help-38

[waxen drift]

Its an explicit formula, not a recusive
If it helps visually its the same as 2n-1

[wooden hamlet]

So when it is explicit n is equal to or greater to is the same as the first step number

[verbal gulch]

I think you're conflating several things together

explicit just means that it's not recursive

[wooden hamlet]

Idk how else to make sense of these equations

[verbal gulch]

it has nothing to do with what n is greater than or equal to

[wooden hamlet]

Yea but when you write the equation you also have to write that

[waxen drift]

Thats right
Since it doesnt need to start on the 2nd step like in a recursive it can start on 1

[verbal gulch]

you have to write it because you're saying that "this formula works for D(n) when n >= 1"

it's not just that you have to write it because someone told you to

[wooden hamlet]

Is this true for both geometric and arthmetic

[verbal gulch]

it's part of the meaning conveyed by writing the formula

[wooden hamlet]

Not for this problem but other problems I have to

[verbal gulch]

again, geometric and arithmetic have nothing to do with what you write n >= somethinng

[wooden hamlet]

But it is apart of the equations I need to know

[verbal gulch]

these are separate parts with separate meanings

okay, but if someone writes three sentences, it doesn't mean that all three have to be together

[verbal gulch]

if you order large purple shoes, it doesn't mean that purple has to be with shoes

[wooden hamlet]

I mean it’s on my math final 😀

[verbal gulch]

okay, but what does that have to do with what I said?

just because I ordered large purple shoes once doesn't mean that red shoes can't exist

or small shoes

[wooden hamlet]

I dont understand what you mean I have to know both versions

[verbal gulch]

when you say "n >= 2 goes with explicit formulae" you're saying that "purple" goes with "shoes"

they're separate concepts enntirely

[waxen drift]

For the most part yeah
Unless theres something else telling you that it isnt allowed, yeah

[wooden hamlet]

They r separate I understand that but I need to know how to write them correctly based on certain equations

[verbal gulch]

All D(1) = 1, D(n) = 2 + D(n-1) for n >= 2 is saying is:

If n = 1, then D(n) = 1,
otherwise, if n >= 2, then D(n) = 2 + D(n-1)

so if you want to evaluate, for example, D(3), you see that 3 >= 2, so you write: D(3) = 2 + D(3-1) = 2 + D(2)

and similarly you can figure out D(2)

[verbal gulch]

if they're separate concepts entirely, then that means you can't simply guess n >= 2 because it's explicit or whatever

that's like guessing purple based on ordering shoes

[wooden hamlet]

Yea I was just confused because I thought when a equations has n>=# it has to always be the second step

[verbal gulch]

what has to be the second step?

your job is to write something for D(n) that accurately expresses the pattern given

[wooden hamlet]

I mean yea but the only way I’m going to be able to write these equations is if I know the differences between them

[waxen drift]

Not necessarily
Thats what sacc is trying to say. It's based on what youre doing not what you have

[verbal gulch]

to know the differences between them you have to know what each of the parts mean, not just try to guess at what to write

[wooden hamlet]

Yea but if I am given a table and the steps are 1,2,3,4,5 I have to know if I put n>=1 or n>=2

[verbal gulch]

yeah and my point is that you can't just guess it off whether it's explicit or recursive or whatever

that's not how it works

it's part of the meaning you need to express

[wooden hamlet]

But when I’m writing a equation I have to know which one to use I can’t just guess

[waxen drift]

Unless the instructions say which one to use you have the freedom to use whatever one is easier for you to use

[verbal gulch]

For example, you can also write:

D(1) = 1
D(2) = 3
D(n) = n + 2 for n = 3
D(n) = 2n -1 for n >= 4

the n >= part is describing for which n the equation you wrote applies

[wooden hamlet]

My tecaher said for recursive ones it has to be the second step

[verbal gulch]

it's not just something you fill in based on something else

what second step

[wooden hamlet]

Based on the table or the sequence

[verbal gulch]

what do you mean by second step

like D(2)?

[wooden hamlet]

Maybe? Ids hard to explain

[verbal gulch]

if it's hard to explain, then it's probably not correct

[wooden hamlet]

Like for this table . When I write a recursive equation it would have n>=1

[verbal gulch]

no? that's not true at all

[waxen drift]

I mean, it can
But it depends on the pattern and where they start you at

[verbal gulch]

1. the table only expresses a subset of values
2. even if you assume the sequence starts at 0, there's nothing wrong with assuming that it has several base cases

[wooden hamlet]

If I did not start at 0 I would just add 1 to the step I started at

[verbal gulch]

take for example the fibonacci sequence

[waxen drift]

Here your "2nd" step is at n=1 because n=0 is the "1st" step

[wooden hamlet]

Yea

[verbal gulch]

like I'm sorry what you're saying about the second step just isn't correct

[wooden hamlet]

I mean it’s the only way I’ve been able to do the work and it’s been correct so far

I just forgot the question above cus I did this packet last week

[verbal gulch]

okay well maybe it lets you write down the correct answers to some problems but it's just wrong in general

[wooden hamlet]

So I don’t remember the rules

[verbal gulch]

there aren't any rules except that you have to write down something that accurately represents the sequence

[waxen drift]

I think theres just a bit of technicalities here on vocabulary and being able to understand the concept

[verbal gulch]

no, the second step thing is just blatantly false

the fibonacci sequence has two base cases

[wooden hamlet]

I mean it works for high-school math I’m not like planning on doing anything begind pre calc

[verbal gulch]

sure, if you just want to pass a test and not understand anything, just memorize every single pattern and match it I guess

[wooden hamlet]

And I don’t really undestand the unit cus it’s from September but I got good grades on those tests in Septembers either A’s or A+

[verbal gulch]

it's more effort than learning it properly, but whatever floats your boat

[waxen drift]

I understand, which is why i disclaim that it depends on the case in my responses to them
But in generality it's likely to be a linear or geometric equation

[verbal gulch]

I mean if you're going to assume that, then why not just tell everyone to check the differences between the terms and just write the memorized formulae for them?

[waxen drift]

Hey thats a bit harsh i think. I get that its better in the long run to learn and understand but not everyone is cut out for higher math. If the goal is just to be able to pass then let it be and get your head out of your ass
You dont have to stay to help if you dont agree

[verbal gulch]

that's not being harsh; I'm saying that's a valid approach

[supple copper]

There’s nothing wrong with studying to pass, other than school is supposedly trying to teach you skills, and maths isn’t the subject that teaches memorisation

[verbal gulch]

like if you're sure that it's a one-off and you're not learning anything that relies on the knowledge, there's nothing wrong with just memorizing all of the patterns and accepting you'll get something wrong

[waxen drift]

I agree it doesnt, but some classes does end up thst way sadly
But right now we're trying to help someone to study
Not bicker about how it should be learned

[verbal gulch]

but you shouldn't reinforce false information like "the second step is right' or whatever; it makes for more confusion

[supple copper]

Maths is supposed to teach you how to reason logically, and that’s the skill you should be focused to practicing when you study maths…

You’re not a computer that just follows predetermined algorithms given to you by a teacher

[waxen drift]

I personally dont agree is false
If youre going from a list thats in numeria order, then its just 1st 2nd 3rd with the placement, not necessarily the n's numerical value

[verbal gulch]

some people just want to pass math class, and I think it's valid to just not understand it and memorize some heuristics

[verbal gulch]

yeah, but it's not "when it's recursive, you start n at the second thing"

[wooden hamlet]

How would you type the equation in number 4 into a calucltor?

