Show that the difference between the isothermal and adiabatic compressibility is:

From the following definitions:

Coefficient of volume expansion -

Adiabatic compressibility -

Isothermal compressibility -

Proof:

For the right side of the equation,

For isothermal compressibility,

For adiabatic compressibility,

Since S is constant it follows that

Then,

Solving first for the derivative of T with respect to P,

We have,

Since

Therefore,

Find a formula for:

My Solution:

Since

We have,

Thus,

Here are some examples of false Mathematical proofs:

Number 1:

Number 2:

Number 3 (the famous one):

let:

Number 4:

(1-1) + (1-1) + … + (1-1) + (1-1) + … = 0

1 + (-1+1) + (-1+1) + … + (-1+1) + … = 0

1 + 0 + 0 +… + 0 + .. = 0

1 = 0

 

Prove that:

Solution:

First,

It follows that:

Then,

Hence,

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5 040 – Greek philosopher Plato gave this number as the population of an ideal city.

This number has several factors, which implies an efficient division of population for the sake of land distribution, taxes, war etc.

In fact,

5, 040 = 7! = (1)(2)(3)(4)(5)(6)(7)

5, 040 = (2⁴)(3²)(5¹)(7¹)

Thus, 5 040 has (4+1)(2+1)(1+1)(1+1) = 60 factors

Compute in (less than) 30 seconds without using calculator.

My technique:

Since 2007 is the arithmetic mean of the numbers 2004, 2006, 2008 and 2010, then: Let x = 2007

So, x-3 = 2004, x-1 = 2006, x+1 = 2008, x+3 = 2010

Hence, the expression is equal to

Simplifying,

= x² – 5 = (2007)² – 5 = 4, 028, 044

When the polynomial x²⁰⁰⁷ + 1 is divided by x² – 3x +2, the remainder can be expressed in the form ax+b. Find the value of a.

My Solution:
Let k be the quotient, then:
x²⁰⁰⁷ + 1 = k(x² – 3x +2) + (ax+b)
Since x² – 3x +2 = (x-2)(x-1),
x²⁰⁰⁷ + 1 = k(x-2)(x-1) + (ax+b)

To omit k and to solve for the value of a easily, set x=2 and x=1:

For x=2
(2)²⁰⁰⁷ + 1= k(2-2)(2-1) + (2a+b)
2²⁰⁰⁷ + 1= 2a+b  (*)

For x=1
(1)²⁰⁰⁷ + 1 = k(1-2)(1-1) + (a+b)
2 = a+b  (**)

To solve for the value of a, perform elimination using (*) and (**).
..Which in turns give: a=2²⁰⁰⁷-1

Compute the number of squares between 4⁹ and 9⁴

My short solution: The technique is to express these two numbers into perfect squares so that it’s easy to compute the number of squares between these two numbers.

4⁹ = (2²)⁹ = (2⁹)² = 512² ;

9⁴ = (3²)⁴ = (3⁴)² = 81²

Therefore, the number of squares between 4⁹ and 9⁴ is (512-81)-1 = 430

Five digit numbers containing all digits from 1-5 are arranged from highest to lowest. What is the 25th number?

(Examples of these numbers are 54 321; 23 451; 41 532)

My Solution:

First, it’s impractical to list these (5)(4)(3)(2)(1)=120 numbers from highest to lowest and then determine the 25th number in the list. So, we can instead do some technique and investigate the distribution of these numbers.

Since it’s from highest to lowest, the first step is to determine the number of these five digit numbers which have a ten thousands digit of 5.

For thousands digit: 4 numbers to choose.

For hundreds digit: 3 numbers to choose.

For tens digit: 2 numbers to choose.

For unit digit: 1 number to choose.

Thus, the number of these 5 digit numbers which have a ten thousands digit of 5 is (4)(3)(2)(1)=24. Then these 24 numbers are the first 24 numbers in the list. Hence, the 25th number is the highest number which has a ten thousands digit of 4. And this 25th number is 45 321.

Prove that:    , for any

Solution: Since     , then in order to produce a short

proof, the expression   should be express in the form  

Doing so,

Hence,

At what angle of projection is the range maximum? Prove it mathematically.

