help-17

[stone timber]

since we need tan on the right side, do we need to convert the bottom back to cos2x or what next 😭

[blazing wigeon]

Break apart the tan2x and rewrite the right side in sinx

[stone timber]

is this what you mean

[blazing wigeon]

Use a different cos2A

And divide out a cosx

Simplify the left side too

[stone timber]

idk how to go about simplifying the left side

[blazing wigeon]

There are like terms

-2A-2A=-4A

1-1=0

Don’t forget the negative on the right

[stone timber]

so is it -4sin^2x on top and bottom?

[blazing wigeon]

Yes but more on the bottom

[stone timber]

elaborate pls

[blazing wigeon]

1+1=2

[stone timber]

oh

-4sin^2x + 1

[blazing wigeon]

Yes and you can divide out a 2

[stone timber]

from the left side?

[blazing wigeon]

Yes

After that you should be done

[stone timber]

i think something got lost here

[blazing wigeon]

You’re missing a1

[blazing wigeon]

^

[stone timber]

a 1 where

[blazing wigeon]

2 steps back

And a negative

[stone timber]

oh ok

[stone timber]

and where’s the negative go 💀

[blazing wigeon]

Just don’t get rid of the negatives

Look at previous steps they’re there

[stone timber]

oh okay yeah but i don’t see how it equals the right side

[blazing wigeon]

You’re so close

The third to last line is all correct

[stone timber]

ive been doing this one problem for 30 minutes now 💀

[blazing wigeon]

Just don’t forget any of the terms

The penultimate line left numerator is correct

That’s the only part that is correct there

[stone timber]

i got the numerator to match but idk about the denominator

[blazing wigeon]

Show me please

[stone timber]

wait no i don’t even think i got the numerator to match 💀

[blazing wigeon]

Left side looks good

For the second to last line

[stone timber]

oh then what am i doing wrong to simplify this

[blazing wigeon]

Divide out a2 from it

[stone timber]

i thought thats what i did

[blazing wigeon]

No

4A/(4A+2) = 2A/(2A+1)

Cosx in the denominator of the right side and a forgotten negative

[stone timber]

are we dividing by a 2 or a -2?

[blazing wigeon]

Just 2

[stone timber]

[blazing wigeon]

Sweet

Good job

[stone timber]

that took way too long

thank you

[blazing wigeon]

Just don’t forget that cosxcos2x = cosx(1-2sin^2x)

It wouldn’t be right to write it as 1-2sin^2xcosx

[stone timber]

oh okay

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[heavy palm]

problem 20

[heavy palm]

i need to split this up into 2 parts right?

2 different frac

how would i do that?

[urban edge]

you dont need to split it

[dull bear]

Do you remember your log rules?

[heavy palm]

uhh

no 😭

cry emoj

[urban edge]

i was thinking of a different method, chartbits method is fine

[dull bear]

You don't remember what log(p/q) turns into?

[heavy palm]

log p-log1

dumb number

Pad

I extend a deep gratitude for your kindness.

[dull bear]

so you did remember

[heavy palm]

Look I have a autofill in for thank you

Actually it’s probably too corny

[dull bear]

hey, it's quite cool

[heavy palm]

I’m able to plug in numbers now right

After splitting

[dull bear]

You could if you wanted to, or however they presumably taught you to deal with telescoping sums 🔭

(also @hushed pewter hiii )

[heavy palm]

Yea you plug in a few

For the pattern

Then Elipse thingy

To the an term you know

That probably didn’t make sense

But you know

[dull bear]

I kinda get what you mean

[heavy palm]

like this

[dull bear]

Ah that's fair yep you can try that here of course, and find out what happens and all

[heavy palm]

I think you are a good cat

https://tenor.com/view/cat-cat-exploding-black-cat-cute-cute-cat-gif-11526660326797842323

[dull bear]

Awwwww

[heavy palm]

It diverge yay

But.

How does it go to negative infinity?

i wrote down the pattern

and every term cancelled each other out

except 0 (the very first term)

[silk osprey]

infinite sums do not work that way

[heavy palm]

Oh

[silk osprey]

infinite sums are the limit of the partial sums

[heavy palm]

Ah right because these like

Forever subtract

And stuff

Right

[silk osprey]

if you stop at any given partial sum you’ll have a -ln(n+1) left over

which diverges to -inf

note that ln(1) is 0 also

[silk osprey]

well what do you think is canceling?

[heavy palm]

Uhh

The

[silk osprey]

stop the sum at any point youd like

what’s canceling

[heavy palm]

The previous term

[silk osprey]

well all the middle terms yea

but the extreme ends remain

ln(1) - ln(N + 1)

this is our sequence of partial sums

then the infinite series is simply the limit of these partial sums

as N -> inf you get -inf

[heavy palm]

Extreme ends remain like the last terms

[silk osprey]

well depends on what you mean by term

[silk osprey]

i meant this^

[heavy palm]

I see

[heavy palm]

This makes sense

[silk osprey]

$\sum_{n = 1}^N \ln \frac{n}{n+1} = \mathcolor{red}{\ln 1} - \cancel{\ln 2} + \dots + \cancel{\ln N} - \mathcolor{cyan}{\ln (N + 1)}$

[twin meteor]

knief

[silk osprey]

then take lim N -> inf

[heavy palm]

Oh I see

Will the last term never cancel?

