0=-1 (Calculus Puzzle)

[RoboDesigners]

Here's a neat puzzle I've come up with/discovered... See if you can find the error/oversight.

∫tan(x) dx = ∫sin(x)/cos(x) dx = ∫sin(x)sec(x) dx

integration by parts:
u = sec(x)
du = sec(x)tan(x) dx

dv = sin(x) dx
v = -cos(x) dx

∫tan(x) dx = -cos(x)sec(x) - ∫-cos(x)sec(x)tan(x)dx

∫tan(x) dx = -1 + ∫tan(x)dx

Now subtract ∫tan(x) dx from both sides and...

0 = -1

[Ether]

∫tan(x) dx = -1 + ∫tan(x)dx

Now subtract ∫tan(x) dx from both sides and...

0 = -1

∫tan(x)dx - ∫tan(x)dx is not equal to 0.

It's an arbitrary constant.

[RoboDesigners]

∫tan(x)dx - ∫tan(x)dx is not equal to 0.

It's an arbitrary constant.
Yep... that's basically it...

I had a feeling you'd be the first one to respond to this thread, and you'd have the answer... :)

//Andrew

[Dusk Star]

Well, and you left off the "+ C" that you always need...
(though that may very well be what Ether meant)

[Peck]

now make 1=2 :)

[Ether]

now make 1=2 :)

0 = -1

add 2 to both sides:

2 = 1

[Peck]

Let a and b be equal non-zero quantities

a = b

2. Multiply through by a

a^2 = ab

3. Subtract b^2

a^2 - b^2 = ab - b^2

4. Factor both sides

(a - b)(a + b) = b(a - b)

5. Divide out (a - b)

a + b = b

6. Observing that a = b

b + b = b

7. Combine like terms on the left

2b = b

8. Divide by the non-zero b

2 = 1

[jblay]

1/3=.33333...
2/3=.66666...
1/3+2/3=.99999...
1=.99999...

[Ether]

Let a and b be equal non-zero quantities
5. Divide out (a - b)

Divide by zero.

[Ether]

1=.99999...

The ellipsis means infinitely repeating decimal so

.99999... means limit(sum(9/10^i,i,1,n),n,inf) = 1

there is no contradiction.

[Peck]

Has this turned into a thread of trying to find a math trick that Ether doesn't know how to work it? If so, i want a chance to disprove it before he dose :).

[Brian Ha]

Peck, sorry to tell you but i doubt ull win haha

Although ether, can you prove or disprove Riemann Sums? Now that would be cool.

[PAR_WIG1350]

167 = -53

No illegal operations were performed, slightly non-standard

[DonRotolo]

How about this one:

13 * 7 = 28

Here is the explanation (http://www.youtube.com/watch?v=rLprXHbn19I) (BTW, the guy in the glasses is Ether as a young man :p )

[Ether]

Although ether, can you prove or disprove Riemann Sums?

fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet

[Peck]

fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet


ingles por favor sinor

[PAR_WIG1350]

ingles por favor sinor

1) *Inglés or favor señor.

2) according to google translate: It is a wonderful demonstration of what share of corruption. Shortness of this margin does not take

but that seems like it was lost a bit in the translation:p

It is actually taken from Fermat who wrote about a theorem (a^x + b^x = c^x has no integral solution set a,b,c for any integer value for x greater than 2) in the margins of a book in 1637. It is actually the description of the proof for this theorem which he came up with and it translates more accurately to I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
unfortunately his proof seems to have disappeared off the face of the earth, assuming he even wrote it down.
Over the years people proved that it held true for specific exponents. It wasn't until 1995 that it all came together in an extremely complex proof that was probably not what Fermat had in mind, but a general proof of the theorem non the less.

[EricH]

1) *Inglés or favor señor.

2) according to google translate: It is a wonderful demonstration of what share of corruption. Shortness of this margin does not take

but that seems like it was lost a bit in the translation:pSounds like the classic "I have discovered a truly marvelous proof of this, which this margin is too narrow to contain" from Fermat. Which, on investigating the Wikipedia entry on Fermat's last theorem, is exactly what it is.

Well played, Ether.

[Ether]

Well played, Ether.

Thank you :)

[Peck]

1) *Inglés or favor señor.



my computer is annoying whenever i try to insert alt symbols. also, i was using spanish, not latin and por was correct.

[RoboDesigners]

Has this turned into a thread of trying to find a math trick that Ether doesn't know how to work it? If so, i want a chance to disprove it before he dose .
I have an article here one of my math professors gave me that has a dozen "proofs" that 1=2. (She gave me this after I showed her my "proof," which happened to be in the article... I guess I didn't come up with it first...)

One that's not so hard to disprove is as follows:

0 = (1 - 1) + (1 - 1) + (1 - 1)...

Rearrange parenthesis:
0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1)...
0 = 1 + 0 + 0 + 0....
0 = 1

(Ether's going to get this oh-so-fast...)

