find the average rate of change of f over the interval -5<=x<=5
#help-49
lavish venture
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lavish venture
$\frac{f(b) - f(a)}{b-a}$
grand pondBOT
knief
lavish venture
is average rate of change of f over (a,b)
upbeat plinth
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lavish venture
slope of secant line
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lavish venture
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never
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fickle oriole
Lock in keno
lavish venture
12th
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fickle oriole
Kenzo
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fickle oriole
Use talk tuah threorom for this problem
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fickle oriole
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Lock in wtf man
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fickle oriole
Do you skip the tests?
lavish venture
we should just close this
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fickle oriole
these help channels are useless
😭🙏
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fickle oriole
You average like 1-4 questions a night
lavish venture
more
fickle oriole
Maybe you need to try solving the problems yourself more
Because what is probs happening is
lavish venture
fickle oriole
You just blank during test
lavish venture
because knief solves them all for him
upbeat plinth
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lavish venture
you probably look at a question, don’t know what to do immediately
then go to a help channel
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lavish venture
when you should bash your head into the wall
trying to solve it
for at least 5-10 minutes of no ideas
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fickle oriole
And also don’t have the answer book right next to you 😭🙏
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lavish venture
man reads the solutions
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fickle oriole
It only gets harder 😭🙏
lavish venture
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lavish venture
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brainrot
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lavish venture
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nah
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lavish venture
doesn’t matter
upbeat plinth
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fickle oriole
More to life than math is equivalent to like a 90 🙏😭
upbeat plinth
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fickle oriole
A 73 is sleeping through the class
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fickle oriole
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L mindset
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fickle oriole
“I decided before I even got to college that math is too hard”
lavish venture
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sure
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flint obsidian
Yes
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“I decided before I even got to college that math is too hard”
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lavish venture
and this is mean value theorem
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flint obsidian
Look at the slope graph and count how many times y=3/5
upbeat plinth
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upbeat plinth
or you’re not up to that yet
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lavish venture
funny rock was probably his friend
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fickle oriole
Bro acting like he isn’t checking the answer key😭🙏
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lavish venture
😹😹😭🙏🏻
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fickle oriole
Bro has to buy a chegg sub now😹😹😭🙏
upbeat plinth
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upbeat plinth
Look at the slope graph and count how many times y=3/5
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lavish venture
legendary set up
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lavish venture
yuh
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lavish venture
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they’re building up to it with this question
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they’re building up to it with this question
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fickle oriole
u dont need mvt
fickle oriole
also mvt doesnt even work
lavish venture
i know
lavish venture
it just guarantees the existence
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fickle oriole
This lowkey should be a light problem
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fickle oriole
After doing derivatives for like 2 months
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lavish venture
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have you been reading our messages
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fickle oriole
All you need to know from geometry is trig
Which you learn in precalc
And all you need for algebra 1 is covered in precalc
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lavish venture
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huh
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lavish venture
yea but if you never did that your fundamentals are probably just shit
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fickle oriole
No you weren’t
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fickle oriole
Precalc just strengthens your algebra skills
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fickle oriole
Because half the time isn’t not the calc that’s hard
It’s the algebra that messes you up
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fickle oriole
Ye you can fix that pretty easily
When you see yourself struggling with something in algebra
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fickle oriole
Take 15 minutes to review
fickle oriole
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You might need it😭🙏😹😭
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fickle oriole
In all seriousness can you factor
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fickle oriole
Beauty of living in the modern age
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fickle oriole
That is factored 🙏😭😹😹😹😹
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fickle oriole
youre mad geeked ✌️😭😭
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midnight plankBOT
midnight plankBOT
Available help channel!
meager ore
Let ( f : I \to \mathbb{R} ) be a function with the intermediate value property. If for every ( c \in \mathbb{R} ) there exists only a finite number of points ( x \in I ) such that ( f(x) = c ), then ( f ) is continuous.
midnight plankBOT
Let \( f : I \to \mathbb{R} \) be a function with the intermediate value propert...
grand pondBOT
Halex
meager ore
what does it mean "only a finite number of points x in I"
how do I express it mathematicaly?
Like how do I use it as an hypothesis
how do I choose a finite number of points in an interval I?
midnight plankBOT
@meager ore Has your question been resolved?
meager ore
how do I choose a finite number of points in an interval I?
<@&286206848099549185>
blissful summit
so a mathematical way to describe a random number generator within an interval?
idk if its possible to have a function written down that gives u a truly random number without any input from the user
i think there are pseudorandom number generators that can be written down on paper, but it needs a seed chosen by the user
rose trout
<@&286206848099549185>
What are you trying to do? Understand the statement above? Prove it?
The idea is that when a function is continuous on an interval, it has the intermediate value property (that's called the Intermediate Value Theorem).
But the converse isn't true, i.e. functions that have the intermediate value property on an interval aren't necessarily continuous.
In particular, the added property above, that the functions only takes any value a finite number of times, ensures that the function is continuous if it has the IVP.
meager ore
What are you trying to do? Understand the statement above? Prove it?
Understand it to prove it
meager ore
The idea is that when a function is continuous on an interval, it has the interm...
Let $\epsilon > 0.$ For every $c \in \mathbb{R}$ there exists only a finite number of points $x \in I$ such that $f(x) = c$ means that $|x - c| < \delta \implies |f(x) - c| < \epsilon$
grand pondBOT
Halex
meager ore
this is not a finite numbers of x tho
rose trout
Well the hypothesis includes the intermediate value property, so you should expect to use it somewhere.
It won't be a one line proof
meager ore
Well the hypothesis includes the intermediate value property, so you should expe...
I'm not even sure how to express the hypothesis in a mathematical way
Might you help me with the proof?
rose trout
It's not an exactly easy proof iirc
meager ore
It's not an exactly easy proof iirc
It would be awesome if you could help me to know how to get started
midnight plankBOT
@meager ore Has your question been resolved?
midnight plankBOT
@meager ore Has your question been resolved?
twilit field
midnight plankBOT
twilit field
Hmm, I think I can do this, but this is just a transformation between two fields
runic hamlet
F^n and F^m arent fields
twilit field
F^n and F^m arent fields
oh right
We preface this by assuming $\mathbf{F}= \R$
\
$T(x_1,\dots,x_n)=T(x_1,0,\dots,0 \dots)+T(0,x_2,\dots,0) +\dots + T(0,0,\dots,x_n)$ By the definition of linearity.
