#help-33
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$\frac{e^{\frac{1}{x}}}{(x+2)^2} \cdot -\frac{x-1}{x^3+2x^2} \cdot e^{\frac{1}{x}}$
alee
but its so strange in this form
and there is a way to get this result
.
but i dont know how
I have to find a simple result, then make f'(x) > 0
you don't get why this works?
this is just simple fraction manipulation
can you help me
pls
we have to start from here:$\frac{1}{(x+2)^2} \cdot e^{\frac{1}{x}} + \frac{x+1}{x+2} \cdot -\frac{1}{x^2} \cdot e^{\frac{1}{x}}$
alee
trying to do that
yeah but make the question clear, what's f(x)? what do you want with f'(x)?
prove that f'(x)>0?
I just want to try to simplify what I sent you as easily as possible
then i will do f'(x) > 0 myself
:$\frac{1}{(x+2)^2} \cdot e^{\frac{1}{x}} + \frac{x+1}{x+2} \cdot -\frac{1}{x^2} \cdot e^{\frac{1}{x}}$
alee
this is correct i think
yes but i didnt did that
just factor e^(1/x)
i know but how i can do that
from here?
ok $$e^{1/x}\left(\frac{1}{(x+2)^2}-\frac{x+1}{x^2(x+2)}\right)$$
moriaritie
then make the denominators same
$$e^{1/x}\left(\frac{x^2}{x^2(x+2)^2}-\frac{(x+1)(x+2)}{x^2(x+2)^2}\right)$$
moriaritie
$$e^{1/x}\left(\frac{x^2-(x+1)(x+2)}{x^2(x+2)^2}\right)$$
moriaritie
$$e^{1/x}\frac{-3x-2}{x^2(x+2)^2}$$
moriaritie
that's what you wanted right?
yes
tell me which step you didn't understand
ok here now i check others
just multiply top and bottom by the same thing
and why we do that?
we can write $\frac{1}{x}=\frac{y}{yx}$
moriaritie
to make the denominators (the bottom) of the two fractions equal
im glad you got it
yeah sure
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Is pi make able through e if you consider all the elementary operators i.e. division addition, adding other rationals?
I am pretty sure we do not know
Oh ok that's pretty interesting
What if you constrain it to e= a+b*pi
Where a and b are rationals
if you could get to here then you could get to pi (take away a, divide by b), and the other way around
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Let $d_b(x,y)=\min(1,|x-y|)$ be a metric on $\mathbb R$. Let $X$ be the set of countable sequences of real numbers and define a metric $\varsigma_b$ on $X$ by $$\varsigma_b((x_n),(y_n))=\sup{d_b(x_n,y_n):n\geq 0}.$$ Let $S\subseteq X$ be the subset of sequences such that $\lim x_n=0$. \textbf{Is $S$ open in $X$?}
Junojuno
Isn't a sequence countable by definition?
So if I’m reading this right, S is the set of sequences whose terms go to 0?
Yes
I just wanted to be sure I understood your problem correctly
Do you remember the definition of open for subsets of a metric space
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i have a questionabout a graph wuick
if i have a decreasing velocity function that represents a car breaking
what is the area under curve represent?
didtsnce traveled while breaking?
yes, since position is the antiderivative of velocity, it represents the change in position over the given time
@twin nebula Has your question been resolved?
can you explain more what the integral is
so is it the output of b-a
of the original function
well it's the output of the antiderivative
where $F$ is an antiderivative of $f$: $\int_a^b f(x) dx = F(b) - F(a)$
tatpoj
oh, that pretty much is the fundamental theorem
yeah
Do you mean like an example of a problem where you might be asked to use it?
Part 1 is like: If $F(x) = \int_a^x f(t) dt$, then $\frac{d}{dx}F(x) = f(x)$
tatpoj
notice the variable x is one of the bounds of the integral
so we basivally put x instead of t?
so that could be any variabel
if i want to take an integral in terms of another variavle
yes, and a could be any constant
ok
im learnign this today actually
so maybe ill uinderstand the use of it
i had a test today so thats why i asked the question
i wanna make sure i wrote the write thing even though it wont change anything lol
thanks tho
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ok thanks
np 👍
i would be here all day
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I just need help finding a closed form expression to solve this q: "Let n be a positive integer. How many different candy bowls can you make if you must include
n pieces of candy, subject to the following conditions:
• The number of Skittles is a multiple of five.
