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@clever sequoia Has your question been resolved?

[clever sequoia]

uh tbh i dont really know integration xd

[analog raptor]

howd you even get this integral?

its not even elementary

to solve it you should know the value of the guassian integral

[clever sequoia]

i was looking for a way to get the area under the curve of a normal distribution

for zscores without looking at a table

[analog raptor]

ig integration si the only method then

the integral of $e^{-x^2}$ from -inf to +inf is $\sqrt \pi$

[glossy valve]

math X meth ✓

[analog raptor]

so try and get it in this form

[full forum]

@clever sequoia Has your question been resolved?

[clever sequoia]

oh well idk integration rlly xd

[full forum]

@clever sequoia Has your question been resolved?

[astral crown]

that specific integral is given by a special function called "erf"

you can't write a standard anti-derivative for it

[clever sequoia]

yea ig it requires series expansion

to get definite integral of it

well thanks anyways

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[astral crown]

people usually have a lookup table, or numerical methods to do this

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[clear swift]

guys how do i factorise x^2 +2x -10

[wide sundial]

Look at it closely, see that 2 and 5 doesn’t make 2

Cry a bit

Then complete the square

[clear swift]

how do i do that

[echo marsh]

add and subtract square of (2/2)^2

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@clear swift Has your question been resolved?

[clear swift]

guys my pp itchy

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[rain oar]

I’m kind of stuck on how to continue this partial fraction decomposition

[plain fox]

first u can let N = 0 and then N-500 = 0

u would get A and B values respectively

idk why this works so pls no ask , but thats how partial fraction decomposition is done

[rain oar]

thank you so much! Yeah, I'm usually okay at solving these, but it's been so long since last time. I should probably do some practice questions before I keep going...

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[raven flax]

How do I do A

[compact zodiac]

First, can you draw a graph

[raven flax]

it is a semi circle

[compact zodiac]

Yeah so you are halfway there

Do you recall how to make it invertible

Or, when it is invertible

[raven flax]

when it passes horizontal line?

[compact zodiac]

Uhh, not sure about that

Why do you think so?

I do think you can relate with horizontal line

[raven flax]

how do u solv e then

[compact zodiac]

I mean, you should have learned about inverse function

Right?

[raven flax]

yea?

[compact zodiac]

Do you recall what you learned about it

[raven flax]

reflex aorund y = x ??

[compact zodiac]

Well, other than that

Like, can you enumerate the things you learned about inverse functions

[raven flax]

can u just tell me how to do it plz

[compact zodiac]

Sorry, I can't do that

I can say, at most, that if you know y=x reflection

Then you can check when the reflection is a function.

[raven flax]

that's what I said

about the horizontal line test

[compact zodiac]

If you reflect an entire half-circle, the resulting shape is not a function, because often there are two result values for same parameter.

So you need to avoid that, that's the problem.

[raven flax]

I got D: 0 <= x <= 2 or -2 <= x <= 0

Is there any way to do this algebraically

[compact zodiac]

Algebraically? I don't think it is doable in general

In this specific case, you can look at how many "roots" of y0 = - sqrt(4-x^2) for fixed y0.

[raven flax]

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[silent quest]

Can anyone explain to me, why the answer is E?

[wide sundial]

Clearly it needs to change it can’t be constant

which does that rule out?

[silent quest]

I just don't understand with the sudden of acceleration increase

[silent quest]

A and D is false

[wide sundial]

now of the remaining 3

all of them start negative

yeah?

[silent quest]

ye

[wide sundial]

that means down is negative

[silent quest]

ye

[wide sundial]

acceleration downwards is the negative direction ok good

[silent quest]

ye

[wide sundial]

now we got gravity going when it falls so it should stay constant at negative since gravity is constant

[silent quest]

ye

[wide sundial]

now as it hits the floor what happens

[silent quest]

a = 0

[wide sundial]

(imagine the ball is squishing)

the ball is inelastic

[silent quest]

ye

[wide sundial]

you should imagine that the ball is like a ball of playdough

[silent quest]

ok

it bounce when hitting the floor

[wide sundial]

it says that it sticks to the ground

no, playdough doesn't bounce

[silent quest]

oh

ok

it sticks

[wide sundial]

ok if it sticks to the floor

clearly it needs to no longer move

no longer accelerate

[silent quest]

ye

[wide sundial]

it must be at a = 0

[silent quest]

ye

[wide sundial]

thats it

there's only 1 answer that starts at negative, ends at 0

[silent quest]

umm

but how the acceleration increase rapid when touches the ground

can you draw a free-body diagram?

to specify all the force include during hitting the ground?

Suppose Weight is the only force during falling

when the ball touches the ground

the ground gives a reaction force to the ball, which is the normal force, N = W

However, the normal force should balance with the weight and makes the ball not falling

So how does it occurs?

[celest pewter]

that's the transition period from -9.8 to 0

[silent quest]

then the range of acceleration should be in -9.8 to 0 and not positive

[celest pewter]

gravity is negative -> to actively decelerate it must be positive

if acceleration is never positive, how would it ever slow down to reach velocity = 0?

[silent quest]

OH

You're right

But how to draw the free-body diagram?

Like this

[celest pewter]

net force ( F ~= a)

[silent quest]

can you label out all the force included

in the process of deceleration?