\
We then have $T(x_1,0,\dots,0 \dots)+T(0,x_2,\dots,0) +\dots + T(0,0,\dots,x_n)=T(1 \cdot x_1,0,\dots,0 \dots)+T(0,1 \cdot x_2,\dots,0) +\dots + T(0,0,\dots,1 \cdot x_n)$
\
We know that both $x_i and A_{j,k}$ are scalars, so
\
$T(x_1,0,\dots,0 \dots)+T(0,x_2,\dots,0) +\dots + T(0,0,\dots,x_n)=T(A_{1,1}x_1,0,\dots,0 \dots)+T(0,A_{2,2}x_2,\dots,0) +\dots + T(0,0,\dots,A_{n,n}x_n)= x_1T(1,0,0 \dots,0)+\dots+ x_nT(0,0,0 \dots,1) $
\
We know that this is a mapping :$\R^n \to \R^m$
\
Thus $T(1,0,\dots,0)= (A_{1,1},A_{1,2} \dots , A_{m,1})$ , so $x_1 T(1,0,\dots,0)=. (x_1A_{1,1},x_1A_{1,2} \dots ,x_1 A_{m,1})$for instance.
\
Repeating this for all the standard basis of $\R^n$.
\
We arrive at the desired result
twilit field
F^n and F^m arent fields
But its the cartesian product ofi a field, right?
runic hamlet
sure
grand pondBOT
A dense set
twilit field
Is this proof fine?
We preface this by assuming $\mathbf{F}= \R$
\
$T(x_1,\dots,x_n)=T(x_1,0,\dots,0 \dots)+T(0,x_2,\dots,0) +\dots + T(0,0,\dots,x_n)$ By the definition of linearity.
\
We then have $T(x_1,0,\dots,0 \dots)+T(0,x_2,\dots,0) +\dots + T(0,0,\dots,x_n)=T(1 \cdot x_1,0,\dots,0 \dots)+T(0,1 \cdot x_2,\dots,0) +\dots + T(0,0,\dots,1 \cdot x_n)$
\
We know that both $x_i and A_{j,k}$ are scalars, so
\
$T(x_1,0,\dots,0 \dots)+T(0,x_2,\dots,0) +\dots + T(0,0,\dots,x_n)=T(A_{1,1}x_1,0,\dots,0 \dots)+T(0,A_{2,2}x_2,\dots,0) +\dots + T(0,0,\dots,A_{n,n}x_n)= x_1T(1,0,0 \dots,0)+\dots+ x_nT(0,0,0 \dots,1) $
\
We know that this is a mapping :$\R^n \to \R^m$
\
Thus $T(1,0,\dots,0)= (A_{1,1},A_{1,2} \dots , A_{m,1})$ , so $x_1 T(1,0,\dots,0)=. (x_1A_{1,1},x_1A_{1,2} \dots ,x_1 A_{m,1})$for instance.
\
Repeating this for all the standard basis of $\R^n$. and then adding them up
\
We arrive at the desired result
grand pondBOT
A dense set
runic hamlet
so you replaced all the 1's by A_ij? why
midnight plankBOT
@twilit field Has your question been resolved?
twilit field
the 1s are transformed to $\R^m$
grand pondBOT
A dense set
twilit field
R^m elemennts
twilit field
We preface this by assuming $\mathbf{F}= \R$
\
$T(x_1,\dots,x_n)=T(x_1,0,\dots,0 \dots)+T(0,x_2,\dots,0) +\dots + T(0,0,\dots,x_n)$ By the definition of linearity.
\
We then have $T(x_1,0,\dots,0 \dots)+T(0,x_2,\dots,0) +\dots + T(0,0,\dots,x_n)=T(1 \cdot x_1,0,\dots,0 \dots)+T(0,1 \cdot x_2,\dots,0) +\dots + T(0,0,\dots,1 \cdot x_n)$
\
We know that both $x_i$ and $A_{j,k}$ are scalars, so
\
$T(x_1,0,\dots,0 \dots)+T(0,x_2,\dots,0) +\dots + T(0,0,\dots,x_n)=T(A_{1,1}x_1,0,\dots,0 \dots)+T(0,A_{2,2}x_2,\dots,0) +\dots + T(0,0,\dots,A_{n,n}x_n)= x_1T(1,0,0 \dots,0)+\dots+ x_nT(0,0,0 \dots,1) $
\
We know that this is a mapping :$\R^n \to \R^m$
\
Thus $T(1,0,\dots,0)= (A_{1,1},A_{1,2} \dots , A_{m,1})$ , so $x_1 T(1,0,\dots,0)=. (x_1A_{1,1},x_1A_{1,2} \dots ,x_1 A_{m,1})$for instance.
\
Repeating this for all the standard basis of $\R^n$. and then adding them up
\
We arrive at the desired result
grand pondBOT
A dense set
twilit field
.close
midnight plankBOT
midnight plankBOT
Available help channel!
last slate
Hi guys
I am new to working with Lambert's W functions and I would like to get my answer checked
this is the question:
$$3^x \cdot e^x = 27$$
midnight plankBOT
Hi guys
I am new to working with Lambert's W functions and I would like to get...
grand pondBOT
Edmund Cloudsley
last slate
I shall post my answer here
upper crag
no lambert W should be needed for this question, standard log will work
last slate
yeah I noticed that
but the question specifically asks for Lambert's W for some reason
dunno
why
upper crag
can you post a picture of the original question
last slate
yeah sure
just gimme a sec
This is from a notion page I found online tho
not from a proper author or something
if we were solving this using logs then
$$(3e)^x = 27$$
$$x = \log_{3e} 27$$
grand pondBOT
Edmund Cloudsley
last slate
this should be the answer right?
upper crag
not from a proper author or something
does this mean chatgpt came up with the question
last slate
I got this from a notion page my friend shared with me
But I have no clue where they got it from
could be given that this doesn't required W functions in the first place
upper crag
$$(3e)^x = 27$$
$$x = \log_{3e} 27$$
yeah that's pretty much it
midnight plankBOT
upper crag
yup you're welcome
midnight plankBOT
Available help channel!
vagrant crest
midnight plankBOT
vagrant crest
f(0)= -1
f(2)= 3
there wouldnt be anything in between( 0-2) x values that would get outside of (-1, 3) y values no?
midnight plankBOT
@vagrant crest Has your question been resolved?
vagrant crest
<@&286206848099549185> 
midnight plankBOT
@vagrant crest Has your question been resolved?
vagrant crest
.close
raven dirge
not sure if this is the right place to ask, but when I have a DFA (deterministic finite automata), can I have 2 accepted states?
midnight plankBOT
not sure if this is the right place to ask, but when I have a DFA (deterministic...
last slate
I forgot nearly all of automata theory but you’ll probably have a lot of luck asking this in #foundations or #discrete-math
robust isle
not sure if this is the right place to ask, but when I have a DFA (deterministic...
yes
twilit field
midnight plankBOT
twilit field
I wish to prove this
Rephrased a bit, this is
Given that $a \in \Z$. $3 \mid (a^3-a)$
grand pondBOT
A dense set
lethal path
factor a^3 - a
twilit field
je sais la, mais merci
It's easy to see that $a^3-a = a(a+1)(a-1)$. We thus have to prove $3 \mid [a(a+1)(a-1)]$
\
It follows that $(a-1)(a)(a+1) = \frac{(a+1)!3!}{(a-2)!3!} = ^{a+1}C_{3}3!$.