[silk osprey]

mhm

the n + 1 is always too far ahead in the race

there ln(n) bit will always need an additional term to cancel any ln(n+1)

so if we stop the sum at say n = N then ln(n) will only ever reach ln(N) meanwhile the ln(N + 1) maintains the lead in the race

[heavy palm]

I see

Thank you

[silk osprey]

you’re welcome

[heavy palm]

So

That makes sense

Only part I’m still confused about

Is the cancelling part still

Apologies

[silk osprey]

if you’re one step ahead of me and for every step i take you also take one step then i’m never going to catch you am i

assuming we take steps of the same length of course

[heavy palm]

Oh so

The point of cancelling is to depict

The end term and terms at the start that don’t cancel

Like

The steps

That

Show the length

bad explanation prob

[silk osprey]

yea idk what you mean?

there was no "point of canceling" it was just a matter of fact

it’s just what happened

think of the ln(n) as pac-man

[heavy palm]

Is the point of canceling these terms to mark the steps from the beginning to end and the steps in the middle are irrelevant to our final series equatuokbthing we get at the end (idk what it’s called)

[silk osprey]

eating up the remains of ln(n+1) from the previous term in the sequence

[silk osprey]

well it just helps to see how it’s behaving

[silk osprey]

the point is to find the sequence of partial sums which for any N is given by this

[heavy palm]

I see so our end product shows its behavior

[silk osprey]

i mean of course you don’t have to write out the terms if you don’t want to

but it helps

[heavy palm]

I see

[silk osprey]

you can reindex a finite sum

,, \begin{align*}
\sum_{n = 1}^N \ln(n) - \ln(n+1) &= \sum_{n = 1}^N \ln(n) - \sum_{n = 1}^N \ln(n+1)\
&= \ln(1) + \sum_{n = 2}^N \ln(n) - \sum_{n = 1}^N \ln(n + 1)\
&= \ln(1) + \sum_{n = 2}^N \left(\ln(n)\right) - \ln(N + 1) - \mathcolor{cyan}{\sum_{n = 1}^{N - 1}\ln(n+1)}\
&= \ln(1) - \ln(N + 1) + \sum_{n = 2}^N \ln(n) - \mathcolor{cyan}{\sum_{n = 2}^N \ln(n)}\
&= \ln(1) - \ln(N + 1)
\end{align*}

[twin meteor]

knief

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[modest wharf]

for this q do I just equate them

this the whole q

oh I found the inverse wrong

but would I just equate them

[outer warren]

setting them equal like that can be quite tedious

note that inverses are a reflection over y=x
and intersections would lie on that line

and your inverse is still wrong

check your work here

[vocal sleet]

@modest wharf Has your question been resolved?

[modest wharf]

wat

so there is an easier way?

[vocal sleet]

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[modest wharf]

.reopen

[vocal sleet]

✅

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[versed pewter]

Prove that there exists number x with 2025 digits that is divisible by 5^2025 with all digits of x being odd.

[marsh jungle]

im not sure exactly how to do it, but i believe you can do that with induction. maybe that'll help a bit but im sure someone else can better help

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@versed pewter Has your question been resolved?

[edgy sapphire]

try looking at n=3 with {000,125,250,...875} and look at which digits are even/odd

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[quick wing]

Question is asking to find H without any cartesian equations. I found N and M using vector projections. I'm completely stuck on how to do this

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@quick wing Has your question been resolved?

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[bleak prawn]

Im stuck on part (b)

Do we take the derivative of c_2 and sub in (0, 0) or something?

I dont really know how to get the derivative of a circle equation

[paper depot]

you can do it like that with implicit differentiation

but in the case of a circle, rather than of a general curve, you can make do without calculus entirely

[bleak prawn]

Im not really sure how

Im completely stuck usually I have a general idea but this time I really dont 😅

[paper depot]

the tangent to a circle is perpendicular to its corresponding radius

[bleak prawn]

Ahhh ok

1 moment I think I can work with that

(x - 1)^2 + (y - 4)^2 = 5

[paper depot]

mmm

arithmetic?

[bleak prawn]

(4 - 0)=m(1 - 0)

4 = m

So the slope of the tangent is -1/4

And the y intercept is obviously 0

So the equation is y = -x/4

[paper depot]

hold your horses!!!

[paper depot]

this equation for c_2 is wrong

it should be (x-1)^2 + (y-2)^2 = 5

[bleak prawn]

Ohhhhh yayayaya 😅

Thanks haha

(x - 1)^2 + (y - 2)^2 = 5
(2 - 0)=m(1 - 0)
m = 2

So the equation is y = -x/2

[paper depot]

indeed

[bleak prawn]

Arctan(-1/2) = -pi/4

[paper depot]

nope!

[bleak prawn]

Ohhh

[paper depot]

arctan(-1) = -pi/4.

[bleak prawn]

-pi/3

[paper depot]

even worse

[bleak prawn]

😭

[paper depot]

arctan(-1/2) can't really be simplified in any way

[bleak prawn]

Ah

[paper depot]

tan(pi/3) = sqrt(3) anyway not 1/2

[bleak prawn]

Hmm im not really sure what to do then

[paper depot]

arctan(-1/2) can't really be simplified in any way

[bleak prawn]

So thats just the answer?

[paper depot]

best you can do is write it as like... -arctan(1/2)

yes it is

if you are curious it is about 0.46 rad

[bleak prawn]

Ahh yeye thats the answer they wanted! 😄

[paper depot]

,calc atan(1/2)

[twin meteor]

Result:

0.46364760900081

[paper depot]

,calc 0.4636476*180/pi

[twin meteor]

Result:

26.56505066137

[paper depot]

or in degrees if you're so inclined

[bleak prawn]

All of these are generally accepted right?

I know it depends but like generally

[paper depot]

generally, if there are angle units mentioned in the question itself, report your answer in the same units

otherwise i would default to radians

but you have to ask your teacher(s) what units they prefer

[bleak prawn]

You guys are my teacher(s) 😝

[bleak prawn]

But yeye thats a good rule to follow haha

Thank you so much!!

❤️

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[bleak prawn]

I thought I knew how to do this one but I got a negative inside a square root

Factor out x from all terms in the numerator and set it equal to x - 2 and x + 2

3x^2 - 3x + a = x - 2

3x^2 - 4x + 2 = -a

-3x^2 + 4x - 2 = a

(-4 +- sqrt(16 - 4(-3)(-2)))/2(-3)

(-4 +- sqrt(-8))/(-6)

-8 inside square root, does that mean its not a solution, right?