[verbal gulch]

depends on what calculator you have

[waxen drift]

And i have said it depends on the pattern

[wooden hamlet]

TI-83

[supple copper]

You shouldn’t think of it that way

[wooden hamlet]

What other way would you do it?

[verbal gulch]

if you have just a scientific calculator, then you have to do it the manual way of figuring out what D(2) is

[supple copper]

Did you do part a

[verbal gulch]

then D(3) and so on

[wooden hamlet]

That’s the part I’m trying to do

[supple copper]

Okay so you have f(1) and now you want to find f(2)

[wooden hamlet]

Yea

[verbal gulch]

Allie doesn't understand what the notation means, so you have to go over that first

[supple copper]

Well it says f(n) = 2 + f(n-1)

[verbal gulch]

that's what I've been trying to get at

[supple copper]

What does this mean for f(1)

[supple copper]

This is really bad, it’s like trying to write an essay in French but you don’t know how to spell in French

[wooden hamlet]

I mean I’ve been going over it for the pass week I just forget it

Over and over

[supple copper]

I’ve been going over notation for the last 15 years of my life

[verbal gulch]

it's fine to not know it, but all that means is that when you do these problems, you have to spend time understanding the notation

[supple copper]

A week is nothing

[waxen drift]

You cant use an actual equation. However
If you follow what it says and type 43 =12
Then multiply ans4
You can keep clicking enter to make it recursive

[verbal gulch]

I'm not saying that's a bad thing; I'm just saying that you have to start from the basics and work your way up

[supple copper]

You don’t learn a new language in a week

Maths is a language

[wooden hamlet]

I mean I don’t really have time to do that 😭

[verbal gulch]

you can get away with using a dictionary and memorizing the whole thing, and tbh if it's the last essay in French you'll ever write, that might not be a bad idea

[wooden hamlet]

I just need to undeatand it in this scenario

Undestand

[verbal gulch]

what I'm saying is that it actually doesn't take nearly as long as you think

the notation is simply a short way of saying something in english

[supple copper]

To me the methodology of doing maths is far more important to the actual maths itself for most people

[wooden hamlet]

I mean like I really don’t have time to learn the thoery of it unless I want to go to bed at five am

[supple copper]

f is a function do you know what a function is?

[wooden hamlet]

I mean not exactly

[verbal gulch]

f(n) = 2 + f(n-1) is literally just saying "if you want to calculate f(n), you take what you have for f(n-1) and add 2 to it"

For example, if you want to calculate f(3), you take f(2) and add 2 to it

[supple copper]

It’s a process that you give it some ingredients, and it gives you some output

For example, the process of baking a cake is a function

[verbal gulch]

it literally takes less time to learn the theory of it than to memorize all of the patterns and the cases

[wooden hamlet]

I tried that it gave me the wrong answer

[supple copper]

You put the ingredients in and you get out a cake

f(n) = 2n means you give it n, you get out 2n

[wooden hamlet]

4 is the gf right?

[supple copper]

The 2n is like the recipe of the cake, it tells you what exactly to do with the input

[verbal gulch]

then you didn't do it correctly, because that's literally what it means

For example, if I want to calculate f(2):

I need to take f(1) and add 2 to it.
what is f(1)?
the problem gives f(1) is -5
therefore, I add 2 to it and I get -3

[supple copper]

The process of turning the problem into this message is what maths is trying to teach you

[wooden hamlet]

But what about the f(n-1)

Where did that go

[waxen drift]

That just means the previous number

[verbal gulch]

The statement was "for any integer n bigger than 1, if you want to calculate f(n), you take what you have for f(n-1) and add 2 to it"

[supple copper]

f(2) = 2 + f(2-1) right?

And 2-1 is just 1

So f(2) = 2 + f(1)

[waxen drift]

So if you using n=3 and trying to find f(3), f(n-1) means use f(2)

[verbal gulch]

since 2 is an integer bigger than 1, if you want to calculate f(2), you take what you have for f(2-1) [this is f(1)] and add 2 to it

[supple copper]

But we know f(1) = 3 because that’s what they told you

[verbal gulch]

it's -5

oh wait nvm that's the next problem

[wooden hamlet]

So for the problem I send does f(2)=12

[supple copper]

Oh oops

[verbal gulch]

yes, f(2) = 12

[supple copper]

Sorry f(n) = 4 * f(n-1)

Yeah

[supple copper]

How did you get 12

[waxen drift]

She send a few problem a few min back

[waxen drift]

@supple copper

[wooden hamlet]

Prob not the right way but I knew the nth term started with the starting value. And that is 3. But we have 4 in the orginal equation so that must be the gf and also it says f(1)=3 which proves it is the starting amount

I do math very weirdly like the way I multiply and divide stuff is crazy

Or the fact I cannot do long devision but I can do polynomial long devision

[supple copper]

What is the gf

[wooden hamlet]

Growth factor

[supple copper]

Hmm

[verbal gulch]

that makes it much more difficult than saying, for example:

the problem gives f(1) = 3
the problem also gives that f(n) = 4 x f(n-1)
so if we plug 2 in for n, f(2) = 4 x f(1) = 4 x 3
Therefore, f(2) = 12

[waxen drift]

If its weird but has mathmatical value youre fine
I learned abacus method a few years back
Its weird but it works

[verbal gulch]

note that nothing in what I just wrote required you to memorize what a "growth factor" was

[supple copper]

So the way we’d normally say is the nth term is the last one * 4

So you can interpret f(n) = 4* f(n-1) as the value quadruples every time

Is that what you mean?

[wooden hamlet]

I mean I just say that cus it’s different equations for arthmetic and geometric sequences and a gf is multiplication while a rate of change is addition or subtraction

[supple copper]

But the fact you know this but not how functions work is strange

[wooden hamlet]

I’m bad at memorizing vocab

And also middle school screwed me up cus it was Covid

And that’s like the fondation of a lot of things

[supple copper]

Yeah well covid sucked

😦

[waxen drift]

Work with how her brain works then
Thats what i've learned when teaching students 1 on 1

[wooden hamlet]

It may help to say I’m also dyslexic

Would this table be correct?

[verbal gulch]

okay but do you see how what I wrote doesn't even require you to know what an arithmetic or a geometric sequence is?

[wooden hamlet]

Really?

[waxen drift]

At n=4 and on is incorrect. You added weird

[verbal gulch]

yeah I never used either of those words

[wooden hamlet]

Whoops

[wooden hamlet]

How do you know how to write the equations then since it switches orders

[verbal gulch]

if you approach it from a logical and mathematical perspective, you don't have to memorize very much

[supple copper]

It might help to actually dissect what you said here, because your reasoning is very strange and doesn’t make sense

[verbal gulch]

what switches orders?

