help-26

[topaz sinew]

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[toxic aspen]

$e^{-|x|}$

[thorny flame]

Merineth

[toxic aspen]

How do i find the derivitive of something like this?

I'm aware that i can't use e^kx = ke^kx

[ruby tree]

Find the derivative when x<0, then when x>0

[toxic aspen]

so basically i put in 1 and -1 into x and see what spits out?

[ruby tree]

No, you rewrite the thing based on whether x is positive or negative

[toxic aspen]

Uhm

so you mean

$x > 0
x < 0$

[wooden osprey]

is the derivative of |x| sgn(x)?

[thorny flame]

Merineth

[toxic aspen]

I'm not sure what you mean by that

[ruby tree]

For some definition of sgn(x), yes

@toxic aspen what is |x| if x > 0?

[wooden osprey]

sgn(x) also includes x=0 though, so i assume its incorrect

[toxic aspen]

negative

[wooden osprey]

but its $\frac{x}{|x|}$, which essentially is sgn without case of x=0

[thorny flame]

FungusDesu

[sullen sentinel]

|x|, when x > 0 is what?

[toxic aspen]

If x > 0 then |x| negative since we have -|x|

[ruby tree]

You're just going to confuse them

[wooden osprey]

right, it was out of my curiosity anyway so yeah, thanks for your answer though

[ruby tree]

I never asked about -|x|, I asked about |x| only

[toxic aspen]

Positive then

[ruby tree]

I don't just want to know whether it's positive or negative, I want to know what it equals

What's the definition of |x|?

[toxic aspen]

I don't remember at all, it's been several years since i did absolute

[sullen sentinel]

@toxic aspen what is the absolute value of a positive number?

[toxic aspen]

a positive number

same with negative

[sullen sentinel]

so |x| = x, when x > 0 right

[toxic aspen]

even if x > 0 and x < 0 it's positve

[ruby tree]

Let's take a few examples I guess

|6| = ?

[toxic aspen]

6

[ruby tree]

|-8| = ?

[toxic aspen]

8

[ruby tree]

Right so it's not just any positive number

|-x| = ?

[toxic aspen]

x

[ruby tree]

Only if x is positive

[toxic aspen]

abs(-8) is 8

|anything| = positive

[ruby tree]

|-(-7)| = 7, even though x is -7

[toxic aspen]

That's the same as doing |-7|

[sullen sentinel]

take x = -7 right, put that into |-x| = ?

[ruby tree]

Well yeah kinda, but |-x| when x = -7 is not x, it's -x

[toxic aspen]

7

[sullen sentinel]

but is x=7?

no right

[toxic aspen]

oh

[sullen sentinel]

boom

[toxic aspen]

That makes sense, x is -7 however - is within the |-(x)|

[sullen sentinel]

yes

[ruby tree]

Anyway, the definition of |x| is piecewise:

• x >= 0 : |x| = x
• x < 0 : |x| = -x

[sullen sentinel]

so using that we know that at 0 the behavior of the absolute value function changes

[toxic aspen]

But doesn't -|x| mean that even if x is positive or negative it's always going to result in a negative value?

[ruby tree]

Yes but I never asked about -|x|

[toxic aspen]

Oki, just making sure

[ruby tree]

Do you agree with this?

[toxic aspen]

Yes

[ruby tree]

So can you rewrite e^{-|x|} when x > 0?

[toxic aspen]

$e^{x}$

[thorny flame]

Merineth

[toxic aspen]

Nvm it should be like this?

[ruby tree]

No

[sullen sentinel]

no be careful

if u input a positive x value into e^(-|x|)

[toxic aspen]

Aahh

[ruby tree]

You just said -|x| is a negative value, and we just took x as a positive value

[toxic aspen]

I think i get what you mean

if i input a positive value on x i get a positive value out. But since we have -|x| it results in being negative?

so

$e^{-x}$

[thorny flame]

Merineth

[sullen sentinel]

yes

when x >= 0

[ruby tree]

Right, now can you rewrite e^{-|x|} when x < 0?

[toxic aspen]

$e^{x}$

[thorny flame]

Merineth

[toxic aspen]

since - - = +

[ruby tree]

Yes, so take the derivative of each and assemble them back into one piecewise function

[toxic aspen]

f(x) = e^x
f'(x) = e^x

g(x) = e^(-x)
g'(x) = -e^(-x)

[sullen sentinel]

u should write it in this format

but for the derivative

[toxic aspen]

aah okay once sec

x >= 0 : e^x
x < 0 : -e^-x

[ruby tree]

No

[sullen sentinel]

f(x) = e^(-|x|)

f'(x) = ?, when x >= 0
f'(x) = ?, when x < 0

fill in the ?, with what you have figured out

[toxic aspen]

f(x) = e^(-|x|)

f'(x) = -e^(-x), when x > 0
f'(x) = e^x, when x < 0
f'(x) = 1 when x = 0

[ruby tree]

Still no

You found e^x when x is positive or negative?

[toxic aspen]

shit

[sullen sentinel]

lol

now it's right

[ruby tree]

Not quite

[toxic aspen]

= should be

?

[ruby tree]

We only took x<0 and x>0, never x=0

Yeah

[sullen sentinel]

oh true

[ruby tree]

Can you imagine what it should be for x=0?

[toxic aspen]

f'(x) = 1 when x = 0

Would be my guess

since |0| is 0 and e^0 is 1

[ruby tree]

That's not how it works

[sullen sentinel]

then what is f'(x), when x = 0

[toxic aspen]

0?

[ruby tree]

That's also not how it works...

You have two pieces for your function, e^x and -e^(-x)

[toxic aspen]

Yes

[ruby tree]

They almost meet at x=0

What would both equal if you plug in x=0?

[toxic aspen]

1 and - 1

[ruby tree]

Right

So there's a jump here

The derivative is discontinuous

More than that, it's undefined at x=0

[toxic aspen]

What do you mean by jump?

