
I was taking about how dldh helped me last week
grand flint
Hello I am working with a boolean function : ~In1 ~In2 ~In3 in4 + in4 in5 + ~In1 In2 In3 in4 + In1 ~In2 In3 in4 + In1 In2 ~In3 in4 (~ = not)
**Is there anyway to further minimize this function or is this the minimal form? **
I was thinking maybe since ~In1 In2 In3 in4 + In1 ~In2 In3 in4 + In1 In2 ~In3 in4 without the in4 could be minimized further with xor such as in1 xor in2 xor in3, but I am not sure if this is possible with in4 being there. Help is appreciated.
lone heartBOT
Hello I am working with a boolean function : ``~In1 ~In2 ~In3 in4 + in4 in5 + ~I...
@grand flint Has your question been resolved?
lone heartBOT
@grand flint Has your question been resolved?
bleak ridge
You could try making a massive kmap
Unless that's what you already did
That's prob more work than it's worth though
As for the xor deal you can factor out the in4
bleak ridge
~In1 In2 In3 in4 + In1 ~In2 In3 in4 + In1 In2 ~In3 in4
In4(~In1 In2 In3 + In1 ~In2 In3 + In1 In2 ~In3)
In4(In3(~In1 In2 + In1 ~In2) + In1 In2 ~In3)
In4(In3(In1 xor In2) + In1 In2 ~In3)
This is the simplest I could get it but
grand flint
Unless that's what you already did
yeah have done it already.
grand flint
~In1 In2 In3 in4 + In1 ~In2 In3 in4 + In1 In2 ~In3 in4
In4(~In1 In2 In3 + In1 ...
ok thanks!
bleak ridge
Well actually one sec
Yeah I think that's about as simple as it gets
I'm not 100% sure ofc but
undone cave
lone heartBOT
undone cave
Hello, I need your help. I already tried to solve the problem, but the answers are wrong even the method I used was wrong. These three problems are the ones that caused me the most conflict when solving them since I did everything wrong. please help me to solve it!!
lone heartBOT
@undone cave Has your question been resolved?
little drum
Hello, I need your help. I already tried to solve the problem, but the answers a...
heyy
$\frac{3}{4-\log x} - \frac{1}{4+\log x} = 1$
ocean sealBOT
Ansh
little drum
This the question !?
pale kestrel
what base logarithm is this..
pale kestrel
Hello, I need your help. I already tried to solve the problem, but the answers a...
well what method did you try
boreal surge
For numbers 0.005875×10^3 and 11 find the product, present it in a normalized form. In your answer, write down the mantissa, separating the integer part from the decimal point.
tacit arch
what have you tried?
tacit arch
For numbers 0.005875×10^3 and 11 find the product, present it in a normalized fo...
is there a typo here or in your image?
5785 != 5875
boreal surge
is there a typo here or in your image?
?
OH
Feeling dumb
WAIT NO
Its typo
Fixed @tacit arch
My teacher said its wrong answer
Idk what to do honestly
<@&286206848099549185>
lone heartBOT
@boreal surge Has your question been resolved?
lone heartBOT
@boreal surge Has your question been resolved?
lone heartBOT
lone heartBOT
Available help channel!
harsh ocean
lone heartBOT
harsh ocean
Id need to confirm whether the area is 2/3 or 22/7
Or maybe neither?
I tried doing $\int_{0}^{2} \abs{(x^{2}-3x)+x^{3}}dx$
ocean sealBOT
JSGF
harsh ocean
Looks like it is 22/3, but I can't seem to get it right
grizzled path
shouldn't it be x^3 - x^2 +3x inside the integral
because the integral of the blue function is negative
harsh ocean
Looks like $\int_{0}^{2} (x^3+\abs{x^{2}-3x})dx$ works, but the 1st one does not. WTF
ocean sealBOT
JSGF
lone heartBOT
@harsh ocean Has your question been resolved?
whole wadi
What is not true of z = 2¹² × 5⁹ × 3²?
a) z is an integer
b) z is evenly divisible by 2
c) z/25 is evenly divisible by 5
d) z/9 is evenly divisible by 3
whole wadi
I am supposed to solve this without a calculator, any advice?
We can cross out a) immediately because we're only multiplying the integers.
