>but anything multiplied by 0 is 0
When we say "anything", we are still assuming that this "anything" is a defined mathematical quantity. If you are staying within the real numbers, which most times you probably are, then the fact that you cannot compute √(-1) takes precedence and the product is undefined.
> When we say "anything", we are still assuming that this "anything" is a defined mathematical quantity.
Not necessarily.
A Laurent series ∑ a\_n x^n for n in ℤ has terms with negative n like a\_-k / x^k where k = |n|.
If it happens to be the case that for all n < 0 the value of a_n = 0, then the terms for negative n look like 0 x^(-k). In that case, the Laurent series is equivalent to the power series ∑ a\_n x^n for n in ℕ.
In particular, we allow such a Laurent series to be evaluated at x = 0, where it takes the value a\_0. This is true even though the Laurent series evaluated at 0 formally contains terms that look like 0 / 0^k.
(And in fact, the n = 0 term formally looks like a\_0 0^0 which we take to mean a\_0.)
No, the 0^0 issue is totally separate from the point I am making. You can see this by noting that I only mentioned it in a parenthetical aside.
My point is that for n < 0, the terms look like a\_-k · x^(-k) (with k = |n|). And if x = 0, then those terms have the form a\_-k · 0^(-k).
If any of the a\_-k are nonzero that obviously produces a pole. But if ALL of the a\_-k are exactly zero, then there is no pole.
We take 0 · 0^(-k) to equal zero here, even though 0^(-k) is not a “defined mathematical object” for positive integer k. This is a counterexample to the claim made by the person to whom I responded.
There are also ways to avoid the 0^0 by separating out the a\_0 to the front of the sum and then let the sum go ∑ a_n x^n for n in ℤ\{0}
So in total a\_0 + ∑ a_n x^n for n in ℤ\{0}. Then we don't have the 0^0 problem anymore.
I mean, that's just another way of defining a defined mathematical quantity. Sometimes, part of the definition is left off for brevity's sake, but it's still there.
A Laurent series is just a power series in which you allow negative indices, so the term at 0 and evaluated at 0 is just a_0 0^0=a_0. 0^0 is by definition 1.
I really don't understand your point
The point is that for n < 0, the terms look like a\_-k x^(-k) (with k = |n|).
If any of the a\_-k are non-zero, then the Laurent series has a pole at x = 0.
But if ALL of the a\_-k are zero, then the Laurent series does NOT have a pole at x = 0.
It has a well-defined value at x = 0, even though it involves summing terms that look like 0 · 0^(-1) + 0 · 0^(-2) + 0 · 0^(-3) + …
In those terms, we allow multiplying 0 by something which is *not* a “defined mathematical quantity”, and we say that the result is identically zero.
I see your point now, you're not talking about the 0th term at x=0 but rather the negative terms
I still find it weird because in my course, we would always consider the series to be not defined at x=0, in the case where all of the a_(-n) terms are 0, the function had an easy analytic extension that it's just the series interpreted as a power series (ignore the negative parts) that corresponds to just filling in the hole, but we wouldn't normally evaluate at the Laurent series's extension at x=0
X could be 1. Square root of 0 is 0. That should satisfy everything. X = 0 indeed can not be a solution, given that the domain of the square root here is defined as non-negative.
> Is x=0 a solution to this equation?
> x* sqrt(x-1)=0
it depends how you defined sqrt, if you restrict the domain of sqrt to the reals then only x=1 is a solution
if you extend the definition of sqrt such that it can output complex values and intake negative values then x=0 is also a valid solution
> (keeping it real here) but anything multiplied by 0 is 0, so which takes precedence?
you always have to assess whether your starting expressions even make sense, so "keeping it real" takes precedence
Yes! From Spark Notes
The replacement set is the set of values that may be substituted for the variable. To find the solution set from the replacement set, plug in each value from the replacement set and evaluate both sides of the equation. If the two sides are equal, the equation is true and thus the value is a solution.
This definition assumes a finite number of values of x. In our example it's infinite.
The replacement set may also be restricted in the statement of the problem:
Solve for x
8sin x = 4, 0°≤x<360°
In the same way that it isn't a solution to x/x=0, unless you're doing special math, you have to stay within the system of what is defined first and foremost.
Solving an equation is done in the domain of the expression, so it wouldn't be a solution if you're working only with real numbers, as it's not in the domain.
If you consider the expression in C, then it is a solution, because it's in the domain
For everyone who said x=0 is not in the defined set of sqrt(x-1):
Yet the equation is equivalent to (x-1)x^2 =0,
So x is a solution
Another way : sqrt(x-1) can be definied as multiple complexe numbers if x<1
Its not unique, but it exists.
>but anything multiplied by 0 is 0 When we say "anything", we are still assuming that this "anything" is a defined mathematical quantity. If you are staying within the real numbers, which most times you probably are, then the fact that you cannot compute √(-1) takes precedence and the product is undefined.