\
It's trivial to verify that $3 \mid (3! ^{a+1}C_{3})$.
\
Attentive readers will notice that this proof is only valid for $\mathbb{W}$.
\
but we have a trick up our sleeve .
\
if $a^3-a<0$, we simply re-write it as $-(a-a^3) = -(a(1-a)(1+a)$.
\
After which the proof is identical, we should note that we've used a previously proven result here.
\
that $b \mid a \implies b \mid -a$
It thus follows that , given $a\in \Z$, $a \equiv a mod(3)$
grand pondBOT
A dense set
junior flower
junior flower
this only proves it for the naturals
weird proof but ig it’s fine for a >= 2 yes
twilit field
It's easy to see that $a^3-a = a(a+1)(a-1)$. We thus have to prove $3 \mid [a(a+1)(a-1)]$
\
It follows that $(a-1)(a)(a+1) = \frac{(a+1)!3!}{(a-2)!3!} = ^{a+1}C_{3}3!$.
\
It's trivial to verify that $3 \mid (3! ^{a+1}C_{3})$.
\
Attentive readers will notice that this proof is only valid for $\mathbb{W}$.
\
but we have a trick up our sleeve .
\
if $a^3-a<0$, we simply re-write it as $-(a-a^3) = -(a(1-a)(1+a)$.
\
After which the proof is identical, we should note that we've used a previously proven result here.
\
that $b \mid a \implies b \mid -a$
It thus follows that , given $a\in \Z$, $a \equiv a mod(3)$
grand pondBOT
A dense set
twilit field
weird proof but ig it’s fine for a >= 2 yes
oh true
junior flower
rather than say “attentive readers will notice…” after the fact, why not just say “suppose a >= 2” first
junior flower
wdym?
grand pondBOT
A dense set
grand pondBOT
A dense set
junior flower
but yes it would be better to write something that doesn’t involve cases
grand pondBOT
A dense set
junior flower
induction would be very unnatural here
twilit field
Or I could prove that the product of any $n$ whole numbers is divisible by 3
grand pondBOT
A dense set
twilit field
and then use the fact that if $b \mid a \implies b \mid |a|$
grand pondBOT
A dense set
junior flower
eh?
twilit field
if$b \mid a \implies b \mid -a$
grand pondBOT
A dense set
junior flower
Or I could prove that the product of any $n$ whole numbers is divisible by 3
what does this mean
twilit field
consider 3 consecutive wholes , $n,n+1,n+2$. We wish to prove $ 3 \mid [(n)(n+1)(n_2)]$. if $n \geq 1$, we can re-write this as $\frac{(n+2)!3!}{n-1)!3!} = ^{n+2}C_{3}3!$.
which is divisible by $3$, If $n,n-1,n+1 =0$, then we have $3 \mid a^3-a$ trivially.
\
We now consider the case wherein $n, n-1,n+1 <0$.
\
In this scenario we consider n(n+1)(n+2). But we've already proven $3 \mid |(n)(n+1)(n+2)|$, thus $3 \mid (n)(n+1)(n+2)$.
grand pondBOT
A dense set
twilit field
what does this mean
Is this fine?
junior flower
your cases are weird
junior flower
**A dense set**
what does this mean?
should this just say “if n = 0 then 3 | n^3 - n”
i also don’t know what’s going on in the last case
fair kayak
"in how many different ways can the 9 letters of the word TELESCOPE be arranged so that there are exactly two letters between the T and the C?"
guys, i only know 7!/3!, i don't know what to do next
midnight plankBOT
"in how many different ways can the 9 letters of the word TELESCOPE be arranged ...
lethal path
"in how many different ways can the 9 letters of the word TELESCOPE be arranged ...
try grouping as (T??C) ?????
so the bracket is 1 object, so 1 object + 5 other letters = 6 objects and hence 6!
strong lava
you could also have C??T
lethal path
ah wait there are repeated letters
strong lava
yes
hence I don't want to participate
I say while working on a combinatorial problem
lethal path
it's an A level question, found the paper
strong lava
I didn't say I can't do it, I merely said I'm not interested
keen saddle
"in how many different ways can the 9 letters of the word TELESCOPE be arranged ...
You should first arrange the letters with a condition
In this case T and C
fair kayak
try grouping as (T??C) ?????
so the bracket is 1 object, so 1 object + 5 other ...
ohhh
fair kayak
you could also have C??T
yes 😦
fair kayak
it's an A level question, found the paper
ure right i am in the cambridge curriculum
fair kayak
You should first arrange the letters with a condition
okiee
visual tiger
"in how many different ways can the 9 letters of the word TELESCOPE be arranged ...
you have two choices for whether T comes before or after C, and you have 6 choices for placing whichever letter comes first
after that, it's about ordering the remaining letters from left to right the way you want...
lethal path
try grouping as (T??C) ?????
so the bracket is 1 object, so 1 object + 5 other ...
I think you need to split into cases, so there can be 0 Es, 1 Es, or 2 Es in the bracket
lethal path
after that, it's about ordering the remaining letters from left to right the way...
or yeah that's easier
keen saddle
you have two choices for whether T comes before or after C, and you have 6 choic...
E is repeated
But you can just do /3! So yeah
visual tiger
E is repeated
so the number of orderings of the remaining letters will be found accordingly
keen saddle
I think you need to split into cases, so there can be 0 Es, 1 Es, or 2 Es in the...
Cases will be harder lol
@fair kayak any progress?
keen saddle
Show your working
fair kayak
the same Es
lethal path
you just need to divide by 3! as you said at the beginning
fair kayak
fair kayak
Can you explain why you multiply 2! twice?
the first 2! is for the position of T and C
like it can be TxxC or CxxT
keen saddle
Yup correct
fair kayak
and the xx can be xx 
keen saddle
Hmm that is actually handled in the 6!
visual tiger
not completely no
keen saddle
Yes it would become 7 but yeah it is
visual tiger
you didn't even choose which letters go in between T and C
fair kayak
Hmm that is actually handled in the 6!
oh 😭
fair kayak
you didn't even choose which letters go in between T and C
how to do that
visual tiger
Yes it would become 7 but yeah it is
and still missing something
visual tiger
and another times 6
visual tiger
how to do that
after placing the T and C
just view placing the remaining letters as in a straight line
like if you have T__C_____
there are 7 letters to arrange in the empty spots
fair kayak
or yeah that's easier
thankyouthankyou!!
fair kayak
and another times 6
what is the times 6 for?
keen saddle
I was wondering that too
keen saddle
but it is correct based on the mark scheme
The into 6 is correct?
fair kayak
The into 6 is correct?
yes
grand pondBOT
visual tiger
you can't place the T and C anywhere you want, since they need to FIT
2 choices for whether T or C is first
6 choices for placing letters T and C
and 7!/3! choices for the other letters
@fair kayak
keen saddle
Ohhh
midnight plankBOT
@fair kayak Has your question been resolved?
fair kayak
you can't place the T and C anywhere you want, since they need to FIT
ohhhh
fair kayak
and 7!/3! choices for the other letters
alrighttt thank you so so much
midnight plankBOT
@fair kayak Has your question been resolved?
midnight plankBOT
midnight plankBOT
Available help channel!
sterile stream
Hi!