We also know that even * odd * odd = odd which means that b) is incorrect, right?
buoyant kayak
careful there
whole wadi
Or is it like this, even*odd = even (2*5 = 10), then even*odd = even (10*3 = 30)
So b is correct?
whole wadi
Yes
So c) & d)
To be evenly disible, the number must end in 5 or 0
For d), the sum of the digits must be evenly disible by 3
Those ones are difficult
If you divide z by 3², we have z/3² = (2¹²×5⁹×3²)/3² => z = 2¹² × 5⁹. So this implies that there is not a number 3 that z is evenly disible by?
pale kestrel
What you seem to be missing is the fact:
If $a$ is prime, $a | bc \implies a | b$ or $a | c$
ocean sealBOT
Shuri2060
whole wadi
I don't understand this, what's the solution if we use this?
whole wadi
**Shuri2060**
I interpret this as ”if z is a prime, and it evenly divides 2*5*3, then z must evenly divide each of these factors.”
glass lichen
what's the prime factorization of z/25?
whole wadi
z/5*5
glass lichen
what's the prime factorization of the number z/25, given you know what z is
whole wadi
idk
glass lichen
do you know what a prime factorization is?
whole wadi
yes but i don't understand the question, i don't know what z is, i know how to prime factorize 25
glass lichen
yes, so that's the prime factorization of z
so what's the prime factorization of z/25?
whole wadi
so 2^12\*5^9*3^2/5\*5
👆 this?
glass lichen
simplified what is it?
whole wadi
2^12 * 5^7 * 3^2
whole wadi
it's true 🙂
glass lichen
yes.
whole wadi
z/3² = (2¹²×5⁹×3²)/3² => z = 2¹² × 5⁹
glass lichen
yes.
whole wadi
therefore there's no 3 for z to evenly divide
lone heartBOT
lone heartBOT
Available help channel!
hardy flume
lone heartBOT
open linden
um idk if im right cause my calc sucks but
For all the parts where x>1 or x<1 it's a polynomial, so it's differentiable, so you don't have to worry about those
You just have to worry about what happens when x=1
obviously you have to have 9x+b = x^2 + ax - 7 otherwise there would be a gap in the graph
So do that
open linden
obviously you have to have 9x+b = x^2 + ax - 7 otherwise there would be a gap in...
at x=1
idk what the next step is
open linden
u get anything
about a and b
wait so 9+b=a-6
b=a-15
Maybe derivative of 9x+b at 1 = derivative of x^2 + ax + 7 at 1
@hardy flume
alpine sable
dont know calc but to make a guess, maybe you have to find the value of a and b that makes the derivitve of both sections the same
alpine sable
at the disconnect point
hardy flume
usually for these questions you can solve for either a or b then imput that into the other one to solve
open linden
derivative of 9x+b at 1 = derivative of x^2 + ax + 7 at 1
Therefore 9 = 2x + ax when x=1
so a =7
probably
alpine sable
i think maybe you have to do
(x^2+ax-7)' = (9x+b)'
if x = 1 what vale of a and b will make this tatment true
im just guessing though, right now you cant diferantiace because theres a jump in the ficntino so if you make the derivites meet then you will make the jump go away?
alpine sable
b+7=a?
coral moss
what do you mean by there is a jump
alpine sable
when you switch from one equation to the other its non coninuous, ie the number jumps from one number to another
coral moss
in any case b could literally take any value because 9x+b will always differentiate to 9 so b would change basically nothing
unless you want both to be equal at 1
alpine sable
$f(x) = x^2+ax-7 // g(x)=9x+b // f'(x) = 2x + a // g'(x) = 9 // f'(1) = g'(1) // 2(1)+a = 9 // a = 7?$
$f(x) = x^2+ax-7 \ g(x)=9x+b \ f'(x) = 2x + a \ g'(x) = 9 \ f'(1) = g'(1) \ 2(1)+a = 9 \ a = 7?$
ocean sealBOT
aspwil
coral moss
i guess you could solve for b then if a =7
alpine sable
but plugging into desmos shows b cant equal just anything
for it to be the same at both points b need to a contant offset from a
coral moss
yeah i know what you mean now i misread the question
alpine sable
desmos seems to say b+7=a, whcih seems to be correct, problem is i dont know how to actauly find that
hardy flume
yeah me neither
wait
hold on
it could be that you have to put that into a for the first one
??