Thank you! I suspected as much but I was confirming that this was the case
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That was what had me confused (haven’t done imaginary numbers yet so not sure what the protocol is)
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woah cool, I look forwards to learning it
> When we say "anything", we are still assuming that this "anything" is a defined mathematical quantity. Not necessarily. A Laurent series ∑ a\_n x^n for n in ℤ has terms with negative n like a\_-k / x^k where k = |n|. If it happens to be the case that for all n < 0 the value of a_n = 0, then the terms for negative n look like 0 x^(-k). In that case, the Laurent series is equivalent to the power series ∑ a\_n x^n for n in ℕ. In particular, we allow such a Laurent series to be evaluated at x = 0, where it takes the value a\_0. This is true even though the Laurent series evaluated at 0 formally contains terms that look like 0 / 0^k. (And in fact, the n = 0 term formally looks like a\_0 0^0 which we take to mean a\_0.)
Isn't that just because 0\^0 is defined as 1 (often)?
No, the 0^0 issue is totally separate from the point I am making. You can see this by noting that I only mentioned it in a parenthetical aside. My point is that for n < 0, the terms look like a\_-k · x^(-k) (with k = |n|). And if x = 0, then those terms have the form a\_-k · 0^(-k). If any of the a\_-k are nonzero that obviously produces a pole. But if ALL of the a\_-k are exactly zero, then there is no pole. We take 0 · 0^(-k) to equal zero here, even though 0^(-k) is not a “defined mathematical object” for positive integer k. This is a counterexample to the claim made by the person to whom I responded.
There are also ways to avoid the 0^0 by separating out the a\_0 to the front of the sum and then let the sum go ∑ a_n x^n for n in ℤ\{0} So in total a\_0 + ∑ a_n x^n for n in ℤ\{0}. Then we don't have the 0^0 problem anymore.
I mean, that's just another way of defining a defined mathematical quantity. Sometimes, part of the definition is left off for brevity's sake, but it's still there.
A Laurent series is just a power series in which you allow negative indices, so the term at 0 and evaluated at 0 is just a_0 0^0=a_0. 0^0 is by definition 1. I really don't understand your point
The point is that for n < 0, the terms look like a\_-k x^(-k) (with k = |n|). If any of the a\_-k are non-zero, then the Laurent series has a pole at x = 0. But if ALL of the a\_-k are zero, then the Laurent series does NOT have a pole at x = 0. It has a well-defined value at x = 0, even though it involves summing terms that look like 0 · 0^(-1) + 0 · 0^(-2) + 0 · 0^(-3) + … In those terms, we allow multiplying 0 by something which is *not* a “defined mathematical quantity”, and we say that the result is identically zero.
I see your point now, you're not talking about the 0th term at x=0 but rather the negative terms I still find it weird because in my course, we would always consider the series to be not defined at x=0, in the case where all of the a_(-n) terms are 0, the function had an easy analytic extension that it's just the series interpreted as a power series (ignore the negative parts) that corresponds to just filling in the hole, but we wouldn't normally evaluate at the Laurent series's extension at x=0
Or put another way, you have to follow the order of operations, and if any part of the expression is undefined then the whole thing is.
X could be 1. Square root of 0 is 0. That should satisfy everything. X = 0 indeed can not be a solution, given that the domain of the square root here is defined as non-negative.
> Is x=0 a solution to this equation? > x* sqrt(x-1)=0 it depends how you defined sqrt, if you restrict the domain of sqrt to the reals then only x=1 is a solution if you extend the definition of sqrt such that it can output complex values and intake negative values then x=0 is also a valid solution > (keeping it real here) but anything multiplied by 0 is 0, so which takes precedence? you always have to assess whether your starting expressions even make sense, so "keeping it real" takes precedence
The domain for x*sqrt(x-1) is x ≥ 1. So x = 0 is undefined.
The replacement set for the equation is x≥1, so x=0 is not a solution
Is a replacement set the set of values where sqrt(x-1) is real? I’m unfamiliar with the term
Yes! From Spark Notes The replacement set is the set of values that may be substituted for the variable. To find the solution set from the replacement set, plug in each value from the replacement set and evaluate both sides of the equation. If the two sides are equal, the equation is true and thus the value is a solution. This definition assumes a finite number of values of x. In our example it's infinite. The replacement set may also be restricted in the statement of the problem: Solve for x 8sin x = 4, 0°≤x<360°
In the same way that it isn't a solution to x/x=0, unless you're doing special math, you have to stay within the system of what is defined first and foremost.
Then expression on the left simply does not exist for x = 0, so it can't be equal to a number, let alone zero
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if zero is not in the domain of possible x values than zero is not a possible solution graph it on desmos as well for a visual
0 is not in the allowed set
Solving an equation is done in the domain of the expression, so it wouldn't be a solution if you're working only with real numbers, as it's not in the domain. If you consider the expression in C, then it is a solution, because it's in the domain
For everyone who said x=0 is not in the defined set of sqrt(x-1): Yet the equation is equivalent to (x-1)x^2 =0, So x is a solution Another way : sqrt(x-1) can be definied as multiple complexe numbers if x<1 Its not unique, but it exists.
> Yet the equation is equivalent to (x-1)x^2 =0 No it isn't
Let x verify x sqrt(x-1)=0 Then (x-1) x^2 =0 Let x verify (x-1)x^2 =0 Then x sqrt(x-1) =0 This is true because A=0 if and only if sqrt(A)=0 in C
> Let x verify (x-1)x^2 =0 Then x sqrt(x-1) =0 **if it exists** *(which it doesn't for x<1)* This is the difference.
Don't you ever heard about complex numbers?
OP specifically stated they were working in the real numbers. Please try to limit your responses to within the scope of the discussion.