[sullen sentinel]

u right

[toxic aspen]

Do we have a bot for a graph?

So i can visualize

[toxic aspen]

Just trying to find a pattern so i can remember.
So if i understand you right, if there is a jump between x = 0 that means it's discontinuous ?

[sullen sentinel]

yes because f'(0) doesn't exist

[ruby tree]

If the limit does not exist at a point, then the function is discontinuous at that point

[toxic aspen]

That makes sense

[ruby tree]

Here you can see the limit from the left and the limit from the right are different

So the actual limit doesn't exist

However it's more than that, like I said, it's undefined

[toxic aspen]

So it's actually not possible to find the derivitive for e^-|x|

[ruby tree]

A discontinuous function could still be defined at the discontinuity (have a value)

[toxic aspen]

Since it's undefined and discontinous

[ruby tree]

Well, not at all points

[toxic aspen]

Because the original question that i had to solve was

[ruby tree]

You still found it for all of R except 0

[toxic aspen]

$(6+4x)e^{-|x|}$

[thorny flame]

Merineth

[toxic aspen]

I was supposed to find the roots, asymptot and max/min points

[ruby tree]

You can still do that

[toxic aspen]

I understand that if i do the derivitive of the function and set it = 0 i get when the graph has 0 incline/decline

However my problem came when i tried to use product rule

and had to find the derivitive of e^-|x|

$4 * e^{-|x|} + (6 * 4x) * (e^{-|x|})'$

[ruby tree]

Right, well the derivative of this is also discontinuous at 0, in pretty much the same way, but that shouldn't stop you from finding the extrema of the function

[sullen sentinel]

u have to find the derivative at x>0 and x<0

[thorny flame]

Merineth

[toxic aspen]

There we go

[sullen sentinel]

this is a piecewise function

[toxic aspen]

peacewise?

[sullen sentinel]

it will have 2 derivatives the same way

[ruby tree]

piecewise

[sullen sentinel]

mb

[ruby tree]

"in multiple pieces"

[toxic aspen]

I've never heard ot piecewise before

[sullen sentinel]

it's when the fucntion is different at different ranges of x

[toxic aspen]

I'm fairly certain we haven't gone through that either

in this case, -1 and 1?

sorry x = 0

[sullen sentinel]

well using what u found for how e^-|x| works you can do the same for this

[ruby tree]

Being piecewise isn't a property of a function, it's a property of how you define (write down) the function

[toxic aspen]

So how do i continue from here considering e^-|x| is undefined

[sullen sentinel]

the derivative of that function is undefined at x = 0

[toxic aspen]

But that means i'll have two derivatives

[sullen sentinel]

it still has all of these

[ruby tree]

It only matters when you want to look a x=0

Everywhere else you can do whatever you usually do

[sullen sentinel]

this is how this function looks, you can still use the derivative to help you find the critical points

[toxic aspen]

I kind of understand however

I'm not sure how to proceed

[ruby tree]

Look at the limits of the derivative on both sides of x=0

[toxic aspen]

If i want to use the product rule i require to do the derivative of e^-|x|

[ruby tree]

No you don't, just split the cases x<0 and x>0 just like before

[toxic aspen]

ooooh

[ruby tree]

The derivative will be in two pieces, but again that doesn't matter

[toxic aspen]

so i rewrite the entire function but instead of doing e^-|x| i do

if x > 0, -e^-x
if x < 0, e^x

allowing me to find the f'(x) = 0

Is that right?

$4 * e^{-|x|} + (6 * 4x) * (e^x)' or 4 * e^{-|x|} + (6 * 4x) * (-e^{-x})'$

[thorny flame]

Merineth

[toxic aspen]

Am i thinking correct?

[ruby tree]

I guess but you're not rigorous

(e^-|x|)' = {x<0 : e^x, x>0 : -e^-x}
In your expression, on the left, (e^x) is already the derivative, drop the '

Same for (-e^-x) on the right

Crap

Using the same notation:

[(6+4x)e^{-|x|}]' = {
x<0 : 4 e^{-|x|} + (6 * 4x) (e^x),
x>0 : 4 e^{-|x|} + (6 * 4x) (-e^{-x})
}

Well my LaTeX skills aren't good enough

Anyway, to find the roots, you don't need the derivative

To find the asymptotes, I guess you can but I don't think you need

[toxic aspen]

It's only for max/min i need it

[ruby tree]

To find the extrema, you just look at each interval separately

[toxic aspen]

i already found the root to be -1,5

[ruby tree]

And then you check what happens at x=0

-1.5? As in -3/2?

[toxic aspen]

Yes

-3/2

[ruby tree]

Ok

[toxic aspen]

I probably should answer exact

[ruby tree]

I mean -1.5 is exact, but only because 10 is divisible by 2

[ruby tree]

So, to do this, you look at the limits of the derivative on both sides of x=0

If it's positive on the left and negative on the right, then you have a maximum; if it's negative on the left and positive on the right, then you have a minimum

[toxic aspen]

Hmm

i'm still trying to find min/max

i'm not sure if i'm doing correctly tbh

[ruby tree]

Show

[toxic aspen]

,rotate

[toxic aspen]

The last part is probably wrong i'd imagine

[ruby tree]

Definitely

[toxic aspen]

But it's right up until that point?

