automatons – find derivative tree for CFG

I need to draw the derived tree for 1-2-3 (3-4) * 5 * 6 of the grammar
How many possibilities are there trees?

Vn = {expr, term, factor, number}

Vt = {(,), -,, 0 … 9}
P = {
expr-> expr-expr | term
term-> term
factor | factor
factor-> number | (expr)
number-> 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
S = expr

The possibilities that I can find are: (1-2) – (((3-4) * 5) * 6), 1- (2- (((3-4) * 5) * 6))

Is there another possibility?

Calculation and analysis. The integral of the derivative does not work.

In 11.3.0 for Microsoft Windows (64 bits) (March 7, 2018) writing:

To integrate[D[ArcTan[Sqrt[1 + x^5]], X], X]

I get:

ArcTan[Sqrt[1 + x^5]]

but if I write:

To integrate[D[ArcTan[x Sqrt[1 + x^5]], X], X]

Mathematics The integral can no longer be solved. Does something escape me?

Real analysis – Explicit expression for derivative.

Leave $ f:[-pi,pi]^ 2 rightarrow [0,1]$ be a smooth function such that $ f (0) = 0 $, $ 0 $ it is the only critical point of $ f $ and the sets of levels $ A_t: = left {z in [pi,pi]^ 2; f (z) = t right } $ They are connected and homeomorphic to a circle to $ t> 0. $

An example of such $ f $ would be $ f (z) = frac { Vert z Vert ^ 2} { Vert ( pi, pi) Vert ^ 2}. $

I would like to know: Can you get an explicit expression for the derivative of line integrals?

$$ F (t): = int_ {A_t} g (z) dz $$

where $ g in C ^ { infty} ([-pi,pi]^ 2). $

It seems that there is no standard expression for the derivative that immediately gives the derivative of this function $ F $.

Multivariable calculation: what is the derivative function?

I would like to illustrate my confusion on this issue by constructing the theme from more or less first principles. Leave $ U subseteq mathbb {R} ^ n $ be an open subset and let $ f: U to mathbb {R} ^ m $. We say that $ f $ it is totally differentiable in $ a in U $ If there is a linear function. $ D_ {a} f: mathbb {R} ^ n a mathbb {R} ^ m $ such that the following holds.

$ lim_ {x to a} frac {|| f (x) -f (a) -D_ {a} f (x-a) ||} {|| x-a ||} = 0 $

We called $ D_ {a} f $ the total derivative of $ f $ to $ a $. My question concerns how we define the derived function. For the family case in which $ n = m = 1 $, this is simple since the total derivative is reduced to only

$ (D_ {a} f) (b) = big ( frac {df} {dx} big { rvert} _a big) b $

for all $ b in mathbb {R} $. Here, $ frac {df} {dx} big { rvert} _a $ Denotes the usual definition of the derivative of $ f $ to $ a $. Then we define the derived function. $ f & # 39 ;: U to mathbb {R} $ as

$ f & # 39; (y): = frac {df} {dx} big { rvert} _y $

for all $ y in U $. Now consider the case when $ m = 1 $. Representing the action of $ D_ {a} f $ through the Jacobian matrix, we have

$ (D_ {a} f) (b) = begin {bmatrix}
frac { partial {f}} { partial {x_1}} big { rvert} _a & cdots & frac { partial {f}} { partial {x_n}} big { rvert} _a
end {bmatrix}
begin {bmatrix}
b_1 \
vdots \
end {bmatrix} $

for all $ b = (b_1, cdots, b_n) in mathbb {R} ^ n $. This suggests that we define the derived function. $ Df: U to mathbb {R} ^ n $ as

$ (Df) (y): = big ( frac { partial {f}} { partial {x_1}} big { rvert} _y, cdots, frac { partial {f}} { partial {x_n}} big { rvert} _y big) $

for all $ y in U $. Maybe we could make a similar argument for the case in which $ n = 1 $. How does this extend to cases where the Jacobian matrix is ​​not a simple column or row vector? What definition of the derived function is more natural in those cases? By natural, I mean that the definition should allow important identities such as the product rule and the chain rule to retain their obvious forms.

tracing: How to manipulate a partial derivative diagram of self-defined bivariate functions?

I have a linear function by parts (spline) that describes a progressive income tax program, with higher marginal rates for higher and higher incomes.

I want to understand the implications of a rule that some tax-exempt income could still increase your average rate in the taxable part. Basically, with X being taxable income and Y being exempt income, your taxes will be f (x + y) / (x + y) * x, where F It is the non-linear scheme that would apply to income without exemptions. I can define F as a linear function by parts, see above.

To calculate the marginal tax rates on taxable and exempt income, I would simply need the partial derivatives of this expression, with the particular function F plugged in, and this I can plot against the two variables. Or trace against one and use the other as a parameter to Handle.

The derivative looks good in Mathematica, although it is a bit difficult to verify with so many cases.