[celest pewter]

it would be some sort of inelastic response force

i'm not sure what you'd call it tbh

[silent quest]

Oh

[celest pewter]

but in the first part, it's just g, and in the third, it's g and normal which cancel

[silent quest]

ye

[silent quest]

Force not found

[wide sundial]

how can my object go from moving downwards to not moving at all

surely it needs to accelerate upwards

[silent quest]

but what's the force?

[wide sundial]

it's hitting the floor

[silent quest]

ye, I miss that point

[wide sundial]

the floor is providing the force

[silent quest]

then the normal and weight balanced

when the object hits the floor

what's the third force

[wide sundial]

wdym third force

[silent quest]

as metioned

there must be another force to decelerate the object

[wide sundial]

well idk this is physics not math anymore

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@silent quest Has your question been resolved?

[silent quest]

oh

ok then

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[west mango]

how do i show that

how do I show that (n^2)^(1/n) is 1?

,w lim n to infinity (n^2)^(1/n)

[glossy valve]

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[sharp flame]

natural

logarithm

[glossy valve]

Sadie Carnot (η > 1)

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@west mango Has your question been resolved?

[west mango]

okay

what now?

how is it 1

[ashen crescent]

$\lim_{n\to\infty} e^{\frac 2n \ln n} = e^{\lim_{n\to\infty} \frac 2n \ln n}$

[glossy valve]

shsgd

[west mango]

okay that makes sense

[ashen crescent]

You know what to do from here?

[west mango]

yeah 0*ln(inf) is 0

because 0 *inf = 0

thanks

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[pale elm]

Hey all! I need help with the following problem: Given a pyramid with a triangular base of sides 13, 14, and 15, and three slant heights of 20, how do you calculate the volume of the pyramid?

What I don’t get is how to find the altitude of the pyramid and how come all the slant heights are equal when the vertices of the base are not equal??

[barren onyx]

why can't all the slant heights be equal even if the vertices of the base are not equal?

[full forum]

@pale elm Has your question been resolved?

[pale elm]

I don’t have a solid argument, I just can’t visualize it in my head that way. I feel like the level of slantedness of each edge would be different.

[barren onyx]

it would be different

and the vertex of the pyramid would not be exactly in the center

[pale elm]

so how do we find the altitude?

[barren onyx]

you will want to find the radius of the circle that circumscribes the base triangle

once you have that, you can calculate the height of the pyramid

just an example so you can better visualize what i'm talking about

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@pale elm Has your question been resolved?

[pale elm]

Thank you. After this, we would need to find the length of the perpendacular from O to each vertice AB, BC and AC (which are equal, lets call it p). And the the altitude h can be found using Pythagorean: 20^2 = p^2 + h^2. Does it sound correct?

[barren onyx]

you don't need to find all 3 lengths from O to each side

just one will be enough

then you use pythagorean theorem to find the altitude

[pale elm]

I get it. That is helpful. Thank you so much!

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[limber flicker]

I actually want to jump off a bridge headfirst , I am getting lost in the maze of notation that are these notes and have never been so confused in my life

so like, in u' = A(x) u

inhomogeneous , and the solution u(x) = psi(x,x0) u(x0)

so

my psi(x,x0) is the 2x2 , e^(Ax)?

and would I form u0(x) = [-1,0] ???

okay so I did that and I got the final answer but the wrong signs on u(x) >_>

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[limber flicker]

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[red valve]

i need someones help

i have had 2 people help me so far and everyone just gets stuck

this question is cursed

[hot herald]

apply inside angles theorem for circle

[red valve]

which would make it 70+70+100

[hot herald]

no

[red valve]

can you elaborate?

[hot herald]

look up inside angles theorem for circle

[wheat ruin]

Basically

A central angle is equal to the angle of the arc

[hot herald]

not really what you want here

also not i'm referring to

[wheat ruin]

It's easier

[red valve]

but we tried it and its wrong

[wheat ruin]

We can find the measure of arc DF

[hot herald]

easier how?

[wheat ruin]

It's

[hot herald]

no

its not

what /2

[red valve]

we already did that and it isnt one of our awnsers

[hot herald]

please...

look up inside angles theorem for circle

no

[red valve]

no

[hot herald]

please delete

don't make me ping mods

[hot herald]

alternative you could apply inscribed angle theorem a few times and do basic angle sums
which is how what i'm referring to is derived

yes. that's the one i'm referring to

[red valve]

inscribed angle theorom on what angle though

[hot herald]

focus on one thing at a time

do you want to first try applying that directly first?

[red valve]

i think so

yes?

[hot herald]

ok, try doing that

here's a colour coded image that may make it easier for you

two chords refer to the red BD and blue CF
measure of each angle refers to the yellow ones I've marked
the intercepted arcs are green and orange. their measures are given to you

[wheat ruin]

But you don't need the orange one, u need minor arc BE not BF to get angle BDE

[red valve]

i dont get it

[hot herald]

with all due respect can you stop interupting

which part of the
theorem
and my breakdown of the image don't you get?