To solve for the required angle of projection ( ), we need first to find a general expression for the range.

To find range, set y = 0 first and use:

,

Implies that,

(non zero value)

Solving for the general expression for range,

Regardless of the value of , the maximum value of range would be attain if is maximum. Since 1 is the maximum value of sine function, then, should be equal to 1. Since sin 90º = 1, therefore (angle of projection) should be equal to 45º

“A physical theory must possess mathematical beauty.”

In the most basic explanation, mathematical beauty is related to simplicity and surprise.
- Paul Dirac, quantum physicist

To celebrate the beauty of our planet on Earth Day, NASA released its 10 favorite photos taken by astronauts on the International Space Station.

Click here: NASA’s top 10 views of Earth

Video by pauleycamerieexodus

Researchers Prove Existence of New Basic Element for Electronic Circuits — ‘Memristor’ from PhysOrg.com

HP today announced that researchers from HP Labs have proven the existence of what had previously been only theorized as the fourth fundamental circuit element in electrical engineering.

This scientific advancement could make it possible to develop computer systems that have memories that do not forget, do not need to be booted up, consume far less power and associate information in a manner similar to that of the human brain.

In a paper published in today’s edition of Nature, four researchers at HP Labs’ Information and Quantum Systems Lab, led by R. Stanley Williams, presented the mathematical model and a physical example of a “memristor” – a blend of “memory resistor” – which has the unique property of retaining a history of the information it has acquired.

Leon Chua, a distinguished faculty member in the Electrical Engineering and Computer Sciences Department of the University of California at Berkeley, initially theorized about and named the element in an academic paper published 37 years ago. Chua argued that the memristor was the fourth fundamental circuit element, along with the resistor, capacitor and inductor, and that it had properties that could not be duplicated by any combination of the other three elements…..

Read More: [...]

(AIME 1994) Find the positive integer n for which
[log₂ 1] + [log₂ 2] + [log₂ 3] +…+ [log₂ n] = 1994
where [x] denotes the greatest integer less than or equal to x (for example [Π] = 3)

Solution:

Observe that:
[log₂ 1] = 0(2⁰)
[log₂ 2] + [log₂ 3] = 1(2¹)
[log₂ 4] + [log₂ 5] + [log₂ 6] + [log₂ 7] = 2(2²)
[log₂ 8] + [log₂ 9] +…+ [log₂ 15] = 3(2³)
[log₂ 16] + [log₂ 17] +…+ [log₂ 31] = 4(2⁴) ……

** [log₂ c] + [log₂ (c+1)] +…+ [log₂ (2c-1)] = b(2^b); where 2^b = c, for positive integers b and c.

So,
0(2⁰) + 1(2¹) + 2(2²) + 3(2³) + 4(2⁴) + 5(2⁵) + 6(2⁶) + 7(2⁷) + 456 = 1994

**7(2⁷) is the sum of [log₂ 128] + [log₂ 129] + [log₂ 130] +…+ [log₂ 255]. Hence, 456 = [log₂ 256] + [log₂ 257] + [log₂ 258] +…+ [log₂ n].

Since [log₂ 256]=[log₂ 257]=[log₂ 258]=…=[log₂ n] = 8
So, 456/8 = 57; Then n = 256 + (57-1).

Therefore, n = 312

There was a young fellow from Trinity (Cambridge University?)…..

…..Who took √∞ ….. But the number of digits….. Gave him the fidgets…..

He dropped Math and took up Divinity…

Hehehehe … ∞

This is the phrase at the bottom of title page of the book One, Two, Three…Infinity by George Gamow.

I enjoyed reading it, since it is about facts and speculations of Science!

Prove that √5 + √3 is irrational

(No.4 of Problem Set 10 in Numbers: Rational and Irrational by Ivan Niven)

Solution:

The argument made here is a proof by contradiction or reductio ad absurdum; That is, assuming that the proposition is false and then derive a contradiction from this assumption.