Why can I not?

We want them to be equal no?

So they cancel out

for the removable discontinuity

Ahhh wait I think I get it, cause the WHOLE thing is not supposed to equal (x + 2) or (x - 2), just one of the factors right?

And set it equal to 0?

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[bleak prawn]

.close

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[sick ivy]

Struggling to solve this DE. I know to use integrating factor method but then I need to integrate $\int \frac{\ln \sin x}{\sin x}$ which I'm unable to do

[twin meteor]

Deepwork

[sick ivy]

Would appreciate any hints to move forward!

[vocal sleet]

@sick ivy Has your question been resolved?

[sick ivy]

no

[sick ivy]

<@&286206848099549185> 🙂

[tame wadi]

Hi

It's a de of form y' +p(x)•y = q(x)

[sick ivy]

yep so I used integrating factor method

[tame wadi]

What's p(x)?

[sick ivy]

-cot(x)

[tame wadi]

So you get d/dx(y•cosecx)= -csc^2x

After multiplying with IF

[sick ivy]

oh hold on I see my mistake

thanks !

.close

!close

how do i close this thing

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[sick ivy]

.close

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[proven bramble]

Hey, I'm working on an inequality for a proof and I got a bit stuck.

Basically, given angles a, b, c each individually less than pi, if I know that:

cos(a+b) < cos(c)

can I prove that:

a + b > c?

[oak magnet]

So they are between 0 and pi ?

[proven bramble]

Yes, individually between 0 and pi

[oak magnet]

The thing is that you need a+b in [0,pi] can you manage to have it ?

[proven bramble]

Yeah that's just the thing, a+b can be more than pi, so I can't really argue it's a decreasing function for both

I'm not very familiar with inequalities like this though, so I wanted to ask if it's still possible

[old sigil]

note that the inequality is false precisely if there exist a, b, c < pi such that cos(a+b) < cos(c), but a+b<=c.

now if a+b<=c, then a+b<=c<pi

[oak magnet]

Its a+b>c isn't it ?

Or am missing something in yo explain

[old sigil]

yes, we're trying to prove that "if cos(a+b) < cos(c), then a+b>c"
so the statement would be false if you could find a, b, c such that cos(a+b) < cos(c) but not a+b>c

a+b>c being false is of course equivalent to a+b<=c being true

it's basically a proof by contradiction, I just tried to phrase it as a hint

[oak magnet]

Oh okay i see

[proven bramble]

Ah so.

Assume a + b <= c

c <= pi ==> a + b <= pi

cos(a+b) < cos(c), since both are less than or equal to pi

a + b > c

Contradiction!

a + b > c

Is that sound?

[old sigil]

yep, that's the idea
essentially, a+b>c always holds under the condition because a+b <= c would put a bound on a+b

[proven bramble]

Yes, thank you so much!

This is the ending to a proof that I had spent a bit of time on and I really feared that I had wasted my time

[proven bramble]

Alright, thank you again.

.close

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[vocal pagoda]

hi

[vocal pagoda]

Am I mistaken to think $\int_a^b \vert \vert F'(t)\vert \vert dt =\vert \vert F(a) \vert \vert - \vert \vert F(b) \vert \vert$

[twin meteor]

Wild123

[paper depot]

very!

[vocal pagoda]

/close

\close

[oak magnet]

.

[vocal pagoda]

forgot how it went

[oak magnet]

.close

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[paper depot]

new coose x2 combo

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[vocal sleet]

@velvet grove Has your question been resolved?

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[broken apex]

for question e i wrote : the ground is level with the bottom of the castle

[broken apex]

would that be right

[rancid ember]

i think they expect smth like neglect air resistance or smthg

[broken apex]

alr

ty

that would apply to all questions asking similar things right (that air resistance is negligible)

.close

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[mild rampart]

Does anyone know if the following applies to the example matrix?

|1 0 0 1|        
|0 1 0 0| = |I F|
|0 0 1 0|           

where F = 
|1|
|0|
|0|

such that
    |-1|
N = | 0|
    | 0|
    | 1|

but in this matrix:

|1 1 0 0|   |I F 0|
|0 0 1 0| = |0 0 I|
|0 0 0 1|

Where F = 

|1|,   

           |-1|
    |-F|   | 1|
N = | I| = | 0| (where is the identity matrix? it doesn't show up in the previous matrix)
           | 0|

this is based on the idea that reducing a matrix to rref returns some form of |I F| where you can easily format the special solutions to result in:

|-F|
| I|

but in the second example, the identity matrix is split down the middle by F, so it doesn't seem to apply properly. is there some form of trick/manipulation of the array that I don't know that makes this trick hold for matrices like the one above?

Also, feel free to ask for elaboration on any of the parts, because i believe i might've missed a detail or two.

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@mild rampart Has your question been resolved?

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@mild rampart Has your question been resolved?

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@mild rampart Has your question been resolved?

[arctic bison]

you understood by now?

[vocal sleet]

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[regal urchin]

is this correct? this is my last attempt

[bronze osprey]

did you use $Pe^{r t}$?

[twin meteor]

south

[bronze osprey]

exponential growth equation

[regal urchin]

i did p=A/e^rt

[bronze osprey]

yep, that's the correct rearrangement

,w 2000000 / (e^(6.75/100 * 22) )

[twin meteor]

Results provided by WolframAlpha

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

Requested by sooouth.

[bronze osprey]

Wolfram rounded but you can click the link to check

[bronze osprey]

yes that's correct

[regal urchin]

got it tysm

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[void stone]

if i have a quadratic like (x+5)(x+8) = 0, then i know that x=-5, x=-8. if the equation's right side wasn't 0, do i have to expand the left side?

e.g., if (x+5)(x+8) = 4, do i need to expand it as x^2 + 13x + 40 = 4, then subtract 4 to make the right side 0?

or is there a trick i can use, knowing that (x+5)(x+8) = 0 will give -5, -8?

