Quartic means that the "dominant" term is proportional to n^4
-66 The sequence can be generated by the following quartic: t(n) = (-3n4 + 42n3 - 177n2 - 358n + 2808)/8 for n = 1, 2, 3, ...
The differences are not the same so the sequence is not arithmetic. The sequence starts with a zero, so it cannot be geometric, or an exponential (power) sequence. The quartic: (2n4 - 19n3 + 64n2 - 83n + 36)/6 fits the 5 points. That gives the next term as 55.
It is possible to find a quartic equation (power 4) so that any fifth number is the next in the sequence. The cubic that fits these is Un = (n3 + 17n + 30)/6 for n = 1, 2, 3, ... and therefore, the next term is 40.
He is known for quartic equations.
No.
It depends on (a) the first five numbers of what and(b) what sort of sequence.ANY 5 numbers can be put into a quartic sequence. So the answer is: every time.
-66 The sequence can be generated by the following quartic: t(n) = (-3n4 + 42n3 - 177n2 - 358n + 2808)/8 for n = 1, 2, 3, ...
A quartic is an algebraic equation or function of the fourth degree.
The differences are not the same so the sequence is not arithmetic. The sequence starts with a zero, so it cannot be geometric, or an exponential (power) sequence. The quartic: (2n4 - 19n3 + 64n2 - 83n + 36)/6 fits the 5 points. That gives the next term as 55.
A quartic.
It is possible to find a quartic equation (power 4) so that any fifth number is the next in the sequence. The cubic that fits these is Un = (n3 + 17n + 30)/6 for n = 1, 2, 3, ... and therefore, the next term is 40.
Luca Pacioli (1445-1515) discussed quartic equations, but did not have a general solution. Lodovico Ferrari (1522-1565) devised a solution.
Leonarda Burke has written: 'On a case of the triangles in-and-circumscribed to a rational quartic curve with a line of symmetry' -- subject(s): Quartic Curves, Triangle
There are many possible answers. But given 5 points, an answer that can be guaranteed is that it is a polynomial of degree 4 (a quartic).In this case, Un = (-13n4 + 166n3 - 719n2 + 1310n - 720)/24There are many possible answers. But given 5 points, an answer that can be guaranteed is that it is a polynomial of degree 4 (a quartic).In this case, Un = (-13n4 + 166n3 - 719n2 + 1310n - 720)/24There are many possible answers. But given 5 points, an answer that can be guaranteed is that it is a polynomial of degree 4 (a quartic).In this case, Un = (-13n4 + 166n3 - 719n2 + 1310n - 720)/24There are many possible answers. But given 5 points, an answer that can be guaranteed is that it is a polynomial of degree 4 (a quartic).In this case, Un = (-13n4 + 166n3 - 719n2 + 1310n - 720)/24
He is known for quartic equations.
No.
Helen Grace Telling has written: 'The rational quartic curve in space of three and four dimensions' -- subject(s): Hyperspace, Quartic Curves