[red valve]

i think its a me problem

i dont think i understand

thanks for the help though

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[hot herald]

its also recommended that you don't just take googles summary from the first hit

[red valve]

no im not

google doenst have the awnsert

[hot herald]

your snapshot is literally the first thing that comes up

when you search what i mentioned

a simplified description of the theorem would be

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[hearty cipher]

This is what I have so far "f the diameter of the moon did doubled in Area then it would amount to 4,318 " Can someone help me with the math on how the volume changes?

[hearty cipher]

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[tired blade]

how did this simplify to this

[hot herald]

prod to sum law

[tired blade]

right I forgot

thank you

[hot herald]

i'm more concerned about the typo on the left side of that line

and missing () in the first line

[tired blade]

ahh yeah the x is an exponent

I'll correct them, thanks again

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[tired blade]

same question

I got the answer as -2.5

but the book got it positive

im not sure which is right

[tired blade]

if we add xlog2 and xlog2 we get 2xlog2

then subtract 4xlog2 from the right side

the answer is -2xlog2

so we end with -2xlog2 = log32

x = -2.5

is that right?

,w 2^((-2.5)^2)

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[pseudo glacier]

Can you help me find the angle x using the cosine rule

[pseudo glacier]

@glossy valve

,help

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[tough hazel]

Are there any more conditions given?

[hot herald]

what have you tried so far?

[pseudo glacier]

just that the angle has to be rounded of to the nearest whole number

[pseudo glacier]

a^2+b^2=c^2 for the left triangle

get the side

[hot herald]

what did you get for the side

[pseudo glacier]

then use the 3 sides to calculate for x

i got

68

[hot herald]

so what's the issue with your calculation for x

[pseudo glacier]

im off by 6

i got 68

[hot herald]

show your work

[pseudo glacier]

its supposed to be 62

[tough hazel]

I knew it

[pseudo glacier]

hold up lemme redo it

cuz i scribbled it

[hot herald]

you didn't set up cos rule correctly

[pseudo glacier]

what

isnt it

[hot herald]

you don't just randomly assign the values to a,b,c

the rule/law follows conventional labelling

[pseudo glacier]

isnt c the highest value

then followed by a and b

[hot herald]

no

[pseudo glacier]

what

[hot herald]

in conventional labelling unless otherwise specified in the question
angle A is opposite side a
B is opposite side b
C is opposite side c

[pseudo glacier]

im only given x tho

and the 90

[hot herald]

the 90° is irrelvant now

[pseudo glacier]

alr

[pseudo glacier]

yea i get that

but like

what do you mean

[hot herald]

in the triangle on the right
which side is opposite x

[pseudo glacier]

79

[hot herald]

yes

[pseudo glacier]

oh wait

so c is 79

[hot herald]

if using the rule that uses C for the angle, yes

[pseudo glacier]

OHH

i see

hold up

lemme try it out

[pseudo glacier]

yooo it worked

thanks so much haha

have a good day

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[analog raptor]

whats the difference?

[analog raptor]

the original q has {x}

but you can rewrite {x} as x - [x]

where {x} is fractional part and [x] is greates integer function

[pliant dagger]

[x] is greatest integer function

[analog raptor]

yes

same thing as floor function

<@&286206848099549185>

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[gilded tapir]

On the drawing MNL triangles biggest side is ML=9 the what can the length of the circle be A)9,4pi B)9pi C)8pi D)7pi E)6pi

[viral jasper]

what do you mean by "length of the circle" Circumference?

[gilded tapir]

Yes

[viral jasper]

Deduce the limits of the radius, then use that to infer the limits of the circumference

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[keen mango]

Can somebody help me with this proof?

[somber niche]

Are you familiar with how to do induction?

[keen mango]

I know the basics

I know I need to prove the base case first

But I don't know if that would be a_k = 7a_k-1
or
a_n = 5*7^(n-1)

[somber niche]

The first is what you are given

the second is what you have to prove

So we have a_1 = 5 and a_k = 7 * a_k_1

what's the base case?

[keen mango]

k = 2?

which would be 35 i think

[somber niche]

What are you proving for?

[keen mango]

n >= 1?

[somber niche]

Read the question carefully

yes

so what's the base case?

[keen mango]

5?

a1 = 5?

[somber niche]

or rather

n = 1

for this we have a_1 = 5 given

if we plug that into the a_n formula, what do we have?

[keen mango]

5 * 7^(n-1)?

[somber niche]

yes

for n = 1 that is?

[keen mango]

5

[somber niche]

good

so the base case is true right?

5 = 5 after all

so far so good

what's the next step?

[keen mango]

simplify this i suppose?

[somber niche]

No...

What's the next step of induction?

[keen mango]

The inductive hypothesis

[somber niche]

Yes

so what's our hypothesis?

[keen mango]

that this is true
a_n = 5*7^(n-1)

for some k, perhaps?

P(n) is true for some integer n = k?

and then plug in k + 1

[somber niche]

Or in other words that a_k = a_n for some n = k right?

[keen mango]

Yes, i suppose

[somber niche]

That's the hypothesis

Rewritten in terms of what we have that would be

[keen mango]

a_k = 5*7^(k-1)

[somber niche]

7 * a_k-1 = 5 * 7 ^ (n-1)

yes

or like this

Now the next step

the case for k+1 and n+1 right?