[wooden hamlet]

Well I had somthing typed out now I’m confused and don’t know how to write a nth term equation for geometric

How do I know what the n is going to be subtracted by

In the exponent spot

Wait

Yea I’m confused

Is the nth term equation wrong should it be n-2 or n-1

[supple copper]

That’s…not right

[waxen drift]

Itd take the form a(k^(n-1)) where a is the starting number, k is the gf (what youre multiplying by to get from one n to the next), and n is your "step" so to sprak

[supple copper]

If 1 is 3 then 2 is 12

On the table

For question 4

[wooden hamlet]

My tecaher checked it 😭

So it is n-1

[supple copper]

I feel like you trying to remember how to do everything by heart is confusing the hell out of you

[waxen drift]

Yes, n-1
So that way when you plug in n=1 the exponent turns to 0

[wooden hamlet]

Yea but the thing is I neeed to knwo everything by heart but I have no idea how to memeorize it if the own keys I keep making are wrong

I had this problem when I first did this also

So now I just don’t know the rules for any of the equations

[supple copper]

The idea is you don’t really need to remember all that much in maths

It’s always been the reason I liked it in school

[wooden hamlet]

But I have to know how to write these equations

[supple copper]

I can always derive the things I need in the test

[wooden hamlet]

So I have to memorize them

[supple copper]

If I forget

[wooden hamlet]

The thing is if I forget if the growth factor goes first or the starting value goes first it messed up the equation and I fail the test

[supple copper]

So why don’t you try to stop remembering and just understand it intuitively

[wooden hamlet]

Idk how I would do that cus the equations don’t make sense when they r together so u have to memorize them separately

WAIR so is it n-2 or n-1

[supple copper]

Okay, let’s take the a geometric progression, where you start with a and every step you multiply the last by b

So first step you have a

[wooden hamlet]

Yea

[supple copper]

Second step is ab

Third step is, take whatever was in step 2 and multiply that by b

Which is (ab)b

Fourth step is abbb

[wooden hamlet]

Yea I undestand that part

[supple copper]

Or rather, first step is ab⁰

Second step is ab¹

Third step is ab²

Fourth step is ab³

So the intuition is that the nth step is ab^(n-1)

[waxen drift]

If you mean if you should have a(k^(n-1)) or (k^(n-1))*a then it doesnt matter. As long as you use proper notation and the exponent goes on the growth factor you'll be fine

[supple copper]

So then the nth term of a geometric progression is ab^(n-1) for starting amount a and growth factor b

[wooden hamlet]

If im starting at the first value

[supple copper]

See how I didn’t need to remember anything

[wooden hamlet]

How do we know it is -1

[supple copper]

Look here

[wooden hamlet]

Because if we started at step 0 it would just be n

[supple copper]

I’m calling that step 1

The first step is just the base amount

[waxen drift]

Its because you need starting number times 1
So to get 1 you need growth factor to the power of 0
Hence, n-1

[supple copper]

If you’d prefer you can call the base amount step 0, then the formula becomes abⁿ

[wooden hamlet]

So for geometric you are never adding 1 to the starting amount for the n-#

[supple copper]

Where does this n-2 come in

[wooden hamlet]

I still don’t see how this would help me know without memorizing the equation

[wooden hamlet]

I’m not sure what u mean

[supple copper]

The idea isn’t the forgo the formula entirely

[wooden hamlet]

Either the second or third step

[supple copper]

It’s to understand how you get the formula vs memorising the formula

[supple copper]

I don’t understand

You ask if it’s n-1 or n-2, why would it be n-2

Oh actually I see your point

[wooden hamlet]

I don’t know the answer for this question so I don’t know how to answer your question

[supple copper]

Suppose you aren’t given the base amount

[wooden hamlet]

Then you would not know

[supple copper]

(Here base amount is step 1)

What if I gave you the value at step 4

And I want you to find the value at step 10

[wooden hamlet]

I would rewrite the equation 😀

Assuming I’m getting at least one other value

In the table

[supple copper]

How would you do that

So let’s say the growth factor is 3 and the 4th term (or fourth step) is 6

The question is, what is the 10th term?

[wooden hamlet]

Is the gf 3 for every 1 step or every 2 steps , every 3 steps etc

[supple copper]

Ahh

I see

It’s every step

[wooden hamlet]

So you just keep multiply

[supple copper]

So how do you find the 10th term

[wooden hamlet]

I would put it into a table and fill it out by multiplying

[supple copper]

Okay what if I want the 27467283th term

You’ll be sitting there for a long time

[wooden hamlet]

I would make an equation by using my table to either get a value at f(1) or f(0) that input that term into f(n)

[supple copper]

Okay good

Right

So you can do that

But what if I asked you, the first term is 6, the growth factor is 3 every step, what is the 7th term?

[wooden hamlet]

One sec

[supple copper]

Going from first to 7th term is 6 steps

[wooden hamlet]

What is the gf?

[supple copper]

3

[wooden hamlet]

So it’s •3 every 3 steps?

[supple copper]

Every step

[wooden hamlet]

Oh

Then u just put it in a table and keep multiplying until 7

[supple copper]

Or…

[supple copper]

.

[wooden hamlet]

Yea but the table way is wayyyy easier

[supple copper]

I’m putting small numbers so it’s easy to write but we should be able to do this with any number

Okay then I ask you what is the 350th term

You’ll be forced to use the formula

[wooden hamlet]

Yea

[supple copper]

I’m trying to explain how the formula works and you’re just going “oh but it’s easier without”

Yes it’s easier for the examples I’m giving you but that’s not the point

[wooden hamlet]

I mean yea but if I was doing a test I would donthe way I’m most confident with and I do understand in order to get larger terms I need a equation

[supple copper]

Going from 4th to 10th term is also 6 steps, and in fact, you will get the same answer as if you went 6 * (3)^(10-4)

This is where the n-# comes in

[waxen drift]

that might be so but it's likely you'll be asked to write the recursive and explicit formulas. you're gonna have to know it anyway. and you might be asked to evaluate when n=15 for example

[supple copper]

We have f(10) = 6 * 3^(10-4), or in general, f(n) = 6 * 3^(n-4)

The reason there is a 4 there is because 6 is the 4th term, not the first time

If 6 was the first term then it would say n-1

If you call the “base amount” the 0th term then it will say n-0 or just n

[wooden hamlet]

Yea

But then for some u have to add 1 to the starting amount

[supple copper]

Everything I have said is to answer this question

[wooden hamlet]

That’s only for nth term equations not recursive

[supple copper]

Right so that’s not the recursive form

This is the equation that brings you from the 4th term to the nth term

[wooden hamlet]

But if it was recursive would we do n-5

[supple copper]

???

Right

Yes

But we wouldn’t write it like that

We’d still say f(n) = 3 * f(n-1)

The n-1 here means “the last term”

The power means to do the multiplication n-4 times

[wooden hamlet]

Yea

Is this correct

[trim jolt]

@wooden hamlet Has your question been resolved?

[waxen drift]

you need to finish simplifying for part c
you can also pull the number from the equation too. Focus on sine, specifically the argument (the stuff inside the parentheses)

[wooden hamlet]

What is it asking me to do here?

Long devision?

[wooden hamlet]

How would simplify? Is it 10pi/pi•t

[waxen drift]

yes, long division, but if i remember correctly there's a formula or a process that tells you the remainder without actually having to do the division

[wooden hamlet]

What would I divide?

