this isn't an equation
#help-10
brave plume
timid silo
[
\map \log{a + b} \c r \ne \map \log a +\map \log b
]
warm shaleBOT
brave plume
What should I do?
timid silo
okay so this is much more elaborate now. You're trying to evaluate[
\lim_{x \to \infty} \parens{4^x + 5^x}^{\f1x} -3
]
warm shaleBOT
timid silo
hmm
brave plume
Here
spice citrus
maybe factor out 5^x?
timid silo
I think so too
spice citrus
It is I think, I can't read everything, but you can also factor out $5^x$, so you would get $$\lim_{x\to\infty} 5\cdot((\frac{4}{5})^{x} + 1)^{\frac{1}{x}}$$
warm shaleBOT
Jelle
brave plume
it looks more complicated
but I can see you just took common 5^x inside what will we do next?
spice citrus
Well, the 5 can go out of the limit
and the inside goes to (0 + 1)^0
which is just 1
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
lucid geode
need help
lucid geode
not sure where to start
tacit scarab
derivative represent rate of change
thus when f'(x)>0, f is increasing. when f'(x)=0, f is stationary and when f'(x)<0, f is decreasing.
tacit scarab
I told you no
lucid geode
bruh then wtf am i supposed to do
tacit scarab
f'(-2)=-1 which is negative
from the graph you can see f'(x)<0 in interval B
thus f is decreasing in B
tacit scarab
yes
tacit scarab
critical point is where f'(x) = 0 or f(x) is not defined
a critaical point can be a local maximum, local minimum or saddle point
tacit scarab
yep
lucid geode
whats relative maximum points
lucid geode
so theres none
tacit scarab
no
for f'(x) there's not
but we want maximum of f, not f'
to figure out if a critical point is max or min, first calculate f''
lucid geode
how
tacit scarab
differentiaite f'
lucid geode
2x+4
tacit scarab
if f''(x)>0 then it's minimum and vice versa
tacit scarab
we know critical points -3 and -1
we just need to find out which is maximum and which is minimum
plug in the critical point to f''(x)
tacit scarab
if f''(x)>0 then it's minimum and vice versa
then ^
tacit scarab
correct
lucid geode
then what
vast musk
obtuse pebbleBOT
vast musk
not sure where i went wrong in this
maybe i got an implication confused with an equiavalence?
vast musk
oh i see, if the curl is zero, there still might not exist a potential for that vector field?
tacit scarab
yes
vast musk
but the rest that i have ticked are true?
obtuse pebbleBOT
Available help channel!
sleek wagon
Both of these questions are confusing me
obtuse pebbleBOT
Both of these questions are confusing me
sleek wagon
For 6 my answer was 5 but it says 2.5 in the answer key(??)
For 7 how do we get the probability when it says nothing about travel insurance in the info
timid silo
,rccw
warm shaleBOT
obtuse pebbleBOT
@sleek wagon Has your question been resolved?
sleek wagon
<@&286206848099549185>
sleek wagon
Yes
severe depot
also the Empirical Rule i'm guessing
sleek wagon
2
severe depot
correct
so let's write this variable rq
X ~ N(mean = 120, standard dev. = 3)
we see "more than 126 minutes" right ?
sleek wagon
Yep
severe depot
that means you need to find the probability of the charging time being more than 126 minutes
P(X > 126)
do u agree?
sleek wagon
Yes
severe depot
Yes
could you use the complement's rule on this?
for normal distribution it's also very good to draw a graph
sleek wagon
Yea so I originally did 100-95 to get 5 but the answer is 2.5
severe depot
you were almost there
the 5 is the extremely low charging times and the extremely high charging times
we only need the extremely high ones, so to speak
so you divide by two
severe depot
always draw a graph, best tip ever
sleek wagon
True
severe depot
got me through normal distribution easily
so you'll see what probabilities you need etc
sleek wagon
Yea
Ill try
Bro how hard is it to make a venn diagram with 4 circles that makes sense
mellow forge
how do i show this
obtuse pebbleBOT
how do i show this
mellow forge
g(x)=f(x^2)
timid silo
what is g and what is f
mellow forge
Its that the taylor expansion
4th order taylor expansion equals the second order taylor expansions
steel blade
Have you already calculated all derivatives of g up to 4th order?
steel blade
Well if you differentiate the first line again, you get two terms: The first one is correct, but the term where you diff. f'(x^2) you have to consider the inner derivative aswell
steel blade
Im not sure what u is, you have to use the chain rule
mellow forge
i am using the chain rule
nvm im not
what two functions do we have here then?
f'(x^2) and x^2?
steel blade
Yeah f'(x) and x^2
mellow forge
so f'(g(x)*g'(x)
steel blade
yes
steel blade
yes but you already have 2x in front
mellow forge
now we have to differentiate this
that was the first term
this will be quite difficult though
steel blade
So what is your second derivative now?