• Any number of peanuts is possible.
• Any number of gummy bears is possible.
• There are at most two Snickers.
• There are at most four KitKat.
• There is at most one chocolate bar.
You answer should be a closed formula."
I have the generating function btw
$\frac{1}{1-x^5} * \frac{1}{1-x} * \frac{1}{1-x} * \frac{1-x^3}{1-x} * \frac{1-x^5}{1-x} * \frac{1-x^2}{1-x}$
nosqldb
<@&286206848099549185>
i think just simplify this (and partial fractions) then you're good
huh but I need the coef
for x^n
😦
huh
I did partial fractions like in high school 💀
you have to help me through this
sure
also i assume that plus sign should be multiplication?
yes
yay
do you know how to factor like (1 - x^3) and (1 - x^2) ?
uhh (1-x)(1-x+x^2)
or sm like that
I forgot
but 1-x^2 is ez
(1-x)(1+x)
ye square diff
,w calc (1-x)(1-x+x^2)
(1 - x^3) factors into (1 - x)(1 + x + x^2)
ah I was pretty close fr
in general, (1 - x^n) factors into (1 - x)(1 + x + x^2 + ... + x^(n - 1))
so for now try to simplify this expression a bit
there's a lot of things you can factor and/or cancel
,w simplify \frac{1}{1-x^5} * \frac{1}{1-x} * \frac{1}{1-x} * \frac{1-x^3}{1-x} * \frac{1-x^5}{1-x} * \frac{1-x^2}{1-x}
what I do with that
you can extract the coefficients
you can consider each term separately
yes
mhm and
okay so we made progress
boys
$\frac{1}{1-x^5} * \frac{1}{1-x} * \frac{1}{1-x} * \frac{1-x^3}{1-x} * \frac{1-x^5}{1-x} * \frac{1-x^2}{1-x}$ = $-\frac{5}{x-1} - \frac{9}{(x-1)^2} - \frac{6}{(x-1)^3} - 1$
nosqldb
but now what to do
oh wait I'm so stupid @sand fable
why not take derivatives
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yeah i was wondering if you were allowed to use calc
well I can use wtv I want
good on figuring that out. that's pretty much how to continue from there
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,w what's the nth derivative of $-\frac{5}{x-1} - \frac{9}{(x-1)^2} - \frac{6}{(x-1)^3} - 1$
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Is my homework answer key erroneous or did I make several calculation mistakes? We're asked to determine general term of arithmetic sequences:
a. -14, -11, -8, -5, ... has answer key 3n-20 and I got 3n-17
b. 0, -3, -6, -9, ... has answer key -3n+6 and I got -3n+3
c. 4, 0, -4, -8, ... has answer key 12-4n and I got 8-4n
can you send a picture of your homework
yeah that was weird like all the other ones are off by 3
teacher probably accidentally calculated tn=a+(n-2)d or smth
thanks!
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why did they change the integral
in the second one
to 0 to pi/3
and not to the first question? - where they kept the integral the same
@upbeat oar Has your question been resolved?
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\begin{figure}[ht]
\centering
\begin{tikzpicture}[node distance={35mm}, thick, main/.style = {draw, circle}]
\node[main] (1) {1};
\node[main] (3) [right of=1] {3};
\node[main] (4) [below right of=3] {4};
\node[main] (2) [below left of=4] {2};
\node[main] (5) [left of=2] {5};
\draw (1) -- (3);
\draw (1) -- (5);
\draw (2) -- (3);
\draw (2) -- (4);
\draw (3) -- (4);
\draw (3) -- (5);
\end{tikzpicture}
\end{figure}
a) Suppose this is an electric circuit in which all edge conductances are unity.
i) By assigning columns/rows to nodes according to the numbering shown, find the graph Laplacian matrix (\mathbold K), the degree matrix (\mathbold D) and the adjacency (\mathbold W).
