Hi, supposed to calculate lim{x-->0} of (x-sin x)² ÷ ((cosh(x)-1)÷2)³ . Tried using Taylor series for sin(x), e^x and e^(-x) which simplified the expression quite a bit but still got the problem of dividing by 0. Any smart suggestions?
#[Analysis] Calculating limit using Taylor series.
celest patio
signal auroraBOT
bitter gorge
Why in your cosh(x) expansion do you still have exp terms ?
celest patio
It is said in the question that cosh(x) = (e^x + e^(-x))÷2
bitter gorge
Yes but
When you expand it
It should be a polynomial
In x
There arent any exp terms inside it
In other words, if you write
Exp(x) = 1+x+x^2/2 + x^3/3! + ....
And exp(-x) = 1-x +x^2/2 - x^3/3! + ...
Adding them up makes all the odd power terms vanish
celest patio
Yes that's also what I got
bitter gorge
No, the last line is the final result
But if you take each exp then yes
Cosh is the hyperbolic cosine
Which is a way we make the the exponential function even
(if you remember the definition of even function)
celest patio
Oh I didn't know
bitter gorge
Take any function f (which is whatever)
If you let
g(x) = (f(x)+f(-x))/2
celest patio
I see, not sure this helps my expression though
bitter gorge
Then g is even
And h(x)
= f(x)-f(-x))/2 is odd
(so hyperbolic sine is
Sinh(x) = (exp(x)-exp(-x))/2)
celest patio
Yeah alright
bitter gorge
Why im telling you this
Because
You can check your Taylor expansions
If a function is odd/even
It will have only odd/even powers of x
Respectively
So sin(x) is odd, it will have only odd powers
In its expansion
(just making sure uve seen this to help you)
What did you find for the expansion of the numerator ?
(without the square)
celest patio
That's kinda cool, see that clearer now
The numerator I got to this: (but also continued with recalculating my expression after what you explained about the hyperbolic function)
Unsure exactly how to handle the rest term (marked in blue) though?
Chose to take the Taylor polynomial to the 2nd degree just to mainly remove the x from the numerator and - 1 from the denominator
(also curious why sinh(x) exists if we said it was to make a function even, if
Sinh(x) = (exp(x)-exp(-x))/2) then it's odd but sin(x) already is odd right? At least it's Taylor polynomial gives odd powered terms 🤔)
celest patio
Think I solved it, at least the answer matches the answer for this problem.
bitter gorge
Writing alpha may be not necessary
It's already clear that you select alpha = 0
So no need to write alpha
celest patio
Oh, alpha I denote as my rest term
Cause you can only do the substitution for the actual function if you include it right?
bitter gorge
You still dont need it
Because these rest terms are negligible
The standard notation is o(x-a)^(k+1)
When using the Taylor Young series
You dont need this rest explicitly
In problems like those
You seem to have some confusion
Regarding this psi_L
What you call Psi_L is actually called Xi_L
This Xi_L is a real number in the interval [a;x]
It is not a ("alpha")
This is just one expression of the remainder on which we have applied the mean value theorem
We know there exists some Xi_L inside the interval [0;x]
That makes the rest Rk equal to this
This is not alpha as you said
Sometimes this form is called the Lagrange form
But this is not necessary here
So it isnt that you wrote alpha instead of Xi_L
They're different
celest patio
okay, yeah I thought Xi was the argument in the restterm I denoted alpha with
but the lagrangre form for R(x) has to be included for the substitution since its definition is literally f(x)-p(x) no? where f(x) is the function you want to replace with your taylor series for calculating the limit and p(x) is the taylorpolynomial
If I don't have to include the rest term when substituting a function for its taylorpolynomial theese problems just got a lot simpler
So far though I have included it and used then fact that as x goes to some value (often the value i create the taylorpolynomial around) then alpha goes to another value and I just plug that value in.
bitter gorge
The Lagrange form is overkill
all you need is to write
F(x) = p(x) + a rest that goes to zero
You dont really need anything about this rest
Not even its explicit form
Just that it goes to zero
Of course in more theoretical problems
This is useful
If we want to estimate errors
celest patio
alright
bitter gorge
Explicitly
celest patio
thats true
bitter gorge
But here we are computing limits
So these rests/errors are not very important
As long as they are small
(aka go to zero)
celest patio
can you say that the rest goes to 0 even when we are creating our taylorpolynomial around a different point then 0 though?
bitter gorge
Yes !