I was trying to use proof by contradiction to proove that [\lim{n \to \infty } \sin(n)] does not exist. I saw some proofs online that started out by stating: [\lim{n \to \infty } \sin(n) = g] => [\lim_{n \to \infty } \sin(n+1) = g]. What is the reasoning behind that implication? Is it that if n approaches infinity then if we increment the argument of sin we still get a sin of an infinitely large number so the limit stays the same? Is there a way to prove it using trig identities?
midnight plankBOT
Hi!
I was trying to use proof by contradiction to proove that [\lim{n \to \infty...
halcyon trellis
It basically follows from the definition of limit
sterile stream
How so? Later on in the proof i got: [\lim{n \to \infty } \cos(n+1/2) = 0] and I implied that [\lim{n \to \infty } \cos(n) = 0]. My professor told me that I need to reason that implication for the proof to be correct
midnight plankBOT
@sterile stream Has your question been resolved?
lofty girder
Sin(x) approaching inf will be just oscillating between 1 and -1
sterile stream
i know that but for the sake of the proof i assumed that the limit does exist to later come to a contradiction proving that the limit does not exist
midnight plankBOT
@sterile stream Has your question been resolved?
umbral scroll
can someone explain to me why if G is the center of gravity of a triangle vector(GA) + vector(GB) + vector(GC) = vector(0)
midnight plankBOT
can someone explain to me why if G is the center of gravity of a triangle vector...
elder zephyr
can someone explain to me why if G is the center of gravity of a triangle vector...
thats the definition, so huh
umbral scroll
thats the definition, so huh
uhm well the definition i have is that it is the point of intersection of the medians
elder zephyr
uhm well the definition i have is that it is the point of intersection of the me...
well for a triangle, yes it is
umbral scroll
well for a triangle, yes it is
yes but how is the litteral definition equivalent to the vector definition
umbral scroll
yes but how is the litteral definition equivalent to the vector definition
<@&286206848099549185> pls ?
elder zephyr
yes but how is the litteral definition equivalent to the vector definition
welp idk
umbral scroll
<@&286206848099549185> 😢
ornate obsidian
@elder zephyr how do I get the role that “does nothing”?
umbral scroll
<@411882877933060109> how do I get the role that “does nothing”?
mind helping me out here pls
ornate obsidian
Oh yeah right mb
ornate obsidian
mind helping me out here pls
Might not be very helpful rn tho
My brain is fogged up
Yeah I can’t help sorry
Gl on ur quest for knowledge
midnight plankBOT
@umbral scroll Has your question been resolved?
midnight plankBOT
@umbral scroll Has your question been resolved?
cerulean igloo
for any two of three vectors, you can add eachone with half of edge which opposite to rest vector, and then add them up. because those two temporary added vector are contrast on orientation and same with length, so the result will be same as just add these two vector up. Result vector will be double on the extension of rest vector to appositive edge. Since we know that this part's length is half of rest vector, so the final result adding them up will be zero vector.
cerulean igloo
yes but how is the litteral definition equivalent to the vector definition
if you're searching for how to get vector definition from medians definition.
hushed hinge
oh it's the bot nevermind
midnight plankBOT
@umbral scroll Has your question been resolved?
umbral scroll
oh it's the bot nevermind
?
umbral scroll
and also your name
is that some sort of problem to you ?
umbral scroll
it's a username i made long time ago
umbral scroll
before i had the mental abiliy to make a username that is not stupid
umbral scroll
like math related
oh nah it's my discord username
i didnt make it specially for this server
hushed hinge
but why is the channel name your username
umbral scroll
but why is the channel name your username
have you ever been in this server ?
hushed hinge
no Im new
umbral scroll
you should take a look at other channels
help channels
and i didnt quite get it when you said nevermind it's the bot
hushed hinge
I wrote a message before that I deleted I thought the bot name was that
doesnt matter tho I was just curious sorry if I bothered you
last slate
i cant find any function which satisfies these conditions
midnight plankBOT
i cant find any function which satisfies these conditions
last slate
are there any? if there are not how do i prove it? do i suppose that a function adheres to all of these and use the definitions to prove that a contradiction arises?
winter elbow
what have you tried
proud island
I would try to sketch the curve
and looking for the function which is similar to the curve (Note that the curve needs to obey the properties)
last slate
why is it clear?
last slate
none, this is a a calculus one course. i either have to state the function and explain why it follows the condtions (doesnt need to be rigorous). if i need to disprove it my idea is to assume a function f satisfies all confitions and show that there is a contradiction
using the defintions of the properties
subtle zinc
i think i have a counterexample
it's a bit crazy
$$f\left(x\right)=\frac{1}{x\left(\sin\left(\frac{1}{x}\right)+1\right)}$$
grand pondBOT
locally ringed klein bottle
last slate
ill graph it
that is a slightly insane graph
but it wouldnt satisfy the third condition because of vertical tangent lines
subtle zinc
the derivative switches signs but gets larger in magnitude
so doesn't actually approach + or - infinity
subtle zinc
the idea is basically ur approaching infinity but going up and down as u do
subtle zinc
but it wouldnt satisfy the third condition because of vertical tangent lines
i misunderstood.. it's differentiable
fuck
it's not defined everywhere
last slate
yep
subtle zinc
i cant find any function which satisfies these conditions
anyways i'm sure the answer is yes
i can imagine such a function
last slate
i cant becuase i dont know how differentiating a function changes it from a visual perspective much. im just spamming random examples but not getting anywhere
subtle zinc
$$f\left(x\right)=\frac{1}{x}-\sin\left(\frac{1}{x}\right)$$
grand pondBOT
locally ringed klein bottle
subtle zinc
i cant becuase i dont know how differentiating a function changes it from a visu...
slope of tangent
last slate
oh right
subtle zinc
this is a good question tho
midnight plankBOT
@last slate Has your question been resolved?
last slate
why unanswered
was the above function your answer?
subtle zinc
yes
subtle zinc
like at which point
subtle zinc
**locally ringed klein bottle**
this is defined for all x>0 and is differentiable
last slate
like at which point
getting closer to x=0
subtle zinc
getting closer to x=0
buddy
the tangent is never vertical
it's just approaching infinity
last slate
it's just approaching infinity
upper crag
that'd be like for g(x)=sqrt(a^2-x^2) which is a semicircle of radius a, so it's actually defined at x=a and has a tangent line at x=a that's vertical.