[waxen drift]

closer. remember that pi is a number, and that you don't need the variable t here
t is the time it takes. just give the number of minutes

[waxen drift]

hold on a minute. I need to look it up to rmember. it's been a while since i used this method

[wooden hamlet]

But I cannot just get rid of t

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[waxen drift]

so the method is called remainder theorem
if you have the polynomial f(x) and you're being asked to divide by (x-a), then you can find the remainder by evaluating f(a)

[wooden hamlet]

.reopen

[trim jolt]

✅

[wooden hamlet]

Oh I regonise that name

Wdym f(a)

[waxen drift]

it's not that you're getting rid of it, it's just not needed

[wooden hamlet]

Do I just input the values

[wooden hamlet]

So what do I do with it

I cant take it away

[waxen drift]

that's right. so if it says divide by (x-4) plug in 4, if it says (x+2) plug in -2

[wooden hamlet]

Kk

[waxen drift]

it's not that you're manipulating it in anyway, you literally just don't need it. you wrote too much. it didn't need to be there to begin with
so inside the sine function you're given (2pi*t/10)
2pi is the length of sine's period
10 is the length of the story's period (in this case how long it takes the ferris wheel to make 1 revolution)
t is the variable that represents time

the question only asks how long it takes the wheel to make 1 complete revolution. so you only need 10 minutes

[wooden hamlet]

Ohhh ty

[trim jolt]

@wooden hamlet Has your question been resolved?

[torpid trellis]

fuck you no one answered my question

[trim jolt]

Channel closed

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[trim jolt]

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[glad lake]

why did he use pythagoras?

[glad lake]

I just did tan(55) x 20 to get height

by using sohcahtoa

[fierce lake]

Why wouldn't you use Pythagoras here

[glad lake]

20 is our only given base

[fierce lake]

Also whatever is that?

[glad lake]

sin o h , cos a h, tan o a

[tawdry bloom]

That's the height of the pyramid at base, not the height of the pyramid at each triangle

[glad lake]

huh

[fierce lake]

This is what X and M are

[glad lake]

So how was i supposed to get real height then because wheres the other side to calc pythagoras

[tawdry bloom]

What do you mean?

All four triangles are equal. Equal area, equal lengths, equal sides

[glad lake]

Also why are X & M being mentioned, it's not even in the question itself

[fierce lake]

Im referring to the solution they've given

[tawdry bloom]

It's to represent midpoints so we can write the formulas simpler

[fierce lake]

Since you have a problem with the solution

[glad lake]

Wait so we calc height from base of square?

By splitting it into triangles

[tawdry bloom]

Which height?

[glad lake]

height of triangle

singular triangle

[tawdry bloom]

Yes

We need to find the height of the pyramid first in order to get the height of each triangle

[glad lake]

but that gives me AC

[tawdry bloom]

Can you draw your explanation please?

[fierce lake]

No?

[fierce lake]

You're not splitting the square into triangles

[tawdry bloom]

No

Just imagine point X as the center of the base and the midpoint of each base side as M. The height of the pyramid is VX, and the height of each triangle is VM.

To find the surface area of each triangle, we need to find VX and VM first.

Do you get it?

[glad lake]

Ah alright

thanks

[tawdry bloom]

No worries

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[red loom]

that is dot product becuase note that BA and BC are vectors not scalars :)

[vocal epoch]

ok

can some1 verify this final answer

also i saw in my answer key that that instead of finding a unit vector of BC for the projection they just square BC, is it fine to do that

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[full willow]

somebody pls help with my calculations

[full willow]

i got a bit stuck here

[trim jolt]

@full willow Has your question been resolved?

[tawdry bloom]

Do you know where the circuit starts and ends?

[full willow]

starts at vin and ends at rl

[tawdry bloom]

Yes

[full willow]

i saw an exampe

but the

i know that

av =

rc/re

and rc = RC/RL

but cant seem to figure out how to find re

[tawdry bloom]

So what's the load resistor here?

Rl, correct?

[full willow]

yes

but it is

generaly find v out

wait wrong

[full willow]

ignore this

[tawdry bloom]

Rc is parallel to Rl, so Av should be - Rc || Rl over Re

[full willow]

oh

is that an actual formula?

then it is

[tawdry bloom]

Pretty sure it is

[full willow]

av =3.6 || 10 / 1

[tawdry bloom]

Don't forget negative

[full willow]

negaitve?

oh

[tawdry bloom]

Av = -(Rc||Rl)/Re

[full willow]

-Rc

why -?

[tawdry bloom]

The voltage decreases across Rc

[full willow]

oh ok

and i then get ac right

[tawdry bloom]

Nope

How much did you get for Av first?

[full willow]

17/45 which flipped is 54/17

45/17

i mean

which is

[tawdry bloom]

Approximately 2.647

Yeah

[full willow]

2.64

yeah so that is my av

[tawdry bloom]

So that's the voltage gain

[full willow]

ye

[tawdry bloom]

Now we just need to calcullate the output voltage

Which uuses the formula of Vout = Av * Vin

[full willow]

ye

because vin is 2mv do i just multiply

[tawdry bloom]

Don't need to overthink

You already have values of Av and Vin

[full willow]

i ask this because in this

this is a different questin

[full willow]

2 mv x 2.64 v is 0.02 x 2.64?

@tawdry bloom sorry for ping

[tawdry bloom]

Don't forget Av is negative

[full willow]

oh ok

0.02 x -2.64

which is

[tawdry bloom]

Approximately -0.053 V

[full willow]

oh ko

yes thx

[tawdry bloom]

There you gio

No worries

[full willow]

can u help me with

one more thing

[tawdry bloom]

Sure

[full willow]

this basically

is re = re'?

@tawdry bloom will u be available after like a few hour maybe 4-5

[tawdry bloom]

Not really

[full willow]

oh

im gonna sleep

[tawdry bloom]

I'm just heading to sleep in a few minutes

[full willow]

that is why

oh ok

got exams

[tawdry bloom]

Next morning maybe

[full willow]

before u go

can u tell me why re' = re?

[tawdry bloom]

The prime (') is just used to distinguish it from other resistances

Both denote the same emitter resistannce

[full willow]

oh

so re' = re? so if re resistance is re 200 ohm or k ohm re' is also 200 k oh

[tawdry bloom]

Yezh

Yeah

[full willow]

oh ok

big thanks

[tawdry bloom]

Or otherwise written as r'e

[tawdry bloom]

No problem

[full willow]

now i hope i remember my boolean algrbea for tmews exam

[tawdry bloom]

Good luck

[full willow]

wait do u think u cantell me how 2 more things work prettly pls

[tawdry bloom]

Mm alr

[full willow]

you know k-map?

[tawdry bloom]

Im afraid not

[full willow]

boolean algebra stuff ukk karnaugh map

then it is fine another thing i wanna ask is op amps

[tawdry bloom]

?

[full willow]

this

op amps

[full willow]

this

[tawdry bloom]

What's your specific need

[full willow]

how did

why is it 0-v1 on top part but then it is

v1-0

ima sleep

[tawdry bloom]

I believe they are different voltages

[full willow]

oh

thx for help

[tawdry bloom]

Np

[full willow]

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[late flume]

I need help to formalize the arguement

I am unable to generalize

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@late flume Has your question been resolved?

