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]

Originally Posted by RoboDesigners

∫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]

Originally Posted by Ether

∫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]

Originally Posted by Peck

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]

Originally Posted by jblay

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

[Ether]

Originally Posted by Peck

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

Divide by zero.