steel blade
Not quite, the 2 f'(x^2) term you had first was already correct, only the second term was missing an additional 2x
mellow forge
you play chess?
mellow forge
Not quite, the 2 f'(x^2) term you had first was already correct, only the second...
rly?
youu sure
steel blade
Very sure
steel blade
Yeah you have to use the product rule first and then apply the chain rule when you differentiate f'(x^2)
steel blade
yes that looks good
mellow forge
yh i get the same now
didnt know you used the product rule first and only the chain rule for the ones that needed it
steel blade
yh i get the same now
Okay nice, now you have to calc. the 3rd and 4th derivative
steel blade
Haha maybe
mellow forge
you high rated?
steel blade
Not really, I have like 1200 online
mellow forge
i might be doing it wrong again
ill try and see if we get the same
1 sec
how do i differentiate f''(x^2)
just 2x*f'''(x^2)?
steel blade
yes
steel blade
Almost, the second term is correct, but the third term should have x^3 and the first term f''
steel blade
You are missing one ' in the first term
mellow forge
Almost, the second term is correct, but the third term should have x^3 and the f...
yeah i see
i move 2x down to 4x^2 and get 8x^3
oops
i didnt write it, but it is supposed to be x^3
steel blade
Haha dw I know what you mean
mellow forge
what do i do after i have done this 4 times
steel blade
You must have some formula for the Taylor expansion right?
steel blade
One should be enough :D
steel blade
Yes but you can simplify that
steel blade
nice
mellow forge
now onto g(x) or?
steel blade
Well you just calculated the derivatives for g(x)
steel blade
Hm maybe, I mean we want to show that T^4(g)(x) = T^2(f)(x^2) right? And if you now plug in the derivatives in the formula for the taylor expansion we have the left side
mellow forge
yh
steel blade
uhm left side sorry
mellow forge
yh but i thought we had to do something to both
but if your method makes it right (which it is supposed to do) im sure its correct
at any rate, lets proceed
steel blade
There is nothing to really calculate for the right side, the left side is the hard part
steel blade
okay so what do you get if you plug in all the derivatives?
steel blade
yes actually
mellow forge
so i do put on both?
steel blade
yes
steel blade
If it is not stated that f(0) = 0, we have to leave it like that
steel blade
Yes you are not supposed to set these terms to zero
mellow forge
in this term for example
it matter which term i apply the x^2 to
or is it supposed to be in parentheses
steel blade
You apply x^2 to both terms in the enumerator yes
steel blade
Looks good
steel blade
We know have to write down the righhand side explicitly
mellow forge
explicitly?
steel blade
What I meant is you have to write down what T_2(f)(x^2) actually is
steel blade
Okay so what we've done so far is the calculation for the lefthand side (T_4(g)(x)). The goal here is to show the equality T_4(g)(x) = T_2(f)(x^2) right? So we know what the lefthand side is, now we use the definition for the taylorexpansion again but for the right hand side and hope that it actually is the same. Note that T_2(f)(x^2) is NOT the taylor expansion for f(x^2), but for f(x) where you plug in x^2 instead of x in the end
steel blade
Yes T is for taylor
The problem is that the way it is written here is a little ambigous. T_2(f(x^2)) means taking the taylor expansion for f up to second order (so f(y) = f(0) + y* f'(0) + 1/2 y^2 * f''(0) ) and then AFTERWARDS, you plug in x^2 in the argument (so y = x^2)
steel blade
The derivative of f(x) is just f'(x), there is now way to further simplfy that since we dont know how f(x) looks like
steel blade
Well normally we have an x^2 term in the second order taylor expansion right? But now what we do is we replace all x with x^2, so the x^2 term suddenly becomes x^4
steel blade
Maybe this helps:
mellow forge
what is mcavs
steel blade
If we compare your result for the left side with what i wrote I would say they are both equal, or what would you say?
steel blade
Yes
steel blade
Ahh then you misinterpreted my bad handwriting
steel blade
the late term has f''
steel blade
f
mellow forge
i thought we were done now
steel blade
Yeah we are done
steel blade
Remember the lefthand side is this right?
