Mathematics marathon

[cmhardw]

I love the idea of this thread. It appears that there is no question to answer right now, so I'll post one.

Problem Goal: Spot the error in the proof
Required Pre-requisite: Up to differential calculus. Probably easy/intermediate for most people on this forum.

Spoiler

Examine the function \( f(x) = x^2 \) on the open interval \( (0,\infty) \)

On this interval we know that the following equation holds true:
\( x^2 = x^2 \)

Now let's rewrite things a bit for fun:

\( x^2 = x + x + x + \ldots + x \)
Where there are \( x \) instances of the variable \( x \). This is by the definition of multiplication as a repeated addition.

\( \frac{d}{dx} \left( x^2 \right) = \frac{d}{dx} \left( x + x + x + \ldots + x \right) \)
Differentiate both sides.

\( 2x = 1 + 1 + 1 + \ldots + 1 \)
Where there are \( x \) instances of the number 1.

\( 2x = x \)
Repeated addition can be simplified to a multiplication

Since we are considering the open interval \( (0,\infty) \), then we know that \( x \) will never have the value 0 anywhere on this interval. This means that dividing by \( x \) will be defined.

\( \frac{2x}{x} = \frac{x}{x} \)

\( 2 = 1 \)
All math is meaningless, because there is really only one number anyway.

 

[Rinfiyks]

Attempt at an answer

Spoiler

When you differentiate x + x + x + . . . + x, you're not taking into account the x that tells you how many xs there are. You're just assuming it's a constant amount of xs, and not a variable amount as it is. Not a very rigorous disproof

Problem

Spoiler

Without evaluating either of them (e.g. with Maclaurin sequences or whatever those things are called), find which number is largest.
\( \sqrt{2} \)

\( \sqrt[3]{3} \)

 

[einstein00]

answer

Spoiler

raise each to the 6th power. \( (\sqrt{2})^6 = 8 \), and \( (\sqrt[3]{3})^6 = 9 \). Clearly, the latter is larger.

problem

Spoiler

Ann and Betty initially each have $50. Ann then receives $10 from her mother, and Betty receives $20 from her mother. Together, they only have $120. They neither gained nor lost any money in any other way. Explain.

 

[cmhardw]

Attempt at an answer

Spoiler

When you differentiate x + x + x + . . . + x, you're not taking into account the x that tells you how many xs there are. You're just assuming it's a constant amount of xs, and not a variable amount as it is. Not a very rigorous disproof

Without giving away too much I can tell you that this is not the first error I made in the proof. Try to find the first error.

 

[einstein00]

another attempt at an answer to cmhardw:

Spoiler

If x is not an integer, how can there be x instances of x. For example, 0.123 instances of 0.123 doesn't make sense (unless you're multiplying 0.123 by 0.123, but that doesn't follow the form x+x+x+...+x, it's x*x)

 

[cmhardw]

another attempt at an answer to cmhardw:

Spoiler

If x is not an integer, how can there be x instances of x. For example, 0.123 instances of 0.123 doesn't make sense (unless you're multiplying 0.123 by 0.123, but that doesn't follow the form x+x+x+...+x, it's x*x)

Yes, this is the first, and largest, error in the proof. The other errors are all consequences of this false assumption. Also one comment to your statement:

Spoiler

If x is not an integer, how can there be x instances of x.

The identity \( x^2 = x + x + x + \ldots + x \) with x instances of x does not hold for all integers. It holds true only for the natural numbers.

 

[einstein00]

Yes, this is the first, and largest, error in the proof. The other errors are all consequences of this false assumption. Also one comment to your statement:

Spoiler

The identity \( x^2 = x + x + x + \ldots + x \) with x instances of x does not hold for all integers. It holds true only for the natural numbers.

probably already obvious..

Spoiler

all integers > 0 are natural numbers.

 

[Baian Liu]

problem

Spoiler

Ann and Betty initially each have $50. Ann then receives $10 from her mother, and Betty receives $20 from her mother. Together, they only have $120. They neither gained nor lost any money in any other way. Explain.

Spoiler

Betty is Ann's mother, and Betty receives $20. They each have $60.

 

[Keroma12]

Problem:

Spoiler

\( n^{(-2)} \int_{-n}^{n} (n-x)^{\frac{1}{2}}(x+n)^{\frac{1}{2}}\,dx \)

Edit: Thanks Chris, I wasn't sure how to write it this way.

 

[Christopher Mowla]

Squared, but the ^2 wasn't there when I answered it:

I noticed that, where I posted the problem, latex makes the supercript 2 so small

The ^2 was always there. I just was implying that since it was made so small by the latex, I could understand why you missed it. When I first edited that post, it was because I wrote the problem
\( \frac{d}{dx}\left[ \frac{1}{\sqrt{1-\sin ^{2}\left( x \right)}} \right] \) instead of \( \frac{d}{dx}\left[ \frac{1}{\sqrt{\sin ^{2}\left( x \right)-1}} \right] \). But you obviously saw that edit before you answered.

No worries. Latex has some weird ways of displaying symbolic logic (although I can't complain about most of the output).

 

[cmhardw]

Problem:

Spoiler

n^(-2) * integral from -n to n [(n-x)^(1/2)(x+n)^(1/2)dx]

@Keroma

Do you mean?

