dnd 3.5e – How Do Prepared Actions Involving a Movement Interact with a Charge?

Situation:

In turn, Alice uses the action ready. It specifies the condition "a creature attacks me" and an action that would take her out of her melee range (like teleporting with dimension door).

At Bob's next turn, he's charging Alice. The attack that he would take at the end of his charge meets the trigger conditions of Alice's prepared action, so she teleports.

What exactly come?

  • Can Bob make his melee attack before Alice is teleported?
  • Does it matter if Alice teleports somewhere that would have been in Bob's original charge range, compared to a place where he could not bill?
  • If Bob fails to attack, does he still lose all of his full round action for charging? Or did he just spend a movement action if he can never attack? What happens if it is already farther than it could be with a single action (since you can load twice as fast)?
  • Would it matter if Alice had chosen a different trigger condition for her prepared action, like "a creature approaching me at 20 feet?"

lee groups – Derivative of expression involving an invariant left-hand connection

I'm trying to understand a calculation made in this article. A little simplified, the configuration is as follows.

Let $ G $ to be a Lie group, and $ varrho $ his Lie algebra.

Let $ nabla $ to be an invariant connection on the left, which means that $ nabla $ evaluated in two left invariant vector fields is itself an invariant vector field on the left.

They calculate then;

$$ left. frac {d} {d}} right | _ {t = 0} nabla_ {u (t)} {u (t)} = nabla _ { left. frac {d} {dt} right | _ {t = 0} u (t)} {u (0)} + nabla_ {u (0)} { left. frac {d} {dt} right | _ {t = 0} u (t)} $$

Any suggestion as to the reason for this situation would be greatly appreciated.

dnd 3.5e – How is this scenario involving the Polymorph and Alter Self episodes?

We had an interesting scenario popping up at our table. A dwarf runesmith (Races of Stone, p.118) level 9, preparing for a great battle, took the following steps:

  1. He put all his equipment on the ground so that nothing was eaten by a polymorph
  2. He threw Polymorph and turned into Behir
  3. He launched the Blessing of Girralon (Compendium of the Kinds, p.106) to obtain a pair of arms.
  4. He launched the Fearless Grappling (Spell Compendium, page 90) to gain 2 pairs of tentacles.
  5. He released Fuse Arms (Spell Compendium p.100) for +20 untyped strings from Behir's 2 additional claw pairs + all his spells brought.
  6. He then launched Alter Self and became a dwarf again.
  7. He put all his armor on and cast Stone Fist (Spell Compendium, 94) to get a +6 bonus to Strength.
  8. He launched Enlarge Person on himself to become big and get a size bonus of +2 to force

In total, he claimed to be a big dwarf with 54 points of strength, 11 dexterity and 21 of constitution.

It works?

The argument was that Alter Self transforms you according to your "normal form" which is a dwarf, but does not explicitly affect your ability scores, which are obscurely high since the Polymorph. And since he's taken the form of this creature (a humanoid dwarf), this has changed his type of magical beast so that he can always launch Enlarge Person. Is it correct?

5th dnd – What would be a balanced Eldritch invocation involving blind vision?

I've tried to create an Eldritch Invocation that values ​​Warlock's blinding, but I'm not sure about the scope of this blindness. I first thought of comparing it to the pre-existing summons, Devil's Sight, which offers 120 feet of dark vision through magical and non-magical darkness, but blind vision is ( obviously) much better than darkvision. I have therefore abandoned this comparison. Any suggestion as to the scope of blind vision should be greatly appreciated.

I was hoping to make sure that this blind Eldritch Summon did not have a Warlock or Pact level requirement, but I would be ready to add a level prerequisite if that is absolutely necessary.

limit involving a harmonic function

Let $ u $ harmonic function in $ mathbb {R} ^ 3 – {0 } $. I know it $$ lim_ {x to0} sqrt {| x |} cdot u (x) = k < infty $$

I'm trying to show that $ k = 0 $. I tried by contradiction, but I failed and I start to suspect that $ k may be different from $ 0.

recursion – Recurrence involving a recurring product

I can solve the following recurrence problem as follows:

    Erase everything[a, b];
a[1] : = 2;
a[2] : = a[1] + b;
b = 2;
a[n_] : = Product[a[i], {i, 1, n - 1}]+ b
list = table[a[i], {i, 1, 5}]

which gives: {2, 4, 10, 82, 6562}.
But when I try:

    f = FindSequenceFunction[list]