If you say it's fine for you then yeah ofc, I just wanted to emphasize that g and f are not equal with this😅
mellow forge
it is just the thing we've just made
I know it is written in danish, but i the math is the same
steel blade
Dw I can partly guess what is written there because I'm German. Looks good overall
mellow forge
Dw I can partly guess what is written there because I'm German. Looks good overa...
actually
have we not made a mistake
when you differentiate the first time you get 2x*f'(x^2)
if u put in 0 here you get 0*f(x^2) which is 0
steel blade
Im not sure to which line yo are referring to
steel blade
Ahh okay
But remember we set a = 0 in your formula for the taylor expansion, but the x stays as it is
Maybe it would be better if you replace the x's with a's when you calculated the derivatives
mellow forge
so how should i write it instead
steel blade
you write 2a f'(a^2) for the first derivate, 2 f'(a^2) + 4a^2 f''(a^2) for the second derivatve and so on
so you replace a with x only were you calculated the derivatives
mellow forge
ok so in the first,second, third and fourth
steel blade
yes
mellow forge
in my pdf
if you scroll up to when i insert a=0
have i not missed some x's
that i havent inserted a=0 on
steel blade
This is exactly why it's better to replace the x's that were used for the derivatives with a's because we set a = 0 and not x = 0
steel blade
yes
mellow forge
my question is though
or tryign to understand rather
when we calculate the left side
we just insert a = 0 and the first term goes away
but on the right side
we suddenly cant do that?
why
@steel blade
steel blade
We played by the same rules on the right side, we used the formula for the taylor expansion for f(x), set a = 0 and then replaced x by x^2
mellow forge
but should the first term not be equal?
steel blade
you mean x^2 f'(0) ?
mellow forge
in the first
we say that the first derivative is 2x*f'(x^2)
or rather 2*a
when we insert 0
entire thing becomes 0
mellow forge
why doesnt the same thing happen on the other side
steel blade
The right side is this f(a) + x^2 f'(a) + 1/2 x^4 f''(a) right? We then only set a=0 but NOT x = 0. So they dont vanish
The reason why it vanished on the left side is because the term 2 a f'(a^2) has an a in front so it vanishes when we set a = 0
steel blade
Yes but one the right side we dont have that. We only have a's in the argument, NOT a^2's, so we dont need to apply the chain rule
steel blade
No problem, I hope it will make 100% sense after some time :D
steel blade
sure why not
mellow forge
i just wrote you just follow server rules
i dont think you can idle cha here
.close
frail depot
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
obtuse pebbleBOT
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) ...
frail depot
I feel like im in an endless loop?
warm shaleBOT
SimonWin
frail depot
I will just keep doing integration by parts forever, by I notice now that I have gone a full "round" i am back at the original integral
Lemme just underbrace what i mean
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \underbrace{\int e^x \sin(x)dx}_{\text{this is the same as the original integral}}
\end{align*}
warm shaleBOT
SimonWin
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
SimonWin
frail depot
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \underbrace{\int e^x \sin(x)dx}_{\text{\Large this is the same as the original integral}}
\end{align*}
warm shaleBOT
SimonWin
frail depot
Don't know if i am supposted to use that?
Otherwise, I feel like I'll keep doing integration by parts forever
grizzled shore
You can call the original integral I
Then say you have a bunch of stuff + I
Pretty sure it should be a different sign though
frail depot
But that seems magical?
grizzled shore
You should probably get - I on the right side
frail depot
Why can I just say okay now we're arrived at the same, so the original integral is now I
violet epoch
Hi
frail depot
Aah..
grizzled shore
Then you add I to both side divide by 2
frail depot
Aaah, yes so I will cancel out
grizzled shore
And you can solve for I explicitly
grizzled shore
Yeah read integration by parts on Wikipedia
grizzled shore
Go to the tabular section
frail depot
I have already done integration by parts
grizzled shore
There’s a method to do this quickly
Yeah there’s a fast method for repeated integration by parts
grizzled shore
Just a tool you might want to add to your tool belt
frail depot
frail depot
Just a tool you might want to add to your tool belt
Yeees!
Definitely
Is it just this you mean?
I will read it after, I am just going to trey this real quick
grizzled shore
Yeah
See the example with x³ cos x
Wait no
The eˣ cos x one
Pretty much the same as your question
frail depot
It's exactrly the same right?
grizzled shore
Have a read if you have time
muted halo
Isnt it integration by parts??
grizzled shore
Just try it
grizzled shore
Yeah
frail depot
thats the whole point we're making right, that they're repetitive
grizzled shore
Just try and see
frail depot
it makes sense!
grizzled shore
Yep
frail depot
yes
will do
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
So going from this
warm shaleBOT
SimonWin
muted halo
This ones a bit complicated one
grizzled shore
It also gives you a better idea what to choose for your derivative
Because the column you derive you want it to go to 0 at the end
warm shaleBOT
SimonWin
frail depot
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
Now we see that we're arrivated at the same original integral, hence we can define $I=\int e^x\sin(x)$
muted halo
Can anyone tell me the question maybe i will be able to help
grizzled shore
There’s not much to do
frail depot
Can anyone tell me the question maybe i will be able to help
a second, i think maaaybe we got it 😂
warm shaleBOT
SimonWin
grizzled shore
Just getting him to try tabular IBP
frail depot
i think i might have it tbh
i see it for me now
im him right now
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
Now we see that we're arrivated at the same original integral, hence we can define $I=\int e^x\sin(x)$ hence,
$$
\int \cos(x)e^x&= \cos(x) e^x + I
$$
\noindent\rule{20cm}{0.4pt}
Now we can start substituing the integrations by parts backwards,
\begin{align*}
I &= \sin(x)e^x - \cos(x) e^x + I \
I - I &= \sin(x)e^x - \cos(x) e^x \
0 &= \sin(x)e^x - \cos(x) e^x \
\end{align*}
warm shaleBOT
SimonWin
LaTeX source sent via direct message.