However, I have empty plots, which, of course, do not make sense to manipulate. Where is this going wrong?

Handle[Plot[D[Tax[x+y]/ (x + y) * x, y], {and, 0,200000}], {x, 0,200000}]

Think about Tax in this way:

Tax [z_]: = By parts[{
                        {0,0    <=z< 11265},
                        {0+ 0.0856(z-11265),11265<=z}

gm. General Mathematics: first derivative with respect to x and the Pell equation.

When you graph the Pell equation:
x = $ sqrt (Dy ^ 2 + 1) $and the first derivative of y with respect to x of the Pell equation, d / dx $ sqrt ((x ^ 2 – 1) / D) $ for D> 1, you will see them intersect at x = 1.618. Is there any evidence that shows that the point of intersection is exactly equal to $ phi $, where $ phi $ = ($ sqrt5 + 1) / 2 $? If not, can someone prove it (or refute it)?

calculation and analysis – Find the derivative of a loss function with respect to the output of a neural network

I have a loss function $ L $, specifically a CrossEntropyLayer["Probabilities"]. I also have a trained neural network $ text {net} $ that outputs a softmax layer, mapping $ out = text {net} (entry) $. I'm experimenting with backward propagation – all I'm trying to find is $ frac {dL (out)} {d (out)} | _ {input = bf {x}} $, and I can not figure out how to do it.

Of course, that derivative is an abbreviated form of a list of partial derivatives for each (4315) of the probabilities returned from the SoftmaxLayer.

I have seen NetPortGradient, but it only seems to be able to calculate the gradient of the output with respect to something, instead of the gradient of the loss with respect to the output.

numerics – Error with NSum: returns NSum :: nsnum: Summand (or its derivative) f[n] it is not numeric at the point n = 17

Consider the following example (I had a lot of trouble finding an example of minimal work, I think it's already compact enough).

Omega0 = 1.
t = 2
nAvg = 10

Omegan[n_] : = Omega0 * Sqrt[n + 1]

F[n_] : = By parts[{{Cos[Omegan[{{Cos[Omegan[{{Cos[Omegan[{{Cos[Omegan[n]* t / 2]^ 2 *
ABS[Exp[-nAvg/2]* Sqrt[nAvg]^ n / Sqrt[Factorial[Factorial[Factorial[Factorial[n]]],
0 <= n <= 20}}, 0]NSum[F[F[F[f[n], {n, 1, 100}]

If you run this short script, you should return it:

NSum :: nsnum: Summand (or its derivative) […] great message […] It is not
numerical at point n = 16

This problem that I am facing occurs only with some specific function. It happens with this complicated looking function that I gave you, but if you try a simpler one, the script can work correctly.

My questions :

First: I would like to understand why I have this error

Second: How to solve it?

Extra question

Is it really more efficient to use NSum?[] what N[Sum[Sum[Suma[Sum[]]. Because I've read (I do not remember where) that when Mathematica sees N[Sum[Sum[Suma[Sum[]]understands that the sum must be done numerically (instead of trying the symbolic method, THEN numerically approximates).

Extra infos:

I've already seen Is this an NSum error?

With some functions it solves the problem to add NSumTerms-> number, with others not. The thing is that I would like to be able to face this problem "in general", so I need to understand what is happening (I read the documentation and I do not).

In summary: how to make numerical sums in general with mathematics? In my specific case I have functions that can be defined in parts. In general, my function can be a product / sum of functions by parts, so it is not obvious at first glance to know the limit of the sum without looking more closely, which I would like to avoid.

External derivative

I have a 3-manifold with Riemannian metric $ g = omega_1 ^ 2 + omega_2 ^ 2 + omega_3 ^ 2 $, where $ omega_i $ They are 1-forms (coframe fields).

I am in this situation:
$ -du wedge omega_2 wedge omega_3 (u ^ {- 1} p + frac {3} {2} u ^ {- 1}) + omega_2 wedge d omega_3 = d omega_2 wedge omega_3 PS, which should be possible to write as:
$ -du wedge omega_2 wedge omega_3 (u ^ {- 1} p + frac {3} {2} u ^ {- 1}) = d ( omega_2 wedge omega_3) $.

How do I get it $ omega_2 $ Y $ omega_3 $?

indexed variable – Derivative that is not evaluated when the function is defined

I have the following code

ExD = Sum[D[y[n, t], t], {n, 0, 1}];
ExS = Sum[y[n, t], {n, 0, 1}];
[Sigma][0]    = 1;
[Sigma]& # 39;[0] = 1;
Y[0, t_] = [Sigma][0]    + t [Sigma]& # 39;[0]

Note that when it is executed, ExD is still general for D[y[0,t], t], while ExS prints and[0,t] how it was defined It seems I can not understand why the first derivative with respect to the second variable of y[0,t] does not evaluate even though[0,t] It is clearly defined and printed correctly. I need to use Evaluate[] here?