coral moss
b = -8
hardy flume
im lost
ocean sealBOT
I can't believe you've done this
hardy flume
wherd you get a = 7 from tho
coral moss
setting the derivatives equal to each other and solving for a
9x + b always differentiates to 9
coral moss
i still don't know if that's right
coral moss
Well that's good to know
The way it works is that you need 1 derivative for every value but at $x = 1$ you'd get a derivative from $x^2 + ax - 7$ and $9x+b$
that means that both must be equal to ensure you get 1 derivative. Since $9x + b$ always differentiates to 9, then the $2x + a = 9$, at $x = 1$. Solving for $a$ gets you $a = 7$ (as the other person showed). You then need to solve for b such that $9x + b$ and $x^2 + 7x -7$ both intersect at $x = 1$
ocean sealBOT
I can't believe you've done this
coral moss
Sorry for any silly mistakes i might have made i've been awake for nearly 20 hours so hopefully they're excusable
lone heartBOT
@hardy flume Has your question been resolved?
true eagle
lone heartBOT
coral moss
the first part of the question tells you to assume the depth at t = 0 is the average of the depth so find the average (midpoint) of 8.1 and 11.9 to get the first part of your equation
wet trellis
think about what you want the amplitude of the graph to be (the middle value up to the max value), what you want the period to be (how often it oscillates), then apply those to the standard sine graph, then how far you need to push that graph up or down to get what you want
if that all is a lot, at least give me the first two things (amplitude, and period) and we can go from there
lone heartBOT
@true eagle Has your question been resolved?
barren nest
lone heartBOT
barren nest
I tried using substitution but I don’t think it’s working
glass lichen
Post your work
barren nest
glass lichen
yeah, cause you didn't square (mx+1)^2 properly
@barren nest
$(mx+1)^2\neq (mx)^2+1^2$
ocean sealBOT
Mosh
lone heartBOT
@barren nest Has your question been resolved?
white jolt
@barren nest u could've made a quadratic like my^2=y-1...and then u can apply D>0
glass lichen
I was getting there........
@white jolt
They also did the correct thing, just messed up squaring.
white jolt
@glass lichen k
glass lichen
You also shouldn't just give complete steps like that.
alpine sable
Could someone please help me with this problem?
I tried everything yet I am still stuck <@&286206848099549185>
lone heartBOT
Available help channel!
proud bobcat
Hello im having an issue knowing how to disprove something not being a subset of the other
proud bobcat
I have my list of numbers and have figured out what number the other set does not have
But
Im not sure how to word it
I have this for the first one proving they are sets but the second one for proving it isnt one is definitely wrong
I know i start with the number so for that, a has 2 in its subset and it is not in 8
But idk what to say
lone heartBOT
@proud bobcat Has your question been resolved?
lone heartBOT
@proud bobcat Has your question been resolved?
naive valley
Your image doesn't even include the full definition of B.
Can you upload a better picture?
proud bobcat
Your image doesn't even include the full definition of B.
Here you go
Can you check 1) and tell me if i worded it right
And then for 2 i just dont know how to start it
wary stream
Is that a test?
proud bobcat
its essentially saying dont copy off other peoples work
im not looking for like a solution im wanting just how to start it up
i had my notes on it but the one for disproving isnt on there for some reason
and her video doesnt have it either
i mean if you feel uncomfortable i understand, just asking cause i dont have it in my notes
and she wants it worded a certain way so i can usually guess if its missing in my notes but oof
dammit the reason theyre not in my notes is cause i was doing them on the board
cucked myself ig
thanks for help though
.close
true eagle
if that all is a lot, at least give me the first two things (amplitude, and peri...
I think the amplitude is 1.9
lone heartBOT
I think the amplitude is 1.9
true eagle
if that all is a lot, at least give me the first two things (amplitude, and peri...
I guess the period is 6
lone heartBOT
@true eagle Has your question been resolved?
lone heartBOT
@true eagle Has your question been resolved?
lapis ore
lone heartBOT
lapis ore
I have absolutely no idea how it is not -2/x^3
lapis ore
so it would just be -2
hasty elk
yeah
lapis ore
makes sense
any idea on this one
I've tried but the multiplication of all the numbers are tripping me up i suppose
This is where i'm at atm
south kraken
lapis ore
I'm sort of confused how that helps
placid zinc
I doubt they've seen the derivative rules, yet
Check your signs. The negative on -4/x should distribute
lapis ore
Right. I'm at the point where my professor is giving us algebra instead of actual calc
I heard it gets easier after this I guess
Thanks for the help, I'm just going to sleep on it. 2 AM my time lol it ain't due till sunday anyways.