```Compilation error:```! Misplaced alignment tab character &.
l.1426 \int \cos(x)e^x&
= \cos(x) e^x + I
I can't figure out why you would want to use a tab mark
here. If you just want an ampersand, the remedy is
simple: Just type `I\&' now. But if some right brace
up above has ended a previous alignment prematurely,
you're probably due for more error messages, and you
might try typing `S' now just to see what is salvageable.```
muted halo
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) ...
How can u even type this😨
frail depot
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
Now we see that we're arrivated at the same original integral, hence we can define $I=\int e^x\sin(x)$ hence,
$$
\int \cos(x)e^x= \cos(x) e^x + I
$$
\noindent\rule{20cm}{0.4pt}
Now we can start substituing the integrations by parts backwards,
\begin{align*}
I &= \sin(x)e^x - \cos(x) e^x + I \
I - I &= \sin(x)e^x - \cos(x) e^x \
0 &= \sin(x)e^x - \cos(x) e^x \
\end{align*}
frail depot
How can u even type this😨
Wdym?
muted halo
Like how
warm shaleBOT
SimonWin
muted halo
Ye
frail depot
or how i know it?
we have like 30 pages reports every week
so you're like forced to just learn it
😂
muted halo
Oh lol
frail depot
- i have invested a lot of time in making my own latex-template
with like cool boxes etc
muted halo
Damn
uneven echo
you made some sign errors
frail depot
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
Now we see that we're arrivated at the same original integral, hence we can define $I=\int e^x\sin(x)$ hence,
$$
\int \cos(x)e^x= \cos(x) e^x + I
$$
\noindent\rule{20cm}{0.4pt}
Now we can start substituing the integrations by parts backwards,
\begin{align*}
I &= \sin(x)e^x - \cos(x) e^x + I \
I - I &= \sin(x)e^x - \cos(x) e^x \
0 &= \sin(x)e^x - \cos(x) e^x \
\end{align*}
frail depot
you made some sign errors
did i? where 😿
warm shaleBOT
SimonWin
muted halo
So u guys actually type it
uneven echo
I = sin(x)e^x - (cos(x)e^x+I)
muted halo
Or some ai thing do it
frail depot
So u guys actually type it
no im literally sitting and typing it rn 😂
its taking forever
muted halo
Aa-
frail depot
and im typing in discord too so no autocompletion pain
muted halo
I shouldnt be here i guess
Coding is not my cup of tea
am fine with learning about medic
frail depot
I = sin(x)e^x - (cos(x)e^x+I)
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
Now we see that we're arrivated at the same original integral, hence we can define $I=\int e^x\sin(x)=\int \sin(x)e^x$ hence,
$$
\int \cos(x)e^x= \cos(x) e^x + I
$$
\noindent\rule{20cm}{0.4pt}
Now we can start substituing the integrations by parts backwards,
\begin{align*}
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx, \qquad \text{ recall $I=\int e^x\sin(x)=\int \sin(x)e^x$ }\
I &= \sin(x)e^x - (\cos(x) e^x + I) \
I &= \sin(x)e^x - \cos(x) e^x - I \
\end{align*}
warm shaleBOT
SimonWin
muted halo
I still dont know what is the question😅
frail depot
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
Now we see that we're arrivated at the same original integral, hence we can define $I=\int e^x\sin(x)=\int \sin(x)e^x$ hence,
$$
\int \cos(x)e^x= \cos(x) e^x + I
$$
\noindent\rule{20cm}{0.4pt}
Now we can start substituing the integrations by parts backwards,
\begin{align*}
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx, \qquad \text{ recall $I=\int e^x\sin(x)=\int \sin(x)e^x$ }\
I &= \sin(x)e^x - (\cos(x) e^x + I) \
I &= \sin(x)e^x - \cos(x) e^x - I \
2I &= \sin(x)e^x - \cos(x) e^x \
I &= \frac{\sin(x)e^x - \cos(x) e^x}{2} \
\end{align*}
So, maybe this is correct?