[ruby tree]

Yeah

Remember you assumed x>0

Also this part is wrong, even if I get what you meant

[toxic aspen]

Yeah that's just the derivative

so i don't forget

Hmm this is getting tricky

i assumed x > 0

[ruby tree]

Ok but don't write down the equal sign then

[toxic aspen]

however i can't be guessing to the answer by putting in positive x values

[ruby tree]

You have 4e^{-|x|} - 6e^{-x} - 4xe^{-x}

[toxic aspen]

I'd like to break out 4e^-x

from 6e^{-x} - 4xe^{-x}

[ruby tree]

You can't really do that

[toxic aspen]

$4e^{-x} -6e^{-x} -4xe^{-x} = 0$

[ruby tree]

Again, you assumed x>0

Look at each term

[toxic aspen]

does that mean i can assume -|x| is -x

[ruby tree]

Yeah of course

[thorny flame]

Merineth

[toxic aspen]

I can break out 4e^-x

[ruby tree]

Don't

Just factorize the e^{-x} part

It's useless to bring the 4 with you

[toxic aspen]

Personally i would bring the 4 with me since i was taught to always break out the largest

[ruby tree]

But 4 doesn't divide 6 very well

[toxic aspen]

$e^{-x}(4 -6 -4x) = 0$

[thorny flame]

Merineth

[ruby tree]

Yes

[toxic aspen]

x = -1/2

[ruby tree]

Right

[toxic aspen]

So that was when x > 0

now i have to do the same when x < 0 also

[ruby tree]

Stop

[toxic aspen]

Ok

[ruby tree]

Think about the solution you just found

[toxic aspen]

The solution means that at x = -1/2 we have either a max or a min

[ruby tree]

It would in another context

What was your assumption to get to that solution?

[toxic aspen]

oh wtf you are right

how can it be -1/2

if we assumed x is positive

[ruby tree]

It can't, so there is no solution

Makes sense?

[toxic aspen]

That makes sense

yeah for sure

[ruby tree]

Now do x<0

[toxic aspen]

i got x to be -5/2

[ruby tree]

Correct

[toxic aspen]

And since we assumed x < 0

we can confirm that it is actually a min or max there

[ruby tree]

Yes

[toxic aspen]

Holy shit it makes sense now

Actual epiphany

[ruby tree]

Is it a min or a max?

[toxic aspen]

Well i can see on the graph that it is a min however, i can prove that by using the derivative of the function when x < 0 and put values below and over -5/2

[ruby tree]

Right

[toxic aspen]

It doesn't need to be tho however

[ruby tree]

Or just use the function itself, you know it's continuous and there is no other extremum on that interval

[toxic aspen]

Yes!

can't i just assume that it's a minimum since e^x is a graph with an incline

Or actually i'll just do a table to prove it so i don't lose points

[ruby tree]

That's not a sufficient condition

Here is f(x)=(1-3x)e^x

You can see it's a max

[toxic aspen]

aah

yeah you are right

i'll just stick with doing tables

Seems safer that way

[ruby tree]

[toxic aspen]

Tysm nel

[ruby tree]

Don't forget to do x=0

[toxic aspen]

Well x=0 is undefined?

[ruby tree]

Not for the function

[toxic aspen]

|x| doesn't have a min/max

only when

x > 0
x < 0

[ruby tree]

It does, at 0

|0| = 0

Only the derivative is undefined

[toxic aspen]

do you mean to find the roots?

x = 0?

[ruby tree]

No I mean |x| has a minimum at x=0

It happens to equal 0, but the point is that it's still an extremum

[toxic aspen]

Wtf i was told |x| doesn't have extreme

[ruby tree]

It doesn't have a maximum but it clearly has a minimum

[toxic aspen]

So in what function do i put in x = 0

not the original one since that results in me getting the roots

which is -3/2

[ruby tree]

Remember the definition of |x|?

[toxic aspen]

Yes that it's only defined when x > 0 or x < 0

[ruby tree]

I wrote x >= 0, not x > 0

[toxic aspen]

[ruby tree]

The derivative of |x| is undefined at 0

But |x| is perfectly well defined at 0, its value here is just 0

This is your function along with its derivative:

The function is defined at 0, it equals (6 + 4*0)e^0 = 6

[toxic aspen]

So you want me to substitute my |x| with 0

[ruby tree]

Not really, you can't do that with the derivative since it's undefined at 0. What you can do is look at the limits of the derivative on both sides of x=0

[toxic aspen]

ooooh

UH

Ahhh

Yeah i get it now

the original function when x = 0 is 6

I understand now

But we don't really need to touch the derivitive at all right?

[ruby tree]

You do, you need to know if the function increases then decreases or decreases then increases, or neither

That's pretty much the definition of a local extremum

[toxic aspen]

It's getting a bit complex

i don't really get how we get that 6 is a max

if we can't use derivative

since it's only defined when x > 0 and x < 0

[ruby tree]

I said you can use limits

[toxic aspen]

oh

when x -> 0

lim x -> 0 of the function

[ruby tree]

You see on the graph of the derivative that it goes to 10 and -2, right?

At x=0

[toxic aspen]

Yes

[ruby tree]

10 is the limit from the left

-2, from the right

You don't need actual values, but you need to know if the derivative is positive or negative

Here it's positive before 0, and negative after 0, right?

[toxic aspen]

lim (6+4x)e^-|x|
x->0

Isn't this what i'm trying to achieve?

[ruby tree]

No that's just the function, you don't need the limit, you can directly plug in x=0

[toxic aspen]

ooooooh

I get what you mean

i just need to know if it's increasing/decreasing when x>0 and x<0 and from where in y it start

and to get where in y it start i plug in x = 0

[ruby tree]

The y part is only if you need the value of the max

[toxic aspen]

oh

true

[ruby tree]

To know if it's a min or a max, you need to know if it's increasing or decreasing before and after 0

[toxic aspen]

But i think i need y since i need to give the location of where it is and if it's a min/max

But yeah i get what you mean now

This assignment was extremely hard, this can't be what is coming on the upcoming test, hopefully..

gonna have to ask tomorrow honestly

[ruby tree]

It's not that hard, it's just new to you

[toxic aspen]

I probably need more practice i guess

[ruby tree]

You lack a bit of rigor but I think you understand the ideas pretty well

[toxic aspen]

I just despise when i don't understand really

[ruby tree]

Get used to it

[toxic aspen]

Yeah i'm trying to cope :P

[ruby tree]

So, just making sure, is it a min or a max and why?

