I have a series of numbers but need to get the formula to calculate the number at position x.
The starting number is always -1396 and every value is a multiplication of 1396. So it starts with -1396 then goes up by (3 * 1396), stays the same for 3 times and then goes up for (3 * 1396) again. This will always be the same pattern, how can I find a formula for it?
The only formula I could get from an online formula creator was:
x = 1396(z^3 + 3z - 1) / (z-1)^2 (z^2 + z + 1)
but this does not seem to get the right answer.
-1396
2792
2792
2792
6980
6980
6980
11168
11168
11168
You can use Math.floor to make it jump each 3 steps.
z = (Math.floor((x - 1) / 3) * 3 + 2) * 1396
Related
I have a very simple (and stupid :) question:
How do I get from this expression:
To this recursion formula?
I do not want any complete solution, but maybe you can help me to understand it with just a few tips.
Thx, and a nice day :)
This is very similar to [Wikipedia]: Fibonacci number, and is a typical [Wikipedia]: Mathematical induction example.
Starting from the assumption that the formula is true for (a given) n (let's call this P(n)):
Jn = 2 * Jn - 1 + (-1)n - 1
then, using the above equality and the function definition, you must prove P(n + 1) - that the formula is also true for (the consecutive) n + 1. This is the one in your question:
Jn + 1 = 2 * Jn + (-1)n
Since this is a programming site (and I'm too lazy calculating the first few values manually), here's some Python code that does that for us:
>>> def j(n):
... if n <= 0:
... return 0
... elif n == 1:
... return 1
... else:
... return j(n - 1) + 2 * j(n - 2)
...
>>>
>>> for i in range(15):
... print("{:2d} - {:4d}".format(i, j(i)))
...
0 - 0
1 - 1
2 - 1
3 - 3
4 - 5
5 - 11
6 - 21
7 - 43
8 - 85
9 - 171
10 - 341
11 - 683
12 - 1365
13 - 2731
14 - 5461
From the above, it's kind of noticeable by the naked eye that the formula is true. Now the induction mechanism should be applied: start from P(n) and using the function definition (3rd branch) get to P(n + 1). That might not be trivial, considering there's a level 2 dependency (most of the terms that will eventually reduce each other, but I didn't try it to see how "visible" that would be). You could check [SO]: Recursive summation of a sequence returning wrong result (#CristiFati's answer), for more details on a simpler problem.
Note:
Given the current coefficients, I must mention recursion's characteristic equation (check [Wikipedia]: Constant-recursive sequence) that would give a non recurring formula for Jn:
Jn = Jn - 1 + 2 * Jn - 2 translates to: x2 = x1 + 2 * x0 -> x2 - x - 2 = 0 (which has -1 and 2 as roots), and from here (using Binet (or Moivre) formula):
Jn = (2n - (-1)n) / 3 (denominator value is: 2 - -1)
Let the computer do some calculations from us (check that previous implementation results match this one's):
>>> def j_simpl(n):
... return (2 ** n - (-1) ** n) / 3
...
>>>
>>> print(all(j(i) == j_simpl(i) for i in range(20)))
True
For n binary digits with base x, the maximum value will be:
x^(n-1) + x^(n-2) + ... + x^1 + x^0
By using geometric progression,
r=1/x
Using formula for sum of n finite numbers i get:
(x^n - 1) / (x - 1).
But my answer should have been x^n - 1.(formula: M= x^n - 1)
You are correct that the sum of the geometric series x0 + x1 + x2 + ... + xn-1 is indeed (xn - 1) / (x - 1). For example, if we pick x = 10 (base 10) and n = 3 (a three-digit base 10 number), we get back
1 + 10 + 100 = (1000 - 1) / 9 = 999 / 9 = 111.
However, the largest three-digit number is 999. And by looking at the above sum, you might get a sense of why we're off by a factor of 9. When writing out numbers in base 10, we'd maximize our number by having each digit be 9, not 1. And more generally, in base x, we'd maximize our value by having each digit be x - 1. That means that the maximum value is
(x - 1)(x0 + x1 + x2 + ... + xn-1) = (x - 1)(xn - 1)/(x - 1) = xn - 1.
Here's another, easier way to see this. What is the smallest number you can make with n+1 digits? That would be xn. Since that's the smallest (n+1)-digit number, the largest n-digit number must be that minus one, giving xn - 1 without needing to discuss geometric series.