Because
(well in general of course)
The taylor polynomial
Is a polynomial that is super close to your function
Near the point x=a
The more terms you add
The closer it will be to your function near point x = a
celest patio
So far I think I understand yes
bitter gorge
So the error (the difference between your function and the taylor polynomial)
Will get smaller the more you add terms
bitter gorge
This also happens in real life
When you iron your clothes
You start by ironing a bit
Then more so the error is closer to zero
Then you add more ironing
At any point you can iron
Iteratively
To get as close as possible
Shitty metaphore
But
celest patio
Thought about like what if x goes to 2 and you create your taylor polynomial around 0 and then ur rest term is between 0 and x, but since x goes to 2 and not 0 your rest term doesnt go to 0. I saw a professor calculate it that way but I guess they were sloppy.
haha
bitter gorge
I think this is the reason
You are confused
You seem to confuse
What is local
With what is global
celest patio
Yeah I dont iron much but the metaphore checks out, I aprove.
oh ok
local and global... 🤔
Like are the polynomial and the limit we are calculating thought of as in seperate scopes?
bitter gorge
When you see integrals and
Xi_P
And all that shit
It means that formula is global
celest patio
yep
bitter gorge
But for computing limits, we need to find errors locally
Near x = a
We dont care about integrating
On intervals
To find global errors
We want just local errors
(this is just an explanation for the intuitive aspect)
Dont write this in exams
So try to use the correct form of Taylors theorem
Depending on the context
celest patio
dammit, already though of writing "I see integral and all that shit, hence I conclude that the proof..."
bitter gorge
But tbh this is a good analogy
When you want to find limits
Stop using integrals
And explicit remainders
This is just sloppy
As you said
You should instead learn and use the small O notation
(which also will come in handy if you do some coding)
Small O, or big O
These are called Landau notations
They allow you to quickly assess the orders of your expansions
And adapt
celest patio
yeah ive only seen big oh notation in code I thought it was just a coincidence they were the same
bitter gorge
They are the exact same !
bitter gorge
Great observation
When you write that this algorithm has running time in o(n)
It means its running time Tn
divided by n goes to zero
What this means is that n increases faster than it
So for instance to make this clearer
e^x = 1+x+x^2/2+o(x^2)
Meaning
e^x = 1 + x + x^2/2 + garbage terms
Such that garbage/x^2 goes to zero
So x^2 is "bigger" and matters more
Near the point we took
celest patio
Well explained
yeah I connect it more now
but it still feels a bit like cheating to ignore the rest term, like if i want to see if e^(-x) approaches some value and i substitute it for 1-x + garbage terms like in this case we are saying that x^2/2 is one of the garbage terms
celest patio
also those should get different results right
celest patio
like for x going to i.e. infty
celest patio
so even though they exist they wont ever change the value that the limit approaches?
bitter gorge
Not x = infinity
bitter gorge
so even though they exist they wont ever change the value that the limit approac...
Well if they stack up
They can change things
This is why we sometimes will need to push
The order of the expansion
celest patio
Well if they stack up
isnt that the thing? the rest term is all the remaining terms stacked up?
bitter gorge
Yes !
But we dont really care theyre so smol
In general
Sometimes we need more of them
Sometimes less
(but this you will learn by doing your exercises)
celest patio
alright
like tiny wrinkles that are negligible as our ironing approaches completeness
true, puzzled by like how you can use an approximation to substitute in for a function whose expression is really sensitive in a way since you are often letting the argument approach infinity or 0
but true, i'll spot the difference clearer by finding exercises to do
it's getting late here but nice chatting a bit
and thank you so much for the help!
bitter gorge
Welcome ! Goodluck
celest patio
Do I tag this with "solved" btw or like... eh..
bitter gorge
You can
celest patio
do I just leave the post as it is
bitter gorge
But no one cares