[toxic aspen]

it's a max since when x > 0 it decreases and when x < 0 it increases

giving us a nice bump

[ruby tree]

And it decreases or increases because...?

[toxic aspen]

Because we figured out the derivate when x > 0 and x < 0.
And if we plug in values on x into the deriviated functions we get it going from increasing to decreasing when x = 0

I prove it with a table such as this

(but not those values)

[ruby tree]

Ok I guess that's fine if you have a full table

The rigorous way would be to say that the limit of the derivative is positive from the left and negative from the right, hence the function is increasing on the left and decreasing on the right

[toxic aspen]

That sounds more elegant

[ruby tree]

A more rigorous way I suppose

[toxic aspen]

[ruby tree]

You have to be careful when you say things like "it increases when x < 0"

That's just not true, it decreases in (-inf, -3/2]

[toxic aspen]

yeah

I probably should get used to doing that

[ruby tree]

[toxic aspen]

It'd time to make dinner :3

thanks again nel

have a good evening

.close

[topaz sinew]

Channel closed

Closed by @toxic aspen

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>.
• 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.

[obtuse trench]

Anyone who can help how to differentiate this?

T is tranpose, and w and x is vectors

[knotty ledge]

With respect to what

Oh W?

[topaz sinew]

@obtuse trench Has your question been resolved?

[obtuse trench]

Exactly^

[sweet shard]

Just find one partial at a time

[obtuse trench]

Is this right for the first term?

This is a bit more clear

[obtuse trench]

Is this true too?

Like when i differentiate with respect to w does $w^T$ equal 1 and the rest is constants so they stay?

[thorny flame]

SimonWin

[obtuse trench]

$\frac{\partial}{\partial \mathbf{w}}(-2\mathbf{w}^T\mathbf{x_n}t_n)&= -2\mathbf{x_n}t_n$

[sweet shard]

Is this your work?

[thorny flame]

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

[obtuse trench]

Yes of course?

[obtuse trench]

Wdym if it's my work?

Is it that wrong? 😂

[sweet shard]

Did you write that or is it the solution?

[obtuse trench]

Uhm, wolfram said (f(x))^2 was this?

[obtuse trench]

So, I assumed I could use that here^

[sweet shard]

Uh can you just answer my question?

[obtuse trench]

It's the solution to that one term?

I think? I am not sure, which is why i am asking?

[sweet shard]

The solution from your teacher

[obtuse trench]

Oh

[sweet shard]

Or is it your work that you wrote?

[obtuse trench]

No no, I don't have a correct solution

Nothing in there is from my teacher

[sweet shard]

This is true for single variable yes

[sweet shard]

You have a multivariate input so you need to calculate the partials

[obtuse trench]

Okay, so how would I partially derive (w^T*x_n)^2

[sweet shard]

.

dL/dw_1 = ?

[sweet shard]

Write the right side of this sum in terms of the individual components of w

[obtuse trench]

Individual components?

That confuses me even more haha?

This is the original equation I am given

And I have to derive the optimal least squares parameter values ˆw for the total
training loss

Which is why I tried to find the partial derivatives and then isolate w

[sweet shard]

w = (w1,w2,...,w_m)

[sweet shard]

Did you take multivariate calculus?

[obtuse trench]

Yeah I think there were some of it there tbh so I can go watch a video and recall it

[sweet shard]

w^T x is a dot product

Write out that sum explicitly in terms of the components of w and x

[obtuse trench]

And you're sure this is needed?

[sweet shard]

Needed for you to understand yes

[obtuse trench]

Because the guide I'm following basically says to partially derive them and set them to zero and isolate "w"

Okay

[sweet shard]

To differentiate

Partially differentiate is exactly what I'm telling you

[sweet shard]

.

[obtuse trench]

Yes?

That is what I am trying to do as well, but now I am trying to "Write out that sum explicitly in terms of the components of w and x" xD

[sweet shard]

So where's your partial derivative?

If you can find your partial derivative without the definition of dot product, I'd like to see it

[obtuse trench]

Well I thought it'd be something like this

[sweet shard]

Those aren't dot products

[obtuse trench]

w1x1+w2x2+w3x3

But is this what you mean?

[sweet shard]

Yes

[obtuse trench]

Ah, okay

[obtuse trench]

Aah, okay I see

[sweet shard]

That's the dot product for m=3

[obtuse trench]

Yes

w1x1+w2x2+w3x3+...+w_mx_m

So, here

We are getting the dot product and squaring that number?

[sweet shard]

Doing the entire calculation for m=3 should be enough for you to guess a general formula

[sweet shard]

yes

[obtuse trench]

partial /partial w (w1x1+w2x2+w3x3) = x_1+x_2+x_3

Wouldn't all w_m just be 1?

When you partially derive them

If that it true what I said above

you'd get sum x_m from 1 to m

[sweet shard]

No

You're trying to do too many steps at a time

[obtuse trench]

$\frac{\partial}{\partial w} (w_1x_1+w_2x_2+w_3x_3)$

[thorny flame]

SimonWin

[obtuse trench]

This is what ur asking me, right?

[sweet shard]

No

[sweet shard]

.

Also you're missing the square

w is a vector. You have to treat it like one

[obtuse trench]

Oh, only differentiate with respect to w_1

Not w2 and w_3 etc..

[obtuse trench]

Yeah, i'll do that one after xD

$\frac{\partial}{\partial w_1} (w_1x_1+w_2x_2+w_3x_3)=x_1+0 +0$

If I have understood you correctly from the dL/dw_1

[sweet shard]

No

[obtuse trench]

I know dL is the whole thing but I am just wondering this

Fukk

[sweet shard]

The first term is correct

The second and third aren't

[obtuse trench]

Oh, they're just zero

[thorny flame]

SimonWin

[obtuse trench]

So like this?