\textcolor{red}{\textit{\textbf{Solution}}} The degree matrix (\mathbold D) is given by a (5\times5) matrix where the (D_{ii}) entry is given by the number of nodes connecting the (i)-th node. Consequently, this is given by
$$
\mathbold D =
\begin{bmatrix}
2&0&0&0&0\
0&2&0&0&0\
0&0&4&0&0\
0&0&0&2&0\
0&0&0&0&2
\end{bmatrix}.
$$
The adjacency (\mathbold W) is given by a (5\times5) matrix where the (\mathbold D_{ij}) entry are (1) if there is an edge between node (i) and (j), and (0) otherwise. Thus, it is given by
$$
\mathbold W =
\begin{bmatrix}
0&0&1&0&1\
0&0&1&1&0\
1&1&0&1&1\
0&1&1&0&0\
1&0&1&0&0
\end{bmatrix}.
$$
The laplacian matrix (\mathbold K) is given by
$$
\mathbold K = \mathbold{D-W} =
\begin{bmatrix}
2&0&-1&0&-1\
0&2&-1&-1&0\
-1&-1&4&-1&-1\
0&-1&-1&2&0\
-1&0&-1&0&2
\end{bmatrix}.
$$
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(You may edit your message to recompile.)
First of all, is this correct?
And how can I go about calculating the voltages in nodes 3-5 given that the voltage in node 1 is unit and node 2 is grounded?
Why did you use \(…\) but $$…$$
Looks right. And just do KV=b to solve for the voltages. Are you familiar with that?
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hi
i am coding a sort of google map with SDL and C but i run into a problem
i want to make a zoom function (on the cursor)
what i do for now is
scale the image.x + 100
scale the image.y + 100
deplace the image.x according to offset.x
deplace the image.y according to offset.y
mx = mouse position x relative to screen
my = mouse position x relative to screen
W = constant Width (of the screen)
H = constant Height (of the screen)
Z = constant Zoom factor (100)
d = direction of zoom (1 or -1)
offset_x = mx / W * Z * d
offset_y = my / H * Z * d
is there a better way ?
You want to learn matrix algebra
i've watched some of 3blue1brown videos
do you have any other links ?
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thyself need help with math problem
can you answer the rest of the questions? this type of quadrilateral has two pairs of parallel sides.
I was absent during the lesson so i have no idea how to start
would be a good idea to learn the basics of quadrilaterals
I just dont get why it seems like all of its angles are same. That would mean its a rectangle
but heres an example
ohhhh ok i understand what they want now
ok so um. the capital letters are referring to the points where two sides meet (the corners)
we call those vertices (plural of vertex)
yeah you'll sometimes hear people say vertexes and you get to be smug if you know the fancy word
the lines in the middle of the shape (like the one going from P to N) are called diagonals because, well, they go diagonally
when we want to talk about a shape as a whole we just list out the vertices in clockwise order, so in that second example the quadrilateral is named ABCD. For yours you can start anywhere, just make sure you go in order clockwise
Is clockwise is going to right?
yes
oh i see then how about the second?
well the vertices are just the capital letters themselves
I mean 2) sides
oh right
well to name a side (or any line) we just say the endpoints
so for instance, one of your sides is NM
or you could say MN, those both refer to the same side
Oh so its two letters connected to each other?
it doesn't, that's kind of just a standard to make it easier to tell when you're talking about the same one
it does always have to be in order though, you can't jump around
but if you went the other way around (anticlockwise) it would be fine
oh
the last thing is angles. To name an angle, use this funny symbol ∠ and then three points that define the angle, you need a start point, a vertex, and an end point. so for example this is ∠ONM
oh yeah when you're talking about edges, you should put a line over top of it so like $\overline{NM}$
hayley!
it's like you're drawing the line that goes from N to M
to be honest i don't know exactly
i guess just to remind your reader that N and M are vertices and you're talking about a line, you're not multiplying N by M
some of this notation goes back to ancient greece
Oh i guess that does make sense
anywys opposite sides means Sides of the Opposites so i understand that but Consecutive Sides is the next im still wondering bout
consecutive means they share a vertex so they're next to each other
like NM and MP are consecutive sides
i mean why does it start at A not B
i think you could start anywhere if i understand your question right, your teacher tried to alphabetize it as much as possible i think
Oh so i can start anywhere then?
yeah it's common to start in the top left though
like CD & DA, CB & BA
yep
this exercise seems like practice in writing the same 4 letters as many ways as possible 
Oh i just understood on why the overlining now
its like a symbol like when looking for angles u put an angle
I think i understand it now tyvm
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there's a symbol for triangles as well (it's just a small triangle)
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ni
hi
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is gaussian elimination the quickest way of solving 3 variable system of equations?
