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seen Dec 17 at 21:12

hey :)

high-schooler with interest in math


Dec
10
awarded  Caucus
Dec
8
awarded  Yearling
Nov
28
accepted Is there a specific name for this set of square-rooted primes?
Nov
22
comment Is there a specific name for this set of square-rooted primes?
Thank you for your answer. How about the invertiblity property. Also, it seems to me that its not exactly like the group of the integers under addition as this prime-root set has certain "fundamental" elements. Is there a formal definition/name for this property that you are aware of? Thanks again :)
Nov
22
asked Is there a specific name for this set of square-rooted primes?
Nov
9
revised How to solve $P=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\left(1+\frac{1}{3^3}\right)\ldots \infty$
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Nov
6
revised Discriminant of quadratic formula
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Nov
6
comment Discriminant of quadratic formula
Just stating, I greatly edited my answer as some things in it were plainly wrong / convoluted.
Nov
6
revised Discriminant of quadratic formula
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Nov
6
revised Discriminant of quadratic formula
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Nov
6
revised Discriminant of quadratic formula
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Nov
6
answered Discriminant of quadratic formula
Oct
1
comment Area Of Triangle . Given two equation and point.
Honestly, I fear that you have not tried hard enough as the answer very readily follows from the hints you have been given. Give it another try :) really think what it means to have squares.
Oct
1
comment Area Of Triangle . Given two equation and point.
intersections: (0,0), (-2,2) and (2,2), so A = 4*2*0.5 = 4 units.
Sep
27
revised Rational and irrational numbers under base pi
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Sep
26
comment Interpolating polynomial given only Y values
Gawd.. now that I wrote so much, I'd really like to know how you got to this question anyways :) You got me curious!
Sep
26
comment Interpolating polynomial given only Y values
Also, in my first comment I said that given any number of points, a lagrange polynomial can be constructed… It should state: "given any number of points, none of which that the same x-coordinate but differing y-coordinates, a Lagrange polynomial can be constructed". Other wise the resultant "thing", if it could be made, wouldn't even be a function. Also I meant "The constraint simply makes sure that at certain x-points, our polynomials must have HEIGHTS y an z" - my bad. Hope all this helped.
Sep
26
comment Interpolating polynomial given only Y values
So this completes the answer. So given 2 sets of y and z heights with the same x coordinates, we can construct infinitely many pairs of polynomials of the same degree (starting with a degree of 2d) which at any of the given x points (from the data sets), will have heights of y and z. As there are no unique polynomials to interpolate the data points, there are no unique roots that can be derived from the given data, we will need more constraints to be able to produce such things.
Sep
26
comment Interpolating polynomial given only Y values
So for both sets of data (with the y and z height sets, we can create infinitely many lagrange polynomials which interpolate the data sets such that the the sum of any 2 lagrange polynomials (one form the z set and one form the y set), when summed will give at a point x from the data set height k (where k is equal to y+z). Essentially, we have shown that an infinite number of polynomials will exist which will satisfy our constraints and hence we cannot recover unique original polynomials. The constraint simply makes sure that at certain x-points, our polynomials must have heats y an z
Sep
26
comment Interpolating polynomial given only Y values
A unique lagrange polynomial will exist for both sets of data and it will interpolate both of them perfectly (look up "lagrange interpolation" to see how this works). However, now consider we choose an x value outside of the ones corresponding to our data set of y's and z's. If we evaluate the f(x) of the lagrange polynomial at this x, we will clearly get some point (x,f(x)). However, we decide that we want a new polynomial that still interpolates all our previous points, but at x has a height of f(x) + k. The virtue of lagrange interpolation will allow up to construct such a polynomial too.