//Andrew

[Dusk Star]

I have an article here one of my math professors gave me that has a dozen "proofs" that 1=2. (She gave me this after I showed her my "proof," which happened to be in the article... I guess I didn't come up with it first...)

One that's not so hard to disprove is as follows:

0 = (1 - 1) + (1 - 1) + (1 - 1)...

Rearrange parenthesis:
0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1)...
0 = 1 + 0 + 0 + 0....
0 = 1

(Ether's going to get this oh-so-fast...)

//Andrew

Not if the other Andrew gets there first!

after you rearrange parenthesis, you should have the one at the end that was freed up, therefore
0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1)... ...-1
0 = 1 + 0 + 0 + 0... ...-1
0 = 0

[Nate Laverdure]

after you rearrange parenthesis, you should have the one at the end that was freed up...
But can't the one at the end just be moved into the next (http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel) (1 - 1) term? :)

[Ether]

0 = (1 - 1) + (1 - 1) + (1 - 1)...

Rearrange parenthesis:
0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1)...
0 = 1 + 0 + 0 + 0....
0 = 1

(Ether's going to get this oh-so-fast...)

Sorry, I was offline all day.

The problem with the above "proof" is that the associative law of arithmetic is not universally valid for infinite sums. So you aren't allowed to re-arrange (or remove) the parentheses.

[Ether]

Not if the other Andrew gets there first!

after you rearrange parenthesis, you should have the one at the end that was freed up,

There is no 1 "at the end". It's an infinite sum.

[Dusk Star]

There is no 1 "at the end". It's an infinite sum.




Ah, ok. Not entirely how I was trying to communicate it, but I guess being an infinite sum invalidates that anyways.

[RoboDesigners]

The problem with the above "proof" is that the associative law of arithmetic is not universally valid for infinite sums. So you aren't allowed to re-arrange (or remove) the parentheses.
Hmmm... that's different than what I was going to say... (the article I have doesn't list the answers...)

I thought that because the series doesn't converge, it doesn't equal zero to begin with. :confused:

//Andrew

[Ether]

I thought that because the series doesn't converge, it doesn't equal zero to begin with. :confused:

The original infinite sum you wrote:

(1 - 1) + (1 - 1) + (1 - 1)...

... does converge. It converges because of the parentheses. The term that is being repeatedly added is (1-1). It is equal to zero.

If you remove all the parentheses, so that you have

1 - 1 + 1 - 1 + 1 - 1 ...

then you are alternately adding plus or minus 1, so the sum never converges: it oscillates between 1 and 0.

The error was re-arranging the parentheses. The associative law does not always hold for an infinite sum: You cannot re-arrange the parentheses in an infinite sum unless certain criteria are satisfied.

[PAR_WIG1350]

my computer is annoying whenever i try to insert alt symbols. also, i was using spanish, not latin and por was correct.

The lack of a "p" was a typo, also, I don't like using alt to insert symbols either, but if you use google translate to enter the text the symbol keys (;',./[]\=-) are used to input special letters like vowels with accents

Also, nobody has attempted to figure out this one yet

167 = -53

No illegal operations were performed, slightly non-standard

[Taylor]

I have an article here one of my math professors gave me that has a dozen "proofs" that 1=2. ...
Would you be willing to share this article?

-----

Back in college years, my two buddies and I decided to go to Las Vegas for spring break. We piled into my Chevy Cavalier station wagon (jealous?) and headed West.
About 4am, we got to the point where we were all falling asleep, none of us were able to drive, and the car was too small to comfortably fit three guys. We found a small motel in Kansas, paid $30 for a room ($10 each), and retired for the night.
The clerk was filling out his books when he realized the motel had a policy of $25 per night after 1am. He took five $1 bills and headed to our room. Along the way, he realized we were not going to be able to split five bucks equally among the three of us. He shoved two of the singles in his pocket, gave us each a dollar, and that was that.
Here's the issue:
We were initially charged ten dollars a piece - total of $30.
After the clerk realized his error, the room charge was $25. We paid $9 apiece ($10 - the $1 that was returned) for a total of $27; he put two dollars in his pocket for a total of $29.

29 = 30 ?

[~Cory~]

Guy 1: $10
Guy 2: $10
Guy 3: $10
Clerk: $0
Hotel: $0

Guy 1: $0
Guy 2: $0
Guy 3: $0
Clerk: $0
Hotel: $30

Guy 1: $0
Guy 2: $0
Guy 3: $0
Clerk: $5
Hotel: $25

Guy 1: $1
Guy 2: $1
Guy 3: $1
Clerk: $2
Hotel: $25

Total: $30. Anywho the error comes into play with the 27. You dont add the 2 to the 27, instead subtract from it to get 25.

[RoboDesigners]

Would you be willing to share this article?
She had it in hard copy to give out... Can't find it for free online. :(
http://www.jstor.org/discover/10.2307/25678202?uid=3739936&uid=2129&uid=2&uid=70&uid=4&uid=3739256&sid=47698805324517

(Yes, it is that old)

//Andrew