What To The What Is One?
Find the number of real solutions of x satisfying x x = 1 .
Note: 0 0 is not defined.
The answer is 1.
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1 solution
This solution is incomplete because it doesn't handle negative x . (Compare: x x − 1 = 1 has two real solutions, but only one is found with this solution.)
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Ya I am quite aware of it , you can upload new solution so I can also learn something
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Why don't you post the solution or at least give some hints to solve your posted problem "spicy 1"?I even posted a report asking help but you do not reply?
The question does not state that x is real number, there are infinitely many solution when x is complex number...
Interestingly, the solution is also consistent with the theorem which states that number of roots of the polynomial equation p ( x ) = 0 is equal to the degree of p ( x ) .
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How so? x x is not a polynomial.
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Yes, it isn't.
What I meant was that we take the specific case of polynomials, i.e., when x takes only positive integral values.
The polynomial x n − 1 = 0 is of degree n and has n roots. When n = 1 , x 1 − 1 = 0 has only one solution which is x = 1 .
So, the number of solutions is equal to the solution only when x = 1 .
It was just an observation that I stumbled upon.
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@Harsh Khatri – Polynomials = when x take only positive integer values. Please review the definition of Polynomials .
This is a bad way of thinking / interpreting the problem. Please do not do so. In fact, it is an "observation" that I want to warn against, because it screws with your understanding of the actual theory.
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@Calvin Lin – Ohh. Will take care not to jumble up concepts again and try to stick to the definitions.
That was really helpful. Thank you.
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@Harsh Khatri – Bonus: Is sin x a polynomial?
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@Nihar Mahajan – in fact sin(x) can be approximated by power series
@Nihar Mahajan – no at all sinx is not a polynomial
Take l o g both sides
x l o g x = l o g 1 = 0
either x = 0 o r l o g x = 0
x = 0 , 1 but x = 0 ( 0 0 is not defined)
Therefore only one solution x = 1