[sweet shard]

Right

[obtuse trench]

$\frac{\partial}{\partial w_1} (w_1x_1+w_2x_2+w_3x_3)^2$

[thorny flame]

SimonWin

[obtuse trench]

$$(w_1x_1+w_2x_2+w_3x_3)^2$$
So, can I just say that the whole thing inside $a=w_1x_1+w_2x_2+w_3x_3$ so $(a^2)'=2a$ and then $2(x_1)$ would be the whole thing?

Or do I have to inside the chain rule?

[thorny flame]

SimonWin

[sweet shard]

No

You gave two definitions of a

[obtuse trench]

Chain rule?

[obtuse trench]

Where?

I say a=w_1x2...

So I substitute a in the original formula (w1x1...)^2 with a^2

But do I have to use chain rule?

[sweet shard]

At the end you say 2a = 2x1

[obtuse trench]

Oh, ahha..

Ur right.

2*(w1x1+w2x2...)

then?

[sweet shard]

No

You forgot chain rule

[obtuse trench]

🤦‍♂️ Sorry man

Yes, I kept asking chain rule man

😂

$2*(w1x1+w2x2...)*x_1$

[thorny flame]

SimonWin

[obtuse trench]

So like this?

[sweet shard]

Yes that's right

[sweet shard]

To this partial

[obtuse trench]

And so, the answer

would be

Maybe?

Oh no!

$2\cdot (wx_n) \cdot x_1$

Maybe?

[thorny flame]

SimonWin

[obtuse trench]

Since $w*x_n$ would be equivalent to the (w1x1+w2x2...) part or dot product?

[thorny flame]

SimonWin

[sweet shard]

You're missing the transpose notation

[obtuse trench]

w^T

$2\cdot (w^Tx_n) \cdot x_1$

[thorny flame]

SimonWin

[sweet shard]

Right

[obtuse trench]

Right

Oh holy shit

I almost arrived at the same thing before

But now I actuallt understand whats going on

Wait

@sweet shard ?

Before I had x_n

$2\cdot (w^Tx_n) \cdot x_n$

[thorny flame]

SimonWin

[obtuse trench]

Now I have $2\cdot (w^Tx_n) \cdot x_1$

[thorny flame]

SimonWin

[obtuse trench]

But we're both 100% sure it's x_1 ?

The only thing I realised is that x_n is a number right? not a vector?

Isn't it?

w is a vector, but x_n is a specific data point

So, that wouldn't be the dot product?

[sweet shard]

Sure what's x1?

[obtuse trench]

A data point I thought?

Like W would represents like the loss function line and x would be like all the data points?

So, x_n would be a specific one

[sweet shard]

What's the dimensions of x1?

[obtuse trench]

A number?

So 1x1

I'm pretty sure it's just a number, so x_n is a number, meaning that (f(x))^2 could be valid right? Since now it's just f(W)=w^Tx_n where x_n is a number

Then i get this, but doing the other method i get 2(w^Tx_n)x_1

so almost the same just x_1 instead of x_n

[sweet shard]

You can't take dot products with vectors and numbers

[obtuse trench]

Yeah?

[sweet shard]

x_n is a vector

[obtuse trench]

No, but you can multiply a vector by a number

This doesn't have to be dot product it could just be scaling the vector the vector by a constant x_n which is the data

Such that, x is a vector, and then x_n would be the n'th number in the vector

[sweet shard]

You're using x_n for different things

W has dimension m

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[storm lance]

can someone help me like actually visualize this because I am having a hard time. what if I have like 3 arbitrary points in the rectable, how do I divy them up into squares. Does this mean I wont be able to like fully cover the rectangle?

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@storm lance Has your question been resolved?

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@storm lance Has your question been resolved?

[storm lance]

guys can someone help me :((((

[polar torrent]

hey @storm lance what exactly are you having trouble with? Maybe I can help

Here is an example of a 5x12 rectangle D that has 15 convex sets that all have area 4, and I've assigned a point in each

[storm lance]

No like the points are already there in the rectangle right. You see how P7, P6, P13 they aren’t exactly contained in a square, that’s what I’m asking

Like isn’t it supposed to be only squares?

[polar torrent]

What do you mean by pony?

so yeah, you can always divide D into all squares of the same size. Squares and rectangles are always convex sets. I think what this is saying is that it's possible to divide D into stuff that's not all the same shape AND size

[storm lance]

Pardon me I meant points

What happens if there’s like only 3 points

How would that division look like, then there would still be leftover area on the bottom?

I guess like how would just squares cover this rectangle

https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=35998f3325f7fe67eae7d6e1ecbe5e4ee67b0ced

this is the paper I am reading and it is section 3 of the paper

[storm lance]

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[heavy jewel]

could anyone help me out on how to prove this

[heavy jewel]

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[strange topaz]

can someone explain the answer here

[mint crescent]

Which part don't you get?

[strange topaz]

i dont really understand how to apply the 30 60 90 triangle rules to a triangle when the square root is not in the form x√3

like in this question the leg is √6

so idk what to do

[mint crescent]

It's the same thing

it doesn't have to be a multiple of sqrt 3

that's what they did here

[strange topaz]

i dont understand

[strange topaz]

this

oh nvm i understand

thanks

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[zinc night]

Hey! I don't even know what I'm looking for so here goes nothing, what would be the name of the formula that estimates the interest of a compound interest? I don't know if that makes sense. Let's say something has an interest of 0.005% daily, if I invest again and it gives me 0.005% again, and so on, what would be the ACTUAL interest by the end of the month? Is this a thing or I'm just making this up?

[clever badge]

There is a formula, but I don't remember the name. I've only heard of it as the "compound interest formula"

It's basically just a particular case of exponential growth/decay

[zinc night]

Yeah, but most of the things I find are based on the invested capital and not the actual interest rate

[neon iron]

Wouldn't that just be P[(1 + r)^n - 1] where r is your interest rate, n is days in this situation and P is your initial sum

[clever badge]

I mean, how can you calculate the interest without the rate?

