I continue the topic of blunders and squiggles, begun in the article "Blunders of textbooks and curiosities of study . " I recall the definitions:

Blooper is a blatant or veiled mistake, which is not, however, of a fundamental nature, so that after suffering, you can fix it.
Zagogulina is a phrase, a theme set forth in such a way that to understand it one needs to smash one’s head (ordinary, not genius and not talent).

1. What is the simplicity of the formula?

Take the book “Probability Theory. Basic concepts. Limit Theorems. Random processes ”, Yu.V. Prokhorov, Yu.A. Rozanov,“ Science ”, 1967. I
quote the text on page 14:
“ Stirling formula. In all the above formulas, the expression

n! = n (n - 1) \dots 1

$n!=n(n-1)…1$ . The direct calculation of such a product for large n is very laborious. There is a relatively simple formula giving an approximate value for n !, called the Stirling formula: for large n

n! ∽ \sqrt{2 π n} n^{n} e^{- n}

$n! \backsim \sqrt{2πn}n^n e^{-n}$

A similar phrase is found in other books and on the Internet.

I don’t understand something than the Stirling formula is simpler than the defining formula

n! = n (n - 1) \dots 1

$n!=n(n-1)…1$ . In what respect is easier? How to arrange formulas by simplicity? Since the key phrase was “Direct calculation of such a product for large n is very laborious”, it is natural to assume greater simplicity in the sense of fewer calculations. Okay, let's come from this side. Compare the formulas from the position of the number of elementary operations (from the point of view of a computer) in one and the other formulas. In the formula

n! = n (n - 1) \dots 1

$n!=n(n-1)…1$ there are n multiplications. In the Stirling formula, we have the operations:

$\sqrt{2πn}$ - two multiplications and one root extraction. Extracting the root is not an elementary operation, but its implementation requires a computational cycle and the longer, the greater the required accuracy of the calculations.
$n^n$ - n multiplications. $n^n=nnn…n$ . After all, we will not dissemble, that this is one operation. In this case, we can say that n! one operation. At least in no computer, neither degree nor exponential function is elementary operation.
$e^{-n}$ - this is also not an elementary operation, but its implementation requires a computational cycle and the longer, the greater the required accuracy of the calculations.

The funniest thing is that

n^{n} = n n \dots n

$n^n=nn…n$ declared easier

n! = n (n - 1) \dots 1

$n!=n(n-1)…1$ . From any point of view on simplicity, after this one can no longer reason further.
So, it is clear that from the point of computational complexity, the Stirling formula is in no way simpler than the defining formula.
So why then do we need the Stirling formula? There is no universal answer. And it does not come down to simplicity of calculations. It all depends on the situation. For example, it is unlikely to simplify the expression

N! / (N - 1)!

$N!/(N-1)!$ you need to apply the Stirling formula. The defining formula immediately gives

N! / (N - 1)! = N

$N!/(N-1)!=N$ .
In general, if we have an identity like formula1 ≡ formula2, then sometimes it is beneficial to replace formula1 with formula2, and sometimes vice versa.
In some situations, the application of the Stirling formula leads to an obvious reduction in the terms of the formula where it enters, which is difficult to see if the defining formula is applied. At least in statistical physics this is so. There, all fundamental quantities are expressed in terms of statistical weight, the formulas for which are flickered with factorials. But entropy, for example, is expressed through the natural logarithm of the number of states. This is where the role of form begins to play, in this case, the representation of the factorial through the degree:

l n (n!) ∽ n l n (n / e)

$ln⁡(n!) \backsim nln(n/e)$

And here is an example of the application of the Stirling formula, taken from Fichtenholtz (v.2):

2. The catch of the notation

Take the textbook “Physics of Elementary Particles”, author N.F. Nelipa, Moscow, “Higher School”, 1977. On page 19, the relation between the momentum and coordinate representations is recorded:

φ (x) = (1 / (2 π)^{2}) \int d p e^{- i p x} φ (p) (1)

$φ(x)=(1/(2π)^2) ∫dpe^{-ipx}φ(p) \ \ \ \ (1)$

We see that the formula contains both φ on the right and φ on the left. If this is an identity, then looking at this formula, a not very sophisticated person can draw these conclusions.
We take φ (x) = 1
Then from (1) we have

1 = (1 / (2 π)^{2}) \int d p e^{- i p x} 1

$1=(1/(2π)^2) ∫dpe^{-ipx}1$

Got a “wonderful” decomposition of unity. Or, similarly,

s i n (x) = (1 / (2 π)^{2}) \int d p e^{- i p x} s i n (p)

$sin(x) =(1/(2π)^2) ∫dpe^{-ipx}sin(p)$

All this is clearly nonsense. So what's the deal?
Or maybe relation (1) should be interpreted as an equation of the type

x^{2} = b x

$x^2=bx$ ?
We look at other books. Here is another “Physics of Elementary Particles”, author Gaziorovich, Moscow, “Science”, 1969. On page 20 we have the formula

φ (x) = (1 / (2 π)^{2}) \int d p e^{- i p x} φ ̃ (p) (2)

$φ(x)=(1/(2π)^2) ∫dpe^{-ipx} φ ̃(p) \ \ \ \ (2)$

I breathe a sigh of relief. This is just a connection between the coordinate and momentum representations and it is given from the point of view of mathematics by the Fourier expansion. Here φ and φ ̃ are different functions. But what about formula (1) Nelips? From the point of view of mathematics, it is incorrect. If φ is a function, then both φ (x) and φ (p) are all one function. I was tormented for a long time (“The author cannot help but notice this. That means the case is trickier”) and then found an excuse from the point of view of physics. Here it is:
φ (x) and φ (p) are not the same functions, they are one and the same field φ taken in different representations. The field is one, but its presentation is different. The view type is specified by a letter in brackets. We focus on the fact that the field does not change. The view is changing. So I reassured myself.But, gentlemen, the authors, you are writing a textbook. explain to mere mortals what’s what. And then the reader has to find excuses for the author .