.close
lone heartBOT
lone heartBOT
Available help channel!
sage tide
lone heartBOT
sage tide
how do we find <BOA
tacit arch
how do we find <BOA
Are you given AD=DB
tacit arch
No idea, just trying to get any info
lone heartBOT
@sage tide Has your question been resolved?
hollow pelican
hi guys does anyone know how to solve 1d
lone heartBOT
hi guys does anyone know how to solve 1d
rich niche
it is not complete right?
pretty sure its incomplete
you need a side relation or something
plain siren
sly comet
lone heartBOT
dull oak
2nd derivative
alpine sable
lone heartBOT
alpine sable
Why does the equality hold..
vale wigeon
$z = y - \sum_{i=1}^k \ang{y,v_i}v_i$ by definition
ocean sealBOT
Ann
vale wigeon
they just plugged that into <z, v_j>
alpine sable
Why does $u=\sum_{i=1}^k \ang{y,v_i}v_i$ hold
vale wigeon
let u be as defined in the preceding equation
ocean sealBOT
ENG learner
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
vale wigeon
(you don't have the \ang{} macro defined in the bot so it won't work for you: you'll have to use \left< ... \right> or \left\langle ... \right\rangle to get the angle brackets)
pulsar aspen
Wait, does that $W^{\perp}$ mean a function from empty set to W?
ocean sealBOT
Xwtek
vale wigeon
no, it means the orthogonal complement of W
alpine sable
Still dont know why it holds
..
vale wigeon
you're overthinking it
the proof says, "let's examine what happens when we subtract that sum from y"
alpine sable
.close
lone heartBOT
alpine sable
.reopen
lone heartBOT
✅
alpine sable
I need more explanation..@vale wigeon
vale wigeon
there's nothing else to explain, is there?
alpine sable
Then why does the equality second from the end hold
alpine sable
The one after the one after the black box@vale wigeon
vale wigeon
$\ang{y, v_j} - \sum_{i=1}^k \ang{y,v_i} \ang{v_i,v_j} = \ang{y,v_j} - \ang{y,v_j}$
ocean sealBOT
Ann
vale wigeon
this one?
alpine sable
Yes
alpine sable
I can't understand these three
alpine sable
That's not a definition..
But that's not a supposition but conclusion ..
So it's a supposition?
@umbral dune
ㅠㅠhow to
lone heartBOT
@alpine sable Has your question been resolved?
lone heartBOT
@alpine sable Has your question been resolved?
alpine sable
my problem is 5a plus 3a
lone heartBOT
@alpine sable Has your question been resolved?
visual plinth
lone heartBOT
visual plinth
Can someone help me with part d please
Im not sure what rules I can apply
i've managed to get 3-6cot^2(4x) but I don't think thats close
alpine sable
Change tan to sin/cos
visual plinth
how would that look?
I tried that but I think what I did was wrong
would it be 2sin^2(4x)/2cos^2(4x)
gray isle
how are you getting
3-6cot^2(4x)
visual plinth
I'll draw it up brb it'll be easier than typing
actually i don't know
lol
it's grossly messy but hopefully you can see what im thinking up till this point
im new to using these identities so there could be a mistake anywhere :/
gray isle
that doesn't tell me how you got
3-6cot^2(4x)
visual plinth
I know
I used cos/sin = cot
on the last expression
I think that doesn't work now
just ignore the 3-6cot that bad
gray isle
from here you could consider factoring 3 out from the numerator
visual plinth
is it correct till that point?
gray isle
handwriting and notation can be improved
visual plinth
yeah sorry i did it with a mouse.
gray isle
the mouse aspect only accounts for half of the issues
visual plinth
like what?
gray isle
ambiguous skyscraper fractions,
missing parentheses
gray isle
assuming you intended to write
$$\frac{3-6\cos^2(4x)}{2\sin^2(4x)}$$
that's ok
ocean sealBOT
ℝamonov
visual plinth
you multiple the ones touching for the denominator and multiply the ones not touching for the numerator
yes
thats where im stuck at
ive factorised out the 3 but im still not sure what identity I could use here
gray isle
you should make fraction lines distinctly longer and or introduce additional parentheses to remove all ambiguity for sky scraper fractions
pythagorean trig identity
visual plinth
I will in the future thanks
also when you say pythagorean trig identity u mean apply it to the numerator
(1- 2cos^2(4x)?
