I got a sequence of numbers
1, 5, 12, 22, 35
How do i derive a formula from this?
best i came up with myself is
a_n+1 = 1+ Σ 3n+1
but this doesn't predict the first term right and a sum isn't very nice
wolfram alpha gives me a_n = 1/2 n (3 n-1)
it works but i don't know where this comes from
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a(s,n)=(s-2)n(n-2)/2 + n
where s is the number of sides
trying to find the number of dots for any figure n
I believe it's the sum of the series 1+4+7+10... but i don't know much about series
Hello, OP. You are looking at a particular case of a "figurate number", which is the general term for when you have a geometric figure with so-many sides, and so-many dots on a side. The math in this link may be over your head but you should still look at this.
mathworld.wolfram.com
the square and triangular cases are more familiar to us, but the larger cases are a bit more abstract.
Now in this particular case, OP, consider the following. What is it that you are building in each case? How can the problem be described? Let's describe it.
-A regular pentagon always has a side of some integer (whole-number) length. It is followed by another, and another, and another... each time, the side is greater by 1.
-The length is the same, all the time. But the pentagons are "generated as in the diagram." What does that mean?
-It means that the cutoff points between the unit length are denoted by dots, as in the diagram. So a side of length 1 has 2 dots on it, a side of length 2 has 3 dots on it, etc. In short, each side of length n always has n+1 dots on it.
-Now of course, each side has those two corner points on it, so we have to be careful not to double-count those. In fact, once you think a moment, you could obviously say that such a pentagon with side length n has 5n dots on it, exactly 5 times as many!
-But then we notice something: /your diagram is not considering each such pentagon by itself, but it adds on in a funny way! So the above doesn't describe what you want, and we have to think a bit harder about this.
im more interested in deriving the formula for the series
the pic is just where there series comes from
Forget the figures and try an ansatz of the form a_n = bn^2 + cn + d. Use the terms in the series to calculate the coefficients. In general you calculate the differences between each term of the series (which is something like a first derivative), then the differences of those terms (second derivative) and so on until the differences become constant. This will tell you to what order of n the ansatz should be.
Doesn't matter where it comes from.
notice how the progression of the sequence is arithmetic (+4,+7,+10,+13 etc).
this should tell you something: "if the "derivative" of the function I'm looking for is linear, then the function itself is quadratic".
So you would be looking for a_n = A*n^2 + B*n +C.
And if you plug in n=1 and 2 and 3, you obtain 3 equations in A, B and C.
If you solve, you find A=3/2 and B=-1/2
which is what wolframalpha gives you: a_n = n*(3n-1)/2
Remembering what has been said above, what is it that's being added, each time, even going from the "0" case, up to the "1" case?
What is added each time, are the /outer-three-full-side-points of the next-biggest pentagon, with the end-corners included!/ So going from "0" to "1", that's adding 4 points, and going from "1" to "2", that adds another 7 points, etc etc. /As in the diagram/, as we say.
But how can we express this algebraically? Here's a trick, let's think a bit different for a moment.
Let's say I ignore the "nesting" of the sums, for a moment. Let's say I just have some outer pentagon, with a side length of 10. In other words, n=10. In this case, the number of dots on this outer shell would just be 5n = 50, as we just discussed. Now, /how many dots are on the very-next shell to be added to this? We should eventually somehow express this in terms of our current n, which is 10.
Well, you're adding the next "shell", which will have a side length of 11. It should be easy to tell, based on what has just been said, that this outer "shell" of three-full-sides of the next biggest pentagon, /including both new end-corners/, will have 3 n' + 1 = 34 dots on it, where n' is just something I made up for "the next" n, which is 11. So 34 is the difference from one particular thing to the next. Keep this in mind.
Now we should generalize this. adding the "next" shell, then, to an already-generated pentagon of side length n looks like this. set n + 1 = n'. Then take 3n' + 1, and add /that number/ to what you've already got.
Let's start at the beginning, when n = 0. In this case, the number of dots is 1. n' = 1. 3n' + 1 = 4. 4 + 1 is 5, which is the number of dots in the new figure!
Does it work again? n = 1, now. the new number of dots, so far, is 5. n' = 2. 3n' + 1 = 7. 7 + 5 = 12. It works again!
Now, to tidy up, we should get rid of this " n' " thing I just made up, and also /prove/ this, using induction.
Here is what we say, to begin with. /The nth Pentagonal number P_n is equal to the number of dots in a figure of side length n, /as shown in the diagram/./ This is not a very good definition, but it together with the picture and what we do next will allow us to give an exact definition later.
We furthermore say that there is a /zeroth/ pentagonal number when n = 0, and then a /first/ pentagonal number (n=1), and so on. Clearly, when n = 0, P_0 = 1. And incidentally, when n = 1, P_1 = 5.
Now another trick. we begin with a /recursion relation/, but we will be able to improve this later. This is what we say, because we already said the same thing before. Either of these two, because they amount to the same thing, where it is left to you to determin appropriate ranges of n in either case:
[math]
P_n = P_{n-1} + (3n+1) \\
P_{n+1} = P_{n} + (3(n+1) + 1) \\
[/math]
Now that I have this recursive definition, it allows me to generate the first few P_n's, and see if there are any /patterns/, so that we can just give any P_n in /general/, which is what you say you want.
Using my recursive definition, and starting from P_0 = 1 and going up through P_10, I get
{1,5,12,22,35,51,70,92,117,145,176}
One more trick. When one consideres elementary series and sequences, it's always a good idea to look at the differences and/or ratios between successive terms. In this case, it's the difference that matters. The differences between successive terms are
{4,7,10,13,16,19,22,25,28,31,...}
/Now/ we have something that is intelligible. /Now we have something that we can conjecture, and prove, for a general formula. Here is my conjecture.