[bronze osprey]

yes, so what you have used here is something called the zero product property (and it has a few other similar names too)

which says if ab = 0, then at least one of a, b must also =0

you could have both a = 0 and b = 0 actually: that indeed makes ab = 0

(well this case doesn't happen in practice, when solving equations)

the only case you can't have is that $a \ne 0$ and $b \ne 0$, cause then $ab \ne 0$ too

[twin meteor]

south

[bronze osprey]

so 0 is special

if you had (x + 5)(x + 8) = 3 you would have to expand the LHS then subtract 3 on both sides, to make the RHS = 0

[quiet echo]

(There is a different case if x is an integer however, it is called a Diophantine equation.)

[bronze osprey]

oh the trick that also works is difference of two squares

if we change your example to (x + 6)(x + 8) = 4

that's just ((x + 7) - 1)((x + 7) + 1), where a = (x + 7) and b = 1

you have (a - b)(a + b) = a^2 - b^2
= (x + 7)^2 - 1

so (x + 7)^2 - 1 = 4 and now you can solve this

don't forget to take plusminus on the RHS after square rooting both sides!

[void stone]

yeah i get that, but im asking if i need to expand the LHS and then subtract the RHS

[quiet echo]

If you would like, I can give an example problem from a competition I just took

[bronze osprey]

no, the difference of two squares method means that you don't need to expand the LHS

[quiet echo]

Not in all cases

[void stone]

oh alright that answers my question

thank you both!

[bronze osprey]

no worries!

[void stone]

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[restive basalt]

Hi there. I have a quick linear algebra concepts question, since I'm bad at visualising maps.

Suppose I am given a basis (B), another basis (C), and a linear transformation ( \left[T\right]{B \to C}), i.e. a linear transformation that takes a vector expressed in basis (B) and expresses it in terms of basis (C). I'm being asked to find ( \left[Tx\right]{B} ) for a given vector (x).

[twin meteor]

RiverineDolphin

[restive basalt]

As I understand, in order to do so I need to take a few steps, right?

Like first, I need to express the vector (x) in terms of basis B, since that's what the linear transformation takes. So my first step is to calculate ( \left[x\right]_{B} )

[twin meteor]

RiverineDolphin

[restive basalt]

And afterwards, I can simply use matrix multiplication to find:

[
\left[T x \right]{C} = \left[T\right]{B \to C} \left[x\right]_{B}
]

[twin meteor]

RiverineDolphin

[restive basalt]

But afterwards, I need to use a change of basis matrix to then change the result of the transformation into basis B, like so:

[
\left[ T x \right]{B} = \left[ I \right]{C \to B} \left[ T x \right]_{C}
]

[twin meteor]

RiverineDolphin

[restive basalt]

Which is another matrix-vector multiplication.

Is this thought process correct?

So putting it all together:

Suppose I am given a basis (B), another basis (C), and a linear transformation ( \left[T\right]_{B \to C}), i.e. a linear transformation that takes a vector expressed in basis (B) and expresses it in terms of basis (C).

In order to find ( \left[Tx\right]_{B} ) for a given vector (x):

First express the vector (x) in terms of basis B, to obtain:

[ \left[x\right]_{B} ]

Next:

[
\left[T x \right]{C} = \left[T\right]{B \to C} \left[x\right]_{B}
]

And:

[
\left[ T x \right]{B} = \left[ I \right]{C \to B} \left[ T x \right]_{C}
]

Thus obtaining what was required.

[twin meteor]

RiverineDolphin

[restive basalt]

Is this correct? 😄

Hurrah!

And just to confirm my understanding, there are other ways to do it, right?

Like, if I wanted to, could I have applied a change of basis matrix directly on the linear transformation itself

for instance?

Ah, gotcha.

[heavy yoke]

yes, you can see how your final solution is [ [Tx]B = [I]{C\to B} ([T]_{B\to C} [x]B) ] but you could just have easily done [ [Tx]B = ([I]{C\to B} [T]{B\to C}) [x]_B = [T]_B [x]_B ]

[twin meteor]

cloud

[restive basalt]

Understood.

@heavy yoke So, here's a part of the concept that I struggle to understand.

When we're given a vector x = (1, 2, 3), I understand that this is implicitly given in terms of the standard basis E = {e_1, e_2, e_3}

And likewise, if I'm given a matrix that represents a linear transformation, it is both written in terms of the standard basis, and yields a vector in terms of the standard basis

So a regular matrix can be seen as [T]_{E \to E}

But what does it mean for a transformation to be expressed as [T]_{B \to B}?

If I think of it as a function in terms of computer science (a poor analogy, I know), I can understands it as a function that takes an vector as input in basis B, and returns a vector in a basis B

But it's still somewhat unclear to me.

Like, is [T]{B} the same as [T]{B \to B}?

[heavy yoke]

$[T]B$ is just shorthand for $[T]{B \to B}$ since if you only write one basis it's understood that you are using the same basis for the domain & cododomain (input and output)

[twin meteor]

cloud

[restive basalt]

Ah, understood.

And when I "look" at the matrix which represents the linear transformation ( [T]_B ), may I understand it's column vectors as the vectors of the linear transformation, expressed in terms of basis B?

[twin meteor]

RiverineDolphin

[heavy yoke]

the ith column is the image of the linear transformation applied to the ith basis vector of B, expressed in the basis B, yes

[restive basalt]

Like, if I wanted to, I could apply a change of basis matrix ( [I]{B \to E} ) upon the linear transformation ( T{B} ), to uncover how the transformation "looks," would I be correct?

[twin meteor]

RiverineDolphin

[restive basalt]

I apologise if such questions seem rather trite, but it does help me develop my intuition on how change-of-basis works.