[clever badge]

not for compound interest

[zinc night]

I know the rate, I just don't know the initial deposit or capital

[neon iron]

Why not?

That is compounding

[clever badge]

I believe the formula is P(1 + r/n)^(nt) where n is the number of times compounded per year, r is the interest rate

[clever badge]

So you don't know the initial or final amount?

[zinc night]

I think I found it: https://www.calculator.net/compound-interest-calculator.html

[waxen flame]

you don't really need to

you can still estimate your rate of return with (1+r)^n

[neon iron]

Right but we were talking about a fixed interest per day so its (1 + r)^n where n is the number of days

[waxen flame]

also for small percentages, this comes out to e^(rt) or something like that

[zinc night]

I'm just trying to estimate how much will I make if I invest, let's say 100 bucks with a 0.005% daily interest rate; I'm trying to find out if it's better than monthly

[waxen flame]

your initial capital doesn't matter for this calculation

[zinc night]

Exactly

[clever badge]

yes, you can just calculate the percentage

[waxen flame]

and surely you will not with such a pitiful interest rate

[zinc night]

The actual interest rate is something like 0.23% haha

[waxen flame]

that's a colossal interest rate per day

what kind of bank is this

[clever badge]

0.23% daily interest sounds like you are being defrauded in some way

[zinc night]

Thirld world country man

[waxen flame]

even a hedge fund only makes like 0.05% per day, and that's with leverage

[zinc night]

Almost 140% inflation here my dudes

[waxen flame]

LOL

that interest rate checks out

0.23% per day is 131% per year

[clever badge]

well, does investing at 0.23% interest overcome that 140% inflation? Otherwise you could put your money into a more stable currency like the US dollar

[neon iron]

Youd be waiting around 38 years to double your money at 0.005% interest

[pseudo bear]

,calc 1.0023^365 - 1

[thorny flame]

Result:

1.3129784187008

[waxen flame]

idk maybe they're going to print money to pay off the debt

[pseudo bear]

So, 131% APY.

[waxen flame]

lines up perfectly with 140% inflation lmfao

[zinc night]

No, I mean, I work as a dev so I'm doing good with my current job haha it's just a tax thing

[neon iron]

Good old banks

[waxen flame]

is this argentina or turkey

[zinc night]

Alright my dudes, all clear. Thanks for your help and laughs, hoping for a better 2024 for my country haha

[zinc night]

Argentina

[clever badge]

I wish you good luck

[waxen flame]

wishing u the best

[zinc night]

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[desert compass]

Could someone please help me find my error?

[desert compass]

when i work to derive f'(x), i do not get the same approximation that is given in the problem

[topaz sinew]

@desert compass Has your question been resolved?

[sweet shard]

what's given in the problem?

are you just trying to get 3h in the denominator instead of 4h?

[desert compass]

kind of. i also have an extra f(x) in the numerator that isn't in the given approximation, and i have 2f(x-h) whereas the given approximation is just f(x-h)

basically, i'm trying to use taylors theorem to derive the approximation

but when i try it, it doesn't come out the same

[tough cargo]

Hmm well susbtracting the third equation by the first works a lot better BUT

How do you know it's the same c?

[desert compass]

ah, good point, i could see maybe multiplying the 2nd equation by 2 and then subtracting might do something, but i imagine it should work out the same

as for the c, iirc, the instructor explained that the c is the center around which we are approximating

[tough cargo]

If you just substract the two you get the nice 3hf'(x)

And the f(x)'s cancel out

[desert compass]

i'll try that real quick, thanks 🙂

[desert compass]

mmm, that didn't work out:

[tough cargo]

Ill just show u (i assumed the constants were diff in the intervals hence why theyre called cw and c3)

,rotate

...

[desert compass]

mmm

you didn't square the 2

[tough cargo]

4/2

[desert compass]

ahhh

interesting, i don't understand why the results are different

it should be the same, yeah?

but yeah, you're right, doing it your way seems to result in the correct approximation. there must be some mistake in my arithmatic/algebra in the method i tried, yeah?

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[neon iron]

I'm a bit curious on a couple things in this picture

• Why is this depicted in degrees instead of radius
• Why in the second equation, it isn't simply just pi as the k value

Mainly the second point, had they had done it in radians it would've been 2pi/2 which simplifies to pi (aka 360 degrees if we did it their way)

because it seems that the period necessaraly changes if you were to depict it in sine (since a period MUST start and end at the midline in a sine function)

[mint crescent]

• Why is this depicted in degrees instead of radius

🤷‍♂️ doesn’t rlly matter

• Why in the second equation, it isn't simply just pi as the k value

The period is 1.6, not 2

[neon iron]

but the period would change if it started at t=0.4

instead of the parent graph, t=0

[mint crescent]

Then you would have 2.0-0.4=1.6

[neon iron]

that's the thing though, if you look at the graph above (normally), it appears to be a -cos function up until the 3.6 second mark

changing it to start at t=0.4 would mean it'd start at mid-line (suggesting it's a +sine function) which would mean 1 full cycle starts at 0.4 and ends at 2.0

[mint crescent]

would mean 1 full cycle starts at 0.4 and ends at 2.0

So what is the length of this cycle then?

[neon iron]

I see now, thanks

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[empty fossil]

if I have the circle plane ${(x,y)|1\leqslant x^2+y^2 \leqslant 2 }$ described by the relation $(\cos{x}, \sin{x} ) $ \text{~}$(\cos{x+\pi}, \sin{x+\pi}))$ what happens? I dont know how to understand this

[thorny flame]

bigpufik

[empty fossil]

Well from what I understand we are dealing with only points on the circle x^2+y^2=1, and what means that they are being rotated by 180 degrees, but how wil this look like?

[worldly nimbus]

If u rotate a circle by 180 , it's still the same circle

[empty fossil]

the plane is described by a relation, they are the same point, It wont be a circle...