I think that's a bit different than having an asymptote at a point where it's undefined
last slate
hm okay
subtle zinc
**locally ringed klein bottle**
do you agree this is defined for all x>0
last slate
yes
subtle zinc
do you agree that it's differentiable for all x>0
last slate
thinking about merosity explanation because of the vertical tangent line thing
but youre 99% correct so i can mark as solved, i dont mind . i can continue thinking
last slate
true but it tends towards infinity
subtle zinc
try $$f\left(x\right)=\frac{1}{x}-\frac{1}{\sqrt{x}}\sin\left(\frac{1}{x}\right)$$ if u want
grand pondBOT
locally ringed klein bottle
subtle zinc
true but it tends towards infinity
what tends to infinity
last slate
derivative of the function
im wrong nvm
i get it now, thanks
!solved
?solved
.close
midnight plankBOT
midnight plankBOT
Available help channel!
chilly cobalt
hi, im confused on one of the steps to the solution of this problem
midnight plankBOT
hi, im confused on one of the steps to the solution of this problem
rigid hollow
in the solution they said A,P,O,R lie on the same circle
All 4 points lie inside the circle?
Only A doesn't lie in the circle
chilly cobalt
on the same circle (they created another cirlce)
rigid hollow
I think u must draw another circle
chilly cobalt
yeah
rigid hollow
But idk I GTG sorry I couldn't help
midnight plankBOT
midnight plankBOT
Available help channel!
twilit field
midnight plankBOT
twilit field
We first prove that if $T$ is a scalar multiple of the identity, then $ST=TS \forall S \in \mathcal{L}(V)$
\
We now attempt to prove $IS=SI$
\
$IS(v) = (I \circ S)(v)=S(v)$ by definition
\
$S(I(v))= (S\circ I)(v)= S(v)$
\
This means that $T= \lambda I ; \lambda \in \R$.
\. However, $TS=ST$. so $\lambda IS= \lambda SI$.
grand pondBOT
A dense set
midnight plankBOT
@twilit field Has your question been resolved?
twilit field
<@&286206848099549185>
I'm not sure I understand how the map is defined
Basically no element in $U$ maps to $0$?
grand pondBOT
A dense set
hard umbra
We first prove that if $T$ is a scalar multiple of the identity, then $ST=TS \fo...
this is not a correct proof for the exercise
twilit field
this is not a correct proof for the exercise
Hmm, what's wrong
hard umbra
you haven't really proven anything
twilit field
oops
I deleted my OG proof
We first prove that if $T = \lambda I$ then $ST=TS$.
\
$\lambda (IS) =\lambda (SI)$
\
But $SI=IS$
Thus they are the same
grand pondBOT
A dense set
hard umbra
you can't begin with your second line
twilit field
It's an iff , right? Can't I prove any direcition first
hard umbra
no but i'm saying that your proof of the <= direction is logically wrong and poorly structured if you begin with the second line ehre
a proof that TS = ST for all S when T = λI needs to show the equality TS = ST
you start out by writing down an equality that you haven't yet shown, and the equality is not even in the form TS = ST
twilit field
so I first have to prove IS=SI
hard umbra
well we can take that as a given
that's fine
but you should show equalities just as in every other proof
show that LHS = RHS
so for example here, you might begin like
TS = (λI)S = λ(IS) = ...
and it doesn't hurt to note why the equality (λI)S = λ(IS) holds as you're doing this
twilit field
$(\lambda I)(v)= \lambda(I(v))
$ by definition
hard umbra
i mean you don't have to go to that level to show it, there are surely plenty of theorems available to you about compositions of linear functions
hard umbra
okay so axler doesn't prove this outright that's a shame
the equality is certainly true but requires a small amount of thought into it
hard umbra
yes but it's not an immediate consequence that you should have things like (λT)S = λ(TS) = T(λS) just from that definition alone
these at least need a small calculation to verify
twilit field
hmm
I'm not sure how I'd go about this
The best I can do is $(\lambda T)S = \lambda(T)(S) = \lambda(TS)$
grand pondBOT
A dense set
hard umbra
i'm not sure what the middle thing is supposed to mean
twilit field
I'm honstly not sure how to do this
twilit field
I've never verified such properties for functions tbh
hard umbra
,, ((\lambda T) \circ S)(v) = (\lambda T)(S(v)) = \lambda \cdot T(S(v)) = (\lambda (T \circ S))(v)
here's one
grand pondBOT
twilit field
I see
so similarly $T \circ ((\lambda S)(v))= T \circ (\lambda (S(v) ) = \lambda (T \circ S) (v)$
hard umbra
you aren't bracketing things correctly
grand pondBOT
A dense set
hard umbra
you only get to abuse notation like this when you're comfortable with all the properties
at this point, i wouldn't accept this notation
twilit field
hmm
Okay, so what I have to prove is
\
$(\lambda I)\circ S(v) = T \circ (\lambda I (v))$
grand pondBOT
A dense set
hard umbra
,, ((\lambda I) \circ S)(v) = (S \circ (\lambda I))(v)
grand pondBOT
hard umbra
proving this is an exercise in understanding the notation and definitions, knowing what you can and can't do
the result itself is not very deep
twilit field
$((\lambda I) \circ S)(v) = (\lambda I) \circ S(v) = \lambda \cdot I \circ S(v) =\lambda ( I \circ S(v))$
grand pondBOT
A dense set
twilit field
But $I \circ S(v) = S \circ I(v)$
grand pondBOT
A dense set
twilit field
$((\lambda I) \circ S)(v) = (\lambda I) \circ S(v) = \lambda \cdot I \circ S(v) =\lambda ( I \circ S(v))= \lambda (S \circ I(v))$
grand pondBOT
A dense set
hard umbra
**A dense set**
no i have no idea what you're two middle things are here
twilit field
$(\lambda I (S(v))$
hard umbra
write it with the proper notation
grand pondBOT
A dense set
twilit field
I have to go for therepy now, will continue this for sure once that's over
sorry
tahnks
hard umbra
twilit field
<:catThumbsUp:662911317090369546>
Ok, I'm back
hard umbra
damn
twilit field
$(\lambda(I(S(v))))$
grand pondBOT
A dense set
twilit field
is this notation fine
From this notation, and the definition of the identity function, it follows this is $\lambda(S(v))$
grand pondBOT
A dense set
twilit field
$S(\lambda I(v))= \lambda(S(I(v))$ by the definition of linearity
grand pondBOT
A dense set
hard umbra
ye
twilit field
However, $S((I(v)) = S(v)$. So this is $\lambda S(v)$
grand pondBOT
A dense set
hard umbra
everything you've said so far is true
twilit field
So this proves the implication in one directiopn
twilit field
Espescially as I don't have matrices on hand yet
hard umbra
yeah
twilit field
Hmm, I want to prove if $ST=TS$ then $T = \lambda I$
grand pondBOT
A dense set
twilit field
Neither do I have inverse functions on hand , I suppose
hard umbra
you need to choose specific S to test the properties of T
twilit field
I don't even know from which vector space to which vector space this is
so I can't
no?