[heavy yoke]

to fully convert you would do something like [ [T]B = [I]{E \to B} [T]E [I]{B \to E} ] and in reverse [ [T]E = [I]{B \to E} [T]B [I]{E \to B} ] this is made a bit easier when you realize that as matrices [ [ I]{E \to B} = [I]\inv{B \to E} ]

[restive basalt]

With respect to such an example, I can imagine a case where some standard linear transformation : like a rotation, for instance, is expressed in a strange basis B. So at a glance, I wouldn't be able to tell what sort of linear transformation it is. But upon applying a change of basis matrix upon the linear transformation, I can see that it is simply a rotation.

[twin meteor]

cloud

[restive basalt]

I am still pondering your example.

I think I am understanding it now.

[heavy yoke]

if you've covered diagonalization yet, that's why it looks the way it does

[restive basalt]

In order to "fully convert" a linear transformation given as ( [T]{B} ), which is really just ( [T]{B \to B} ), we must apply not one, but two change-of-basis matrices.

[twin meteor]

RiverineDolphin

[restive basalt]

One to transform the "input vector" into the right basis, and then another to transform the "output vector" into the right basis

Hence the little sandwhich that we made there.

[heavy yoke]

yes

[restive basalt]

Ah, gotcha.

Thank you. I really appreciate your help, it does solidfy my understanding quite substantially.

[heavy yoke]

this is the motivation for ``matrix similarity'' which is when two matrices represent the same linear transformations in different bases, i.e. matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ such that [ B = P\inv A P \iff A = P B P\inv ]
which we can understand that $P$ is just the change-of-basis between the relevant bases

[twin meteor]

cloud

[restive basalt]

Ah, I understand now.

[heavy yoke]

an importand example of which being diagonalization, which is where we find a basis of eigenvectors (i.e. a basis in which the linear transformation will have a diagonal matrix)

[restive basalt]

Which is useful, because diagonal matrices are easy to compute with.

Gotcha.

Hmmm.

Can bases be mutually incompatible with each other?

Like, if I have a vector space V, and some subspace U

Would it be possible to find a basis for V that contains no elements of subspace U?

I'm not sure if I am mis-understanding my intuition of span here.

I understand it is very possible to have a vector space like R^3, and then a set of vectors in R^3 which nonetheless span a smaller dimension, like R^2.

[heavy yoke]

certainly, no individual vector might be in the subspace, because we only require that a linear combination of them is in the subspace

[restive basalt]

But in such an example, the smaller set of vectors would not be a basis for R^3...

[restive basalt]

I'm not sure if I understand that sentence. Can you elaborate?

[heavy yoke]

for example the line y = x is a subspace of R^2, and a basis of R^2 is {(1,0), (0,1)}. neither (1,0) nor (0,1) is on the line y = x

[restive basalt]

Ahhhh

I think I understand.

So you're saying. Let us have a vector space V = R^2. Let us define U as a subspace of V, where U is the set of all vectors where y = x.

Now, we can see that there exists a basis for V, which is {(1, 0), (0,1)}

Which nonetheless does not contain any elements of the subspace U.

Is my understanding correct?

[heavy yoke]

yes

on the other hand, given a basis for the subspace U, it's always possible to extend it to a basis of V (i.e. there exists a basis of V containing the basis vectors of U)

[restive basalt]

Ah, I understand. And intuitively, that makes sense because

If our parent vector space V is R^3, i.e. a volume, and within it we have a subspace U, which is a plane

It is easy to add another vector to "stretch out" the plane so that it spans the volume.

Is that intuition correct?

[heavy yoke]

yes, it's always possible to pick a vector outside the plane which completes the basis of R^3

[restive basalt]

Alright, and to go back to the earlier example, but in R^3. Let us suppose we have the vector space V = R^3, which is a volume, and within it we have U = R^2, a plane.

Now, the plane must intersect the origin, because if it doesn't, it wouldn't be a vector space anymore, since all vector spaces must possess the zero-vector.

However, no matter how weirdly tilted the plane of U = R^2 is

[heavy yoke]

well strictly speaking R^2 is not a subpace of R^3. but something like {(x,y,0)} [the xy-plane] is a subspace of R^3 which has some correspondence to R^2

[restive basalt]

We may still find three vectors in R^3, starting at the origin, which nonetheless is not in subspace U.

[restive basalt]

Ah, what you are referring to is that in order for a plane to be a subspace, it must intersect the origin, as I was speaking of earlier, is that correct?

[heavy yoke]

moreso R^2 is (strictly speaking) the set of all vectors with two entries, whereas R^3 is the set of vectors with 3 entries. so no vector in R^3 can be a member of R^2

of course there are many 2D subspaces of R^3, which are the planes through the origin (and any 2D vector space is isomorphic to R^2),

[restive basalt]

Ah, I understand.

Thank you very much for your help tonight.

I'll go ponder on these in my own time ❤️

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[vestal lynx]

Can someone explain to me how to simplify this step by step? I can only think of DeMorgan. Thank you

[twin meteor]

Nicolas Matheisen

[dark kiln]

A → B means (~A or B)

you don't need anything else, just demorgan

[vestal lynx]

Ok, I’ll try to write down the implication differently first?

[dark kiln]

yes

[vestal lynx]

Ok let me try I need 5 min of thinking

[dark kiln]

maybe you need more complex thingas like distribution near the end, i haven't tried

but that's the start

[vestal lynx]

Hmm ok I am confused

I understand what

[twin meteor]

Nicolas Matheisen

[dark kiln]

ok, i solved it, there's no demorgan

[vestal lynx]

As I understood DeMorgan, one thinks that an A becomes a \neg A

[dark kiln]

this exercise has no de morgan

[vestal lynx]

This?

Is it possible to solve this completely once and then I have even more tasks here that I have to solve?

I hope I can understand that then

[dark kiln]

if you want the answer, ask wolfram alpha

i can't help

[vestal lynx]

@wanton elk if u have time can u help me

[scenic ravine]

I think frog meant the site

https://www.wolframalpha.com/

[vestal lynx]

:0

Ty

I think this help

Then I can try to understand the steps

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[scenic ravine]

Is this a test

[lost yarrow]

.close

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[stone oyster]

i dont understand

pls can someone help me 🙂

[pale perch]

whats the problemo

[stone oyster]

so for question a

does that mean it has to equal to 0??

