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Unread 21-03-2012, 16:05
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0=-1 (Calculus Puzzle)

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
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Unread 21-03-2012, 16:19
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Re: 0=-1 (Calculus Puzzle)

Quote:
Originally Posted by RoboDesigners View Post
∫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.


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Unread 21-03-2012, 18:32
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Re: 0=-1 (Calculus Puzzle)

Quote:
Originally Posted by Ether View Post
∫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
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Unread 21-03-2012, 18:51
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Re: 0=-1 (Calculus Puzzle)

Well, and you left off the "+ C" that you always need...
(though that may very well be what Ether meant)
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Unread 21-03-2012, 19:07
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Re: 0=-1 (Calculus Puzzle)

now make 1=2
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Unread 21-03-2012, 19:12
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Re: 0=-1 (Calculus Puzzle)

Quote:
Originally Posted by Peck View Post
now make 1=2
0 = -1

add 2 to both sides:

2 = 1


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Unread 21-03-2012, 19:39
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Re: 0=-1 (Calculus Puzzle)

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
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Unread 21-03-2012, 20:09
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Re: 0=-1 (Calculus Puzzle)

1/3=.33333...
2/3=.66666...
1/3+2/3=.99999...
1=.99999...
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Unread 21-03-2012, 20:23
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Re: 0=-1 (Calculus Puzzle)

Quote:
Originally Posted by Peck View Post
Let a and b be equal non-zero quantities
5. Divide out (a - b)
Divide by zero.


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Unread 21-03-2012, 20:31
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Re: 0=-1 (Calculus Puzzle)

Quote:
Originally Posted by jblay View Post
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.


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Unread 21-03-2012, 20:58
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Re: 0=-1 (Calculus Puzzle)

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 .
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Unread 21-03-2012, 21:44
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Re: 0=-1 (Calculus Puzzle)

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.
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Unread 21-03-2012, 21:44
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Re: 0=-1 (Calculus Puzzle)

167 = -53

No illegal operations were performed, slightly non-standard
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Unread 21-03-2012, 22:13
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Re: 0=-1 (Calculus Puzzle)

How about this one:

13 * 7 = 28

Here is the explanation (BTW, the guy in the glasses is Ether as a young man )
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Unread 21-03-2012, 22:21
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Re: 0=-1 (Calculus Puzzle)

Quote:
Originally Posted by Brian Ha View Post
Although ether, can you prove or disprove Riemann Sums?
fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet


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