frail depot
I still dont know what is the question😅
repeated integration by parts
muted halo
repeated integration by parts
Oh i zee
muted halo
Differentiate it to check it
frail depot
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
Now we see that we're arrivated at the same original integral, hence we can define $I=\int e^x\sin(x)=\int \sin(x)e^x$ hence,
$$
\int \cos(x)e^x= \cos(x) e^x + I
$$
\noindent\rule{20cm}{0.4pt}
Now we can start substituing the integrations by parts backwards,
\begin{align*}
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx, \qquad \text{ recall $I=\int e^x\sin(x)=\int \sin(x)e^x$ }\
I &= \sin(x)e^x - (\cos(x) e^x + I) \
I &= \sin(x)e^x - \cos(x) e^x - I \
2I &= \sin(x)e^x - \cos(x) e^x \
I &= \frac{\sin(x)e^x - \cos(x) e^x}{2} \
I &= \frac{1}{2}(\sin(x)e^x - \cos(x) e^x) \
I &= \frac{1}{2}e^x(\sin(x) - \cos(x)) \
\end{align*}
warm shaleBOT
SimonWin
frail depot
,w integrate e^xsin(x)
warm shaleBOT
frail depot
I did it!
💪
and plus constant
😂
Let $f(x)=e^x\sin(x)$, then integrate $f(x)$,
$$
\int f(x)dx = \int e^x\sin(x) dx
$$
For this, we will use integration by parts,
$$
\int udv = uv - \int vdu
$$
\noindent\rule{20cm}{0.4pt}
Let,
\begin{align*}
&u = \sin(x) \implies \frac{du}{dx}=\cos(x) \implies du=\cos(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
where $e^x$ is the family of functions $\mathbf{F}$, and C is any abritrary constant $C \in \mathbb{R}$
Integration by parts:
\begin{align*}
\int udv &= uv - \int vdu \
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx
\end{align*}
\noindent\rule{20cm}{0.4pt}
Again, we notice we have to apply integration by parts on the $\int e^x cos(x) dx$ term. Let,
\begin{align*}
&u = \cos(x) \implies du=-\sin(x) , dx \
&dv=e^x \implies v=\int e^x dx = e^x + C
\end{align*}
So,
\begin{align*}
\int udv &= uv - \int vdu \
\int \cos(x)e^x&= \cos(x) e^x - \int -e^x \sin(x)dx \
\int \cos(x)e^x&= \cos(x) e^x + \int e^x \sin(x)dx
\end{align*}
Now we see that we're arrivated at the same original integral, hence we can define $I=\int e^x\sin(x)=\int \sin(x)e^x$ hence,
$$
\int \cos(x)e^x= \cos(x) e^x + I
$$
\noindent\rule{20cm}{0.4pt}
Now we can start substituing the integrations by parts backwards,
\begin{align*}
\int \sin(x)e^x &= \sin(x)e^x - \int e^x cos(x) dx, \qquad \text{ recall $I=\int e^x\sin(x)=\int \sin(x)e^x$ }\
I &= \sin(x)e^x - (\cos(x) e^x + I) \
I &= \sin(x)e^x - \cos(x) e^x - I \
2I &= \sin(x)e^x - \cos(x) e^x \
I &= \frac{\sin(x)e^x - \cos(x) e^x}{2} \
I &= \frac{1}{2}(\sin(x)e^x - \cos(x) e^x) \
I &= \frac{1}{2}e^x(\sin(x) - \cos(x)) + C \
\end{align*}
daring rock
after all that work you can't forget the constant 😮
muted halo
Damn bro is a pro
warm shaleBOT
SimonWin
frail depot
after all that work you can't forget the constant 😮
I know at the exam i would
and then id fail
id get 1/2 point for it because i miss the + C
daring rock
Lmao yeah it happens
muted halo
Forgetting constant is a sin in integration😭
frail depot
Forgetting constant is a sin in integration😭
Instant reason to fail them 😂
ggs guys
thanks
❤️
.close
jade summit
I'm stuck here, my professor talked about getting the product of z1 and z2 to find z but there's nothing in my notes about finding another solution out of an equation and z1
obtuse pebbleBOT
I'm stuck here, my professor talked about getting the product of z1 and z2 to fi...
jade summit
I was thinking of finding z from the equation which would be sqrt(i) but im not sure where to go from there
obtuse pebbleBOT
@jade summit Has your question been resolved?
jade summit
<@&286206848099549185>
obtuse pebbleBOT
@jade summit Has your question been resolved?
jade summit
<@&286206848099549185>
obtuse pebbleBOT
@jade summit Has your question been resolved?
obtuse pebbleBOT
alpine bison
the second root must stay in the unit circle
obtuse pebbleBOT
Available help channel!
jade summit
reopening
jade summit
I'm stuck here, my professor talked about getting the product of z1 and z2 to fi...