[wary hull]

If f(2)=0, and f(4)=0, what does that imply about 2 and 4?

[wary hull]

Not at all.

[stone oyster]

can we do a first please

[stone oyster]

i thought 1 root = 0

[pale perch]

it would mean there is only one value that makes f=0

[wary hull]

That's what I'm doing.

Wait.

[stone oyster]

isn’t that b??

[wary hull]

Yeah. But still, your reasoning is incorrect.

[stone oyster]

can u explain pls

[wary hull]

No. If the quadratic has one root, it has a root of multiplicity 2. Meaning, it occurs twice in the factorization.

So, what can you deduce about a?

[stone oyster]

hmm]

a has a root multiplicity of 2??

[wary hull]

No, the quadratic does.

[stone oyster]

im not to sure

[wary hull]

Okay. Do you know what a perfect square trinomial is?

[stone oyster]

nope

[wary hull]

It is in the form (x-a)^2.

[stone oyster]

vertex form??

completed the square form???

[wary hull]

It has a root, x=a, with multiplicity 2.

[wary hull]

I guess, the vertex is (a, 0).

[stone oyster]

tp??

[wary hull]

Hm?

[stone oyster]

is a just -2

[wary hull]

Yep.

[stone oyster]

i just guessed that what’s the real reasoning

OHHH

wait im dumb

if it has 1 root only

and x-2=0

that means there’s only 1 answer

ok got it

lets do b

[wary hull]

Yes

[wary hull]

^

[stone oyster]

what does f(2)=f(4)=0 actually mean tho

[wary hull]

What does f(2)=0 mean to you?

[stone oyster]

no idea at all

[wary hull]

It means that 2 is a root of the equation.

What about f(4)=0?

[stone oyster]

4 is a root of the equation??

[wary hull]

Yep.

Therefore, a=?

[stone oyster]

wait

(2-2)(4+a)

is that what its saying??

[wary hull]

In a sense.

(2-2)(4+a)=0.

So a=?

[stone oyster]

4+a = 0

a = -4

[wary hull]

Yep.

[stone oyster]

the wording confuses me ngl

[wary hull]

Can you try part c)?

[stone oyster]

so (0-2)(0+a) = 10

is that what its saying??

a=-5??

[wary hull]

Yep.

Now try to use the reasoning that the other guy told you to do part d).

[stone oyster]

which guy

[wary hull]

The one who helped you in the other thread.

[stone oyster]

line of symmetry = -b/2a

x = -b/2a = 3

???

[wary hull]

Yes

Or, actually, you can just use the fact that the average of the roots is the x-coord of the turning point.

[stone oyster]

hm?

[wary hull]

So, $\frac{2-a}{2}=3$.

[twin meteor]

mathisfun

[wary hull]

It only works if the leading coefficient is 1.

[stone oyster]

what is that

a=-4

but what’s that formula

[wary hull]

^

[stone oyster]

what’s average of the roots??

[wary hull]

Like, the average of the 2 x-coordinates of the roots.

[stone oyster]

so 4 and 2??

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[tepid kiln]

wondering where they got the highlighted term from

[tepid kiln]

.

[scenic ravine]

You can prove that unequlity is true

*inequality

so they used it

[tepid kiln]

im so confused though they just brought it from no-where @scenic ravine

[scenic ravine]

jmm

not really out of nowhere

you have to sort of "reverse engineer" in such problems

[tepid kiln]

where do i even start in this case😭😭

@scenic ravine

[scenic ravine]

okay

so Let's consider $lim_{x \to \infty} \frac{ x+ \sqrt{x^2-1}}{x}$

[twin meteor]

What a wonderful world !

[scenic ravine]

Can you first off, simplify this for me

[tepid kiln]

its simplified

in the

solution

@scenic ravine

line 3

from where the solution begins

[scenic ravine]

From here?

[tepid kiln]

yes

[scenic ravine]

okay, do you get how the first step came to be

[tepid kiln]

yes but

im confused about the second line

in the solution

i dont really understand it

[scenic ravine]

This part?

[tepid kiln]

line 2

@scenic ravine

[scenic ravine]

1-1/x^2 ≥ 1 +1/x^2 -2/x

can you solve this inequality

[tepid kiln]

no no no its not that but

the highlighted bit

they just got it from thin air

i dont really know what they did to come up with the inequality in the first place

@scenic ravine

[scenic ravine]

They probably just thought this function is squeezed between these two functions as we got to infty

That's teh best I cna think of

[tepid kiln]

oh alright

thanks ig

[tepid kiln]

is anyone else able to help me with this

thanks in advance

.

[scenic ravine]

ping helpers

[tepid kiln]

<@&286206848099549185>

ts

pmo

https://tenor.com/view/me-fr-gif-3258695062595711768

.

.

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@tepid kiln Has your question been resolved?

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[orchid osprey]

particularly confused with question c and d

[rare shell]

Wat is 34×39

[halcyon ice]

!occupied

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@orchid osprey Has your question been resolved?

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[dreamy kindle]

Can someone help me with these

[paper depot]

which one first, 6 or 7?

...hold on, the idea i had for 6 isn't cooking like i thought it would...

[paper depot]

the phrasing of #6 is a bit unclear.

are we supposed to prove f' = 0 everywhere between alpha and beta?

or just that there exists xi between alpha and beta such that f'(xi) = 0?

also this xi is a bit sad

OP's disappeared.

@dreamy kindle are you here?

[dreamy kindle]

Yea

[paper depot]

ok right

can you clarify what was meant for #6?

is xi meant to be under a for all quantifier or a there exists quantifier?

it is indeed even simpler

rolle's theorem.

