Taking the limit of the derivatives of the Beta function

I tried using Limit[D[Beta[a,b],{a,1},{b,1}],{a -> 1},{b -> 1}] and Limit[D[Beta[a,b],{a,1},{b,1}],{a , 1},{b , 1}]
but it doesnt seem to work, why?

matrices – matrix derivative’s simplification

I would like to ask about the steps of the following Simplification.
Can anyone show me that how Step 1 simplify to Step 2 ?

Step 1
Step 2

X , Theta = Matrix
y = Vector

And is there any special meaning of a Matrix multiply by it’s Tranpose ?
Thank you.

real analysis – Analyzing the decay rate of taylor series cofficients when high-order derivatives are intractable

This could be a soft question. I am trying to show that the $n$-th Taylor series coefficients of a function is $O(n^{-alpha})$ for some $alpha>1$. However, because the function is a function composition of another function with itself, it seems intractable to compute high-order derivatives. I was wondering if there are methods that can bound the asymptotic decay rate of Taylor series coefficients without obtaining the exact coefficients. For example, can complex analysis help here?

Thank you so much!

The function that I am trying to analyze is $f(x)=g(g(x))$, where $g(x) = frac{1}{pi}left( xcdot (pi-arccos(x)) + sqrt{1-x^2} right)$. I conjecture that its $n$-th Taylor coefficient about $x=0$ is $O(n^{-5/2})$. I have shown that the $n$-th Taylor coefficient of $$g(x)= frac{1}{pi} + frac{x}{2} + sum_{n=1}^infty frac{(2n-3)!!}{(2n-1)n!2^n pi} x^{2n}$$ is $O(n^{-5/2})$.

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general topology – Properties need to define Derivatives on Topological space

I just started learning topology and was curious about defining derivatives on general topological spaces.
Since we can define continuous functions on Topological spaces, my question is what additional properties one would need to define derivatives on Topological spaces.

I guessed one might only need converging sequences to define derivative, so space must have,

  1. metrizablity: to define some sort of distance between points so one can define converging sequences on space.
  2. Hausdorff property: so convergence would be unique.

but on the other hand, these properties are conserved under homeomorphism, where differentiability does not.
so, there should be some more properties (maybe other than topological properties) one would need to define derivatives, and I’m not sure what kind of property it would be that conserves differentiability.

I hope my question makes sense.

calculus – Trying to understand the chain rule for partial derivatives

So I’ve been studying the chain rule for partial derivatives recently and I’m having an extremely difficult time wrapping my head around it as I’m having an incredibly hard time understanding the formulation of the chain rule for partial derivatives in my textbook. In an attempt to help me understannd how it works I’ve been going over some of the exercises in the back of the book to see if I could at least apply it. I came to an exercise that looks rather simple, but I’m not sure how to solve it. The exercise is as follows:

Consider a differentiable function function $f:mathbb{R^2}rightarrowmathbb{R}$, now find the partial derivative of the function:
$F:(x,y)rightarrow f(2x,3y)$ (State the result in terms of the partial derivatives of $f$)

Now in terms of what I want to find I’m having some doubts around one of the deriavtives. What I find is:

$frac{partial F(x,y)}{partial x}=frac{partial f}{partial y_1}(2x,3y)frac{partial(2x)}{partial x}+frac{partial f}{partial y_2}(2x,3y)+frac{partial(3y)}{partial y}$

My issue here is that I can’t really figure out what $y_1$ and $y_2$ are supposed to be, and the formulation in my textbook is really confusing, I’d really appreciate if anyone could clarify this for me.

Fixed Income Derivatives – Swaps and Swaptions

I’m trying to calculate the forward contract on a zero coupon bond where the forward contract matures at t=4.

The zero coupon bond matures at t=10 and has a face value of 100. The price of that bond is 61.62

n = 10−period binomial model for the short-rate

The lattice parameters are: r(0,0)=5% ,u=1.1 ,d=0.9d ,q=0.5 ,1−q=0.5

It seems that i am a bit stuck.

derivatives – Why Are We Able To “Break Up” A Differential Operator? Example With Method of Integrating Factor

Background: Just relearning some stuff from ODE’s on my own time, and part of a derivation isn’t clicking with me but is probably easily explained. I will use a very simple case study of deriving the method of integrating factors for a first order ODE to illustrate my case, but the attention is not on the ODE itself, just a particular mechanic:

$$frac{dy}{dx} + p(t)y = g(t) tag{1}$$

Multiplying by some function $mu(t): $

$$ mu(t)frac{dy}{dx} + mu(t)p(t)y = mu(t)g(t) tag{2}$$

Assume that $frac{dmu(t)}{dt} = mu(t)p(t)$ …. then we have ….

$$ int frac{1}{mu(t)}dmu(t) = int p(t) dt tag{***}$$

The crux of my question has to do with $(***)$ and specifically why we’re able to separate the $dmu(t)$ from $dt$.

I can’t exactly recall what the concept is that enables us to make the separation possible here, but from what I remember I have had professors say that we cannot do the separation because $frac{d}{dt}$ is a differential operator.

I’m just hoping somebody could help me to patch up this conceptual problem, and I recognize that the vocabulary i’m using (e.g ‘separate’ ) here is very informal, so I apologize.

calculus – Proof of derivatives of exponential function

I am learning Calculus and I ran into the proof regarding the exponential derivative, I haven’t quite understood it though, could you guys please help me out with that? The proof goes like that:

y = a^x
ln(y) = ln(a^x)

Here comes the part I haven’t gotten yet, why is y’/y = lna?

y´/y = lna
y´= lna(y)
y´= lna(a)(a^x)

Thanks in advance!

derivatives – How can $z = xa + x$ be differentiated with only chain rule?

I am trying to put some rigour to my understanding of the Chain Rule (with Leibniz Notation). I came across this question and the second answer there (David K’s) states,

$frac{dz}{dx} = a + 1 + xfrac{da}{dx}$, as you surmised, though you could also have gotten that last result by considering $a$ as a function of $x$ and applying the Chain Rule.

I understand how could one evaluate $frac{dz}{dx}$ using the product rule,

frac{dz}{dx} & = frac{d(xa + x)}{dx} \
& = frac{d(xa)}{dx} + frac{d(x)}{dx} \
& = Bigg(frac{d(x)}{dx}a + xfrac{d(a)}{dx}Bigg) + 1 && text{Product Rule}\
& = a + xfrac{da}{dx} + 1

Though, I seem to be getting something wrong when using the Chain Rule,

frac{dz}{dx} & = frac{d(xa + x)}{dx} \
& = frac{d(xa)}{dx} + frac{d(x)}{dx} \
& = Bigg(frac{d(ax)}{d(ax)} times frac{d(ax)}{d(a)} times frac{d(a)}{d(x)} Bigg) + 1 && text{Chain Rule} \
& = Bigg(1 times x times frac{d(a)}{d(x)} Bigg) + 1 \
& = xfrac{d(a)}{d(x)} + 1

I am missing the $a$ in the chain rule variant. Did I expand the chain rule correctly in terms of the Leibniz notation? What am I missing here?