this
@alpine bison sorry i went afk could you explain more
alpine bison
ok
is it clear to you?
geometrically multiply some unit complex number by itself give to you the complex number with a doubled angle
jade summit
is this something i need to do w/ drawing a circle or am i able to do it mathematically
alpine bison
yes it comes from that $(e^{i\theta})^2=e^{2i\theta}$
warm shaleBOT
everg
jade summit
okay
alpine bison
now in your case z is a unit vector
alpine bison
so it has that form $e^{i\theta}$
warm shaleBOT
everg
jade summit
i was guessing it would have to do with the fact that sqrts have plus or minus
so like the opposite
or something
alpine bison
z = sqrt(i)
don t use sqrt in with complex number ... it isn t a well defined function here
jade summit
it isn't?
alpine bison
you have to chose which sqrt you are using
so which is angle such that $e^{i2\theta}=i=e^{-i\frac{\pi}{2}}$
warm shaleBOT
everg
jade summit
can i ask what that is in relation to the z in the equation
alpine bison
because $|z|=1$ in this case.. are you ok with that ?
warm shaleBOT
everg
jade summit
im not sure how 1 is the modulus of z no could you go into that a bit more
warm shaleBOT
everg
warm shaleBOT
everg
alpine bison
beacuse if you use the expation of $e^z$ you get that $e^{i\theta}=cos(\theta)+i\sin (\theta)$
warm shaleBOT
everg
alpine bison
from the last line its obvious that you can write every unit vector of the plane
ok?
jade summit
mm alright
jade summit
polar form and exponential form
alpine bison
now in polar form its much easier find the behavior of complex multiplication ...its just doubled the angle
so you have only to find the only two angle such that when dubled you get the angle associated with -i $(-i=e^{-\frac{\pi}{2}i})$
warm shaleBOT
everg
alpine bison
first solution
second solution
the first one is given from the problem
the second one its easy ...it is in a midway between 1 e -i so its $\frac{-i+1}{\sqrt 2}$
warm shaleBOT
everg
alpine bison
have you some doubts ?
jade summit
ill try to make some more sense of it and review it for the question
thanks for the help
alpine bison
np
jade summit
.close
woeful talon
obtuse pebbleBOT
woeful talon
If x = (1+t)e^5t, prove the following
wild bough
What is it you are stuck on?
woeful talon
i dont know how to start
alpine bison
just do derivatives
woeful talon
do i just derive x = (1+t)e^5t twice, minus 10*deriving it once and add 25 of it to prive its 0?
alpine bison
yes
woeful talon
yes
ok so
so I derive x = (1+t)e^5t once
using the product rule
giving me
t * e^5t + (1+t)*5e^5t
which is
te^5t+5e^5t+5te^5t
alpine bison
finish what you started
woeful talon
derivative of (1+t) = 1
oh thanks
woeful talon
derivative of (1+t) = 1
after deriving it again
i got
35e^5t + 25te^5t
is this correct?
late sequoia
no
late sequoia
derivative of (1+t) = 1
again (1+t)' = 1
late sequoia
e^5t + (1+t)*5e^5t
derivative of this is 5e^5t + 25e^5t(1+t) + 5e^5t
woeful talon
derivative of this is 5e^5t + 25e^5t(1+t) + 5e^5t
im confused
if we try and simplify this
or say expand it
its 10e^5t + 25e^5t(1+t)
expand the second term
25e^5t+25te^5t+10e^5t
would this be considered correct?
becuase u can make that 25e^5t + 25te^5t
if thats correct
late sequoia
its 10e^5t + 25e^5t(1+t)
this is your second derivative now you replace it in your formula
woeful talon
alright
late sequoia
10e^5t + 25e^5t(1+t) - 10 * (e^5t + (1+t)*5e^5t) + 25 * ((1+t)e^5t)
10e^5t - 10e^5t = 0
25(1+t)e^5t - 50(1+t)e^5t(1+t) = -25(1+t)e^5t
woeful talon
thank you!
late sequoia
woeful talon
thanks
woeful talon
I have another question how would i do this
After n years, the value (V) of a principle of P dollars that is invested at a rate of r% per year (with r expressed as a decimal) and compounded continuously is given by V=Pe^(rn)
show the dV/dn = Vr
how do i approach this
late sequoia
After n years, the value (V) of a principle of P dollars that is invested at a r...
derivate it but your variable n
late sequoia
do you know formulas of derivatives?
V is function that depends from n. It mean you need to use same derivatives formulas but instead of x there will be n
woeful talon
do you know formulas of derivatives?
yse
woeful talon
im still confsed
so fiest you dervie e^rn
to become re^rx
but then what does the other part have to do with this?
+6544
late sequoia
but then what does the other part have to do with this?