[late flume]

<@&286206848099549185>

[bright jacinth]

hello

nvm im dum

[limpid belfry]

what have you done/tried?

[late flume]

I have tried for 20 21 and 22

And I have found a way for those

But I am not sure how to generalize it

For 20 I formed pairs

[limpid belfry]

hmm

[late flume]

We had an extra 2 x 4 x 6 .... 20

Since there is an even number of 2s

We can neglect that

[limpid belfry]

well on first thought seeing this the only reason the product 1!2!...n! is not gonna be a square is that it has an odd amount of some prime

[late flume]

We get 1x2x ....10

Hence we divide by 10! And 1!

[limpid belfry]

and then maybe prove that there exists some j such that it makes the prime count even

(probably some logic holes but you get the idea)

[late flume]

I have no idea how to generalize though

[limpid belfry]

oh yea the qn said to make cases

let n = 4k, 4k + 1, 4k + 2 and prove for each case i guess

you should start from n = 4k

[late flume]

I tried 20

But I am not sure if this works in the specific case or is the general idea

[limpid belfry]

your idea works yea!

(or maybe theres some other way to solve this that i dont see)

[late flume]

How do I create a general arguement out of this

Like a proof

Or something

With complicated symbol manipulations

[limpid belfry]

^(maybe)

sorry i have to go😓

gl on this tho

[late flume]

Have a nice day

[late flume]

<@&286206848099549185>

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@late flume Has your question been resolved?

[raw estuary]

Not a solution but some ideas:

1. $n^2$ mod 4 is 0 or 1 (for an integer n)

2. $1! * 2! * 3! * ... * n! = n^1 * (n-1)^2 * (n-2)^3 * ... * 2^{(n-1)}$

3. For the product to be a square, the exponents in its prime factorization must be even

[solid kiln]

whiskyheat

[late flume]

I will give it a try in the morning

I still have no ideas 2 mod 4

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[raw estuary]

.reopen

@late flume Here is my proofsketch. It turned out to be quite easy. I hope everything is clear and I did no mistakes.

For shorter notation, let $sf(n) = n! (n-1)! ... 2!$ be the superfactorial.
Notice that $sf(n) = n (n-1)^2 (n-2)^3 ... 2^(n-1)$.

We make cases for n=4k, 4k+1, 4k+2.

Case I: n = 4k
Prove by induction that sf(4k)/(2k)! is a square.

Base case: k = 1: sf(4) = 4 * 3^2 * 2^3 and sf(4)/2! is a square.

Assume the statement holds for some k, e.g. sf(4k)/(2k)! is a square.

k -> k+1:
sf(4(k+1))
= sf(4k+4)
= (4k+4) (4k+3)^2 (4k+2)^3 (4k+1)^4 (4k)^4 ... 2^4 sf(4k)
= 2(2k+2) (4k+3)^2 (2(2k+1))^3 (4k+1)^4 (4k)^4 ... 2^4 sf(4k)

Dividing by (2k+2)! gives a square.

Case II: n = 4k+1
sf(4k+1) = (4k+1)! sf(4k)

Dividing by (4k+1)! and (2k)! gives a square

Case III: n = 4k + 2
sf(4k+2)
= (4k + 2)! (4k+1)! sf(4k)
= (4k + 2) (4k+1)^2 (4k)^2 ... 2^2 sf(4k)
= 2(2k+1) (4k+1)^2 (4k)^2 ... 2^2 sf(4k)

Dividing by (2k+1)! and 2! gives a square.

This completes the proof of the exercise.
As the exercise states, this does not work for 4k+3. Indeed, one cannot use the same trick as in the steps above. This is because 4k+1 and 4k+3 may be prime twins (for example, 5 and 7, 17 and 19, 41 and 43) of which there are infintly many.

[solid kiln]

whiskyheat
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[raw estuary]

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[full willow]

is rc = (-RC//RL)/ RE or (-RC//RL)/re

if i dont have re or re' do i use Re?

this

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[rough turret]

for this function f the teacher wants me to give an alegebraic analysis on it

i understand everything but how to describe the behavior

i know that the end behavior is the quotient of the leading terms so 2/1= 2

but i dont know how to describe why this function increases on its intervals

no derivative given

[bright quarry]

partial fractions

(just split the fraction appropriately)

[rough turret]

i dont understand that

[bright quarry]

$2\left(\frac{w + 4 - \frac{5}{2}}{w + 4}\right)$

[solid kiln]

knief

[bright quarry]

now split the fraction

[rough turret]

as in split the numerator and denominator?

[bright quarry]

what

$\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$

[solid kiln]

knief

[bright quarry]

choose a and b appropriately

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[upbeat fulcrum]

Hello, could someone show me how they pulled out a 1/3, when I distribute the 3^n/n^3 im not seeing how there's a 1/3 in there

[upbeat fulcrum]

or uhhhh

[left oriole]

3^(n+1) is 3 times 3^n

[upbeat fulcrum]

ohh right okay

ok that's all thxxxx

!

[left oriole]

yw

[upbeat fulcrum]

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[jagged wharf]

I've stumbled upon this quite seemingly basic idea while studying rational canonical form, yet I'm in doubt. Consider p(t)^m where p(t) is a monic, irreducible factor and m a positive integer. Also consider q(t)^n where q(t) is a monic irreducible factor and n a positive integer. Now in a proof I'm reading that proceeds by contradiction, we have that p(t) and q(t) are relatively prime, yet p(t)^m = q(t)^n. The book says this is impossible. Somehow I can't see why this is false. Can someone elucidate this?

[nimble niche]

Well, let's say a and b are two relatively prime numbers. That means they will have no prime factors in common.
Eg. 12 and 19
12 = 2^2 * 3
19= 1* 19
Now let's raise a and b to some powers m and n.
a^m, b^n. Do you see that even when we raise them to some power, they still will have no prime factors in common?

[jagged wharf]

Yeah, makes sense. I guess we can say something similar for irreducible, monic polynomials too then?

[nimble niche]

Correct.

You can type .close if your doubt is resolved btw.

[jagged wharf]

.close

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[frank rain]

Is the working out enough reasoning for part i

[marble wharf]

yes

Q'(x) should be there tho

check your product rule

also last line P'(a) dont forget ()

[frank rain]

oh yeh

i meant to put the fash

for the second Q(a)

[frank rain]

yeh, thank u

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[olive cargo]

help

[olive cargo]

dont know where to start

ive drawn the triangle but like

what...

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[frank rain]

.reopen

[frank rain]

.close

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[nimble niche]

Need to explain your thought process, can't just give out answers.

[frank rain]

nah i am writing on paper

i am going to take a pic

[frank rain]

but i think im wrong with my working out

[nimble niche]

Why give them the answer then? That'll just cause uneeded confusion.

[frank rain]

if u multiplied the vector by itself, is it 0

[frank rain]

sorry mb

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[long bridge]

hey, i am a student for entrance exams. Can someone help me with the probability content?

