Lambda Calculus – Equivalent terms in call-by-name, but not call-by-value

When working with the untyped lambda calculation, I am asked to specify two terms that are equivalent in call-by-name semantics, but not in call-by-value semantics.

Call $ text {fls} = lambda x. lambda y. y $ and $ Omega = ( lambda x. X x) ( lambda x. X x) $ , I was suggested to look at these terms:

$ text {fls} ( lambda x. omega) $ which is reduced to in both semantics $ lambda y. y $

$ text {fls} ( lambda x. omega x) $ which is reduced to call-by-name $ lambda y. y $ but in call-by-value, the evaluation of the argument diverges $ text {fls} $".

I do not understand how they diverge unless I assume $ text {eta} $Conversion that was not accepted in my course. In addition, I do not see how the argument can be distinguished and evaluated $ text {fls} $, Does that make sense to you?

Aside

I suggested the conditions $ ( lambda f. omega) $ and $ ( lambda t. lambda f. omega $ I think that's a valid example …

Calculus – Mean Value Contradiction

Question: Let f (x) = 2 – | 2x – 1 |. Show that there is no value of c such that f (3) – f (0) =
f & # 39; (c) (3 – 0). Why does this not contradict the mean value theorem?

Note: I am not asking how to solve this problem. I ask a question about this problem.

EX1: To answer this question, x is not differentiable at 1/2 because the left slope (2) is not equal to the right slope (2). So there is no value c, where f; (c) has a slope of -4/3 in the interval (0,3).


EX2: My question is, what happens if you have restricted the domain to (3,9)? Then the function would be continuous as before, but this time also DIFFERENT. However, there still is no value c at which f & # 39; (c) has a slope of -4/3.

I know that my question sounds confusing, but in a nutshell, what does it really mean to say that "the mean is contradicted"? Is EX2 not contradictory to MVT and EX1 because EX1 has an additional condition where it is indistinguishable on the domain (0,3)?

Calculus – How can I evaluate $ frac {d} {dx} int_ {0} ^ {x} dy ? f (y) int_ {0} ^ {y} dz ; f (z) $ right?

Naively, I thought that the second factor does not depend on it $ x $, I can use
$$ frac {d} {dx} int_0 ^ x f (y) dy = f (x) $$
Find
$$ frac {d} {dx} int_ {0} ^ {x} dy ; f (y) int_ {0} ^ {y} dz ; f (z) = f (x) int_ {0} ^ {y} dz ; f (z) ,. $$
However, this obviously can not be correct, since the second integral still depends on the integration variable of the first integral, which is no longer present in the formula.

How can I rate $ frac {d} {dx} int_ {0} ^ {x} dy ; f (y) int_ {0} ^ {y} dz ; f (z) $ correctly?

(I suspect the correct answer is $ f (x) int_ {0} ^ {x} dz ; f (z) $ but I'm not sure how to show it.)

Calculus and Analysis – Find a constant with which an integral can finally be

I have a function that is not so complicated polynomial, depending on a parameter $ c $:

$$ f (r, c) = frac {-2r ​​^ 4-4r ^ 3-6r ^ 2-6r-3} {8r ^ 3} – frac {c left (-2r ^ 2-2 r-1 right)} {4 r ^ 3} $$

What I am looking for is that a function of this function has a finite integral. I want the following expression to exist:

$$ int_0 ^ { infty} 4 pi r ^ 2 left ( exp {(- r)} (1
+ f (r, c) right) ^ 2 mathrm {d} r $$

If $ c = -3 / 2 $, this integral is finite, as I can check with Mathematica:

f(r_, c_) := ((-3 - 6 r - 6 r^2 - 4 r^3 - 2 r^4)/(8 r^3) + ((-1 - 2 r - 2 r^2) c)/(4 r^3))
Integrate((Exp(-r) (1 + f(r, -3/2)))^2*4 (Pi) r^2, {r, 0, (Infinity)})

The result is $ pi / 16 $as it should be. For other values ​​of $ c $ (as far as I know) the integral is not finite. But I want to check this Mathematica,

The simplest idea that I had is simply to ask for the integral as a function of $ c $:

Integrate((Exp(-r) (1 + f(r, c)))^2*4 (Pi) r^2, {r, 0, (Infinity)})

The result, however, is that the integral does not converge.

How do I find all values ​​of $ c $ Where is this integral finally?