+6544
you have V = P*e^rn
and you replace P*e^rn with V
late sequoia
V is function that depends of n
late sequoia
this is just for clarity
woeful talon
this is just for clarity
ok so can we do this step by step
V = Pe^rn
we start with the equation
then we derive the equation
e^rn
giving us re^rn
late sequoia
yes
woeful talon
and then dV/Dn = P x e^rn
and then this can be rewritten as
r * pe^rn
which is r * v
but why dont we diff the P
at the start
nvm im stupid
late sequoia
P is a const that multiply e^rn
you can use formula of derivative of multiplication and see.
woeful talon
i understand now
late sequoia
obtuse pebbleBOT
@woeful talon Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
buoyant crag
I'm working on this problem:
obtuse pebbleBOT
I'm working on this problem:
buoyant crag
I have this as my total arrangements :
and this when the 2 red balls are next to each other
Is it correct to interpret it in this way
or would total just be 10! (I'm not sure when to apply which one)
I treated the red balls being together as a group, hence the 9!, and multiplied by 2 since there are 2! ways to arranging the two red balls
drifting wraith
you can't say for sure, it's a guessing game
10! is the less likely one
in this case it should give the same answer anyway
drifting wraith
the x2 is an error
buoyant crag
so iss this the correct setup for the problem
why
isnt there 2 possible arrangements of red balls (could be left or right)
drifting wraith
yes, and you should count them as the same thing
everything else you do tries to count arrangements that looks the same once
if you count arrangements "as many times as possible"
if you count them once only
tawdry vortex
obtuse pebbleBOT
tawdry vortex
hi i am really having trouble with these odd power graphs
like x^5 and x^3
How does a horizontal shift to the left make it look like any of those options
ALso if anyone knows a video that can help explain this concept for these odd power functions plz
Thanks very much
obtuse pebbleBOT
@tawdry vortex Has your question been resolved?
latent walrus
if its just x^n
the even n's all have the same shape
and the odd n's all have the same shape,
they just get flatter in the region |x|<1 and steeper in |x|>1 as n gets larger
why are you confused on the horizontal shift?
is it because the example graphs are all quite zoomed in compared to the options?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
versed ferry
hey did i do this correctly? this is for precalc
obtuse pebbleBOT
hey did i do this correctly? this is for precalc
versed ferry
obtuse pebbleBOT
@versed ferry Has your question been resolved?
versed ferry
<@&286206848099549185>
obtuse pebbleBOT
@versed ferry Has your question been resolved?
versed ferry
.close
mighty pilot
obtuse pebbleBOT
tepid mountain
It is better to add it to clear any confusion if something weird happened without it being there
tepid mountain
Weird way to do long division
But it is correct
tepid mountain
I didn't meant easier way
I just said that the way itself is a bit weird
mighty pilot
oh ok
mighty pilot
yea synthetic division is easier for me, but i wanted to practice long division as well
tepid mountain
Oh no problem
Long division in my opinion is better than synthetic division
As it can be done whatever the dominator is
mighty pilot
i think i did? 4x-3
tepid mountain
What does it mean to say
D(x) • Q(X)
mighty pilot
multiplied?
junior flame
i dont know how its graded but my teacher woudl remove a few marks because you didn’t move the negative sign to the front for the remainder
and some software are very picky
im just assuming
tepid mountain
You need to Express the function P(x) using the function D(x) and the division result Q(x) + the reminder R(x)
In the way mentioned above
P(x) = D(x) • Q(x) + R(x)
You just need to say
(4X-3)(X²-1)+(-3)/(4X-3)
@mighty pilot understood?
mighty pilot
okay yea i think i understand that now
tepid mountain
But I think there is something off here
mighty pilot
im just tryna backtrack it and figure out how to start doin it myself, im a bit slow today lol
mighty pilot
the +R(x) wouldnt that just be -3
tepid mountain
You put the reminder like that
mighty pilot
the remainder
tepid mountain
the +R(x) wouldnt that just be -3
Yeah
mighty pilot
yea thats what kinda confused me a bit
tepid mountain
yea thats what kinda confused me a bit
Is it correct?
tepid mountain
No problem
It is normal though to be slow
Nothing weird in it
You just take your time to understand and to solve the questions correctly
Don't forget to close the channel using .close
obtuse pebbleBOT
tepid mountain
ima power through the rest now, got 3 hours before all is due lol
Lol solving things before deadline is something all students love LOL
mighty pilot
haha fr, youtube all day so far
obtuse pebbleBOT
Available help channel!
timid silo
obtuse pebbleBOT
wooden wharf
Aren't these the same and if not why so?
obtuse pebbleBOT
Aren't these the same and if not why so?
worn yoke
try expanding both and see
mighty pilot
any idea why its coming out wrong?
obtuse pebbleBOT
any idea why its coming out wrong?
mighty pilot
im sure its correct, i double checked with a calculator
subtle sinew
any idea why its coming out wrong?