I believe so
do you mean for a human or for a computer?
human
I guess it depends on the person and system then
I think for most people, it will be the fastest method
for me personally though, I find substitution quicker for most nice systems
can u help me with this problem? I learned gaussian elimination today, but I want to learn how to apply it. I simplified the bottom so that it is
$(\alpha -10)z= \beta - 16$
$2x+y+6z=0 \$
$-y-2z=-4 \$
$(\alpha -10)z=\beta -16$
This is what I have from gaussian elimination
You know when you get no solution ??
idk
i forgot
when there are no intersection points between the 3 equations
i dont have a graphic calculator tho
I got a=10 and B doesnt equal 16 for the no solution
idk if this is correct
You know matrix form of it
Like putting in a matrix
You get no solution if your last row is like [0 0 0 b]
See there is no solution if coff. Of you last equation is 0
And after equal part is non zero
i found a similar problem here. can u confirm the solution is correct??
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i am making a weighted average calculator for my bio class, did i use the correct weight for each unit
what i did is i took 32% of each unit and put it in the row with the the blue label that matched the unit name. then i took 48% of each unit and put it in the row with the purple label that matched the unit name
then i took 20% of each unit (in total 16%) and added it to the final exam row with the red label. i added 20% to that 16%, because the final is worth 20%
@stoic drum Has your question been resolved?
Looks right, except that I’d expect the final exam to have a weight of 20 out of the total 100
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hi
Need help with this
So for this question to figure out the work do I need to calculate the force first?
you could do it using potential energy, if you know about that
well this is calc 2
so idk how to do that
ah ok
can't you use work = force * distance (to avoid explictly using energy calculations)
and force is mass * gravity (I don't know what this is in imperial units)
@pearl granite Has your question been resolved?
so I would do 1,500 times 40?
that what I thought initally but wasnt sure
what gravitational acceleration?
yes, it's the same formula
but you need force
you don't have force, you have mass
so use F=ma to find force
we're not working with springs
well this was how we were taught
so i think if I just do 1,500 times 40 I should get it
the answer was 1500 times 40
thats all I needed to do
now I am not sure how to do this
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felt like this one should be simple but can't seem to get it done
I tried solving for y and got y = 12 - 6x
then because its rotated about the y-axis I solved it in terms of x and got y -12 = -6x so (y-12)/-6 = x
then since I knew that for the volume of a solid region I needed to do PI * integral from a-b of [R(x)]^2
I took the (y-12)/-6 and squared it to get
Then I solved for the zero of the function, which is given by setting y to 0, so 0 = 12 - 6x so -6x = -12 so x = 2
And so if x = 0 is a bound and y = 0 is a bound and x = 2 is our zero, we can integrate from 0 to 2
Leaving us with 61/9
But the PI on the outside needs to be remembered, so 61*PI/9
Can anyone help me with this? Where is my logic wrong?
It's not three functions
Oh I see what you mean
nvm
Sorry I misinterpreted
Yeah you get a triangle
use it to find its volume
Yeah but I am not meant to do it by the triangle
I am supposed to integrate, that's what will be on my test
pretty much the same, except with volume now you use the volume equation, so PI * integral of [R(x)]^2 where R(x) is your function that creates the graph
In the case of 2 functions creating a region in between them, it's [R(x)]^2 - [r(x)]^2 where R(x) is the function on top and r(x) is the function that lies below
im thinking we should get the area of the triangle and multiply by 2pi
other than that im stumped :/
you gotta wait for someone else to come help you im sorry
Just use the formula for area of a cone
I can't
Why?
It's a requirement that I integrate
Oh
bounds are wrong
you're rotating around the y-axis with disk's, / integrating wrt y
the bounds should be the y-coords
yes
ok
you are 100% right
I didn't realize that
I only flipped the function
not the bounds lol
ty
I still get it wrong lol
I get -4 doing it from 0 to 12
And then I multiplied it by PI to get -4PI
but it's not correct
can you take a pic of your work on paper
@trail flint Has your question been resolved?
sure 1 sec
ok here
That's the work
work on paper seems fine
you plugged your bounds into the integrand instead of the antiderivative in mathway
Oh I see I plugged it in wrong
my work was right I worked it out by hand and got 16 * PI and it was accepted
ok one last question it's of the same type
I am guessing I messed up along the way somewhere again
I will post work
This feels good to me
But I'm not sure
is this right?
@trail flint Has your question been resolved?
