Is x an L-Tuple or a Member of an L-Tuple in ℝL+ Notation?

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Homework Statement


Let's say we have a notation:
gif.gif


Does it mean that x is an L-tuple? Also, if it does, does it have to have L members?
If it x isn't an L-tuple, does it mean that x belongs to an L-tuple? That is, x is inside of an L-tuple: (1, x, ... L)? And again, does it mean that that L-tuple has to have exactly L members?

Homework Equations


gif.gif

The Attempt at a Solution


Googling
 
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Cinitiator said:

Homework Statement


Let's say we have a notation:
gif.gif


Does it mean that x is an L-tuple?

Yes.

Also, if it does, does it have to have L members?

Yes.

For example, typical members of \mathbb{R}^6 are x=(1,2,3,6,4,2). An element like (1,2,3) is not an element of \mathbb{R}^6.
 
Hi Cinitiator! :smile:
Cinitiator said:
Does it mean that x is an L-tuple?

Yes. :smile:
Also, if it does, does it have to have L members?

Yes!

A Cartesian product space, of L individual spaces, only has elements that are L-tuples.

If the individual spaces are all different, that's fairly obvious … it's the same even if the individual spaces are identical. :wink:
 
Hi Cinitiator,

Does it mean that x is an L-tuple?

No, x \in ℝL+ usually means that x is a column vector with L rows with real entries where the subscript + requires that all its entries are positive.

http://en.wikipedia.org/wiki/Column_vector

A tuple is also a kind of array but such that the listed objects are ordered horizontally. It is usually notated like this:
L-Tupel: (n1,n2,...,nL)

http://en.wikipedia.org/wiki/Tuple

Also, if it does, does it have to have L members?

Yes, exactly L entries (or alternatevely members)
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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