and ur talking about this identity
?
gray isle
wait
visual plinth
im struggling to apply the identity here
gray isle
1 sec
visual plinth
ok
gray isle
didn't catch the mistake yuo made
$2\tan^2(4x)\neq \frac{2\sin^2(4x)}{\red{2}\cos^2(4x)}$
ocean sealBOT
ℝamonov
gray isle
yes
ocean sealBOT
kenzo
visual plinth
ok im here now
im going to try to use the Pythagorean identity in terms of sine
is it 1.5?
gray isle
yeh
visual plinth
thank you , i feel like I learnt alot from this one question
gray isle
alternative you can first factor out the 3/2
and consider csc and cot
visual plinth
.close
alpine sable
Can anyone help me why do i get square root of 1+t^2 when i need just 1+t^2 thanks
tacit arch
What are you even trying to show
alpine sable
trying to get the t formula for sin theta using sin/cos = tan and sin^2 + cos^2 = 1
if this helps
tacit arch
trying to get the t formula for sin theta using sin/cos = tan and sin^2 + cos^2 ...
What is t formula
warped ferry
general question but for euler's theorem and Fermat's little theorem, do both numbers need to be natural numbers?
pale kestrel
its Number Theory.. well yes???
Divisibility makes no sense for decimals
If you're asking if they could be integers... look up the statement of the theorems. They say.
warped ferry
ok ty
sorry i'm not familiar with number theory so i'm not aware of obvious things in it
.close
lone heartBOT
lone heartBOT
Available help channel!
raw sleet
lone heartBOT
raw sleet
i still dont know how to start this
sin
its in spanish
yep
same
calc 1
thats the thing
im not allowed to use lhopital
:0
oof
nope lol
to solve it with trig. identities
:0
i dont knowwwwwwwww
:(
let me try it
<@&286206848099549185>
blissful whale
You can ignore the bottom right cos(x) as it is just 1.
blissful whale
Other than that it looks like a prime use case for Hôptial's 😄
raw sleet
but i cant use ittt
blissful whale
😦
raw sleet
yep
blissful whale
How about Taylor expansion? Is that allowed?
raw sleet
How about Taylor expansion? Is that allowed?
do i have to derivate with it?
blissful whale
Kinda - kinda not - it's already derivated if you wish.
You find the taylor approximations using derivates, but they may be given to you already.
raw sleet
idk if i can use it
blissful whale
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \ldots$$
ocean sealBOT
M8732
blissful whale
Without either Taylor nor Hôptials this will be very difficult.
blissful whale
Disbelieve 😮
blissful whale
Are we allowed to use $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ and similar things?
ocean sealBOT
M8732
raw sleet
yep
blissful whale
Okay that's derivates in disguise, but let's not tell the professor lol
raw sleet
lmao
blissful whale
Then I suggest
$$ = \lim_{x \to 0} \frac{\sin(1 - \cos(3x))}{1 - \cos(3x)} \frac{1 - \cos(3x)}{x \tan \left( \frac{x}{4} \right)} \frac{1}{\cos(x)}$$
$$ = \underbrace{\lim_{x \to 0} \frac{\sin(1 - \cos(3x))}{1 - \cos(3x)}}{ = 1} \lim{x \to 0} \frac{1 - \cos(3x)}{x \tan \left( \frac{x}{4} \right)} \underbrace{\frac{1}{\lim_{x \to 0} \cos(x)}}_{ = 1}$$
ocean sealBOT
M8732
blissful whale
wow this looks ugly 😄
blissful whale
yeah if all limits exist that's possible
you just need to find a "split"
that does not get you to indeterminate things like
infinity times 0, infinity minus infinity or something like that
raw sleet
raw sleet
uhhh
blissful whale
yeah
because that's just a 1.
an ugly way to write 1, but it helps us recognize sin(x)/x form
because 1 - cos(3x) is the argument of the sin
blissful whale
yes
blissful whale
Which minus sign and why do you want to change it?
blissful whale
no I am not trying to do conjugates
blissful whale
Maybe try simpler example to see, let's say
$$\lim_{x \to 0} \frac{\sin(f(x))}{x}$$ with some $f(x)$.