hard umbra
so you can just construct a few S
grand pondBOT
A dense set
grand pondBOT
A dense set
twilit field
what am I doing wrong
hard umbra
well these are certainly true statements
but you've just picked S = αI, which we already know commutes with every other linear function
so this choice of S doesn't tell us anything useful
twilit field
Well, $S$ ia linear function
grand pondBOT
A dense set
twilit field
and every linear function from $V \to V$ is of this form
grand pondBOT
A dense set
hard umbra
thats false
twilit field
wait
right
the basis can be transformed in different ways
let the basis of $V$ be $e_1,e_2, \dots, e_n$
\
we have $S(v) = \alpha_1e_1+\alpha_2e_2 + \dots+ \alpha _n e_n)$
\
So $T(S(v)) = T(\alpha_1e_1+\alpha_2e_2 + \dots+ \alpha _n e_n)$
\
And $S(T(v)) = \alpha_1T(e_1)+ \alpha_2T(e_2) + \dots \alpha_nT(e_n)$
grand pondBOT
A dense set
twilit field
thats false
Is this fine
twilit field
real numbers
hard umbra
thats not the point of the question
twilit field
Is the idea right though
hard umbra
im asking you what they are because you just wrote down this expression
and you haven't told me how the alphas and the v relate at all
are the alphas constants? do they vary with v?
twilit field
let the basis of $V$ be $e_1,e_2, \dots, e_n$
\
The transformation can be described by just following where the basis vectors end up
\
so we have $T(e_1)+T(e_2)+ \dots + T(e_n)$
\
This is $\alpha_1e_1+\alpha_2e_2+ \dots+ \alpha_ne_n$; \alpha_i \in \R$
grand pondBOT
A dense set
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
hard umbra
hmm
like
i can see that you're trying to express the function using a basis, but this isnt the right way to express it
ill give you a starting point
let $e_1, \dots, e_n$ be a basis of $V$, and define $S$ in the following way:
[ S(\alpha_1 e_1 + \dots + \alpha_n e_n) = \alpha_1 e_1 ]
we can also write
[ T(e_i) = \beta_{i, 1} e_1 + \dots + \beta_{i, n} e_n ]
for numbers $\beta_{i, j}$ because $e_1, \dots, e_n$ form a basis.
\medskip
we have
[ T(S(\alpha_1 e_1 + \dots + \alpha_n e_n)) = T(\alpha_1 e_1) = \alpha_1 \beta_{1, 1} e_1 + \dots + \alpha_1 \beta_{1, n} e_n ]
on the other hand, we have
[ S(T(\alpha_1 e_1 + \dots + \alpha_n e_n)) = S \parens [\bigg] {\sum_{i = 1}^n \sum_{j = 1}^n \alpha_i \beta_{i, j} e_j} = \sum_{i = 1}^n \alpha_i \beta_{i, 1} e_1 ]
grand pondBOT
twilit field
why are you describing it to be a multiple of just a single vector
*single basis vector
hard umbra
im defining S
grand pondBOT
A dense set
twilit field
let ${e_1,e_2,\dots, e_n}$ form a basis of V
We then define
$S(e_i) = \sum_{j=1}^{n} \beta_{i,j} e_j$..
\
We define $v\in V$ to be $\sum_{i=1}^{n} \alpha_i e_i$
\
We thus have $S(v)= S( \sum_{i=1}^{n} \alpha_i e_i) = \sum_{k=1}^{n} \sum_{j=1}^{n} \alpha_{k} \beta_{k,j} e_j$
Something feels off about the notation I've used
This feels better
grand pondBOT
A dense set
twilit field
what am I doing wrong here
let ${e_1,e_2,\dots, e_n}$ form a basis of V
We then define
$S(e_i) = \sum_{j=1}^{n} \beta_{i,j} e_j$..
\
We define $v\in V$ to be $\sum_{i=1}^{n} \alpha_i e_i$
\
We thus have $S(v)= S( \sum_{i=1}^{n} \alpha_i e_i) = \sum_{k=1}^{n} \sum_{j=1}^{n} \alpha_{k} \beta_{k,j} e_j$
\
\
We now define $T(v)= T( \sum_{i=1}^{n} \alpha_i e_i) = \sum_{l=1}^{n} \sum_{j=1}^{n} \alpha_{l} \beta_{l,j} e_j$
\
\
We now look as $T(S(V))= T(\sum_{k=1}^{n} \sum_{j=1}^{n} \alpha_{k} \beta_{k,j} e_j)$.
\
We can define
$S(T(v))=. S(\sum_{l=1}^{n} \sum_{j=1}^{n} \alpha_{l} \beta_{l,j} e_j)$
How do I look at $S(T(v))$
grand pondBOT
A dense set
twilit field
feels like it wiil be very cumbersome
grand pondBOT
A dense set
twilit field
Now what
twilit field
let $e_1, \dots, e_n$ be a basis of $V$, and define $S$ in the following way:
\[...
Can I have some help , pelase
okay, gtg for my class now
midnight plankBOT
@twilit field Has your question been resolved?
violet root
ok so i asked a question about the rule of fifteenths for predicting tidal heights and was told that it follows the same general rules as the rule of twelfths. i was wondering what pattern is follows/what fractions it uses each time frame.
midnight plankBOT
ok so i asked a question about the rule of fifteenths for predicting tidal heigh...
violet root
.close
midnight plankBOT
midnight plankBOT
Available help channel!
twilit field
midnight plankBOT
twilit field
$T(v)=\begin{cases} S(v) & v \in U \ 0 & v \in V \land v \notin U\end{cases}$
grand pondBOT
A dense set
midnight plankBOT
@twilit field Has your question been resolved?
twilit field
We assume $t \in U$ and $r \in V \setminus U$. We now have $t+r=p$ say. where $t,r,p \neq 0$
\
We now have two choices for $p$.
\
Case 1:- $p$ is in $V \setminus U$.
\
We then have $T(r+t) =T(p)= T(r)+T(t) =T(p)$. This implies that $S(t)=0$.