What's the graph/equation given?
mighty pilot
$3x^3+x^2-7x-5$
warm shaleBOT
x927373
subtle sinew
And what about the graph? The graph should assist you with the zeros
subtle sinew
Those are correct
subtle sinew
It doesn't say to list the repeated roots
mighty pilot
it does
subtle sinew
You can
sterile inlet
Hey guys, so I'm stuck on how to factor this expression, specifically that first step where the two middle terms are split. I understand how to factor by grouping, but I don't understand how I would figure out how to split the middle terms so I can factor by grouping.
obtuse pebbleBOT
Hey guys, so I'm stuck on how to factor this expression, specifically that first...
timid silo
Hey guys, so I'm stuck on how to factor this expression, specifically that first...
i usually do synthetic division not sure if grouping would work for all cases
like 3x^3 + 12x^2 + 5x + 3
sterile inlet
how would you use synthetic division on this expression? im familar with the concept but only when i have a factor for the expression
timid silo
Hey guys, so I'm stuck on how to factor this expression, specifically that first...
in this case c=-1 works out
obtuse pebbleBOT
@sterile inlet Has your question been resolved?
spare mortar
my brain fried, need help
obtuse pebbleBOT
my brain fried, need help
spare mortar
warm shaleBOT
Ratatatat
spare mortar
a tautology is supposed to be true for all possible values no??
obtuse pebbleBOT
@spare mortar Has your question been resolved?
spare mortar
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
native portal
can someone explain how to solve these?
obtuse pebbleBOT
can someone explain how to solve these?
wraith forge
@native portal option 1: what is the least number of basis vectors you need to represent matrices of the form $\begin{bmatrix} a & b \end{bmatrix}, \quad a,b \in \mathbb{R}$?
warm shaleBOT
Kaiser
wraith forge
clearly, you need two vectors, one for each entry, ${ \begin{bmatrix} 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \end{bmatrix} }$ and thus $\dim(M_{1 \times 2} (\mathbb{R})) = 2$
warm shaleBOT
Kaiser
obtuse pebbleBOT
@native portal Has your question been resolved?
native portal
clearly, you need two vectors, one for each entry, $\{ \begin{bmatrix} 1 & 0 \en...
then ig for M(3X3) we would have dimension 9 ? instead of 3
wraith forge
then ig for M(3X3) we would have dimension 9 ? instead of 3
yes
obtuse pebbleBOT
tender tusk
bro elaborate
ancient iris
💀
timid silo
lmfao
hot hazel
.close
obtuse pebbleBOT
wraith forge
<@&268886789983436800>
daring elbow
ugh...
next reef
<@&268886789983436800>
tender tusk
bro that was disgusting thank god it got removed... send it again.
next reef
tender tusk
yeah looked ai af
daring elbow
nami it was on screen for like 3 seconds 
obtuse pebbleBOT
Available help channel!
tepid yoke
Could anyone point out my issue here? I’m trying to use integration by parts but my answer is wrong
obtuse pebbleBOT
Could anyone point out my issue here? I’m trying to use integration by parts but...
thorny drum
What general formula can be for this sequence? I found 60*36^(n-1)/2 only for odd "n", but can someone help me with finding out general formula for all natural "n" with recursion and without recursion?
obtuse pebbleBOT
What general formula can be for this sequence? I found 60*36^(n-1)/2 only for od...
warm shaleBOT
brazen gorge
,w factor 504
warm shaleBOT
brazen gorge
,w factor 2160
warm shaleBOT
brazen gorge
,w factor 18144
warm shaleBOT
brazen gorge
so it's $2^{n+1} \cdot 3^{n} \cdot$ {{5 if n is odd, 7 if n is even}}
warm shaleBOT
artemetra
thorny drum
I thought task requires general formula without such conditions of if n is odd or even, like formula for any n. Is it possible, or only your formula is the final possible one?
wraith forge
you can get rid of even/odd branches in general by playing with (-1)^n, if you want
but i don't think its particularly helpful to do so
wraith forge
for example
if you want a function $f(n)$ to produce 5 for odd $n$ and 7 for even $n$, you can achieve this by $f(n) = 6+(-1)^n$
warm shaleBOT
Kaiser
royal basin
!status
obtuse pebbleBOT
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
@daring salmon
steel urchin
UU^tx
Whats the order of multiplication where x is a vector and U is a matrix
Do I get dot of u^tx or outter uu^t first?
obtuse pebbleBOT
UU^tx
Whats the order of multiplication where x is a vector and U is a matrix
...
obtuse pebbleBOT