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Maths help......

Can anyone explain to e what exactly a harmonica sequence is?

Also, what is the shortest way of working out a cubic sequence?
Reply 1
Avatar for mqb2766
mqb2766
21
Original post by Lacrimosa1
Can anyone explain to e what exactly a harmonica sequence is?

Also, what is the shortest way of working out a cubic sequence?

1, 1/2, 1/3, 1/4, ....
https://en.wikipedia.org/wiki/Harmonic_progression_(mathematics)
What do you mean "to e"?

What is the cubic, what have you tried?
(edited 2 years ago)
Reply 2
Avatar for davros
davros
16
Original post by mqb2766
1, 1/2, 1/3, 1/4, ....
https://en.wikipedia.org/wiki/Harmonic_progression_(mathematics)
What do you mean "to e"?

What is the cubic, what have you tried?

Perhaps they just meant "to me" ?
Reply 3
Avatar for Lacrimosa1
Lacrimosa1
OP
16
Original post by davros
Perhaps they just meant "to me" ?

Thanks. Sorry, I always rush-type
Reply 4
Avatar for Lacrimosa1
Lacrimosa1
OP
16
So, in a harmonica sequence, the denominator gradually increases?
Reply 5
Avatar for mqb2766
mqb2766
21
Original post by Lacrimosa1
So, in a harmonica sequence, the denominator gradually increases?

Its clearly defined in the wiki page and there are some good examples.
Reply 6
Avatar for Lacrimosa1
Lacrimosa1
OP
16
Original post by mqb2766
1, 1/2, 1/3, 1/4, ....
https://en.wikipedia.org/wiki/Harmonic_progression_(mathematics)
What do you mean "to e"?

What is the cubic, what have you tried?

There is a formula for the cubic.
= an3+bn2+cn+d
Reply 7
Avatar for mqb2766
mqb2766
21
Original post by Lacrimosa1
There is a formula for the cubic.
= an3+bn2+cn+d

Either compute the terms like you do for a quadratic by differencing (3 times( or write it as the sum of the first four terms in the sequence each multiplied by a cubic in n so something like
x_1*(n-2)*(n-3)*(n-4)
and expand.

Not really sure what you mean about working out a cubic sequence. Posting an actual question helps / with some ideas/working

Again, a quick google gives
https://www.youtube.com/watch?v=ckH7MaF_a8Y&ab_channel=RonBarrow
(edited 2 years ago)

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