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This question is plaguing me
I can post my work so far
I only have 2 trains of thought for it
Yes that looks alright to me
this one up here
this ones start and stops are still in terms of x and need to change to terms of y
so 0-32?
yeah
ok so 0-32 I tried both 0-32 and 2-6 but neither worked for me
I got (-32^2)/2+36(32) so 640
and * PI
so 640*PI
but it was not accepted
What if you multiplied the pi and 640? does it typically want pi left byitself?
oh I got it 🤦♂️
It is 640
but...
-4(32)
?
Because the bounds are x = 2 and x = 6, but we rotate around x = 0
so we don't want from x = 0 to x = 2
therefore we subtract x = 2 function from the given function
ah, yeah. right minus left
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in this form of u substitution, the sub taken is 3-x. How is that affecting the x in the original equation if the u sub is 3-x and not just x. If this is some kind of rule, can someone explain how it works?
right i get this, but should this be done in every situation of x? for example, if there was another x somewhere in the qeuation somehow would i do it then to? do i do it to every x every time?
you are changing your integration variable from x to u
so you should always accordingly change every x to be in terms of u yes
You want to pick something that makes it easy for you. If you made u = x, the du = 1 dx, but that doesn't make it easy with the the 3 - x part
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Suppose that you want to have a $39,000 retirement fund after 39 years. How much will you need to deposit now if you can obtain an APR of 11.3%, compounded daily? Assume that no additional deposits are to be made to the account.
$472.47
$639.14
$475.80
$411.67
None of the above.
Since irl nobody would tell you which angle is the right angle, how to identify the hypotenuse, opposite and adjacent without knowing the right angle? What if the theta is placed on right angle and the hypotenuse is oppositing the theta?
!occupied
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What have you tried?
Identify the possible outcomes when two dice are rolled to get a sum of 7.
my work
@lavish kernel this is myy work could you please check it
https://www.youtube.com/watch?app=desktop&v=H02B3aMNKzE this video is an okay intro
Tutorial on how to calculate conditional probability (Bayes Theorem) for two events P(A), P(B), P(B|A) with two examples using
Playlist on Probability
http://www.youtube.com/course?list=EC482E6C6B5F0A30E7
http://www.facebook.com/PartyMoreStudyLess
Created by David Longstreet, Professor of the Universe, MyBookSucks
http://www.linkedin.com/in/...
Yeah, you corrected calculated it using the formula for conditional probability
thank you
thank you...
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Can I have some help?
Oh
ITS YOU
?
waddup
If x² - 2ax + a²=0, find the value of x
-
a
Yes
yea
I realised I forgot all my maths which isn't good, so revising it all again hahaa....
To find the value of $\frac{x}{a}$, we will solve the given quadratic equation $x^2 - 2ax + a^2 = 0$ and then divide the solution $x$ by $a$.
David K.
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I have tried converting the x denominator into x*sinx/sinx and put it in the exponent, I just got ln(1+sinx)^sinx and sat like a duck, the answer is supposed to be 1 but I kept getting 0
I honestly thought there'd be a form of some sort for ln(1+x)^x, but nope, it is always x and 1/x and vice versa
God I hope to learn that tomorrow because it seems really helpful. I've been asking limits in the past 3 days and I always encountered that name.
Which I did, I had x*sinx/sin, which is sin x/ x sin x when flipped, right?
you want to use the result (1+sin(x))/sin(x) at 0 is 1
Ohhhhh, and I disect x and sin x and place eachother accordingly.
Yeah I got it, 1*1
Morning limits. haha. Thanks for the help!
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Hello
is there any theorem saying
how to quickly find roots of a polynomial
(write as a product of polynomials)
approach with rational root theorem i suppose
what is it
way of searching for candidates for rational roots
moreover, what does it say? like how can i use it
it'll be easier for you to look it up
i did
try applying it here
after you find one of the roots, you can use long division
then repeat as needed
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Is n+1,n+2,n+3 also correct?
right ty
np
Do i have to state the root 11 thing
or is it implied
I never seen this before 😵💫
do you know why the root thing works?
nope
say you needed the factors of 100
you usually have a small number * a big number = 100
you meet in the middle with 10 * 10 = 100
then have a big number * a small number = 100
when you go through the factors in order
1 * 100 = 100
2 * 50 = 100
...
10 * 10 = 100
...