[worldly nimbus]

Cosx+pi=-cosx

Sinx+pi=-sinx

[empty fossil]

Yes.

But these points dont just go to the place, they bend. They become the same point

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[mortal rose]

“the function f is given in the form f(x)= (x+a)/(bx+c); x ≠ -c/b. Given that the graph contains asymptotes in x = -4 and y = -2, and that the point (2/3,1) exists in the graph, find the values of a, b, and c.”

[mortal rose]

i know i can solve it with the information given i just don’t know how 😭

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@mortal rose Has your question been resolved?

[mortal rose]

<@&286206848099549185> 🥲🥲🥲

[scenic crow]

Hey

[mortal rose]

hi!

[scenic crow]

I can help you

[mortal rose]

thanks 😭

[scenic crow]

Ok so the first thing we have is the two asymptotes

Yk that the vertical asymptote is x=-c/d

And it says that that vertical asymptote is x=-4

[mortal rose]

so -c/b = -4?

[scenic crow]

Yes

And like for now you can just pick c=4 and b=1 and multiply them by another number if you need to (like for example you could have c=8 and b=2, or smth like that with the same proportion)

[mortal rose]

so just any set of values that will give me -4

[scenic crow]

Yes

And then the horizontal asymptote is 1/b

You know that right ?

[mortal rose]

no i forgot all of this 😭 i’m v sorry

is that a general rule?

so 1/b has to give me -2

[scenic crow]

Well yeah, usually when you have f(x)=(ax+b)/(cx+d) the horizontal asymptote is y=a/c

[scenic crow]

Yes

[mortal rose]

so b has to be -2

no

[scenic crow]

-½

[mortal rose]

okay! yeah got it

so yhen i can plug that into the previous one

-c/(-1/2) = -4

[scenic crow]

Yeah exactly

And then c equals?

[mortal rose]

-2

so would it be

-(-2)/(-1/2)=-4?

[scenic crow]

Yes exactly, now we have c=-2 and b=-½ and we need to find a

[mortal rose]

i can just plug the values into the original equation

and replace f(x) and x with

the coordinate values

right?

[scenic crow]

Yes exactly

[mortal rose]

okay! awesome!

i was so stumped 😭 but you really cleared it up for me

thanks a lot!

[scenic crow]

No problem 🙂

[mortal rose]

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[arctic rock]

WHERE DO I BEGIN

b i), ii), and iii)

<@&286206848099549185>

[mild trail]

do u know the triangle of Pascal?

@arctic rock

basically, on the top of the triangle, it's (a+b)^0

so on top of it u put 1, then on the vertical and diagonal sides u always put 1, and then u do the addition of the numbers as u can see in the pic

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[snow forum]

How did they get rid of squareroot I forgot how to deal with that lol

[snow forum]

You can ignore the mesurments C m v etc..

[strange whale]

,calc -5/sqrt(2)

[thorny flame]

Result:

-3.5355339059327

[strange whale]

yeah -3.54 checks out

@snow forum

[snow forum]

Yeah but im suppose to do that manually

That is the problem

So is there a way to solve it elegantly

[strange whale]

well just keep 5/sqrt(2) then if you don't have a calc

or if you're a madlad compute 5/1.414 by hand

[snow forum]

Yeah lol that is possible XD

Well thqnk you glat to know that there is nothing im missing , I have to write it something like 3.53

Or just leave it

Tnx

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[cinder owl]

i have some questions about svd

[cinder owl]

so SVD is a transformation with orthonormal vectors right?

also, im confused aboutt his

[neon venture]

you know ? not everyone knows what svd stands for so call the thing by its name

[cinder owl]

sorry

uhhh

singular value decomposition

[neon venture]

well its a generalization of spectral theorem

so the eigenvectors are orthonormal

[cinder owl]

spectral theorem is just diagonalizing orthogonal matricies right

[neon venture]

yea yea if im not wrong

[cinder owl]

so how is svd related to spectral theorem them

[neon venture]

its almost the same thing, its just that it can work with any matrices that are not only square

spectral theorem works for only square matrices

[cinder owl]

but isnt that just regular diagonalization

where you have the matrix of eigenvalues, and then the matrix of eigenvectors

[cinder owl]

also do you know what this is about

oh shoot so svd is breaking down matrix vector multiplication

[neon venture]

it is
the whole thing about diagonalization is to write a matrix in a basis where that matrix is simple to manipulate

but spectral theorem and svd works with orthogonal basis

[cinder owl]

right because U and V are both symmetric

[waxen flame]

U and V are orthogonal, not necessarily symmetric

especially if you're dealing with the compact SVD, in which case U and V are not square

[topaz sinew]

@cinder owl Has your question been resolved?

[cinder owl]

do you know about the svd in rank r format

and also how you find the basis of different spaces

[waxen flame]

the thing you've posted is a compact svd

[cinder owl]

oh interesting

[waxen flame]

you can get U, V, and Sigma from taking eigenvalues/eigenvectors of AA^T and A^T A

[cinder owl]

man i dont know why svd is so bad for me

right that makes sense

U and V are orthogonal, which means they are linearly independent

so that is already the basis

what is this talking about

[waxen flame]

what about it

[cinder owl]

Shoot i think im overthinking this

left null space is the same as null space but with u right

[waxen flame]

yes I believe so

[cinder owl]

what does it mean when it says like the first k

[waxen flame]

A has rank k

so you pick the first k things

this is assuming that S is sorted in descending order

well, the weaker condition is that the nonzero singular values are grouped together at the beginning, but whatever

[cinder owl]

oh and Vn is like the last one right

also does S need to be in descending horder

[waxen flame]

not to make the statements given here; you can actually just say that the nonzero singular values are the first k

but the svd is defined that way so it's more convenient

[cinder owl]

i think in class singular values were defined to never be 0

[waxen flame]

I think you need some zero singular values for rank-deficient matrices...

for the compact svd, you leave the zero singular values out

[cinder owl]

i dont know, what exactly is the point of compact svd

[waxen flame]

you get to ignore the whole min(m, n) thing

and the other eigenvectors really don't matter anyway

[cinder owl]

oh honestly idk my linear alg class doesnt rlly talk about time complexities in computation

that is really interesthing tho

[waxen flame]

not time complexity

like the size of Sigma is kxk instead of min(m, n) x min(m, n)

and yes there's some computational advantage to not dealing with all of the extra eigenvectors

[cinder owl]

thanks this is so helpful

can i also ask about best fit subspace for svd

also kinda confuses me

[topaz sinew]

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[sonic bone]

help i needa find common denominator and solve and simplify to one term

[proper valve]

well, what have you tried, what would be the common denominator?