Note: My solutions are also parallel to Prof. Niven’s

First, suppose that √5 + √3 is rational, say a: √5 + √3 = a

By squaring, so that radicals would reduce and make the proof shorter:

5 + 2√15 + 3 = a² ….. √15 = (a² -8)/2

—————————————————————————————-

Note: Rational numbers are closed under the four fundamental operations. But…..

…..proving that √15 is a rational should be done first to verify the assumption (√5 + √3 is rational)

—————————————————————————————–

Second, assume that √15 = x/y, where x and y are integers having no common factor other than 1; That is, in lowest term

Squaring, we would obtain: 15 = (x²)/(y²) ….. x² = 15y²

15y² is divisible by 15, so x² is divisible by 15, and hence x is also divisible by 15. Let’s say x = 15w

Then, x² = 15y² ….. (15w)² = 15y² ….. 15w² = y²

It implies that y² is divisible by 15, so y is also divisible by 15.

—————————————————————————————-

Now, it tells that x and y have another common factor, 15; Which is contradict to the second assumption that x/y is a rational fraction in lowest term.. Hence, √15 can’t be represented as a ratio of two integers; An irrational number.

Therefore, it follows that √5 + √3 is also an irrational number.

——————————————————————————————

Prove that cube root of 3 is irrational

Solution: Let cube root of 3 = r/s, where r/s is a rational fraction in lowest term

By cubing, to eliminate radical sign: 3 = (r³)/(s³) ….. 3s³ = r³

It implies that r³ is divisible by 3, so r is also divisible by 3; Supposing r = 3c

Then, 3s³ = (3c)³ ….. s³ = 9c³

It tells that s³ is divisible by 9, so s is also divisible by 9. Since it is divisible by 9, s is divisible by 3.

Since r and s are both divisible by 3, hence it contradicts the assumption that r/s is a fraction in lowest term.

Therefore this contradiction lead us to conclude that cube root of 3 is irrational

Quote:

“Reductio ad absurdum, which Euclid loved so much, is one of a mathematician’s finest weapons. It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game.” – Godfrey Harold Hardy

Let P and P’ denote points inside rectangles ABCD and A’B'C’D', respectively. If PA = a+b, PB=a+c, PC = c+d, PD =b+d, P’A=ab, P’B'=ac, P’C'=cd, prove that P’D'=bd.

(Pi Mu Epsilon Journal – 4(Spring 1967)258, proposed by Stanley Rabinowitz)

————————————————————–

*Let sides AB and CD are the horizontal sides, and sides BC and AD are the vertical sides.

*Draw 2 perpendicular lines (vertical and horizontal) inside rectangle ABCD such that their intersection is point P. It follows that the the vertical line segment is perpendicular to sides AB and CD, and the horizontal line segment is perpendicular to sides BC and AD.

*Let:

point a₁ – intersection of the v.l.segment and side AB

point a₂ - intersection of the h.l.segment and side BC

point a₃ – intersection of the v.l.segment and side CD

point a₄ – intersection of the h.l.segment and side AD

*Let: u=distance bet. A and a₁= distance bet. D and a₃

w=distance bet. B and a₁= distance bet. C and a₃

x=distance bet. A and a₄= distance bet. B and a₂

y=distance bet. D and a₄= distance bet. C and a₂

*Then, by Pythagorean theorem

(PA)² = x² + u², (PB)² = w² + x² , (PC)² = w² + y², (PD)² = y² + u²

Since (y² + u²) + (w² + x²) = (x² + u²) + (w² +y²).

Therefore, (PD)² + (PB)² = (PA)² + (PC)²,

(b+d)² + (a+c)² = (a+b)² + (c+d)²,

(ac)² + (bd)² = (ab)² + (cd)² – - – (1)

*Same procedure to A’B'C’D',

(P’D')² = (P’A')² + (P’C')² – (P’B')²,

(P’D')² = (ab)² + (cd)² – (ac)²

*Substitute (1):

(P’D')² = [(ac)² + (bd)²] – (ac)²

Therefore, P’D’ = bd

Quote: “As long as algebra and geometry have been separated, their progress have been slow and their uses limited; but when these two sciences have been united, they have lent each mutual forces, and have marched together towards perfection” - Joseph-Louis Lagrange