My math is pretty weak, and I'm having confusion over the differences. I'm trying to find out the midi formula, to output frequency when I have midi value
MidiNumber = 69+12* Log2(440/Frequency)
So I derived
Frequency = (-69 + 5280 Log2 + MidiNumber)/(12 Log2)
If I plugin things this works correctly
440 = (-69 + 5280 Log2 + 69)/(12 Log2)
If I do this though things do not work correctly
(-69 + Log[2, 5280.] + 69)/Log[2, 12.]
This is the output I get in my programming, I don't know exactly the difference between the two equations. Maybe it's 12*Log2, but is that 12*Log2[1] or, ...? No idea.
Part of your confusion seems to be treating Log2(n) as Log2 * n. Log2 is actually a function, the inverse of which is 2^x.
So your derivation should go something as follows:
MidiNumber = 69 + 12 * Log2(440 / Frequency)
MidiNumber - 69 = 12 * Log2(440 / Frequency)
(MidiNumber - 69) / 12 = Log2(440 / Frequency)
2^((MidiNumber - 69) / 12) = 440 / Frequency
Frequency = 440 / 2^((MidiNumber - 69) / 12)
I have this recursive function:
f(n) = 2 * f(n-1) + 3 * f(n-2) + 4
f(1) = 2
f(2) = 8
I know from experience that explicit form of it would be:
f(n) = 3 ^ n - 1 // pow(3, n) - 1
I wanna know if there's any way to prove that. I googled a bit, yet didn't find anything simple to understand. I already know that generation functions probably solve it, they're too complex, I'd rather not get into them. I'm looking for a simpler way.
P.S.
If it helps I remember something like this solved it:
f(n) = 2 * f(n-1) + 3 * f(n-2) + 4
// consider f(n) = x ^ n
x ^ n = 2 * x ^ (n-1) + 3 * x ^ (n-2) + 4
And then you somehow computed x that lead to explicit form of the recursive formula, yet I can't quite remember
f(n) = 2 * f(n-1) + 3 * f(n-2) + 4
f(n+1) = 2 * f(n) + 3 * f(n-1) + 4
f(n+1)-f(n) = 2 * f(n) - 2 * f(n-1) + 3 * f(n-1) - 3 * f(n-2)
f(n+1) = 3 * f(n) + f(n-1) - 3 * f(n-2)
Now the 4 is gone.
As you said the next step is letting f(n) = x ^ n
x^(n+1) = 3 * x^n + x^(n-1) - 3 * x^(n-2)
divide by x^(n-2)
x^3 = 3 * x^2 + x - 3
x^3 - 3 * x^2 - x + 3 = 0
factorise to find x
(x-3)(x-1)(x+1) = 0
x = -1 or 1 or 3
f(n) = A * (-1)^n + B * 1^n + C * 3^n
f(n) = A * (-1)^n + B + C * 3^n
Now find A,B and C using the values you have
f(1) = 2; f(2) = 8; f(3) = 26
f(1) = 2 = -A + B + 3C
f(2) = 8 = A + B + 9C
f(3) = 26 = -A + B + 27C
solving for A,B and C:
f(3)-f(1) = 24 = 24C => C = 1
f(2)-f(1) = 6 = 2A + 6 => A = 0
2 = B + 3 => B = -1
Finally
f(n) = 3^n - 1
Ok, I know you didn't want generating functions (GF from now on) and all the complicated stuff, but my problem turned out to be nonlinear and simple linear methods didn't seem to work. So after a full day of searching, I found the answer and hopefully these findings will be of help to others.
My problem: a[n+1]= a[n]/(1+a[n]) (i.e. not linear (nor polynomial), but also not completely nonlinear - it is a rational difference equation)
if your recurrence is linear (or polynomial), wikihow has step-by-step instructions (with and without GF)
if you want to read something about GF, go to this wiki, but I didn't get it till I started doing examples (see next)
GF usage example on Fibonacci
if the previous example didn't make sense, download GF book and read the simplest GF example (section 1.1, ie a[n+1]= 2 a[n]+1, then 1.2, a[n+1]= 2 a[n]+1, then 1.3 - Fibonacci)
(while I'm on the book topic) templatetypedef mentioned Concrete Mathematics, download here, but I don't know much about it except it has a recurrence, sums, and GF chapter (among others) and a table of simple GFs on page 335
as I dove deeper for nonlinear stuff, I saw this page, using which I failed at z-transforms approach and didn't try linear algebra, but the link to rational difference eqn was the best (see next step)
so as per this page, rational functions are nice because you can transform them into polynomials and use linear methods of step 1. 3. and 4. above, which I wrote out by hand and probably made some mistake, because (see 8)
Mathematica (or even the free WolframAlpha) has a recurrence solver, which with RSolve[{a[n + 1] == a[n]/(1 + a[n]), a[1] == A}, a[n], n] got me a simple {{a[n] -> A/(1 - A + A n)}}. So I guess I'll go back and look for mistake in hand-calculations (they are good for understanding how the whole conversion process works).