[dreamy kindle]

Rolles theorem

[paper depot]

which kind of spoils it

f(alpha) = f(beta) = 0 is quite obvious yes

@dreamy kindle do you understand what to do for 6?

[dreamy kindle]

Yeah thanks I got it now

[paper depot]

ok

and 7?

any progress so far?

[dreamy kindle]

What about the other one

Nah those were the only 2 I couldn't solve

[paper depot]

this one seems like you can just rawdog ∂^2u/∂x^2 and then use the fact that u is a symmetric function of x, y and z

to cut down the amount of calculation you have to do by a factor of three

[dreamy kindle]

Let me try

[paper depot]

maybe factorizing the stuff inside the log could help a bit

[dreamy kindle]

@paper depot

What do I do after this

[paper depot]

your numerator for ∂^2u/∂x^2 looks wrong

what was your ∂u/∂x?

[dreamy kindle]

Oh ye wait

(3x^2-3yz)/(x^3+y^3+z^3-3xyz)

I got it thanks for the help

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@dreamy kindle Has your question been resolved?

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[vast shale]

I do not know anything about the topic? May you suggest any videos or serve discussions about this?

[flat python]

linear regression?

https://youtu.be/wRZwrcPnmc4?si=Mk9vBoyNRlcwHSj5

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

Thank you

[vast shale]

I'm sorry. This isn't close to the photos.

[stray shard]

You might consider specifically "Correlation Analysis" as the term.

[vast shale]

Is this close to the photos? https://www.youtube.com/watch?v=11c9cs6WpJU

[rapid anvil]

yes

[vast shale]

ok

After watching that video, I still do not understand. May someone help me understand?

[stray shard]

Any specifics here you don't understand?

[vast shale]

I do not know the step by step process

I'm very organized so The photo is quite confusing for me

[vast shale]

<@&286206848099549185>

[vast shale]

<@&286206848099549185>

[stray shard]

Is it the numerator or denominator you have trouble understanding here?

The numerator contains the sums of error in relation to the mean.

The denominator contains "corrected sample standard deviation" for both x and y.

This is called "sampled correlation coefficient". Does this term help?

The formula looks just like that.

!status

[vocal sleet]

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

[golden ice]

@everyone

Need help

[vast shale]

Like I do not know what is the first step

whether it be the table first or the formula

[stray shard]

Usually you need the samples first, therefore the table.

The formula comes in as a tool/function for you to input the calculations towards r.

Everything to the left of the board is the theory, describing the coefficient.

[vast shale]

The x table is the observations here from what I recall. What is the y?

[vocal sleet]

@vast shale Has your question been resolved?

[stray shard]

It would be the corresponding value to the x.

Even if the table is not sorted, there are still data points in the sample.

[vocal sleet]

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[stone stream]

I need the solution for this pls(my homework due tomorrow)

[silk osprey]

x = 2, y = 5

[stone stream]

i know

but i need the solution

[silk osprey]

then what is confusing

what is confusing

[stone stream]

ohhh now ok i know how to

sqrt 4(2) + 15/2 = 2 + 15/2 = 9.5

[silk osprey]

nope

not sqrt

[stone stream]

oh my bad cube root

[silk osprey]

but it seems like you knew that

[stone stream]

i just dont remember

[civic otter]

There's nothing to remember actually

[stone stream]

my brain isnt braining at that time

[silk osprey]

have a donut

[vast shale]

too sugary. have just nuts instead

[silk osprey]

have a nut

[night mortar]

Have a do

[vocal sleet]

@stone stream Has your question been resolved?

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[distant perch]

A careless mathematician didn’t wash his hands properly and ended up catching a cold. To recover, he had to take all 15 pills from a blister pack of cold medicine. The doctor's recommendation is that he should take either 2 or 3 pills each day. Assuming the mathematician decided to take his health seriously and follow the doctor’s advice precisely, in how many different ways can he consume the 15 pills?
(A) 21 (B) 28 (C) 36 (D) 45 (E) 55

[distant perch]

I already know the answer because I tried to brute force it, counting all the possible ways. But there’s definitely an easier and faster way to solve this — I just don’t know what it is.

I found 28 possible ways

[dark kiln]

there's two cases, depending on the last day

e.g. 3 pills on the last day, there's as many ways as you would count in 12 days pills without restricition

or the last day had two, then it's 13 days

so 2 pills is 1 and 3 is 1 and 4 is 1

[distant perch]

<@&268886789983436800>

If the last day was 3 pills, so in the other days he had 12 pills, so I have to find how many ways he could take these 12 pills

[dark kiln]

yeah

[distant perch]

Alright

[dark kiln]

then 5 pills is 2: maybe it ends in 3, then there's one way, or it ends in 2, then there's one way

then 6 pills is something, and eventually you get to 15

[distant perch]

I didn't understand this

[dark kiln]

that's the base

you can eat 3 pills if you eat 3 pills

there's no second way

4 can only be 2+2

[distant perch]

Alright

So I have to count how many ways I can get 2, 3, 4, 5, 6 ... ways

Right?

[dark kiln]

yeas

[distant perch]

Great

Thank you very much

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[digital glen]

How do we do the last part?

[digital glen]

Assume the dice is fair and six sided

[zinc quail]

two events are independent if P(A and B) = P(A) * P(B)

therefore determine P(A and B), P(A) and P(B)

[digital glen]

Is the probability for both equal to 1/6 in this case?

P(A) × P(B) and P(A and B)

[wary hull]

For getting both at the same time, or for getting either?

[digital glen]

Both at the same time

[zinc quail]

ys both RHS and LHS are 1/6

[wary hull]

Then you are right

[digital glen]

Thanks

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[peak shore]

Hello

[peak shore]

Hello

How to get S(a)?

I get stuck

@vocal sleet

[hexed crown]

Can you find the equation of the tangent at x = a?

[vast shale]

integrate it mate

do you know how to integrate?

[wary hull]

This is confusing.

Why did you equate the derivative with the function?