[keen mango]

Yes

I don't know how we can do that with a disagreement of variables

[somber niche]

k+1 = n+1

basically

[keen mango]

so this is true?
7 * a_k-1 = 5 * 7 ^ (k-1)

[somber niche]

yeah that should be ok

or rather make them both n

[keen mango]

something like that

[somber niche]

ye, that's the assumption

[keen mango]

so now we can plug in (n + 1)?

[somber niche]

yes

[keen mango]

well that should just be this i think

[somber niche]

ok

now what could you do to the left side of that?

avoid changing the right side

hint : we know the formula for a_n right?

[somber niche]

basically a_n = a_k don't let the letter confuse you

[keen mango]

?

[somber niche]

yes

[somber niche]

you should write n here

[keen mango]

n instead

[somber niche]

yeah

and that second term

what is that?

have you seen that before?

somewhere

[keen mango]

it equals this i guess

maybe sub it in for that?

[somber niche]

You're just backtracking now

[somber niche]

We have this right?

[keen mango]

but then it just equals itself

[somber niche]

Now look carefully at our hypothesis

[somber niche]

No, don't change the right side

don't touch it at all

This is our hypothesis, we know this to be true

This is what we currently have right? We are trying to prove it

[torn jolt]

Is it strong induction tho?

[somber niche]

10 amps

[keen mango]

it only asks for "induction"

is this true

[somber niche]

No, how'd you get that?

Pay mind that there are 2 sevens on the left side

[keen mango]

ah, yes

i forgot that...

[somber niche]

So what would you have now?

[keen mango]

im still looking at these two formulas trying to figure it out

[somber niche]

Look at this

couldn't you substitute the hypothesis in here?

[keen mango]

...?

[somber niche]

Good

[keen mango]

or

[somber niche]

now

That left seven?

should it be there?

You had 7 * 5 * 7^(n-1) right

and you simplified it down to

5 * 7^n

that left hand 7 is gone now

and you have

5 * 7^n = 5*7^n right?

this is a true statement

as such the proof is complete

[keen mango]

but this is just like saying 1 = 1 isnt it

don't we need to prove 5 * 7^(n+1) = 5*7^(n) ?

[somber niche]

no

we need to get 1 = 1

because 1 = 1 is a true statement

so with the base case being true

under the assumption of the hypothesis

we got a true statement

as such the (albeit weak) induction is complete

[keen mango]

how does this differ from strong induction

and does induction always imply assuming the conclusion, and then proving it "backwards"?

[somber niche]

You're actually proving it forwards since you assume the case for n and then prove for n+1

as for strong induction, it's a bit different, in this case probably not necessary

[keen mango]

but we're only given this for k >= 2

we do not know that this is true

we must show it

and so from what im gathering, we are assuming it to be true in order to make our argument

which sounds a bit out of whack

[somber niche]

Is this the first induction problem you're doing?

[keen mango]

it's the first of this kind i suppose

with sequences

[somber niche]

To be frank I think this one can be proven with deduction in a way since it's literally just "take 5 now multiply iz by 7 over and over" lol

So induction has 3 basic steps

So ominous btw

[keen mango]

text-smoothing turned off is even more ominous imo

[somber niche]

so the claim of this problem is that for any k = n >= 1, a_k = a_n holds

right

Step 1 : Check the base case

For n = 1

In this case it might actually be more correct to check n = 2 as well

since a_1 is given and a_k has a formula

If this step is true (it was) move on to the next step

Step 2 : Assume that for some n = k the statement holds, this is our hypothesis

so we assume a_k = a_n

now this might seem counterproductive, but it's just how induction works

and then step 3

Step 3 : Using the assumption of step 2 prove that a_k+1 = a_n+1

If you get a true statement at step 3

you're done

now the whole k and n thing, it's I guess more correct to keep k on left side and n on right

[somber niche]

and then only plug in n when you get here

I'm too sleepy to care about that detail :d

Think of this as a game of dominoes

[torn jolt]

I mean isn't the question sort of imply that it already did strong induction?

If k>1 for n>1 then by letting n=k+1 we have it for k>2
Then
ak+1=5*7^k
Then use ak+1=7ak
And then use n=k
?

Ah whatever I hate induction notation and here we have 2 different variables I have no idea how to write it by strong induction

[keen mango]

are we proving the base case for this?

[somber niche]

We are proving the base case for the statement that a_k = a_n

for k = n = 1

[keen mango]

But we have to make that statement

because the problem doesn't make that claim

[somber niche]

Honestly, I don't think it matters in this case, it's clearly an entry level induction problem so they probably mean weak

The problem does, the claim is prove a_k = a_n

and then we check for n = 1

is it true?

we get 5 = 5

it is

hooray

[keen mango]

interesting

[torn jolt]

Yeah but I have seen a couple of questions where it matters what inequality you establish.
I mean k>2
Then to use k=1
We can only do it if k=k+1
And Therefor ak+1=7ak

[keen mango]

i guess i'm reading it in english and not in math

because to me, "show that..." kinda sounds like its asking you to prove something independent of the original claim from the ground up

to make that connection all by yourself

[somber niche]

Or we can just use n=2 as the base case, ignoring the initial statement and then prove sepperately n=1 and just say "there it works" lol

[keen mango]

so how would i go about explaining the first step formally

"Let P(k): ..."
or something like that?

[somber niche]

Step 1 : Does it hold for n = 1? Let's try it! By plugging in the numbers we get 5 = 5, hooray!