ocean sealBOT
M8732
blissful whale
then
$$\lim_{x \to 0} \frac{\sin(f(x))}{x} = \lim_{x \to 0} \frac{\sin(f(x))}{x} \frac{f(x)}{f(x)} = \lim_{x \to 0} \frac{\sin(f(x))}{f(x)} \frac{f(x)}{x}$$ $$ = \lim_{x \to 0} \frac{\sin(f(x))}{f(x)} \lim_{x \to 0} \frac{f(x)}{x}$$
ocean sealBOT
M8732
blissful whale
for first limes we can use sin(x)/x thing since we just changed the argument.
So it's one.
The second limes remains without extra sin.
blissful whale
If we also allow ourselves to use the facts
$$\lim_{x -> 0} \frac{\tan(x)}{x} = 1$$
ocean sealBOT
M8732
ocean sealBOT
M8732
blissful whale
we can do the limit.
You can use these in a similar fashion like I got rid of the sin.
raw sleet
so how can I use that to solve the limit?
blissful whale
Do you get how I got rid of the sin?
blissful whale
yes
raw sleet
uhh
kinds
you multiplied by f(x)/f(x)
and then you got two different limits
right?
raw sleet
yep
blissful whale
Let's do another limes as an example together.
$$\lim_{x \to 0} \frac{\tan(x)^2}{1 - (\cos(x))^2}$$
ocean sealBOT
M8732
blissful whale
then
$$\lim_{x \to 0} \frac{\tan(x)^2}{1 - \cos(x)} = \lim_{x \to 0} \frac{\tan(x)^2}{x^2} \lim_{x \to 0} \frac{x^2}{1 - \cos(x)}$$
$$ = \left( \lim_{x \to 0} \frac{\tan(x)}{x} \right)^2 \frac{1}{\lim_{x \to 0} \frac{1 - \cos(x)}{x^2}} = 1^2 \frac{1}{\frac{1}{2}} = 2$$
blissful whale
tan becomes sinx/cosx
Yeah that is true.
Using that you can proof my two "facts" I suggested above, if you are not allowed to use them directly.
blissful whale
I corrected some typos in my calculation, I hope they caused not too much confusion.
ocean sealBOT
M8732
lone heartBOT
@raw sleet Has your question been resolved?
lone heartBOT
@raw sleet Has your question been resolved?
raw sleet
isn't this the answer?
um
i don't think so
this is the limit
and im stuck
here
tacit arch
,w plot sin(1-cos(3x))/(x tan(x/4) cos(x))
ocean sealBOT
tacit arch
the limit is negative infinity?
blissful whale
isn't this the answer?
No, I was doing an example to show the technique.
,wolf limes sin(1-cos(3x))/(x tan(x/4) cos(x)), x -> 0
ocean sealBOT
blissful whale
Not infinite.
raw sleet
;-;
blissful whale
We already got to the point where we said that we can drop the sin in the top part
and the cos in the bottom
Then it's just a product of limits with known facts of tan and cos.
See my above example for how to proceed.
raw sleet
We already got to the point where we said that we can drop the sin in the top pa...
i can eliminate the sin?
blissful whale
Yes, see my explanations in the beginning
blissful whale
Yes, later I also did an example to show the idea in general with a "f" instead of the convoluted mess.
raw sleet
raw sleet
Yes, later I also did an example to show the idea in general with a "f" instead ...
like that?
blissful whale
Yes, though you are missing brackets in the third line.
raw sleet
Yes, though you are missing brackets in the third line.
woops
what do i have to do after that?
blissful whale
Yeah the answer should be 18
though if Hôptal's is not allowed, I doubt Wolfy is allwoed either 😮
blissful whale
Wolfram Alpha
blissful whale
what do i have to do after that?
The left fraction is 1, do you agree?
blissful whale
This means another trigonometric function elimanted.
As the right fraction is now without sin.
blissful whale
yeah
blissful whale
Do you understand my comptation there?
It's veeery similar, if you understand it it should be very easy to transfer.
Just trying to "Likewise, if you are providing help to others, try your best to explain and elaborate instead of simply giving away the answer."
as it says in the rules.
raw sleet
i kinda understand it
blissful whale
I am just grouping/balancing the things that go to zero with simple x'es.
blissful whale
You have 1 - cos(x)
what thing do we need at the other side of the fraction
to apply what we know about it?