Spoiler

\( n^{(-2)} \int_{-n}^{n} (n-x)^{\frac{1}{2}}(x+n)^{\frac{1}{2}}\,dx \)

 

[oprah62]

Spoiler

Every month, a girl gets allowance. Assume last year she had no money, and kept it up to now. Then she spends 1/2 of her money on clothes, then 1/3 of the remaining money on games, and then 1/4 of the remaining money on toys. After she bought all of that, she had $7777 left. Assuming she only gets money by allowance, how much money does she earn every month?

 

[qqwref]

Spoiler

\( n^{(-2)} \int_{-n}^{n} (n-x)^{\frac{1}{2}}(x+n)^{\frac{1}{2}}\,dx \)

Spoiler

= \( n^{-2} \int_{-n}^{n} ((n-x)(x+n))^{\frac{1}{2}} dx \)
= \( n^{-2} \int_{-n}^{n} (n^2 - x^2)^{\frac{1}{2}} dx \)
= \( 2 n^{-2} \int_{0}^{n} (n^2 - x^2)^{\frac{1}{2}} dx \)
= \( 2 n^{-2} \) * area under quarter-circle of radius n
= \( 2 n^{-2} * \frac{\pi n^2}{4} \)
= \( \frac{\pi}{2} \)

New problem:

Spoiler

Let x be some real number from 0 to 1. Imagine we take a square ABCD, of side length 1 (and area 1), and draw a line segment from A to a point x of the way from B to C, then draw a line segment from B to a point x of the way from C to D, and so on, until there are 4 line segments drawn. Let y be the area of the smaller square that is now formed.

If x is 1/2, what is y? If y is 1/2, what is x? If x is rational, does y have to be rational, and vice versa?

 

[Keroma12]

Spoiler

= \( n^{-2} \int_{-n}^{n} ((n-x)(x+n))^{\frac{1}{2}} dx \)
= \( n^{-2} \int_{-n}^{n} (n^2 - x^2)^{\frac{1}{2}} dx \)
= \( 2 n^{-2} \int_{0}^{n} (n^2 - x^2)^{\frac{1}{2}} dx \)
= \( 2 n^{-2} \) * area under quarter-circle of radius n
= \( 2 n^{-2} * \frac{\pi n^2}{4} \)
= \( \frac{\pi}{2} \)

Spoiler

What if n<0?

 

[qqwref]

Spoiler

What if n<0?

Spoiler

Fair enough. If n<0, then after the second line we replace n with m (defined as -n) which doesn't change the look of the problem at all, so we can continue as normal.

 

[Keroma12]

Spoiler

Fair enough. If n<0, then after the second line we replace n with m (defined as -n) which doesn't change the look of the problem at all, so we can continue as normal.

Spoiler

But if you you replace -n with m in the first step:
\( n^{-2} \int_{-n}^{n} (n^2 - x^2)^{\frac{1}{2}} dx \)
\( m^{-2} \int_{m}^{-m} (m^2 - x^2)^{\frac{1}{2}} dx \)
\( -m^{-2} \int_{-m}^{m} (m^2 - x^2)^{\frac{1}{2}} dx \)
And if you continue from there don't you get -\( \frac{\pi}{2} \)

If I'm wrong, where am I making my mistake?

 

[Keroma12]

New problem:

Spoiler

Let x be some real number from 0 to 1. Imagine we take a square ABCD, of side length 1 (and area 1), and draw a line segment from A to a point x of the way from B to C, then draw a line segment from B to a point x of the way from C to D, and so on, until there are 4 line segments drawn. Let y be the area of the smaller square that is now formed.

If x is 1/2, what is y? If y is 1/2, what is x? If x is rational, does y have to be rational, and vice versa?

Answer to qqwref

Spoiler

y = \( \frac{x^2-2x+1}{x^2+1} \)
If x = 1/2, y = 1/5. If y = 1/2, x = \( \frac{2-3^{1/2}}{2} \)
If x is rational, y must be also, but not vice versa.
Correct?

 

[y3k9]

Bold your answers and number your problem!

 

[qqwref]

Spoiler

\( m^{-2} \int_{m}^{-m} (m^2 - x^2)^{\frac{1}{2}} dx \)
\( -m^{-2} \int_{-m}^{m} (m^2 - x^2)^{\frac{1}{2}} dx \)[/math]

Good point.

Answer to qqwref

Spoiler

y = \( \frac{x^2-2x+1}{x^2+1} \)
If x = 1/2, y = 1/5. If y = 1/2, x = \( \frac{2-3^{1/2}}{2} \)
If x is rational, y must be also, but not vice versa.
Correct?

Spoiler

How did you derive the first equation?

 

[JBCM627]

Here are a couple tricky but useful little integrals:

Problem JBCM-1:

Spoiler

This is neatly related to Keroma12's integral:
\( \int_{a}^{b} \frac{1}{x} (x-a)^{1/2}(b-x)^{1/2} dx \)
Where \( a \) and \( b \) are arbitrary real constants. Hint: complete the square. Though I'd be interested if you can evaluate it using Mathematica, too

Problem JBCM-2:

Spoiler

\( \int_{-\infty}^{\infty} \frac{e^{\imath\omega\tau}}{1+(\frac{t}{\tau})^2} dt \)
Where \( \imath = \sqrt{-1} \), and \( \omega \) and \( \tau \) are arbitrary constants. Hint: contour integral.