Anyways, hope this helps.
In general, there is no algorithm for converting a recursive form into an iterative one. This problem is undecidable. As an example, consider this recursive function definition, which defines the Collatz sequence:
f(1) = 0
f(2n) = 1 + f(n)
f(2n + 1) = 1 + f(6n + 4)
It's not known whether or not this is even a well-defined function or not. Were an algorithm to exist that could convert this into a closed-form, we could decide whether or not it was well-defined.
However, for many common cases, it is possible to convert a recursive definition into an iterative one. The excellent textbook Concrete Mathematics spends much of its pages showing how to do this. One common technique that works quite well when you have a guess of what the answer is is to use induction. As an example for your case, suppose that you believe that your recursive definition does indeed give 3^n - 1. To prove this, try proving that it holds true for the base cases, then show that this knowledge lets you generalize the solution upward. You didn't put a base case in your post, but I'm assuming that
f(0) = 0
f(1) = 2
Given this, let's see whether your hunch is correct. For the specific inputs of 0 and 1, you can verify by inspection that the function does compute 3^n - 1. For the inductive step, let's assume that for all n' < n that f(n) = 3^n - 1. Then we have that
f(n) = 2f(n - 1) + 3f(n - 2) + 4
= 2 * (3^{n-1} - 1) + 3 * (3^{n-2} - 1) + 4
= 2 * 3^{n-1} - 2 + 3^{n-1} - 3 + 4
= 3 * 3^{n-1} - 5 + 4
= 3^n - 1
So we have just proven that this recursive function does indeed produce 3^n - 1.
I'm trying to use WolframAlpha to solve for a variable.
I have
u(k, r) = (900-3k)r^(k-1)
and
s(n, r) = sum u(k, r), k=1 to n
and I want to solve for r with
s(5000, r) = -600000000000
I've tried various incantations, but can't seem to get it working. I can't even get s defined to evaluate it.
If you care, it is to solve this problem : http://projecteuler.net/index.php?section=problems&id=235
WARNING: Spoiler below!
You should ask WA to FullSimplify the expression of s(n,r) after you substitute u(k,r) into it. It should give
(3 (299 - 300 r + r^n (-299 + n + 300 r - n r)))/(-1 + r)^2
Solving the final equality is then just finding the root of a (high degree) polynomial:
299 + 200000000000 (-1 + r)^2 + (4701 - 4700 r) r^5000 == 300 r
where r != 1 since that was a pole of the original expression. Note that r must be positive so that the positive quadratic gets negated by the high-degree term. Plotting the function shows that It is positive for r < 1, and negative for r >~ 1, so the solution is somewhere past r=1. Now change variables so that x = r-1 and look near x=0:
200000000000 x^2 + (1 + x)^5000 (1 - 4700 x) - 1 - 300 x == 0
This should be enlightnening:
Plot[200000000000 x^2 + (1 + x)^5000 (1 - 4700 x) - 1 - 300 x, {x, 0, 0.003}]
Using FindRoot with a good guess gives x=0.002322108633 or r=1.002322108633.
The WA commands follow.
First I used
FullSimplify[Sum[(900-3k)r^(k-1),{k,1,n]]
Then you would have to retype the expression it spits out:
Plot[(3 (299 - 300 r + r^5000 (-299 + 5000 + 300 r - 5000 r)))/(-1 + r)^2 + 6000000000,{r,-2,2}]
At this point I manually replaced r with x+1:
Plot[200000000000 x^2 + (1 + x)^5000 (1 - 4700 x) - 1 - 300 x, {x, 0, 0.003}]
And solving for the root:
FindRoot[200000000000 x^2 + (1 + x)^5000 (1 - 4700 x) - 1 - 300 x, {x, 0.0023}]
Which doesn't give enough precision, and this is as far as you can go using only WA. You can try to subtract off the first few digits that WA gives you, and do another substitution with y = x + 0.00232211 to get the next few digits, but that is too tedious for me to try.