[vocal sleet]

@peak shore Has your question been resolved?

[vocal sleet]

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[white horizon]

Prove 1/x + 1/y = 1/z

[wary hull]

Finally.

Ok, what do you know about similar triangles?

[white horizon]

Ok nvm I just accidentally forgot to reciprocate both sides

My bad

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[analog marlin]

Logic problem, trying to make a latch that only updates when the bit labled clock switches from 1->0, and uses a D latch as normal, outputting the stored result when clock is 0

[analog marlin]

to use the same tool, go to https://nandgame.com and then go to the settings tab, click import, input the attatched text file, click import, back, and you have the level with my current solution

please ping me if responding

[vocal sleet]

@analog marlin Has your question been resolved?

[analog marlin]

<@&286206848099549185>

[vocal sleet]

@analog marlin Has your question been resolved?

[vocal sleet]

@analog marlin Has your question been resolved?

[vocal sleet]

@analog marlin Has your question been resolved?

[analog marlin]

the error might be in the program, if this seems correct please ignore

[marsh jungle]

might be more of a CS question, you can also go to #old-network and join the computer science server. im sure more people will be able to help you with your problem there.

[analog marlin]

that's awesome, didn't know we had one of those

ty

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[marsh jungle]

of course!

[vocal sleet]

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[finite hatch]

would it be dy/dx = 2x + 4

i dont get it when it says f(x) = x^2 + 4x

is it like d/dx = 2x + 4

[river kettle]

bum chicken

[river kettle]

wdym?

[finite hatch]

olh nvm

i gya thtis

[river kettle]

hello bum chicken

blud is finally doing calc!!

vectors

[finite hatch]

I DID IT LAST YTEAR

[river kettle]

whats ur syllabus

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[tender gale]

How do I put sec into a calculator

[hard atlas]

depends on the calculator

otherwise just do 1/cos

[tender gale]

Cos-1?

[hard atlas]

1/cos

fraction

[paper depot]

1 divided by cosine

[tender gale]

Isn’t that cos to the -1

[paper depot]

no

[hard atlas]

no

[tender gale]

Bc it goes into a fraction 1/cos

[paper depot]

the notation is confusing but cos^-1 does not mean the same thing as 1/cos

[tender gale]

Like x^-1 is 1/x

[paper depot]

cos^-1 means the inverse cosine, ie "cosine of what angle equals this number?"

[hard atlas]

welcome to the clusterfuck that is trig notation

[paper depot]

yeah, but with functions an exponent of -1 conventionally refers to the functional inverse

[tender gale]

Yay

The first of many I assume

Ty guys

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[tender gale]

.reopen

[vocal sleet]

✅

[tender gale]

Wait so do I still do hyp/adj

Or is it sohcahtoa

[rugged orchid]

Cos = adj/hyp

So if you do cos^-1(adj/hyp) you get the angle in that triangle

You get the angle that’s adjacent to the side that you’ve put in

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[vast shale]

why couldn't this be:

"doing it on both sides of the equation"

[paper depot]

you... just multiplied some stuff ONLY on the RHS?

[vast shale]

?

[paper depot]

you multiplied by (x+3)(x+4) but only on the right

without also doing it on the left as you should have

[vast shale]

i did do that

[paper depot]

nope

[vast shale]

5 is being multiplied by x+4

and 3x by x + 3

[paper depot]

the denominator would have vanished

but it didn't

[vast shale]

u saying x=4 shouldcut out

[paper depot]

no

[vast shale]

then

[paper depot]

easier to show what i mean than say it in words

[vast shale]

jsut write the better version of it

[paper depot]

one moment please

[vast shale]

ok

np

[paper depot]

i've shown the thing i multiply both sides by in red

and the cancellation that follows, in green

[vast shale]

the question

[paper depot]

ok sure

[vast shale]

second step

[paper depot]

i could have guessed you started with this

[paper depot]

and then got to this

[vast shale]

ok

[paper depot]

do you understand this? yes/no

[vast shale]

what

i do

[paper depot]

ok

[vast shale]

, do you mean this?

the yes

?

[vast shale]

?

.close

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[vocal sleet]

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[vocal pagoda]

$\Omega \subset \mathbb{R}^n$ is an open set, $A, B$ differentiable functions, $A : \Omega \rightarrow M_{mn}(\mathbb{R}), B : \Omega \rightarrow M_{nk}(\mathbb{R}), u : \Omega \rightarrow \mathbb{R}^n$. $\forall x \in \Omega, V \in T_x\mathbb{R}^n$, we define $V(A), V(u)$ as the differential of A and u along the vector field V.
I try to see if $V(AB)=V(A)B(x) +A(x)V(B)$.

[twin meteor]

Wild123

[vocal pagoda]

Hi

Firstly, I'd like to see if I am correct in writing that for $x=(x_1,...x_n), V=\Sigma_{i=1}^n V_i\partial_i$ and $A=(a_{ij}(x)){i, j= \overline{1,n}}$, then $V(A)=\Sigma{i=1}^n V_i(\partial_iA)$

[twin meteor]

Wild123

[vocal pagoda]

If the above is correct, When proving V(AB) is what I wrote above, given it's a matrix, can I just look at an arbitrary component and prove the equality stands?

Here's what I came up with:
looking at element jk (line j, column k), note $A_j$ the line j of matrix A, $B_k$ the column k of B. We have:

$\Sigma_{i=1}^nV_i (\partial_i \langle A_j, B_k\rangle )=\Sigma_{i=1}^n \langle V_i(\partial_iA_j), B_k(x) \rangle +\Sigma_{i=1}^n \langle A_j(x),V_i(\partial_iB_k) \rangle$

[twin meteor]

Wild123

[vocal sleet]

@vocal pagoda Has your question been resolved?

[vocal sleet]

@vocal pagoda Has your question been resolved?

[vocal sleet]

@vocal pagoda Has your question been resolved?

[vocal pagoda]

!helpers

[vocal sleet]

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