50 * 2 = 100
100 * 1 = 100
youll notice the big * small part of the list is just a repeat of the small * big part of the list
you already went through all the factors on the small side
so they have to reappear on the big side
that means the big side is redundant
the separation between them is that 10 * 10 = 100, where both are equal
and so you can stop after √100 = 10
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np
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i don't know how to start this, the method i used was to try and isolate R and then take the partial derivative of r1 but I got stuck when I saw dr/dr1 on both sides
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I don't know where to exactly start for b
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can somone explain why as x goes to infinite
it became ln 4
from what i see its should jsut be ln(1) as infinite cancel out
Try writting L=ln(f(x) )so e^L=f(x)
Evaluate it now
And then Take the log of both sides
ok
Seeing the second is equivalent, may be easier to see if you write
[
\ln\qty( \frac{2x^2 + 3}{(x + 1)^2 } ) = \ln\qty( \frac{2x^2 + 3}{x^2 + 2x + 1} ) = \ln\qty( \frac{2 + \frac3{x^2}}{1 + \frac2x + \frac1{x^2}} )
]
@glass silo
oh
that make more sense
cause what i done was infinite +3 =infinite
yes exactly
so top bottom both becames infinite
but it actully 3/infinite
so it becames 0
yeah
thgat make so much sense
I think we forgot to square it
but it does not matter
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I have been stuck on this problem for a few hours and found an online solution that I think is wrong. I've transcribed it out to make it easier to read.
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@gilded sierra Has your question been resolved?
@gilded sierra Has your question been resolved?
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Why 2xy=12
divide both sides by 2x? (assuming x is not zero)
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im not sure
Are you familiar with the fundemental theorem of calculus
well
This is FTOC Part 1
i know them but i dont konw how to apply them. i just learn the steps to solve problems
well okay its really two things that you need:
FTC and chain rule
so first of all we establish FTC
,, \dv x\int_a^x \m ft \dd t = \m fx
for reference this is what i learned
yeah the first one is all you need
when do you use the second one?
the second one is basically the first in disguise
Yes! Use this first one here
,, \int_{\m hx}^{\m gx} \m ft \dd t = \int_{\m hx}^a \m ft \dd t+ \int_a^{\m gx} \m ft \dd t = -\int_{a}^{\m hx} \m ft \dd t + \int_a^{\m gx} \m ft \dd t
im not good at understandn letter notation. it seems like the first intergral n's should be a to a.
since it goes from
a to hx and then gx to a
oh
i read it wrong
yes
it makes sense
i dont understand how to apply it. or rather, i dont understand what tells us to apply it. what part in the problem indicates to us we should use this theorem. what should i look for to know to use this
anyone??
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can someone explain me why 1/1 * 2 + 1/2 * 3 is equal to 1/1 - 1/2 + 1/2 - 1/3 + 1/3
work backwards
hmm?
@brazen flame Has your question been resolved?
$\frac{1}{2} - \frac{1}{3} = \frac{3}{2\cdot 3} - \frac{2}{2\cdot 3} = \frac{3-2}{2\cdot 3} = \frac{1}{2\cdot 3}$
artemetra
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how do you do this problem?
split it up first
idk how to apply this to this problem
,, \sum_{k=1}^n a_k + b_k = \sum_{k=1}^n a_k + \sum_{k=1}^n b_k
well we can do it a step at a time
but do you understand how to do this
no 😦 so i got n(n+1)(2n+1)/6 + 5n(n+1)/2 + 8(n) so far but idk what the next step is
okay so you do know how to do it since you implicitly did exactly that
anyways that's correct
the rest is just algebra
ok yayy
like for the next part i see that they got 9n^2+9n and 24n and idk what number they mulitiplied it by?
you dont even need to do anything else. Your answer suffices. Although simplification is possible
ive had several attempts at this question and they want the answer to be in the same 1/3n(n^2+...+...) form
are you familiar with the GCF (greatest common factor)
ok so you have [
\f{n(n+1)(2n+1)}6+5\cd\f{n(n+1)}2 + 8n
]
you want to get rid of the fractions. To do that, you can multiply and divide by the least common factor between 6 and 2 (the denominators)
sooo im multiplying by 2...?
that's not the LCM
oh is it 6
yeah
so like now
[
\f{n(n+1)(2n+1)}6+5\cd\f{n(n+1)}2 + 8n = \c b{\f66\cd}\bs{\f{n(n+1)(2n+1)}6+5\cd\f{n(n+1)}2 + 8n}
]
try distributing in like, that 6 in the numerator while keeping out the 1/6
so 1/6[2n^3+3n^2+n]+(30n^2+30n)+48n ... ?