I like to write my proofs as borderline shitposts

[keen mango]

so 5 * 7^(1-1) = 5

[somber niche]

Yes

[keen mango]

yes but we must first state the claim

Let P(n) represent a_n = 5*7^(n-1)

[torn jolt]

Let me give you one induction question :

Q5

[keen mango]

zzzzggggaaa, you are imperializing my help forum

[torn jolt]

Ah yeah sorry

[keen mango]

im trying to understand what the hell induction is

[torn jolt]

Is your question have any other parts in it?

[torn jolt]

Yeah I am also not sure either.

[torn jolt]

Do you have anything in your book about induction or something?

[keen mango]

there's a chapter on it

[torn jolt]

Have you tried to use notation?
I mean you can just start with doing questions with listing all necessary notation and then as you get used to it you can get rid of it

[somber niche]

Basically forget everything you know about proofs, induction is it's own beast

[torn jolt]

[keen mango]

Why does k = 1 and k = 2 not need to be proven for this part

k = 1 is given, but k = 2 kind of isn't

am i crazy

[somber niche]

Welcome to induction baby

Think of it like dominoes, you pushed over the first one, now you have to prove that if any one is pushed the next one is pushed as well, from that, without looking, you know they're all going to fall

[cobalt path]

k=2 is equal to 35

[gloomy minnow]

what does induction mean

[cobalt path]

its when you induct something

[keen mango]

thats right, but why isn't that a necessary basis step

[gloomy minnow]

idk what that means

[somber niche]

Ah yes, here come the shitpost

https://www.youtube.com/watch?v=JTgWbq-S6Zc

Here this might help you

The voice actor is annoying af though

[gloomy minnow]

thx

[torn jolt]

For me it looks like the example problem but n and k swapped and also k>n>1
So you reduce intitial formula by one to get to n>1
By letting k=n then just do k+1 case for ak+1=7ak=57^k
Then by using ak=57^k-1
You are done
So just check the file I attached and try to mimic strong proof by induction from their example

[cobalt path]

because you're supposed to find it yourself

essentially induction is saying that "thing holds for k=1, thing holds for k=2, thing seems tol hold for k, thing is real"

[torn jolt]

Any help with series my question?
Thanks

[cobalt path]

what

get your channel

[gloomy minnow]

is it like

like

prove something with n

u use what they gave for n

to find n+1

[somber niche]

This became "what the heck is induction" the channel

real quick

[cobalt path]

yes

something like that

[gloomy minnow]

sorry moderator

[cobalt path]

god my eyes burn jesus

[cobalt path]

you, do you know higher statistics

[gloomy minnow]

who want these

[somber niche]

It's when you get heat from an oven with electricity running in it

[gloomy minnow]

bruh

[cobalt path]

you seem smart

not you nrf

[gloomy minnow]

okay buddy

[somber niche]

I barely know normal statistics, statistics are evil I stay away from that

[cobalt path]

goddamn it

[keen mango]

To prove the basis step is true for this example, don't you have to prove that the caluclated values for P(n) match the sequence values P(k)?

[somber niche]

You pick a basis number

n = 1

you plug it into a_k

you plug it into a_n

you compare

are they the same?

Hooray

[keen mango]

okay thats what i was looking for

because we are given P(k) as the truth

so P(n), the hypothesis, must match the sequence for at least the first 1 or 2 steps to be considered true for the base cases

[somber niche]

yes... I think... Weird wording but you got it.... maybe....

[keen mango]

also i think using P(n) and P(k) to represent the sequence, and the statement to be proven is confusing

it should be something like S(n) and P(k) i think

[torn jolt]

Yeah that's why you want to take first basic result as ak+1 so it will be for k>1 and then say k=n. I guess.

Ideally we want to prove nth case from k not vice versa

[keen mango]

yep, im completely lost

[somber niche]

[keen mango]

im trying to write it formally

with A(k) representing the given sequence

the statement to be shown matches the sequence for n=1, k=1 and n=2, k=2

so how do i make the hypothesis now

should i make a new variable like this?

i don't even see how 'n' and 'k' are related at this point

[torn jolt]

By hypothesis?

[keen mango]

i dont see that

[cobalt path]

i wont lie, chegg has this solved

[keen mango]

checked and could not find it

[keen mango]

how does this work again

[somber niche]

a * a^(n-1) = a^n

[keen mango]

i've got this written down so far

i don't see how we go from here to there

anybody have the slightest clue how

i don't think there's any equality that allows for it

anybody have the slightest inkling

[full forum]

@keen mango Has your question been resolved?

[full forum]

@keen mango Has your question been resolved?

[alpine forge]

this is your question?

@keen mango

@keen mango @keen mango @keen mango @keen mango @keen mango @keen mango @keen mango @keen mango

@keen mango

[keen mango]

yeah

i want to write an at least somewhat formally written proof for it

but i keep getting lost in the variables

my work is above

[keen mango]

if you understand mathematical induction i would appreciate the help

[alpine forge]

ok

wait a minute, ok?

[keen mango]

yes sir

[alpine forge]

Fine.