Calculus – Compute $ lim_ {x to 0 ^ {-}} frac { sqrt {x ^ 2 + x + 4} – 2} { ln (1 + x + x ^ 2) – x} $

I tried to rate

begin {equation *}
lim_ {x to 0 ^ {-}} frac { sqrt {x ^ 2 + x + 4} – 2} { ln (1 + x + x ^ 2) – x}.
end {equation *}

We have
begin {equation *}
frac { sqrt {x ^ 2 + x + 4} – 2} { ln (1 + x + x ^ 2) – x} = frac {x ^ 2 + x} { ln (1 + x + x ^ 2) – x} cdot frac {1} { sqrt {x ^ 2 + x + 4} + 2} quad text {for all} x in mathbb {R},
end {equation *}

So I think my practice is boiled down to that
begin {equation *}
lim_ {x to 0 ^ {-}} frac {x ^ 2 + x} { ln (1 + x + x ^ 2) – x},
end {equation *}

and this limit is the same $ – infty $ from De L & # 39; Hôpitals set.

Can I rate this limit without De L & # 39; Hôpital's theorem?

Lambda calculus and church numbers

I study a book about lambda calculus and have met church numbers and the successor function.

I think I understand the basics of beta reduction and I understand how the book implements the numbers

It thus defines a general coding of a value n as

n = 𝜆f.𝜆x.f^n x

^ n is f increased to n, I do not know how to format here superscript / subscript …

Then

0 = 𝜆f.𝜆x.x
1 = 𝜆f.𝜆x.fx
2 = 𝜆f.𝜆x.f(fx)

What I get as a number is just a count of the encapsulation of a function f about x,

It then defines SUC as a way to add 1 n as:

SUC = 𝜆n.𝜆f.𝜆x.nf(fx)

So, apply n to SUC:

SUC_n = (𝜆n.𝜆f.𝜆x.nf(fx))(𝜆f.𝜆x.f^n x)
         (n := (𝜆f.𝜆x.f^n x))
      = 𝜆f.𝜆x.(𝜆f.𝜆x.f^n x)f(fx)
               (f := f)
      = 𝜆f.𝜆x.(𝜆x.f^n x)(fx)
               (x := (fx))
      = 𝜆f.𝜆x.f^n(fx)
      = 𝜆f.𝜆x.f^(n+1)x        This step I get, we are increasing n as an encapsulation of f over x
      = n + 1                 There is step here I don't get. How are we losing f^(n+1)x to get
                              just n+1

So my question is. The last step, what's next? 𝜆f.𝜆x.f^(n+1)x too straight n+1,

Is it a beta reduction that I do not understand / miss here?

Calculus and Analysis – Indefinite The integration of the real variable function gives a complex logarithm

Integrate real variable function in Mathematica

 Integrate(1/Sqrt(2 x - x^2), x)

outputs the following output with complex logarithm
$$
frac {2 sqrt {x-2} sqrt {x} log left ( sqrt {x-2} + sqrt {x} right)} { sqrt {- (x-2) x} },
$$

while I expected $ arcsin (x-1) $ as a result of integration.

I tried to give some assumptions like that

Integrate(1/Sqrt(2 x - x^2), x, Assumptions -> x > 0 && x < 2)

but that does not help.

Is it possible to bypass this complex logarithm and instruct Mathematica to integrate only via Real?

Calculus – Two curves for the orthogonal trajectory of $ y = mx $

Find the orthogonal trajectory of $ y = mx $,

We know,
$$ frac {dy} {dx} = m $$

For the orthogonal trajectory
$$ – frac {dx} {dy} = m $$

We integrate both sides,
$$ y = – frac {x} {m} $$

But if I then use $ m $ and then integrate, I get a completely different curve.
$$ x ^ 2 + y ^ 2 = c $$

Are both curves considered as orthogonal curves to our original equation? I've seen books that only answer the second turn. But is not the first curve orthogonal? If not, where do I go wrong?

Any help would be appreciated.

Calculus – I'm trying to prove the derivation of $ sqrt {x} $ with geometry.

I try to prove the derivation from $ sqrt {x} $ with geometry.

So far I have created a square with area $ x $ and side lengths $ sqrt {x} $,

The derivative of the function is $ frac {d sqrt {x}} {dx} $ With $ dx $ the enlargement of the area.

I have set up the equations for the change in f,

$ df = 2 ( sqrt {x}) ( text {} d sqrt {x}) text {} + text {} d sqrt {x} text {} d sqrt {x} $

$ df = 2 ( sqrt {x}) ( text {} d sqrt {x}) text {} + text {} dx $

I should end with $ frac {df} {dx} $ = $ frac {1} {2 sqrt {x}} $

What I get in the end is $ frac {df} {dx} $ = $ frac {1} {1-2 sqrt {x}} $

What am I doing wrong?

Thank you very much