\
grand pondBOT
A dense set
dreamy lichen
Did you assume T is linear?
twilit field
This is however a contradiction, as $S \neq 0$
grand pondBOT
A dense set
twilit field
My contradiction is $S=0$, but that does't prove $T$ isn't a linear map
grand pondBOT
A dense set
twilit field
I assumed t \in V \setminus U
dreamy lichen
I assumed t \in V \setminus U
do you mean r?
grand pondBOT
A dense set
twilit field
Let $r \in V \setminus U st T(r+v) \neq 0$
grand pondBOT
A dense set
twilit field
can I assume atht
dreamy lichen
yes
and if you want to, you can let r be in V \ U
because you know it's non-empty
twilit field
yup
dreamy lichen
**A dense set**
this wasnt entirely wrong btw, it's just missing one case and the implication in case 1 is unclear. But that doesnt mean it cant be made clear
twilit field
Let $v \in U$ st $S(v) \neq 0$. We also assume $r \in V \setminus U, $ so $T(r)=0$.
\
We now assume $r+v= p $ ; $p \in V \setminus U$.
\
We then have assuming linearity
\
$T(r)+ T(v) = T(p)$.
\
We thus have
$S(v)=0$.
\
but this is a contradiction, as we know there is atelast one $v \in U$ st $S(v) \neq 0$
\
Thus clearly such a p can't exist in $V \setminus U$
dreamy lichen
it's a contradiction bc S(v) != 0
grand pondBOT
A dense set
dreamy lichen
the fact that there is at least one v in U with S(v) != 0 was used when you said "Let v in U st S(v) != 0"
the contradiction arises just because of this
dreamy lichen
or that T is non-linear...
dreamy lichen
you are nesting too many assumptions in an order, that makes it really hard to follow
twilit field
\
We now assume $r+v= p $ ; $p \in V \setminus U$.
\
We then have assuming linearity
\
$T(r)+ T(v) = T(p)$.
\
We thus have
$S(v)=0$.
\
but this is a contradiction, as we know there is atelast one $v \in U$ st $S(v) \neq 0$
\
We now let P \in U$.
\
thus $T(v)= T(p)$
\
This would mean $T(v-p)=0$
dreamy lichen
**A dense set**
you essentially proved that either T is non-linear, which would close the proof, or that p is in U, which is a case you have to deal with
grand pondBOT
A dense set
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
dreamy lichen
you're not far from the end, the other case is quite easy to deal with
you can finish this later
sinful trout
uh, I have to go fo therepy now
therapy? what?
i won't ask
dreamy lichen
therapy? what?
i won't ask
what?
midnight plankBOT
@twilit field Has your question been resolved?
twilit field
therapy? what?
i won't ask
Counselling
twilit field
you're not far from the end, the other case is quite easy to deal with
okay
so I have $T(r)+T(v)=T(p)$
\
I however, know that T(r)=0
\
We thus have $S(v)=S(p)$.
\
We thus know that $S(v-p)=0_v$
\
It thus follows that $v-p \in Ker(T)$
grand pondBOT
A dense set
dreamy lichen
This is way too complicated
the other case is
v + r is in U
v is something in U, r is something not in U
it's kinda similar to saying that rational + irrational can be rational
dreamy lichen
v is something in U, r is something not in U
you wont need anything other than this to reach a contradiction
twilit field
This would mean that $r \in U$
grand pondBOT
A dense set
dreamy lichen
This would mean that $r \in U$
Yes, but why?
twilit field
as $r$ is in span (v,p)
grand pondBOT
A dense set
dreamy lichen
yeah, that works
or you can simply add -v to v + r
and by closure under addition, you get that r is in U
dreamy lichen
**A dense set**
and this is a contradiction
Now just try structuring the proof in such a way it's readable
and note that this 2nd case was so simple, that you could start by proving that r + v can't be in U, and then doing the 1st case to reach a contradiction
so the proof doesnt even have to be broken down into cases
dreamy lichen
yeah, it is
twilit field
Let $v \in U$ st $S(v) \neq 0$. We also assume $r \in V \setminus U, $ so $T(r)=0$.
\
We now assume $r+v= p $ ; $p \in V \setminus U$.
\
We then have assuming linearity
\
$T(r)+ T(v) = T(p)$.
\
We thus have
$S(v)=0$.
\
but this is a contradiction, as we know there is atelast one $v \in U$ st $S(v) \neq 0$
\
We now conisider the case where $ p \in U$..
\
It $p \in U$, it follows that $r \in span(v,p)$, which is a contradiction to what we assumed.
\
Therefore $T$ is not a linear map
grand pondBOT
A dense set
dreamy lichen
It's less readable tho
twilit field
wdym
dreamy lichen
you made 2 assumptions
how do I know which one gets contradicted?
and then you mention "the other case"
when no cases were mentioned before
twilit field
Ah
dreamy lichen
It could be made more readable in 2 ways:
- first proving that r + v is in V \ U and then doing case 1 without that one assumption
- moving the assumption of linearity above the assumption of p in V \ U (and explicitly mentioning case 1 and case 2)
twilit field
Okay, will do
dreamy lichen
and also, I'd replace the assume on first line with let
because there certainly is such r, you just need to pick one to perform the proof on it
and it's also the first time you mention r
so you are essentially bringing it into existence
twilit field
It could be made more readable in 2 ways:
1) first proving that r + v is in V \ ...
This first case doesn't make sense to me
first pointer
Let $v \in U$ st $S(v) \neq 0$. We also assume $r \in V \setminus U, $ so $T(r)=0$.
\
We assume linearity throughout this proof by contradiction
\
Case 1
\
We now assume $r+v= p $ ; $p \in V \setminus U$.
\
We then have assuming linearity
\
$T(r)+ T(v) = T(p)$.
\
We thus have
$S(v)=0$.
\
but this is a contradiction, as we know there is atelast one $v \in U$ st $S(v) \neq 0$. Thus either there is no such $p$ or the transformation isn't linear
\
Case 2:-
We now conisider the case where $ p \in U$..
\
It $p \in U$, it follows that $r \in span(v,p)$, which is a contradiction to what we assumed.
\
Therefore $T$ is not a linear map
grand pondBOT
A dense set
dreamy lichen
Let $v \in U$ s.t. $S(v) \neq 0$. Let $r \in V \setminus U$.
Then $v + r \not \in U$, lest $v + r - v = r$ would be in $U$.