Get this form by introducing factors.
blissful whale
nah then we are not mkaing progress
raw sleet
oof
blissful whale
This is a fact you need.
raw sleet
This is a fact you need.
:0
blissful whale
We can try to prove this fact
raw sleet
We can try to prove this fact
how
blissful whale
It changes fundamentally with how you define cos and what you are allowed to use.
I think one can't really do it without derivatives.
raw sleet
I think one can't really do it without derivatives.
aaaaaaaaaaa
raw sleet
It changes fundamentally with how you define cos and what you are allowed to use...
oooo
pain
blissful whale
Just checked the internet and got an idea, we can reduce it to the sin case.
raw sleet
:0
blissful whale
1 - cos(x) = 2sin²(x/2)
blissful whale
the 3 you can just substitute out
raw sleet
and use u/3 in denominator.
i dont understant this
blissful whale
$$\lim_{x \to 0} \frac{1 - \cos(3x)}{x^2} = \lim_{x \to 0} \frac{1 - \cos(3x)}{\left( \frac{3x}{3} \right)^2} = \lim_{u \to 0} \frac{1 - \cos(u)}{\left( \frac{u}{3} \right)^2}$$
ocean sealBOT
M8732
blissful whale
This corresponds to just 9 times the thing from before.
raw sleet
<@&286206848099549185>
alpine sable
Relatable
raw sleet
aghhhhhhhhhhhhhhhhhhhhhhhhhh
tacit arch
Last line is almost there!
raw sleet
no
its not almost there
i had to learn triple angle identities
and double angle identities
tacit arch
Don't need that
tacit arch
Multiply top and bottom by 3x and multiply top and bottom by x/4
And use definition of tan
raw sleet
w h a
tacit arch
Have you proven $sin(x) / x$ approaches 1 as x goes to zero
ocean sealBOT
riemann
raw sleet
yea
tacit arch
So what about $\sin(kx)/(kx)$ for some nonzero k?
ocean sealBOT
riemann
tacit arch
As the limit x goes to zero
tacit arch
Ok this is tougher, but still easer than fucking triple angle identities
raw sleet
yea lmaoo
tacit arch
As x goes to zero, how does kx behave?
tacit arch
You have to do some things to make it more obvious and apparent
But they're legal things
tacit arch
Multiply top and bottom by 3x and multiply top and bottom by x/4
Write this out
raw sleet
ok
tacit arch
So far so good
raw sleet
raw sleet
So far so good
there?
tacit arch
**riemann**
You want to manipulate your expression so this pops out and you can take the limit at the end
raw sleet
okay...
tacit arch
You did the same thing here with sin(x)/x on the second to last line
little drum
scrweem
little drum
what's the question even?
little drum
ehhhh
little drum
$\lim_{x \to 0} \frac{\sin (1-\cos 3x)}{x\cdot \tan \frac{x}{4}\cdot \cos x}$
ocean sealBOT
Ansh
little drum
$\frac{4(4\sin^2 x)}{x^2}$
ocean sealBOT
Ansh
little drum
Isn't the answer just 16?
little drum
but I'm sure there's an easier way
wdym easier way... what you did is the easiest possible way except if you're proficient with this way, in which case you can directly write the limit as:
$$\lim_{x \to 0} \frac{4\textcolor{red}{\cancel{\color{black}{\qty(\frac{\sin (1-\cos 3x)}{1-\cos 3x})}}} \cdot (1-\cos 3x)}{x^2 \cdot \frac{\tan \frac{x}{4}}{\frac{x}{4}} \cdot \cos 0}$$
tacit arch
Isn't the answer just 16?
,w 16!
ocean sealBOT
ocean sealBOT
Requested by Ansh#3502
bleak ridge
How do you do a red cancel
little drum
oof
bleak ridge
Tell me these secrets
raw sleet
Isn't the answer just 16?
its 18
ocean sealBOT
Ansh
little drum
I don't see why in the slightest would the limit be 18
,w limit (sin(1-cos 3x))/(x tan(x/4) cos x) as x to 0
ocean sealBOT
little drum
whaaa
raw sleet
indeed
little drum
miscalc !? 🤔
frozen python
my god
little drum
cos 3x = 4cos^3x - 3cos x... right!?