I receive: "FindSequenceFunction[{2, 4, 10, 82, 6562}]when I expect the function: 3 ^ 2 ^ n + 1, since:

    Table[3^2^n + 1, {n, 0, 3}]

gives: {4, 10, 82, 6562}. Frustrated; I tried other approaches that did not work either. For example:

(1.) recurrence table approach:

    Erase everything[a, b];
With[{b = 2},
 RecurrenceTable[{a[n] == Product[a[i], {i, 1, n - 1}]+ b, a[1] == 2,
a[2] == a[1] + b}, a, {n, 1, 5}]]

(2.) RSolve approach:

    Erase everything[a, b];
With[{b = 2}, 
 RSolve[{a[n] == Product[a[i], {i, 1, n - 1}]+ b, a[1] == 2,
a[2] == a[1] + b}, a[n], not]]

(3.) Modular approach:

    Erase everything[a, b];
prod[n_] : = Module[{a},
a[1] : = 2; b: = 2; a[2] : = a[1] + b; k: = (n - 1);
a[i_] : = Product[a[j], {j, 1, k}]+ b;
a[n]
  ]

When I evaluate:

    prod[1]

I receive 2. Likewise[2] give 4 and prod[3] gives 10. But, when I try:

    prod[4]

I get the following message: "$ RecursionLimit :: A recursion depth of 1024 was exceeded when …"

I would appreciate any help with the three approaches above. Thank you!

probability – Markov chain involving gerrymandering in Pennsylvania.

I am currently working on a project to analyze the 18 districts of Pennsylvania and use the results of the 2018 legislative elections of the House of Representatives. I understand that the transition matrices for Markov chains must be square; However, I do not know how to do it because I currently have an 18×3 matrix (18 districts, 3 parties (Republican, Democrat, Independent).

Here is a more in-depth summary of my project:

we will build a Markov chain X_t whose state space consists of partitions of a real state and whose initial state X_0 is a totally random partition. let f (X_t) denote the
number of seats R for the partition X_t. then 1/1000 * (f (X_1) + … +
f (X_1000)) would be our sense of the number of R seats for this state. Then see where is the fairness version of our model in the graph of the efficiency gap.

sequence and series – An iterative argument involving $ f (n + 1) – f (n) $

I work with an argument implying an inequality of form:
$$ f (n + 1) leq f (n) + C (f (n)) ^ {1 – frac {1} { gamma}} qquad ( ast) $$
or $ f $ is a positive function, $ gamma> 0 $ and $ C> $ 0. It is known (but not explicitly proven) that $ ( ast) $ leads to the terminal
$$ f (n) leq n ^ { gamma} quad ( forall n> n_0) qquad ( ast ast) $$ for a certain $ n_0 $ to be chosen.

My question is: how to prove $ ( ast ast) $, being that we have $ ( ast) $ ? My failed attempt was to use a telescopic sum
get $$ f (n + ell) – f (n) leq C sum_ {k = 0} ^ { ell – 1} (f (n + k)) ^ {1 – frac {1} { gamma}} $$ but that does not lead to $ ( ast ast) $ straight ahead.

point-bound involving fractional Laplacians on the unit ball

Let $ B $ denote the unit ball in $ mathbb R ^ n $ and $ u in C_0 ^ infty (B) $ to be a radial function such as $ | (- Delta) ^ { frac s2} u | _ {L ^ 2 ( mathbb R ^ n)} leq 1 $ with $ s in (0,1) $. Do we have the estimate
$$ | u (x) | leq C | x | ^ {- frac {n-2s} 2} $$
for a constant $ C $ independent of $ u $.

Solving an initial value problem involving the nonlinear ordinary second order differential equation $ y & # 39; & # 39; (x) = y (x) cdot y # (x) + (y # (x)) ^ 2 $

I have the following equation $ y & # 39; (x) = y (x) cdot y # (x) + (y # (x)) ^ 2 $ with the initial values $ y (1) = 0 $ and $ y (1) = -1 $.

I am looking for advice on the best way to tackle this particular problem.

This article suggests that a substitution v can reduce the problem in a certain way. So I leave $ v = y $ (x) as following:

begin {align *}
y & # 39; & # 39; (x) & = y (x) cdot y (x) + (y (x)) ^ 2 & (1) \
y & # 39; & # 39; (x) & = y (x) cdot v + v ^ 2 & (2) \
end {align *}

When I enter (2) in wolfram, the engine says that the equation is second-rate linear ordinary differential equation.

Can we use substitution from here to solve the problem? Or is there a better way to do that?