[long bridge]

A math teacher gives his students a list of 10 problems and tells them that 5 of them will be chosen at random to be on the final exam. If a student was able to solve only 7 of the 10 problems correctly, the probability that he will get all the problems on the exam right is:

[ancient trout]

What is the probability of 1 question being put

That's the first step

[main sigil]

Alternatively, you could just calculate how many different groups of 5 questions could be formed of the 7 and divide it by number of groups of 5 questions out of all 10

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[long bridge]

ok, thanks

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[dusky thunder]

open

[dusky thunder]

@nocturne mural

here is ur channel

[nocturne mural]

oh thanks

[dusky thunder]

share ur question and ping me once u wanna close the channel

[nocturne mural]

I needed help on this question and I can't think of a way to approach this

[nimble niche]

well, I can tell you the short method

since it's (x+1)^1/2, it would mean that after squaring both sides you'll get a polynomial of degree 1 and some sqrt term

[nocturne mural]

also x is a real number by the way

[nimble niche]

which means there's only one solution

put 0 in

that's your solution

[nocturne mural]

it's the fourth root

not the square root

[nimble niche]

ik

oh wait

lol

[nocturne mural]

THAT'S WHY I ASKED 😭

[nimble niche]

it's 1/8th power then

[nocturne mural]

i can't just square two times and do my own stuffs 😭

[nocturne mural]

the short method is to waste an hour to square both sides twice 😭

[nimble niche]

Well you can still visualise the thing tho, if you raise to power 8 and can imagine the binomial exp but atp maybe it's just better to solve it properly if udk binomial well

[nocturne mural]

the pascal triangle thing?

i know (a + b)^2 = a^2 + 2ab + b^2

(a - b)^2 = a^2 - 2ab + b^2

a^2 - b^2 = (a - b)(a + b)

[nimble niche]

Yeah, except this is a binomial to power 8

So you need to use binomial expansion

[nocturne mural]

alright

[nimble niche]

But nm that now

[nocturne mural]

but i think we need to solve it properly

3x^2 - 2x + 1, 2x^2 - x + 1, x^2 + 1

there's literally no way to factorize those

i tried a = fourth root of x + 1 and got x = a^4 - 1

but substituting won't work (even if i tried to expand)

[nimble niche]

also, this was wrong so nm i can see two roots for this.

[nocturne mural]

0 and 1 right?

[nimble niche]

Ye

[nocturne mural]

that's literally the two roots

some says to assume x^2 = x

and x^2 != x

of course that x^2 = x leads to x = 1 and 0 being the correct roots

[nimble niche]

Why would you do that

That's such a random assumption lol

[nocturne mural]

it's not me to come up with that

it's someone else to come up with that

i don't even know why lol

ai told me to subsitute each 4th root into their own variable

like a + b = c + d

a - d = b - c

and assume a = d then b = c

[nimble niche]

I'm trynna brute force it with binomial rn since I cannot think of any way to simplify

[nocturne mural]

no 😭

idk how that slid into my revision 😭

[nimble niche]

I can think of one other way

[nocturne mural]

me either

[nimble niche]

We can see one root x= 0, then assume LHS as a function f and RHS as a function g. Then check if they are increasing and decreasing, draw their graphs

That's the best I can come up with except binomial

[nocturne mural]

yeah then i'd rather do binomial

BECAUSE MY SYSTEM HASN'T LEARNED HOW TO GRAPH YET

[nimble niche]

Lmao. Maybe I'm hallucinating too and this is just a one line solution idk.

I can't think of any perfect squares or anything.

[nocturne mural]

wait

can you help me with another one?

this is easier

[nimble niche]

K

[nocturne mural]

i tried to set t = x^2 + x + 6

then i got t - x + cbrt(t) = 8x^3 + x

t + cbrt(t) = (2x)^3 + 2x

so cbrt(t) could possibly be 2x

or cbrt(t) can't be 2x

[nimble niche]

yes

[nocturne mural]

but what if cbrt(t) != 2x

[nimble niche]

man algebra is a bad way to these imo, just setting the opposite sides of the equality as functions and checking their behaviour is so much simpler

[nocturne mural]

I DON'T KNOW HOW THAT SLID INTO MY REVISION

[nimble niche]

Do you know what rational root theorem is?

I think that's what you're supposed to use for both these questions

It's basically a method for cubics and higher deg polys where we just guess all integers from [-3, 3] and put them in to check

[nimble niche]

Then we have no solutions from that case, so we can just discard it btw

[nimble niche]

Can just move forward w this

[nocturne mural]

well then that's easy

[nimble niche]

But this still stands, especially for the previous qn.

[nocturne mural]

can you prove it btw?

because it's a theorem

[nocturne mural]

how can i use the rational root theorem to solve these?

[nimble niche]

btw the definition i just said is just a general way, it's not really the rational root theorem

I'll explain it first

then if you want can prove it

Ok so let f(x) = px^3 + rx^2 + zx + q

The rational root theorem says that we need to find all the divisors of p and q

And our roots will be from all the possible combinations of (p/q)

Do you get that@nocturne mural ?

That's the general format for a cubic, but it is also applicable for all polynomials

[nimble niche]

So in this question we can definitely see it's a cubic (or a higher deg poly if we cube it- we'll get extra solutions then)

and too really apply the rational root theorem we would have to do painstaking things like cubing and stuff, but since we know it's valid for all polynomials

we can just put x's from [-3,3] in to check

[nocturne mural]

yes

[trim jolt]

@dusky thunder Has your question been resolved?

[nimble niche]

Maybe there's an easier way of solving this idk

Cause nothing is a perfect square in the first one, and the second if you cube you get a degree 9 poly, so you're gonna have to solve with the taking 't' method.

[nocturne mural]

@dusky thunder im sorry but the question is too hard for someone to solve 😭

[lime sphinx]

https://www.wolframalpha.com/input?i=(x%2B1)^(1%2F4)%2B(3x^2-2x%2B1)^(1%2F4)%3D(x^2%2B1)^(1%2F4)%2B(2x^2-x%2B1)^(1%2F4)

at this point paste i would paste this equation in wolfram

apparently x=1 is a solution

[nimble niche]

We know the solutions lol but the method is behind a paywall here

[lime sphinx]

but how to deduce this idk

[nimble niche]

Rational root theorem is the best i can come up with ngl

Or binomial

[trim jolt]

@dusky thunder Has your question been resolved?

[burnt mulch]

This is one of those stupid trick questions that I hate. \ \ Let $k=\sqrt[3]{x^2+x+6}$. Then, $k^3=x^2+x+6 \implies x^2+6=k^3-x$. So, we can rewrite the equation as $$k^3-x+k-8x^3-x=0$$ at which the factorisation is trivial upon observing that the left hand side is zero when $k=2x$.

[solid kiln]

Civil Service Pigeon

[dusky thunder]

im gonna go offline for a while so the channel may auto close

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[limpid dawn]

you may auto close

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[dense fractal]

Hey! need some help on the following problem:
For n>1 verify that $$1^2+3^2+...+(2n-1)^2=\binom{2n+1}{3}.$$
I tried induction but I can't get to something simple (which tells me it's probably not the way to go except if I made a mistake. What I get on the right hand side after adding $(2n+1)^2$ and some calculating is: $$(2n+1)\frac{(4n^2+10n+1)}{6}$$ which doesnt seem to simplify well.

[dense fractal]

Any inputs on how you would apporach this/any mistakes I made would be more than welcome!