[sonic bone]

i set the first one as cos/cos

cuz 1-sinx=cosx right

[proper valve]

No, sorry. We know that (sinx)^2 + (cosx)^2 = 1, but it's not generally true that sinx+cosx=1

[upbeat crypt]

multiply cos(x) / (1 - sinx) by 1 + sin(x)

and cosx / 1 + sinx by 1 - sinx

[sonic bone]

ohh ok

[proper valve]

Imagine these are just your normal Fractions. How would you add 3/4 + 5/7

You would do: (3*7 + 5*4) / (4*7)

[upbeat crypt]

u will have something like:
[cosx(1 + sinx) + cosx(1-sinx) ] / 1 - (sinx)^2

so u will then have :
2cosx / (1 - sinx^2)

[sonic bone]

are u able to show this

[upbeat crypt]

1 - sinx ^ 2 = cosx^ 2

[sonic bone]

in the latex thing

[upbeat crypt]

so in the end:
-> 2/cosx

is the answer

[sonic bone]

yeah ur going a lil too fast

[upbeat crypt]

why?

oh

u didnt understand?

[sonic bone]

u want me to do this first

?

[upbeat crypt]

@sonic bone

here

@sonic bone

[sonic bone]

oh ty

thats more helpful

where did the two on top come from

confused here

[topaz sinew]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

@sonic bone Has your question been resolved?

[sonic bone]

<@&286206848099549185>

[lone pier]

Sorry, What is your question now?

He/she multiplied the first fraction by 1 = (1+sinx)/(1+sinx)
and the second fraction by 1 = (1-sinx)/(1-sinx)
which gives you two fractions with the same demoninator.

Seems a possible way to go.

@sonic bone Still interested?

Sure, as you say, the sinx * cosx - sinx * cosx will = 0

[sonic bone]

@lone pier im just wonder how they got 2 cos

Does cos+cos = 2cos and not cos^2

[hollow drum]

cos^2 is cos * cos

[sonic bone]

oh is that why cos+cos is 2cos

[hollow drum]

Same logic with x, x + x = 2x not x^2

[sonic bone]

Ohh okay

Ik the answer is 2sec

But im just tryna find out how to get there

[hollow drum]

What do you have now?

[lone pier]

y + y = 2*y for any y

[hollow drum]

Like what step are you on, in your work

[sonic bone]

?

[lone pier]

Stepped away - willleave it with you

[sonic bone]

cos^2

the bottom

that would be 1/cos?

[hollow drum]

What happened to the 2?

[sonic bone]

2/cos?

[hollow drum]

Yes

Then recall that sec = 1/cos

[sonic bone]

yeah dats a lil confusing to me

how it just turns into 2sec

[hollow drum]

It's an identity

sec = 1/cos

[sonic bone]

ik 1/cos is sec

but now its 2/cos

[hollow drum]

$\frac{2}{\cos} = 2 \cdot \frac{1}{\cos}$

[thorny flame]

CaptainNova22

[sonic bone]

okay

i see how that would be

2sec

now

i also need help wit wherever i went wrong wit this

[hollow drum]

You applied the wrong identity

It's sin^2 + cos^2 = 1

Not sin + cos = 1

[sonic bone]

do i multiply everything by sin

to get rid of the fraction

[hollow drum]

You don't need to because the sin cancels in sin * cos/sin

[sonic bone]

?

[sonic bone]

<@&286206848099549185>

[sweet shard]

right

so find when sin(x) + cos(x) = 0

moving one term to the other side may help

[sonic bone]

cos=-sin

or what

[sonic bone]

sorry for ping idk if thats allowed but still tryna figure this problem out

are u saying do this

and find when cos is postive and sin is negative

like (0,-1) 3pi/2

and if i were to subtract cos so itd be sin=-cos

(-1,0) pi

?

[unreal stream]

yes

3pi/2 and 5pi/2

[unreal stream]

times any integer number

fuck

[sonic bone]

I dont even know how to start this tbh if anyone can help

[unreal stream]

i would try something like this

[sonic bone]

whered u get the 6 from

[unreal stream]

that 6 multipling the cos

[sonic bone]

oh

thats

not a number in there

thats just problem number 6

[unreal stream]

oh

[sonic bone]

the question is just cos(2theta-pi/2)=-1

[unreal stream]

you can start taking what's inside that cos as what would be cosx = -1

[sonic bone]

so when cos = -1?

its pi

?

[unreal stream]

yes

then 2theta - pi/2 = pi

[sonic bone]

do i add pi/2 to pi

[unreal stream]

yes

[sonic bone]

?

what now

[unreal stream]

you can divide both sides by 2

[sonic bone]

3pi/4

[unreal stream]

yes

[sonic bone]

why is it also 7pi/4

also i did 3pi/2 divided by 2 on a calculator to get the answer but why does 2 turn into a 4

idk how to do that without a calculator

[unreal stream]

honestly idk

[sonic bone]

lol

[unreal stream]

cause when you divide something you are multiplying by it's inverse, so divide by 2 is the same as multiply by 1/2