The nth pentagonal number is given by...
youtube.com
this video helped a lot
got a_n = 1.5n^2 - .5n
not sure how it turns into the wolfram formula
...
1.5n^2-0.5n = n*(1.5n-0.5) = n*(3n-1)/2
you got bigger issues user
nigga it's 11 pm i didn't even try
[math]
\displaystyle P_n = \sum\limits_{k=0}^{n} 3k+1
[/math]
The proof goes like this. when n = 0, everything goes fine, and so a basis case is established. Then you can go ahead and assume the above, and use it to push things around. If you can use that plus what else we know, to conclude the analogous case for n+1, that proves it for all n. This is left an exercise.
Finally, we must consider the orignal form that wolfram spat out, to get away from series altogether. This suggestion amounts to
[math]
\displaystyle P_n = \sum\limits_{k=0}^{n} 3k+1 = \frac{1}{2} (n+1) (3(n+1)-1)
[/math]
Notice that I made a change of index here, because I set P_0 = 1 as opposed to P_1 = 1, which is what wolfram really does. I've accounted for this. My version amounts to the same as your formula, but I'm far enough along on my train of thought that I'll stick with mine, and it's good practice for you too.
Anyway, when n = 0, again all is well and good. So how do you prove it this time, again? /Use what you know, and add it in to get the end result!/
specifically, use the above to know how to add the same thing to both sides, to get P_n+1, in my usage (add 3(n+1)+1 to both sides). Then rearrange until you get what you expect out of your RHS, in the n+1 case.
A hint about working induction problems: I find it useful to state what I expect to end up with, and work it backwards, literally working the problem from both ends. I just did this, in my check. If the two ends meet and are equal, then your thing is true. If not, then it's not. In this case, it is. the details are left as a further exercise.
A final question becomes this. We just went with Wolfram's algebraic expression for each P_n, after an index change (which we can easily undo), but a deeper issue arises: /how does one see that supposition to begin with, to even test it by induction, as we just did/?. This is where cleanup is needed.
I therefore adjust my above and no longer speak of a "zeroth" pentagonal number; I make a change of index and say that the /first/ pentagonal number is equal to one, etc. This entails that...
[math]
\displaystyle P_n = \sum\limits_{k=1}^{n} 3k-2 = \frac{1}{2} n(3n-1)
[/math]
The double-check of this canonical form is left as yet another exercise. The impetus to even consider the "final" RHS in this case, arises from the following:
find the appropriate geometric argument for yourself! the triangular and square numbers also offer hints. I'm blanking atm.
This seems like an appropriate thread for this.
This isn't homework, but is work related.
When bolting up manways and flanges at work, you typically follow a torque pattern (star pattern so to speak) to get the shit tightened down correctly. You always start by numbering the very top bolt as one, the one opposite it on the bottom a 2, the left side a 3, and the right a 4. You then put a 5 on the bolt that's to the right of the 1, a 6 opposite of the 5 bolt to the left of the 2 bolt , then a 7 above the 3, and an 8 below the 4. You always go in a clockwise fashion.
It's a simple enough thing to do when its 16 or so bolts, but somewhat annoying when you have a 40+ pattern as you typically have to number them.
I got to thinking today while closing a manway that took 80 bolts; Is there some kind of formula that I could use to know what each bolt number would be while going in a clockwise fashion around the flange to avoid having to do this star pattern shit?
Every flange that I tighten up has a number of bolts that is a multiple of 4. No exceptions.
This is my recursive representation:
If n=0
T(n)=1
Else
T(n)=(5*n)+T(n-1)-( 2n-1)
Or
T(n)=3n+1 +T(n-1)
Looks like you can just count up by multiplies of 4 starting with 1 until your next number would exceed the number of bolts, then start over at 4, then 2, then 3.
So for 80 it would be
1,5,...,77
4,8,...,80
2,6,...,78
3,7,...,79
The clue to this motivation that we wish to re-discover in a round-about way, is to consider what the above recursive series gives us, for a useful number of terms of investigation and once more to look for a pattern. We thus compare the sequence n
{1,2,3,4,5,6,7,8,9}
with the sequence P_n
{1,5,12,22,35,51,70,92,117}
and ask whether we can see any useful general patterns, in terms of differences, ratios, etc, as I advised before.
A moment's check confirms that the odd ratios 1/1, 12/3, 35/5 etc are precisely the integers 1,4,7... a familiar progression from before. This invites us to express the whole lot as ratios, and with rearrangement in terms of division by two, a natural choice to make the whole thing the same since the other numerators are odd, we get
2/2, 5/2, 8/2, 11/2, 14/2, 17/2, 20/2, 23/2, 26/2...
Again, incrementing with three! When we investigate in this low scenario, /this is what makes it possible for us to conjecture, and above (prove) the formula/:
-take n, and do something to it. Whatever you end up doing, take that and divide it by two. (or leave it alone, depending on the treatment).
-what is it that you do to n? /you add the corresponding odd number/: 1,3,5...
-e.g. n + (2n-1)
-You then take this that you get, and divide it by two: (n+(2n-1))/2 = (3n-1)/2.
-But what does it mean in this context? /it is the ratio of P_n to n/. So, to find P_n, simply multiply through by n.
THIS IS WHAT GIVES OP'S ORIGINAL FORMULA. And we have considered this single case of the figurate number, every which-way.