Base Case (n=1): We start by checking the base case, which is ( n = 1 ). The sequence is defined as ( a_1 = 5 ), and according to the formula we need to prove, ( a_n = 5 \cdot 7^{n-1} ), it should also yield ( a_1 = 5 ) when ( n = 1 ). Since both the definition and the formula give us the same result, the base case is true.

Inductive Step: Now, we assume that the formula holds for some integer ( k ), which means ( a_k = 5 \cdot 7^{k-1} ). This is our inductive hypothesis.

Next, we need to prove that the formula also holds for ( k+1 ), which means showing that ( a_{k+1} = 5 \cdot 7^k ).

Using the recursive definition of the sequence, we have: [ a_{k+1} = 7a_k ]

Substituting the inductive hypothesis into this equation gives us: [ a_{k+1} = 7 \cdot (5 \cdot 7^{k-1}) ] [ a_{k+1} = 5 \cdot 7^k ]

This confirms that the formula is true for ( k+1 ) if it’s true for ( k ), completing the inductive step.

Therefore, by mathematical induction, the formula ( a_n = 5 \cdot 7^{n-1} ) holds for every integer ( n \geq 1 ).

i cant put an " L " again ;-;

it was to be reactions saying " LOL "

[keen mango]

how can i see this correctly formatted?

i don't know latex so i don't know what \cdot refers to

[alpine forge]

In mathematics, the symbol \cdot is used to represent multiplication. It’s particularly useful in written expressions to avoid ambiguity, especially when dealing with numbers or variables that could be mistaken for one another if written consecutively. For example, to clearly distinguish the multiplication of 2 and 3, you would write it as 2⋅3
instead of just 23, which could be confused with the number twenty-three.
The \cdot symbol is also used to denote the dot product in vector algebra, where it represents the multiplication of two vectors to result in a scalar. However, in higher mathematics, explicit symbols for multiplication are often omitted unless necessary to prevent confusion

[keen mango]

why are you copy pasting from AI

[alpine forge]

because im dumb 😄

[keen mango]

do you have an understanding of the mathematics here

[alpine forge]

yea

1+1=2 LOL

[keen mango]

you are definitely not pending post-graduate

[alpine forge]

2+2=5

134+834=12

[keen mango]

<@&286206848099549185>
Can somebody help me on my mathematical induction proof?

[alpine forge]

me

[high patrol]

Your pending post college?

[alpine forge]

im brazilian, do you really think i can write all this on english?

[high patrol]

Ah then help that dude

[alpine forge]

i alrdy helped

[high patrol]

Why is his math so hard

[alpine forge]

i already said how to solve it

[keen mango]

chat bots are not capable of solving high level math problems

and you're breaking the rules by spamming AI responses

[high patrol]

Why don’t u just translate ur work to English

[patent pelican]

Look at it a little bit longer

rhs is still the same

So the step you are struggling with basically says the following:
7 * 5 * 7^k-1 = 5 * 7^k

[keen mango]

i've been looking at it for 2 hours 😭

[patent pelican]

If you understand this, you are basically done

[keen mango]

do i need to pull from a previously esablished equality?

[patent pelican]

No

[keen mango]

oh.

7^1

(k-1) + 1 = k

[patent pelican]

Exactly

[keen mango]

you know what was confusing me, the mix of superscript and subscript

[patent pelican]

It's just getting used to the notation

You will get it with time

[keen mango]

so now, does this mean it's proven?

[patent pelican]

Yes, and that is a tricky thing to notice too

[keen mango]

I started the proof like this

and then i go on to prove P(k + 1) is true as well

[patent pelican]

Yes

[keen mango]

is my logic correct do you think, for the first part?

is it necessary to differentiate P(n) and A(k), or could i use P(n) and P(k)

[patent pelican]

Yes. You will find with practice that there are better ways to write it, and I don't think that overthinking those things is actually of any help at this stage

[keen mango]

i appreciate the words of encouragement

i just try to be a perfectionist with my assignments

try to get the book-keeping type skills out of the way, like formatting and structure

so i can express the more difficult mathematics (when i get to them) to the most accurate degree

[patent pelican]

You will write better induction proofs when you use the pure logic behind it

I believe you get the idea that makes it work

You can state the base case as P(1), and nothing more, because you want to prove that statement

A(1) = 5 by definition. Make sure to include that when you introduce the sequence

[keen mango]

actually i included that verbatim but deleted it in confusion a while ago

[patent pelican]

That said, you can skip the P(2) part. The induction step will work when k=1 so there is no need to do that. It's not wrong, as I said before, but it can be skipped

[keen mango]

is that because, for the basis step, we need to prove the base cases for the statement that is to be proven, and not the sequence that is assumed to be true?

i got caught up early on deciding whether or not to prove the base case either for the sequence, or for the formula statement

[patent pelican]

The sequence is an object that is just there

The sequence is there and it exists, and is initially defined recursively

And you want to prove a statement for every n

and that statement gives another way of describing the sequence

A priori we don't know if that new expression is correct, so we will find out that it is when we finish our induction proof

The first step is checking that the proposition is true for 1

So write the proposition, and verify that it is implied by something that is true

The statement says that a_1 = 5, and we know that is true, by definition

So that proves the base case

Now the inductive step lets us assume that the proposition holds for some natural k, and we want to show that it holds for k+1

We want to prove that a_k+1 = 5*7^k+1

So we know by definition that a_k+1 = 7*a_k, right?