Further, I turn to the Soviet quantum classics “Introduction to the Theory of Quantized Fields” by NN Bogolyubov and DV Shirkov, Nauka, Moscow, 1973. Page 28.

φ (x) = (1 / (2 π)^{3 / 2}) \int d k e^{- i k x} φ ̃ (k) (3)

$φ(x)=(1/(2π)^{3/2}) ∫dke^{-ikx}φ ̃ (k) \ \ \ \ (3)$

Fine. And the aforementioned Gaziorovich is similar. But, look further. I quote:
“... we find that the function φ ̃ (k) satisfies the equation

(k^{2} - m^{2}) φ ̃ (k) = 0

$(k^2-m^2 ) φ ̃ (k)=0$

And, therefore, can be represented as

φ ̃ (k) = δ (k^{2} - m^{2}) φ (k)

$φ ̃ (k)=δ(k^2-m^2 )φ(k)$

And, further, it is said that with this in mind, decomposition (3) will take the form

φ (x) = (1 / (2 π)^{3 / 2}) \int d k e^{- i k x} δ (k^{2} - m^{2}) φ (k) (4)

$φ(x)=(1/(2π)^{3/2}) ∫dke^{-ikx} δ(k^2-m^2 )φ(k) \ \ \ \ (4)$

Again we returned to the representation of φ through φ. From understandable Gaziorovich came to the incomprehensible Nelipe. What does it mean? Is this an equation? I think that here too it is necessary to give an interpretation similar to the case of Nelipa given above.
This is really a squiggle. For some reason, this squiggle met me only in the books of Soviet authors.
And I will return to the author of Nelip. We take his book “Physics of elementary particles. Gauge Fields. ” The book is recommended as a study guide. This obliges the textbook to make all kinds of defaults as little as possible. No need to force the student to think long about solving the defaults. It’s already difficult for him. However, we take the formula (1.2.14) and the preceding text:
Matrices

λ_{k}

$λ_k$ satisfy the following commutation relations (Lie algebra):

[λ_{k}, λ_{j}]_{-} = 2 i f_{k j n} λ_{n}, S p λ_{i} λ j_{k} = 2 δ_{i j} (1.2.14)

$[λ_k,λ_j ]_-=2if_{kjn} λ_n, Spλ_i λj_k=2δ_{ij} \ \ \ \ (1.2.14)$

And not a word is said that summing is meant by n. This is neither in the preface nor in the main text. And there are a lot of such formulas in the book.
Moreover, the introduction says:
“The scalar product of two four-dimensional vectors is written as

Bloopers and squiggles. 2

1. What is the simplicity of the formula?

2. The catch of the notation

3. Large squiggle measurements

$\ \ \$ 3.1. Dirac

$\ \ \$ 3.2. Landau

$\ \ \$ 3.3. Blokhintsev

$\ \ \$ 3.4. Susskind

$\ \ \$ 3.5.

$\ \ \$ 3.6. Wihman

$\ \ \$ 3.7.

$\ \ \ \ \ \$ 3.7.1. Denial of the projection postulate

$\ \ \ \ \ \$ 3.7.2. Denial of non-deterministic reduction

$\ \ \$ 3.8. Summary

$\ \ \ \ \ \$ 3.8.1. Cooking

$\ \ \ \ \ \$ 3.8.2. Measurement

More articles:

Bloopers and squiggles. 2

1. What is the simplicity of the formula?

2. The catch of the notation

3. Large squiggle measurements

\ \ \ 3.1. Dirac

\ \ \ 3.2. Landau

\ \ \ 3.3. Blokhintsev

\ \ \ 3.4. Susskind

\ \ \ 3.5.

\ \ \ 3.6. Wihman

\ \ \ 3.7.

\ \ \ \ \ \ 3.7.1. Denial of the projection postulate

\ \ \ \ \ \ 3.7.2. Denial of non-deterministic reduction

\ \ \ 3.8. Summary

\ \ \ \ \ \ 3.8.1. Cooking

\ \ \ \ \ \ 3.8.2. Measurement

More articles:

$\ \ \$ 3.1. Dirac

$\ \ \$ 3.2. Landau

$\ \ \$ 3.3. Blokhintsev

$\ \ \$ 3.4. Susskind

$\ \ \$ 3.5.

$\ \ \$ 3.6. Wihman

$\ \ \$ 3.7.

$\ \ \ \ \ \$ 3.7.1. Denial of the projection postulate

$\ \ \ \ \ \$ 3.7.2. Denial of non-deterministic reduction

$\ \ \$ 3.8. Summary

$\ \ \ \ \ \$ 3.8.1. Cooking

$\ \ \ \ \ \$ 3.8.2. Measurement