oh sorry im really lost 
Gamer Sans The Gamer Skeleton
it's $\frac{1}[6} \cdot n(n+1)(2n+1) + 15 \cdot n(n+1) + 48n$
if im not stoobid
```Compilation error:```! Extra }, or forgotten $.
l.49 it's $\frac{1}[6}
\cdot n(n+1)(2n+1) + 15 \cdot n(n+1) + 48n$
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l```
Gamer Sans The Gamer Skeleton
,align
\f{n(n+1)(2n+1)}6+5\cd\f{n(n+1)}2 + 8n &= \c b{\f66\cd}\bs{\f{n(n+1)(2n+1)}6+5\cd\f{n(n+1)}2 + 8n}\
&=\f16\bs{6\cd\f{n(n+1)(2n+1)}6+6\cd5\cd\f{n(n+1)}2 + 6\cd8n}
@harsh vigil this is what you have currently. Do you agree yes or no?
yes
okay, so for now, don't distribute the parentheses
simplify the numbers only
what do you get?
1/6[6n(n+1)(2n+1)/6 + 30n(n+1)/2 + 48n] ??
oh 15
1
so, your updated version?
1/6[1n(n+1)(2n+1) + 15n(n+1) + 48n] ?
15n^2+15
nuh uh
yeah
okk
so, your updated version now?
1/6[n(n+1)(2n+1) + 15n^2+15n + 48n]
1/6[n(n+1)(2n+1) + 15n^2+63n]
yeah
so now going to the left
distribute out all three parentheses
or do you want me to guide you with that?
wait so do i multiply (n+1)(2n+1) by n first then 1/6?
n(n^2+n)(2n^2+n)...
pause
isnt 1n = n??
n goes only once
oh what
,, a(b\cd c) \c r{\ne} ab \cd ac
im sorry how do u distribute it then? i forgot most of my algebra
so like
distribute out n(n+1) what do you get
n^2+n...?
ohhhh thats what you meant
yes
i thought the n goes into both parenthesis
no like
for multiplication its only once
you are confusong it with addition
for addition yes it does distribute
a(b+c) = ab + ac
a(b*c) != ab * ac
!= means not equal to
ohh ok
2n^3+3n^2+n
FINALLY
so your updated version?
we were here btw
we are not done yet, but almost
1/6[(2n^3+3n^2+n)+15n^2+63n]
64n?
18n^2
and, at last, your updated final version is?
1/6(2n^3+18n^2+64n)
lmaoo were almost there at least
what's the greatest common factor between 2n^3, 18n^2 and 64n
2?
i mean, yeah, but you are missing something
2n
1/6 2n(n^2+9n+32)
yess and whats 2 divided by 6 (in fraction)
1/3
so 1/3n(n^2+9n+32) ?!!?!?!
so your last thing is[
\f n3(n^2 +9n +32)
]
yeah although i hope you didnt mean that n to be in the denominator because it isn't
no cus this is my last attempt and i don't want to get the format wrong :(((!!
i mean
we can check for you lol
let's see
,w sum of i^2 + 5i + 8 from 1 to n =n/3(n^2+9n+32)
lifesaver fr fr
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I just want to understand the part why it can assume lambda to be an specific thing, and then prove the unequally for all lambda in R
It’s in spanish but ill explain the important parts:
It is trying to prove the cauchy inequality
Starting with the square of the norm of x+lambda * y, which is always > 0 because its an square
And then, it choses that one specific lambda, which will be enough to make everything work out and prove the inequality
My question is that
If you chose that one specific lambda, does it mean that the inequality will only work if the lambda is only the one you chose?
well the first inequality is valid for all lambdas
the actual inequality you want to prove doesnt depend on lambda at all
I see
So it does work
Now if there was a lambda in the inequality (like the cauchy one)
We shouldn’t assume it works for all lambdas in R?
well depends on the statement of the inequality
it could be "for all lambda this inequality holds"
or "there exists a lambda for which the inequality holds"
the inequality is true
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Can you not use the same tehod you used for a and c?
Here's a hint that may or may not help (depends on what method you use), there's one eigenvector you can kind of tell by inspection
@minor verge Has your question been resolved?
i figured out b the same way, but idk with a 3x3 matrix
Whats the method you've been taught to use?
this is what i did for part a