And now the inductive hypothesis says that a_k = 5*7^k

replace above and you get:
a_k+1 = 7 * a_k = 7 * 5 * 7^k = 5*7^k+1

And that is explicitly the p(k+1) you wanted to prove

and that finishes the proof

[keen mango]

So, the goal is to sort of mold what you know to be true into what you want to prove?

I kind of looked at induction as sort of assuming the conclusion, and working backwards, but people in this chat have told me I'm wrong on that.

But I suppose the most essential part of your conclusion is proving that P(n + 1) actually follows P(n), and less so proving P(n) by itself

[patent pelican]

It's not wrong but you have to notice that all your steps are an "if and only if", so your initial line is implied by something that is always true, like 0=0

And that is trickier than doing what I did above

[keen mango]

Okay I think I see

If you still have time, do you mind if I ask you about one more question

[patent pelican]

Hmm go ahead. I might have to go, but surely someone will answer

[keen mango]

I think this is a lot shorter

I'm trying to find the flaw in this proof

off the top of my head, I noticed that P(0) doesn't seem to be the accurate base case

because 'r' must be any nonzero real number

so it doesn't seem right that P(0) implies any sort of truth

but i'd be curious if you agree on that

[patent pelican]

The flaw is that you want to prove it for k+1, and you are assuming P(i) for all i from 0 to k.

And notice the proof uses that r^k-1 is 1

But if k+1 is 1 (that is k = 0) then when replacing r^k-1 by 1 you should assume P(-1) but that is not true

Since the proof relies on r^k and r^k-1 being true, for every k, the proof should have two base cases

but you will only ever have one base case

Do you see it?

[keen mango]

maybe... k = 1 and k = 0?

[patent pelican]

What do you mean?

[keen mango]

for the two possible base cases

[patent pelican]

k=1 won't work

The equation is only true for r=1

[keen mango]

i just dont understand P(0) being a valid base case

if the goal is to prove that 'r' to any positive, nonzero integer is 1

there is no reason to prove P(0)

your base case should be 1 i think

because its the lowest positive nonzero integer

[patent pelican]

P(0) is true, but no other P(k) is, and the inductive step will always need at least two statements to hold

That's why it fails

The base case holds, but nothing else does

[keen mango]

So how can you determine if your base case is 'bad'

or that it doesn't prove anything past itself

[patent pelican]

It's a case-by-case thing, but in the proof you presented the flaw is what I mentioned

The inductive step relies on the two previous statements being true, but if you started with only one then the induction never "starts"

[keen mango]

So, P(i) being true from 0 through k is not necessarily true?

Or, it's incorrect to assume it's true

because you're essentially assuming (k+1), (k+2), (k+3)... is true?

although I don't really see how that leads to a material problem in the proof

[patent pelican]

That's not the problem. The problem is that even assuming all of that, there exists one case (k=0) that is not guaranteed by all those statements

[keen mango]

I don't understand that

isn't k = 0 proven in the basis step?

[patent pelican]

Yes

But when proving P(1) (I say k=0 because I want to prove P(k+1)) using the inductive step, I will require P(0) and P(-1), but only P(0) is assumed

[keen mango]

I don't see why P(0) and P(-1) are required

[patent pelican]

Notice that proving P(k+1) requires P(k-1) being true

[keen mango]

I guess I just don't understand what base cases are for, or how they're determined
because I thought the base case represents literally the "base case". So if you're accounting for k >= 5, your base case should be k = 5

[patent pelican]

Yes, that is correct

[patent pelican]

Because the r^k-1 is being replaced by 1. That is the same as assuming P(k-1)

But what I say is that the proof doesn't work because if you let k=0, you will be trying to prove P(k+1) = P(1), right?

And you will be requiring P(k-1) = P(-1)

Because that is being used in the inductive step

But the inductive hypothesis is not guaranteeing that specific statement. IH does not guarantee P(-1)

[keen mango]

this jump in logic, i dont see

this whole thing just simplifies to r^(k+1)

isn't it just algebra

[patent pelican]

Okey I have to go now. I'd try giving it some time alone and give it another look when you have more experience with induction. I struggled with it a lot when it was introduced, but practice and time did the trick

[patent pelican]

Yes it does, but the proof will try to rewrite it in a clever way so that it can invoke the inductive hypothesis to simplify the expression

A big part of math is rewriting things in clever ways so that is becomes easier to see some arguments

But those tricks should be used correctly, unlike this proof hehe

[keen mango]

is it from the first denominator to the second denominator?

i think i see that

[patent pelican]

it uses the IH claiming that P(k) and P(k-1) are true

[keen mango]

...but it only proved that P(0) is true, and not that P(k) or P(k-1) = 1

i know you have to go, but thank you for your help thus far

your explanations have been helpful

[full forum]

@keen mango Has your question been resolved?

Channel closed

Closed by @keen mango

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[idle shale]

If (i)¹² = ω¹²
, ω is one of the roots of 1
So why can't i say i = ω as it's wrong?

[tidal folio]

what grade is that?