One thing which I might have done wrong is simplifying the numerator/denominator since I do need to get back to a binomial coef, though it also seems like there is no other easy way to do this by induction 🤔

[solid kiln]

levpromano

[trim jolt]

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

[quick tusk]

I honestly wouldn't even use induction here

[burnt mulch]

This is wrong

So you made an algebra mistake

Hence why I’m asking to see

[quick tusk]

You just expanded it wrong, multiply out the parens and just apply the formula for a sum to n

[tropic moth]

Omg it’s u again I remember u from a. Long time ago

[vestal summit]

Try using the following identity:
C(N+1, k) = C(N, k) + C(N, k-1)

[quick tusk]

whoops

[dense fractal]

how else would you approach this problem?

[wintry stag]

If you already know that the formula is true, it is going to be 2(n+1)+1 choose 3, and the denominator is 6, you already know the numerator is going to look like (2n+3)(2n+2)(2n+1) and it's just a matter of doing the algebra correctly.

[dense fractal]

ah right. Im getting 4n^2+10n+6 now, which if I am not mistaken does simplify to (2n+3)(2n+2). In which case that solves the problem.
Sorry, stupid mistake, and I had verified my work too 😅

[wintry stag]

When you get stuck it is helpful to start with (2(n+1)+1) choose 3 and expand it to see what you are actually working with.

[dense fractal]

right, that's why I tend to do induction without thinking on these. for some reason my mind couldnt see that mistake..

[dense fractal]

yeah, ill try that, thanks!

[dense fractal]

could you expand a little on how that would work?

like I know the identity, but what would you use it for? to get the C(2n+3,3) on the right hand side?

[vestal summit]

C(2n + 3, 3) = C(2n + 2, 3) + C(2n + 2, 2) = C(2n + 1, 3) + 2C(2n + 1, 2) + C(2n + 1, 1) = C(2n + 1, 3) + 2n(2n + 1) + (2n + 1)

[dense fractal]

I see

thanks for the help! (and sorry for bothering for such a mistake..)

.close

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[glad pumice]

I've worked on this question for multiple hours now and can't seem to solve it

[glad pumice]

I've tried doing casework and stuff but it's still not right

nvm i js solved it

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[wraith hinge]

Hi

[grim sparrow]

hello!

do you have a question?

[nimble niche]

Well they're trolling in #discussion lmao, so prolly not

[spiral kettle]

.close

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[urban copper]

all this exercises sucks ass

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@urban copper Has your question been resolved?

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@urban copper Has your question been resolved?

[thorn bay]

Here is the translation of the text in the image:

---

2. Let $f : \mathbb{R}^4 \to \mathbb{R}^4$ be the linear transformation defined by

$$
f(x_1, x_2, x_3, x_4) = (x_1 - x_2 + x_3 + 2x_4,\ x_1 - x_2 + x_4,\ x_3 + x_4,\ x_1 - x_2 - x_3),
$$

and let

$$
S = \left{ \mathbf{x} \in \mathbb{R}^4\ \middle|\ x_1 - 2x_2 = x_2 - x_3 + x_4 = 0 \right}.
$$

Define, if possible, a linear transformation $g : \mathbb{R}^4 \to \mathbb{R}^4$ that simultaneously satisfies:

• $g \circ f = 0$,
• $(f \circ g)(S) = 0$,
• $\dim(\text{Im}(g)) = 2$,
• $(0, 0, 0, 1) \in \text{Im}(g)$.

[solid kiln]

gfauxpas

[thorn bay]

ugh i cant even read the notation like this its too messy, one moment

Let $f : \mathbb{R}^4 \to \mathbb{R}^4$ be the linear transformation defined by

$$
f\begin{pmatrix}
x \ y \ z \ w
\end{pmatrix}

\begin{bmatrix}
x - y + z + 2w \
x - y + w \
z + w \
x - y - z
\end{bmatrix},
$$

and let

$$
S = \left{ \begin{pmatrix}
x \ y \ z \ w
\end{pmatrix} \in \mathbb{R}^4\ \middle|\ x - 2y = 0,\quad y - z + w = 0 \right}.
$$

Define, if possible, a linear transformation $g : \mathbb{R}^4 \to \mathbb{R}^4$ that simultaneously satisfies:

$$
\left{
\begin{array}{l}
g \circ f = 0 \[4pt]
(f \circ g)(S) = 0 \[4pt]
\dim(\operatorname{Im}(g)) = 2 \[4pt]
(0,\ 0,\ 0,\ 1) \in \operatorname{Im}(g)
\end{array}
\right.
$$

[solid kiln]

gfauxpas

[thorn bay]

there now i can actually read it

[urban copper]

yeah

[thorn bay]

this looks annoying

[fleet bear]

$$ (f \circ g)(S) = 0 \implies g(S) \subset \operatorname{ker}(f) $$

[solid kiln]

StrangeQuarkAL

[fleet bear]

oh wait

could also mean that S is a subset of ker(g) I guess

[urban copper]

yes

[urban copper]

ker g is contained in ker f

wait... it's getting confusing

[urban copper]

ty for the translation gfaux

[thorn bay]

img(f)⊆ker(g)
g(S)⊆ker(f)
rho(g)=2
nu(g)=2
follows from the problem

i didnt solve anything im just observing, along the lines of what strange said

np

[urban copper]

rho?

[thorn bay]

rho rank, nu nullity

[urban copper]

ok i didn't know

[thorn bay]

its okay usually people just write rank and nullity i think

[urban copper]

which vectors are in ker g

g o f = 0
and
(fog)(S)=0

f(g(S)) = 0

g(f(x))=0

Im f is in ker g

g(S) in ker f

g(f(g(S)))= g(0) = 0

g o f = 0, (fog)(S) = 0
(g o f)(g(S)) = g(f(g(S))) = g((f o g)(S)) = g(0) = 0

(g o f o g)(S) = g o (f o g)(S) = g(f(g(S)))

(f o g)(S) = 0

idk what i am doing

[fleet bear]

Maybe we should take a guess based on what we know.

g

g(S) c ker(f) so Im(g) should at least partially lie in ker(f). If we assume it lies fully in ker(f), perhaps we can find a g that satisifes the constraints

[urban copper]

but we are still talking about f , but what about g

what about ker g

[urban copper]

yeah, dim im g =2 = dim ker g

[fleet bear]

Still not sure

[urban copper]

im being told ker(g) = im(f)?

but idk

this is what you meant above?

[fleet bear]

I just tried this. Found ker(f) and (0,0,0,1) doesn't lie in it so Im(g) doesn't lie entirely in ker(f)

So I think we can set (0,0,0,1) to one of the basis vectors and find the other in ker(f) with the constraints in mind

[fleet bear]

I think that's right

[urban copper]

how??

[fleet bear]

Since they have the same dimension I think but I might be wrong

If g(v1) = 0 and g(v2) = 0 then surely {v1,v2} should form a basis for ker(g) if they are linearly independent yeah?