raw sleet
frozen python
yes but how
raw sleet
even with the stuff we did
frozen python
what in gods name of trig identities is that
raw sleet
we got 18
little drum
$$\lim_{x \to 0} \frac{4\textcolor{red}{\cancel{\color{black}{\qty(\frac{\sin (1-\cos 3x)}{1-\cos 3x})}}} \cdot (1-\cos 3x)}{(3x)^2 \cdot \textcolor{red}{\cancel{\color{black}{\frac{\tan \frac{x}{4}}{\frac{x}{4}}}}} \cdot \cos 0} \cdot 9$$
okay I got it smh
ocean sealBOT
Ansh
little drum
do uou understand the above!?
raw sleet
yea
little drum
100% sure!?
little drum
cuz sinx / x form
little drum
and cancel the tan cuz tanx/x
raw sleet
yep
raw sleet
yepp
blissful whale
This is all the things I said 8 hours ago or something.
frozen python
can u break it into the products of limits?
raw sleet
This is all the things I said 8 hours ago or something.
i take my time to understant
xd
blissful whale
can u break it into the products of limits?
Yes
little drum
$$\lim_{x \to 0} \frac{4\textcolor{red}{\cancel{\color{black}{\qty(\frac{\sin (1-\cos 3x)}{1-\cos 3x})}}} \cdot (1-\cos 3x)}{\textcolor{red}{\cancel{\color{black}{\qty(\frac{\tan \frac{x}{4}}{\frac{x}{4}})}}} \cdot \cos 0 \cdot (3x)^2} \cdot 9$$
ocean sealBOT
Ansh
frozen python
$$\lim_{x\to 0}\frac{\sin\left(1-\cos\left(3x\right)\right)}{x}$$
ocean sealBOT
MattDog_222
little drum
....
blissful whale
You need to divide by x²
little drum
Matt can you...
frozen python
(stop talking?)
little drum
yepp
so we're left with only a couple terms right!?
$$\lim_{x \to 0} \frac{4\textcolor{red}{\cancel{\color{black}{\qty(\frac{\sin (1-\cos 3x)}{1-\cos 3x})}}} \cdot (1-\cos 3x)}{\textcolor{red}{\cancel{\color{black}{\qty(\frac{\tan \frac{x}{4}}{\frac{x}{4}})}}} \cdot \cos 0 \cdot (3x)^2} \cdot 9$$
ocean sealBOT
Ansh
raw sleet
yea
little drum
$36\lim_{x \to 0} \frac{1-\cos (3x)}{(3x)^2}$
blissful whale
why 36?
ocean sealBOT
Ansh
little drum
because brain ._.
blissful whale
Ah you grouped it this way, okay
little drum
from where u got all of this
hmmm?
You said you got it =_= don't tell me you did not
little drum
wdym from where?
blissful whale
yeah
little drum
nope, I wrote that right from the first step
blissful whale
I think it is better to do it a longer winded way though if there is confusion.
little drum
$$\lim_{x \to 0} \frac{\color{green}{4} \textcolor{red}{\cancel{\color{black}{\qty(\frac{\sin (1-\cos 3x)}{\color{yellow}{1-\cos 3x}})}}} \cdot \color{yellow}{(1-\cos 3x)}}{\textcolor{red}{\cancel{\color{black}{\qty(\frac{\tan \frac{x}{4}}{\frac{x}{\color{green}{4}}})}}} \cdot \cos 0 \cdot (3x)^2} \cdot 9$$
raw sleet
yep im doing it
little drum
hmm, makes more sense
ocean sealBOT
Ansh
raw sleet
then u cancel that
ocean sealBOT
Ansh
little drum
Agreed!?
raw sleet
yep
little drum
next you use your favorite family limit
$\lim_{3x \to 0} \frac{1-\cos (3x)}{(3x)^2} = \frac{1}{2}$
ocean sealBOT
Ansh
raw sleet
aha
little drum
$\lim_{x\to 0} \frac{4(1-\cos (3 x))}{(3x)^2} \cdot 9$
And this becomes: $$\lim_{x\to 0} \frac{4\color{red}{\cancel{\color{yellow}{(1-\cos (3 x))}}}}{\color{red}{\cancel{\color{yellow}{(3 x)^2}}}} \cdot 9 \cdot \color{yellow}{\frac{1}{2}}$$
ocean sealBOT
Ansh