[idle shale]

It's complex numbers

I don't talk about grades

[tidal folio]

-_-

[idle shale]

I don't mean to be rude, just it doesn't matter

[umbral dome]

i is one of the 12 different 12th roots of unity, but it's not equal to any of the others

[thick hedge]

the roots of unity are in a circle in the unit plane

equally spaced

[idle shale]

More explain please

[thick hedge]

try drawing them for yourself

[idle shale]

The 12 roots?

[thick hedge]

yeah

[idle shale]

Bruh

[granite torrent]

i don't see how that helps

cloud's message is enough

[tidal folio]

I'm a helper of grade 0-7

[gloomy minnow]

what if you take w=1

[idle shale]

Can't i even understand it Algebraically?

[gloomy minnow]

1≠i

w¹²=i¹²

[thick hedge]

$e^{\frac{i n \pi}{12}$

[granite torrent]

do you understand cloud's message

[glossy valve]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thick hedge]

are the roots

where n ranges from 0 to 12

[idle shale]

Aha

I got it

Yeah yeah

[granite torrent]

also this is not true

[idle shale]

There are different roots , they don't equal themselves

[thick hedge]

it isn't?

[idle shale]

Alright thanks

.close

[full forum]

Channel closed

Closed by @idle shale

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[full cradle]

Hi. I want to count from 100 to 0 without ever actually hitting 0. For example, 100 - 1 = 99, but then the next time 99 - (number smaller than 1). Can someone help me with a math function that can do this?

[granite torrent]

you can do this with like a geometric series or something

[hardy jewel]

Keep on dividing by 2

100/2 = 50

50/2 = 25

25/2 = 12.5

And so on

You won't ever hit 0

[full forum]

@full cradle Has your question been resolved?

[full cradle]

thanks. I'm trying to do it by only affecting the decrement number. for example, 100 - 1 = 99, 99 - (1 / 2) = 99.5, but it gets stuck in the 90s. I want it to get as close to 0 as possible before it starts to slow down in any way

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@full cradle Has your question been resolved?

[hardy jewel]

100 - 50 = 50

50 - 50/2 = 25

o nvm

ill think smthn

[full cradle]

thanks guys

resolved

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[rough oyster]

when expanding this logarithm:

[rough oyster]

would the final answer be this? and if so why is it not like this,

[slow canopy]

Those are the same after simplifying

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[rough oyster]

thanks

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[fast sluice]

given $a \geq 6$ Find the smalelst value of $S = a^2 + \frac{18}{\sqrt{a}}$

[glossy valve]

cffex

[fast sluice]

I tried using AM GM $a^2 + \frac{18}{\sqrt{a}} \geq 2 \cdot \sqrt{\frac{18a^2}{\sqrt{a}}}$

[glossy valve]

cffex

[fast sluice]

simplified and dont know what to do next

[low echo]

Hii

[fast sluice]

hello

[low echo]

Use calculus

[fast sluice]

im still in middle school

[low echo]

differentiate and equate to 0

You know differntiation?

[fast sluice]

im still in middle school I cant use calculus

[low echo]

Ok wait

[fast sluice]

also afk for a moment sorry

ok im back @low echo

[low echo]

Hm

See this is an increasing f(n)

So as its specified a>=6

[fast sluice]

yea

[low echo]

Least value should be at 6

[fast sluice]

yea

[low echo]

Wait few mins lemme think

[fast sluice]

brb

[low echo]

.

i dont see alternative soln for middle school

Scwartz inequality may work

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[plain scroll]

I’m really confused whether i should distribute or solve the parenthesis first

[plain scroll]

im trying to differentiate between them

[fathom saddle]

Can you elaborate on "solve the parentheses"?

[plain scroll]

uh the numbers in the parentheses

i mean the expression

[fathom saddle]

You only have the option to distribute. Nothing in the parentheses simplifies

[reef dust]

do you mean like reducing it into a single fraction or something

[glossy valve]

faiyrose

faiyrose

[reef dust]

either way itll be the same

[neon geyser]

[plain scroll]

ah so i guess distribute 2 into the expression is the only easier option

[neon geyser]

anything else you think/want to do?

[plain scroll]

I think i can solve it now, Thanks guys

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[hollow loom]

Can someone verify b

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@hollow loom Has your question been resolved?

[ashen crescent]

Check OQ

Nevermind, didn’t see the c

Looks correct

[hollow loom]

Alr thx

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[limber fossil]

why is the answer for the other root 3, not 2?

[hot herald]

how are you getting 2 for the other root

[limber fossil]

solved it

(x-3)(x-2)=0

why is 2 rejected?

[shadow silo]

Because they asked other root

And 2 is given in ques

[naive gale]

k=2 I think

[limber fossil]

oh yeah

im dumb

thanks

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[vast furnace]

howto draw a

[vast furnace]

and this

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@vast furnace Has your question been resolved?

[pallid ivy]

First equate the equations

so uall get 10 - 2x = 4/x + k

then further simplify it

so ull get 2x² + x(k-10) + 4 = 0

now use the b² - 4ac = 0 formula

since it's a tangent we use b² - 4ac = 0

tell what u got after doing this much

@vast furnace ?

[vast furnace]

no i just

want to know how to draw

[vast furnace]

this

[pallid ivy]

oh ok

one sec

see substitute the values of x into rhe equations

start by -3 to 3

but when u take x = 2

ull get y value undefined

So x = 2 is ur asymtote

[vast furnace]

okkk

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