I calculated the upper derivative of the modified Bessel function of the first kind and order alpha j_{alpha} with respect to the variable with the maple program, but I could not show it for example by induction.

May you help me? Give me another key to show it?

Upper derivative of j_alpha

# Tag: kind

## dnd 5e – Do you have to specify what kind of attack or weapon you will Ready?

It seems logical that you can use the Ready action to prepare to grapple or shove an opponent if triggered. But can you just Ready an attack, and then decide when it’s triggered whether to attack as normal or try to grapple or shove?

Similarly, you could certainly Ready a specific weapon attack like “if a creature moves within 5′ of me, I swing my mace” or “if any of them draw a weapon, I shoot an arrow at them”. But could you just say “I Ready an attack, if any of the creatures move close to us?” and then, when it is triggered, decide which weapon to attack with, whether to attack with melee or range, or (as above) whether to use a weapon or grapple/shove? Or disarm?

A similar situation is Help – you could be planning to help one character do something, but by the time of the trigger, prefer help someone else do something else. Could you just say that you are Ready to Help whoever might need it when (trigger)?

Essentially, how specific do you have to be about the Action you are Readying? Can you be as general as saying you will Ready a move, an Attack, a Help, etc.? Or do you have to say who you will Attack, how (melee/range) and with what weapon, or who you will Help and with which of their Actions, etc.?

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## What kind of change of basis keeps the matrix of a linear operator unchanged?

Let $V$ be a finite-dimensional vector space in which the set $mathcal B = {e_i}_{iin{1,cdots,dim V}}$ is a basis. Consider also the dual space $V^*$ and the dual basis $mathcal B^* = {e^i}_{iin{1,cdots,dim V}}$. Now let $A:V to V$ be any linear operator. The matrix elements of this operator are given by

$$

A^i_j = e^i(Ae_j)

$$

Then the question is: if one makes a change of basis by a (-n invertible) linear operator $alpha:V to V, mathcal B mapsto mathcal B’$, what conditions should $alpha$ satisfy such that the matrix elements in the new basis are as those in the old, i.e.

$$

A^i_j = e^i(Ae_j) = e^{i’}(Ae_{j’}) = A^{i’}_{j’}

$$

where $mathcal B’ = {alpha e_{i} = e_{i’}}_{iin{1,cdots,dim V}}$ is the new basis and $mathcal B’^* = {e^{i’}}_{iin{1,cdots,dim V}}$ is dual of the new basis. Also the superscript denotes the row and subscript denotes the column.

**My Attempt:**

By the operator $alpha:V to V$ the new basis vectors in terms of the old are $e_{i’} = alpha^j_{i’} e_j$ where $alpha^j_{i’}$ are matrix elements of $alpha$ wrt $mathcal B$. The dual of the new basis vectors are $e^{i’} = alpha_j^{i’} e^j$, where $alpha_j^{i’}$ are matrix elements of the inverse of $alpha$ wrt $mathcal B’$.

The transformed matrix elements are (in general)

$$

e^{i’}(A e_{j’}) = alpha_k^{i’} e^k(A alpha^l_{j’} e_l) = alpha_k^{i’} alpha^l_{j’} e^k(A e_l)

$$

requiring that the new matrix elements are the same as the old ones means

$$

e^{i}(A e_{j}) = alpha_k^{i’} alpha^l_{j’} e^k(A e_l)

$$

Now I write the last equation in matrix notation as

$$

(A)_mathcal{B} = (alpha^{-1})_mathcal{B’} (A)_mathcal{B} (alpha)_mathcal{B}

$$

Now I use $(alpha^{-1})_mathcal{B’} = (alpha^{-1})_mathcal{B} = (alpha)_mathcal{B}^{-1}$ (but I’m not sure of it and it is the reason for stating this question in general) which gives

$$

(A)_mathcal{B} = (alpha)_mathcal{B}^{-1} (A)_mathcal{B} (alpha)_mathcal{B} qquad iff qquad

(alpha)_mathcal{B}(A)_mathcal{B} – (A)_mathcal{B} (alpha)_mathcal{B} = 0

$$

Which is to say that in oreder to keep the matrix elements of $A$ in the new basis unchanged one has to change the basis by a(-n invertible linear) operator $alpha$ that commutes with $A$.

$$

(alpha, A) = 0

$$

The last result has the following implication: If $mathcal B$ is a basis then $alphamathcal B$ is as well but also are $alpha^nmathcal B$, for $n in mathbb Z$ and wrt all these bases the matrix of $A$ is the same!

It is good now to ask if there is something wrong with the conclusion $(alpha, A) = 0$

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