Solve for X

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You make an easy thing hard.

Lift the denominators by cross multiplication to get

nx = dx - dk which gives

nx - dx = -dk

then by the distributive law

x(n-d) = -dk

then divide by (n - d) to get

x = -dk/( n -d )

ta! da!

ruveyn

Actually, his step 2 is perfectly correct; it comes from the arithmetic of fractions - in particular


(IOW, a quarter plus two quarters is three quarters). You mightn't find this as obvious a method, but it is just as valid as multiplying by dx.

[quote="Orwell]Calculus is fun.[/quote]

I hate calculus !

I think that the best thing ever in physics / maths is dimensional analysis.

For instance for a LC circuit

f = 1 / [2pi square root (LC)]

So

Hz x Hz = 1 / (4 x pi x pi x H x F)

Remove the 4 pi term as it has no units

Hz x Hz = 1 / ( H x F)

Expand out farads

Hz x Hz = 1 / (H x C V-1)

Hz x Hz = 1 / (H x A x s x V-1)

Next expand out the henrys

Hz x Hz = 1 / (V x s x A-1 x A x s x V-1)

Remove most of the terms in the bracket by multiplication

Remove volts first

Hz x Hz = 1 / (s x A-1 x A x s)

Then Amps

Hz x Hz = 1 / (s x s)

Then square root both sides

Hz = 1 / s

Perfect !

it is correct but harder and longer

Here is a different way. 4 steps and I could toss in 1-2 more steps to make it look just like yours but it is simpler my way.

See if you can work out what step is wrong in this:

if a = b
1. a^2 = ab
2. a^2 - b^2 = ab - b^2
3. (a-b)(a+b) = b(a-b)
4. /(a+b) = a+ b = b
5. b + b = b
6. 2b = b
2 = 1

division by 0

Correct, step 4.

Alrighty, speaking of bogus ways to prove 1=0
(1-1)+(1-1)+(1-1)+(1-1)+...=0+0+0...=0 Right?
Now, this obviuhsleh is equal to
1+(-1+1)+(-1+1)+(-1+1)+...=1+0+0+0...=1
∴0=1

your conclusion is false. the rest is accurate

ok


Well, not quite. You've left out what happens at the *end* of the sequence:
(1-1) + (1-1) + ... + (1-1) = 1 + (-1 + 1) ... + ( -1 + 1) + -1

Which is still zero.

Your series does not converge absolutely therefore rearranging the terms may change the result.

There is no *end* to an infinite series. It is perfectly possible to change the limit of an infinite series by rearranging the terms, unless the series converges absolutely. The error is in assuming "this obviuhsleh is equal to ", which it is not. A series converges absolutely if the sum of the absolute value of the summand is finite.
In this case, the sum of the absolute values would be 1+1+1+1+1+... which is infinite.

Slight flaw in your reasoning - my series most certainly does converge absolutely.
For my sum to be absolutely as you say it would need to be something like,
Σ(-1)^(i+1)
which doesn't equal 0 because it doesn't converge at all. No, indeed, my series is in fact
Σ(1-1)
and |1-1| = |0| = 0 and hence my series converges absolutely to 0.

You almost remembered that in the right order. Just for clarification its PEMDAS. You got the order for Exponents and Parenthesis mixed up. Just thought you might be interested to know that. May save you some trouble at some point, I dunno..

No.

One of your series is as you quoted above.

The second was
1+Σ(-1+1)
Which is absolutely convergent to unity.

However, you try to suggest you can remove the parentheses, rearrange terms, etc, but in the process, you transition through the series
Σ((-1)^i)
which is, as nudel correctly pointed out, NOT absolutely convergent.

I'm not really sure I see at all why we need to invoke anything about the Riemann rearrangement theorem in order to resolve this. Obviously, regrouping is based on a naive assumption that ignores what it means for a series to converge, so it seems completely unnecessary to say that I transition through anything, pardon me if I'm still failing to grasp this. If I take,
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?