[urban copper]

sure

[urban copper]

I just tried this. Found ker(f) and (0,0,0,1) doesn't lie in it so Im(g) doesn't lie entirely in ker(f)
so Im(g) not Ker(f)

[fleet bear]

So we know ker(g). I noticed that one of the basis vectors in S lies in Im(f)

[fleet bear]

Yes

But it does partially lie in ker(f)

[urban copper]

wait a second

,w nullspace {{1,-1,1,2},{1,-1,0,1},{0,0,1,1},{1,-1,-1,0}}

[solid kiln]

Results provided by WolframAlpha

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

Requested by renatoxbz

[urban copper]

Ker(f) = <(1,1,0,0),(-1,0,-1,1)>

,w rank {{1,1,0,0},{-1,0,-1,1},{0,0,0,1}}

[solid kiln]

Results provided by WolframAlpha

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

Requested by renatoxbz

[urban copper]

ok

[fleet bear]

We know ker(g)

For Im(g), (0,0,0,1) can be one of basis vectors and the other must lie in ker(f). Let's call it v

Since:

• g(S) c ker(f)
• (0,0,0,1) not in ker(f)

(*) g must map S to scalar multiples of v

Basis for S: { (2,1,1,0), (0,0,1,1) }

So g will map both to scalar multiple of v

I think from here we get solve for the matrix representation of g. We just need to guess an appropriate v in ker(f)

Since we can definitely represent the matrix for g using (0,0,0,1) and v as columns, I would just try placing (0,0,0,1) and v in different column positions and testing against (*). For what v should be, just try one of the basis vectors for ker(f)

[urban copper]

g o f = 0 ==> im(f) subseteq ker(g)
but dim(Im(g)) = 2 ==> dim(ker(g)) = 2
so if dim im(f) = 2 and dim ker (g) = 2
and im(f) subseteq ker(g) then im(F) = ker(g)

[urban copper]

g o f = 0 <=> g(f(x))=0 <=> Im(f) ⊂ ker(g)

Im(f) = <(1,1,0,1),(1,0,1,-1)>

dim(Im(g)) = 2 <=> dim(ker(g)) = 2

and if Im(f) ⊂ ker(g) and dim(ker(g)) = dim(Im(f)) = 2

Im(f) = ker(g)

So Ker(g) = <(1,1,0,1),(1,0,1,-1)>

(f o g)(S) = 0
<=> f(g(S)) = 0
<=> g(S) ⊂ Ker(f)

but g(S) ⊂ Im(g)

so g(S) ⊂ Ker(f) ∩ Im(g)

S = <(2,1,1,0),(-2,-1,0,1)>

g(S) = <g(2,1,1,0),g(-2,-1,0,1)>

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[limpid belfry]

is this proof correct?

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[fallow meadow]

how do i solve this

[hallow spruce]

,rcw

[harsh cypress]

Descartes rule of signs, rational zero theorem, then synthetic division?

[trim lichen]

do you see the instructions in the problem?

have you followed them?

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[tired shuttle]

$\sum_{k=0}^{9} {\left(k + \left(-9\right)^{k}\right) \cdot 8^{9-k} \cdot \binom{9}{k}}$ has exactly 6 prime factors. Compute the sum of the 6 primes

[solid kiln]

acgn

[tired shuttle]

how can i even evaluate that sum 😭

[jagged phoenix]

pattern recognition the binomial theorem out

then consider $\frac{d}{dx}(x+a)^9$

[solid kiln]

Element118

[tired shuttle]

wait

so

$\sum_{k=0}^{9} {\left(k + \left(-9\right)^{k}\right) \cdot 8^{9-k} \cdot \binom{9}{k}} = $\sum_{k=0}^{9} {k \cdot 8^{9-k} \cdot \binom{9}{k}} + \sum_{k=0}^{9} {\left(-9\right)^k \cdot 8^{9-k} \cdot \binom{9}{k}}$

[solid kiln]

acgn
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[jagged phoenix]

yeah

[tired shuttle]

$\sum_{k=0}^{9} {\left(-9\right)^k \cdot 8^{9-k} \cdot \binom{9}{k}} = \left(8 - 9\right)^9$

[solid kiln]

acgn

[jagged phoenix]

that simplifies down

[tired shuttle]

how about $\sum{k=0}^{9} {k \cdot 8^{9-k} \cdot \binom{9}{k}}

[jagged phoenix]

consider this

binomial expand first then differentiate

[tired shuttle]

alr wait

so $\left(x + a\right)^{9} = \sum_{k=0}^{9} {\binom{9}{k} \cdot x^{9-k} \cdot a^{k}}$

[solid kiln]

acgn

[jagged phoenix]

maybe x^k instead so you get the k out when you differentiate

[tired shuttle]

$\sum_{k=0}^{9} {\binom{9}{k} \cdot x^{k} \cdot a^{9 - k}}$

[solid kiln]

acgn

[tired shuttle]

using the constant rule then the derivative is

$\sum_{k=0}^{9} {\binom{9}{k} \cdot a^{9 - k}} \cdot kx^{k-1}$

[solid kiln]

acgn

[jagged phoenix]

how would you set x and a to match what you have?

[tired shuttle]

a = 8
x = 1

[jagged phoenix]

you can also deal with the expression another way, just differentiate without applying binomial

[tired shuttle]

$\sum_{k=0}^{9} {k \cdot 8^{9-k} \cdot \binom{9}{k}} = \dfrac{d}{dx} (8 + 1)^{9}$ ?

[solid kiln]

acgn

[tired shuttle]

nvm

so $\sum_{k=0}^{n} {k \cdot a^{n-k} \cdot \binom{n}{k}} = n \cdot \left(a + 1\right)^{n-1}$

[solid kiln]

acgn

[tired shuttle]

so the entire sum is just 9^9 - 1

now how do i factorize it

so 9^9 - 1^9 = (9 - 1)(9^8 + 9^7 + ... + 9 + 1)

[jagged phoenix]

yeah you need to differentiate before plugging in

[jagged phoenix]

9^9-1

yeah

but you might want to read it as a difference of cubes first

[tired shuttle]

ohh

alr

[jagged phoenix]

or you can read it as $3^{18}-1$

[tired shuttle]

that gives 2^3 . 7 . 13 (729^3 + 729 + 1)

[solid kiln]

Element118

[jagged phoenix]

and do the difference of squares first, since that gives you smaller factors

[tired shuttle]

(3^9 - 1)(3^9 + 1)

then by difference of cubes

(3^9 - 1) = 27^3 - 1 = (26)(27^2 + 27 + 1)

or 2 . 13 . 757 and 757 is a prime number by division until a <= sqrt 757 which is around 27

got it

$9^{9} - 1 = 2^{3} \cdot 7 \cdot 13 \cdot 19 \cdot 37 \cdot 757$

[solid kiln]

acgn

[tired shuttle]

tysm man

.close

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[balmy hornet]

So like

It looks like f is a single variable function

So would the b-1 be f'(x+y) and b-2 be none of the above -- 0 since there is no y based function?

[pallid lake]

the answer depends on the function, so I think it's gonna be none of the above

but it's really unclearly given

[fleet torrent]

Multivariabel chain rule:

$dz/dx = df/du \cdot du/dx$ when you have $z=f(u)$

[solid kiln]

Oliver

[fleet torrent]

Pretend all those d's are actually the partial derivative sign, i cba to do good latex unfortunately

[balmy hornet]

I wanna assume that it's f(x) based on the first part

[fleet torrent]

actually that's really bad lemme try again $\frac{dz}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \quad \text{where} \quad z = f(u)$

[solid kiln]

Oliver