A really bad joke

Soroban · #1 $Old$ November 14th, 2006, 10:31 AM

What is the next number in this sequence? . $1,\;3,\;5,\;7,\;\_$

Awful punchline . The next number is $8$

I had this function in mind: . $f(n) \:=\:-\frac{1}{24}\left(n^4 - 10n^3 + 35n^2 - 98n + 48\right)$

If anyone interested in a simple way to create this function,
. . I'd be delighted to share the secret.

Worst punchline . The next number is $8$

These are positive integers whose (English) names contain the letter "e".

Quick · #2 $Old$ November 14th, 2006, 01:07 PM

Quote:

Originally Posted by Soroban $View Post$

If anyone interested in a simple way to create this function, I'd be delighted to share the secret.

me!

Soroban · #3 $Old$ November 14th, 2006, 02:22 PM

I'm so glad you asked!

I learned this trick many years ago from one the books in my vast library.
. . [My wife said it's "half-vast" . . . well, I think that's what she said.]

We want a function $f(n)$ such that: . $\begin{array}{cccc}f(1) = 1 \\ f(2) = 3 \\ f(3) = 5 \\ f(4) = 7 \\ f(5) = 8\end{array}$

It has the form: . $f(n) \;=\;2n - 1 + (n-1)(n-2)(n-3)(n-4)k$

Note that for $n = 1,2,3,4$ , the second portion vanishes
. . and we get the first four odd numbers.

All we have to do is to determine $k$ so that $f(5) = 8$

We have: . $f(5) \;= \;2(5) - 1 + (4)(3)(2)(1)k \;= \;8$

. . and we get: . $9 + 24k\:=\:8\quad\Rightarrow\quad k = -\frac{1}{24}$

Hence, the function is: . $f(n) \;=\;2n - 1 -\frac{1}{24}(n-1)(n-2)(n-3)(n-4)$

. . which simplifies to: . $f(n) \:=\:-\frac{1}{24}\left(n^4 - 10n^3 + 35n^2 - 98n + 48\right)$ . . . ta-DAA!

Quick · #4 $Old$ November 14th, 2006, 02:33 PM

Quote:

Originally Posted by Soroban $View Post$

I'm so glad you asked!

I learned this trick many years ago from one the books in my vast library.
. . [My wife said it's "half-vast" . . . well, I think that's what she said.]

We want a function $f(n)$ such that: . $\begin{array}{cccc}f(1) = 1 \\ f(2) = 3 \\ f(3) = 5 \\ f(4) = 7 \\ f(5) = 8\end{array}$

It has the form: . $f(n) \;=\;2n - 1 + (n-1)(n-2)(n-3)(n-4)k$

Note that for $n = 1,2,3,4$ , the second portion vanishes
. . and we get the first four odd numbers.

All we have to do is to determine $k$ so that $f(5) = 8$

We have: . $f(5) \;= \;2(5) - 1 + (4)(3)(2)(1)k \;= \;8$

. . and we get: . $9 + 24k\:=\:8\quad\Rightarrow\quad k = -\frac{1}{24}$

Hence, the function is: . $f(n) \;=\;2n - 1 -\frac{1}{24}(n-1)(n-2)(n-3)(n-4)$

. . which simplifies to: . $f(n) \:=\:-\frac{1}{24}\left(n^4 - 10n^3 + 35n^2 - 98n + 48\right)$ . . . ta-DAA!

surprisingly simple

topsquark · #5 $Old$ November 14th, 2006, 07:42 PM

Clever. I haven't seen that trick in a long time.

-Dan

TriKri · #6 $Old$ November 15th, 2006, 02:21 PM

Yeah, real cleverness.

Hey, does anyone know the trick to get a polynomial function for a general series? I think Pythagoras had some method; you took the difference between all pairs of adjacent elements, resulting in a new series with one less elements. Then you continued so until you got one element left. Then you did something to create a formula from all the first elements in the series I think, but I don't remember how ... anyone else who does?