For sake of contradiction, assume that $T$ is linear. Then $0 = T(v + r) = T(v) + T(r) = S(v) + 0 = S(v)$, which is the wanted contradiction.
grand pondBOT
MæthIsAlwaysRight
dreamy lichen
this is what I meant
strong lava
**A dense set**
this has poor wording overall
dreamy lichen
**A dense set**
This is not wrong, there is just so many assumptions that it's hard to read
And also it's not clear how T(r) + T(v) = T(p) implies S(v) = 0
strong lava
**MæthIsAlwaysRight**
this has good wording
twilit field
Let me rewrite it
dreamy lichen
A rule of thumb is to not make assumptions, unless it's necessary
and when you need to make them, do it one by one
dreamy lichen
**A dense set**
This has 2 proofs by contradiction, nested in each other in a weird way
dreamy lichen
**MæthIsAlwaysRight**
This also has 2 proofs by contradiciton, that however follow one by one. And one of them was even reworded to just a simple "lest..."
dreamy lichen
**A dense set**
- define v and r, as previously. Also say what p is, don't delay it until case 1. You'll use it in case 2 as well
- For sake of contradiction, assume T is linear
- Case 1: p is in V \ U. Reach a contradiction
- Case 2: p is in U. Reach a contradiction
- Conclude the proof
this would also be a fine schema
twilit field
We now wish to prove that the mapping that $T$ describes isn't linear. We wish to use proof by contradiction for the same.
\
Let $v\in U$ be a vector such that $S(v) \neq 0$. Let $ u \in V \setminus U$ We also assume $v+u=p$
\
For the sake of contradiction, we assume that $T$ is a linear map
\
Case 1: $p \in V \setminus U$
\
We then have $T(v+u)=T(p)$
\
By linearity
\
$T(v)+T(u)=T(p)$
\
$S(v)+0=0$
\
$S(v)=0$
\
However, $v$ was chosen such that $S(v)$ would be non-zero. However if $p \in V \setminus U$, such a construction is clearly not possible.
\
\
alternatively , if $v+u=p$, it follows that $v \in V \setminus U$, which is clearly a contradiction
\
\
Case 2- $p \in U$.
\
If this were the case $p-v=u$
\
From this it is evident that $u \in span (p,v)$
\
Which clearly isn't possible as the sets we've declared $u$ and, $p,v$ to belong to ,only intersect at ${0}$.
\
Therefore as $T(v+u)$ isn't defined when $v,u$ come from different subspaces of $V$, This isn't a linear transformation
grand pondBOT
A dense set
twilit field
How does this look?
dreamy lichen
u in V \ U sounds kinda strange lol
but why not
I'd change "We also assume v + u = p" to "Let p = v + u"
but that's a minor detail
I also don't see what you mean by this
S(v) = 0, boom, contradiction, case concluded
you can move to case = 2
this is false
it doesnt follow that v is in V \ U
twilit field
it doesnt follow that v is in V \ U
Oh, I have to explain it
dreamy lichen
i actually think it's false
if you clip whole that out and replace it with something that says "Hey, look, contradiction", case 1 will be fine
they dont intersect at all
U intersection V \ U is empty
T(v + u) is defined for all vectors in V
also V \ U is not a subspace (e.g. bc it doesnt have the 0 vector, and it usually fails to satisfy the remaining conditions as well)
twilit field
i actually think it's false
I don't follow
dreamy lichen
this is false
This whole alternative approach seems to be wrong
I agree with v + u = p, that's how p was defined. How does it follow that v is in V \ U though?
you probably assued that V \ U is a subspace and it's closed under vector addition
which it's not
twilit field
I agree with v + u = p, that's how p was defined. How does it follow that v is i...
$v = p-u$ thus $v\in span(p-u)$
grand pondBOT
A dense set
dreamy lichen
v is in span(p, u) you mean?
twilit field
ye
dreamy lichen
okay, and?
p is in V \ U and u is also in V \ U. That doesn't mean p - u is in V \ U tho
grand pondBOT
A dense set
twilit field
How did I miss that
dreamy lichen
they dont intersect at all
e.g. this is wrong, V \ U isn't a vector space, it doesnt have the 0 vector
the intersection is empty
twilit field
We now wish to prove that the mapping that $T$ describes isn't linear. We wish to use proof by contradiction for the same.
\
Let $v\in U$ be a vector such that $S(v) \neq 0$. Let $ u \in V \setminus U$ We also assume $v+u=p$
\
For the sake of contradiction, we assume that $T$ is a linear map
\
Case 1: $p \in V \setminus U$
\
We then have $T(v+u)=T(p)$
\
By linearity
\
$T(v)+T(u)=T(p)$
\
$S(v)+0=0$
\
$S(v)=0$
\
However, $v$ was chosen such that $S(v)$ would be non-zero. We've thus arrived at a contradiction, which arises from the assumption that the function is linear
\
\
Case 2- $p \in U$.
\
If this were the case $p-v=u$
\
From this it is evident that $u \in span (p,v)$
\
Which clearly isn't possible as this means that $u \in U$. But that's not true by assumption.
\
this contradiction arose from the assumption $p$ could lie in $U$
\
\
Thus clearly our initial assumption that the transformation was linear is wrong, and this transformation isn't linear.
dreamy lichen
I also don't get the sentence "However if p in V \ U, such a construction is clearly not possible"
what construction?
twilit field
what construction?
This construction
dreamy lichen
that's not really construction
and existence of v is guaranteed
the contradiction arises mainly because it was assumed that T is linear
while you make it seem like it arises because p is in V \ U
just remove that sentence as say "so we get a contradiction" or sth like that
okay, this looks better
you dont have to mention "which arises ..." but why not
case 2 looks slightly broken now
u is in span(p, v) and thus also in U
grand pondBOT
A dense set
twilit field
Is this better?
twilit field
We now wish to prove that the mapping that $T$ describes isn't linear. We wish to use proof by contradiction for the same.
\
Let $v\in U$ be a vector such that $S(v) \neq 0$. Let $ u \in V \setminus U$ We also assume $v+u=p$
\
For the sake of contradiction, we assume that $T$ is a linear map
\
Case 1: $p \in V \setminus U$
\
We then have $T(v+u)=T(p)$
\
By linearity
\
$T(v)+T(u)=T(p)$
\
$S(v)+0=0$
\
$S(v)=0$
\
However, $v$ was chosen such that $S(v)$ would be non-zero. We've thus arrived at a contradiction, which arises from the assumption that the function is linear
\
\
Case 2- $p \in U$.
\
If this were the case $p-v=u$
\
From this it is evident that $u \in span (p,v)$
\
Which clearly isn't possible as this means that $u \in U$. But that's not true by assumption.
\
this contradiction arose from the assumption $p$ could lie in $U$
\
\
Thus clearly our initial assumption that the transformation was linear is wrong, and this transformation isn't linear.
Is this better
grand pondBOT
A dense set
dreamy lichen
yeah, this is better
midnight plankBOT
twilit field
Sorry, need to get a shave at the barber, I'm done with this, so I closed it
midnight plankBOT
Available help channel!
twilit field
midnight plankBOT