Number series
- Thread starter Saumyojit
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@JeffM
Can you please in simple way just like explaining to layman , please tell me how did you derive the formula.
Why / how did you inserted 1/5 factorial * all those numbers and the power of n.
What was your thought process ?
I Send this to one of my friends who is good in maths but he has no clue (awestruck) that how you made the formula.
I am also Curious extremely.
What are you talking about simultaneousl equations??
What do you mean by the above para.
I already smoke 10 cigarettes I order to find out how did you make up the formula but couldn't
I have been thinking about the quoted post.
You are correct that the answer I gave to your original problem is not one that would occur to many students. But that is exactly the flaw in this kind of puzzle. It teaches that there is only one right answer. But, as has been said and now proven, the data provided may be susceptible to different answers, each of which satisfies the conditions. The practical lesson is not to insist that the most intuitive answer must be correct. People’s intuitions differ. Furthermore, the universe is under no duty to behave in accordance with human intuition. The practical lesson when the data provided are consistent with multiple explanations is TO GET MORE DATA. That is why this kind of puzzle is worse than useless; it teaches to accept a superficial answer rather than seeking information
This magic number [MATH]338[/MATH] is the main reason that JeffM formula is valid.
If you know how to create a formula for [MATH]10, 20, 36, 58, 86, 120, 160,.........[/MATH], you can create JeffM formula easily. It follows the same logic. All you have to do is to find the magic number. The magic number is going to tell you what polynomial degree you need for your formula.
When you find a row that has similar numbers, count the number of rows until that row (skip the original row). For example, the below numbers were found similar in the 5th row.
This means you need a 5th polynomial degree.
[MATH]a_n =An^5 + Bn^4 + Cn^3 + Dn^2 + En + F[/MATH]
Finding the constants is just a matter of setting [MATH]6[/MATH] equations and solving them. Of course, you will not solve them by hand, but this is how JeffM formula was created.
This means you need a 5th polynomial degree.
[MATH]a_n =An^5 + Bn^4 + Cn^3 + Dn^2 + En + F[/MATH]
Finding the constants is just a matter of setting [MATH]6[/MATH] equations and solving them. Of course, you will not solve them by hand, but this is how JeffM formula was created.
A real polynomial of degree n (n being a positive integer) is defined as
[MATH]P_n(x) = \sum_{j=0}^n a_j x^j \text { where } a_n \ne 0 \text { and all a’s are real numbers.}[/MATH]
A polynomial of degree 1 is [MATH]a_0 + a_1x.[/MATH] A polynomial of degree 2 is [MATH]a_0 + a_1x + a_2x^2[/MATH].
Notice that the number of a’s is always one greater than the degree of the polynomial.
Now suppose I know n + 1 pairs of numbers. I can find a polynomial of degree n or lower that will relate them. How?
Well, in this case , I had (1, 7), (2, 9), (3, 12), (4, 48), (5, 190), and (6, 890). So
[MATH]7 = a_0(1^0) + a_1(1^1) + a_2(1^2) + a_3(1^3) + a_4(1^4) + a_5(1^5);\\ 9 = a_0(2^0) + a_1(2^1) + a_2(2^2) + a_3(2^3) + a_4(2^4) + a_5(2^5);\\ 12 = a_0(3^0) + a_1(3^1) + a_2(3^2) + a_3(3^3) + a_4(3^4) + a_5(3^5);\\ 48 = a_0(4^0) + a_1(4^1) + a_2(4^2) + a_3(4^3) + a_4(4^4) + a_5(4^5);\\ 190 = a_0(5^0) + a_1(5^1) + a_2(5^2) + a_3(5^3) + a_4(5^4) + a_5(5^5); \text { and } \\ 890 = a_0(6^0) + a_1(6^1) + a_2(6^2) + a_3(6^3) + a_4(6^4) + a_5(6^5).[/MATH]Now if we substitute variables to eliminate subscripts and do the arithmetic, this transforms to
[MATH]7 = p + q + r + s + t + u;\\ 9 = p + 2q + 4r + 8s + 16t + 32u;\\ 12= p + 3q + 9r + 27s + 81t + 243u;\\ 48 = p + 4q + 16r + 64s + 256t + 1024u;\\ 190 = p + 5q + 25r + 125s + 625t + 3125u; \text { and}\\ 890 = p + 6q + 36r + 216s + 1296t + 7776u.[/MATH]That is a system of six linear equations in six unknowns. We learn in principle how to solve such systems with paper and pencil during first year algebra. Solving that by hand is boring but possible. Solving it matrix style is easy in excel.
Otis
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Hi. Google keywords gaussian elimination tutorial. If you scroll through the links, you'll find both textbook and video-lecture resources.
?
Last edited:
a. Going back to the sequence: 7, 9, 12, 48, 190...
The following quintic polynomial will produce this sequence, followed by any number 'A' you care to enter. (An uncountably infinite set of answers!)
[MATH]\frac{1}{120}[\hspace1ex A (n^5 - 15 n^4 + 85 n^3 - 225 n^2 + 274 n - 120)-552n^5+8485n^4-48330n^3+127595n^2-154398n+68040)\hspace1ex][/MATH]
b. In post #45 I gave the simplest way (today) to get a polynomial to fit the data !
c. However, you can always write down 'the polynomial' directly, without formulating or solving equations:
say it is to go through (1,7) (2,9) (3,12) (4,48) then:
[MATH]p(x)=7 \frac{(x-2)(x-3)(x-4)}{(1-2)(1-3)(1-4)} + 9 \frac{(x-1)(x-3)(x-4)}{(2-1)(2-3)(2-4)}+12 \frac{(x-1)(x-2)(x-4)}{(3-1)(3-2)(3-4)}+48 \frac{(x-1)(x-2)(x-3)}{(4-1)(4-2)(4-3)}[/MATH]
works, for obvious reasons!
If we want p(x) to go through [MATH](i, a_i), i=1...n[/MATH], then
[MATH]p(x) = \sum_{i=1}^n a_i \prod_{j=1, i≠j}^n \frac{x-j}{i-j} \hspace1ex \text{ does it.}[/MATH]
The following quintic polynomial will produce this sequence, followed by any number 'A' you care to enter. (An uncountably infinite set of answers!)
[MATH]\frac{1}{120}[\hspace1ex A (n^5 - 15 n^4 + 85 n^3 - 225 n^2 + 274 n - 120)-552n^5+8485n^4-48330n^3+127595n^2-154398n+68040)\hspace1ex][/MATH]
b. In post #45 I gave the simplest way (today) to get a polynomial to fit the data !
c. However, you can always write down 'the polynomial' directly, without formulating or solving equations:
say it is to go through (1,7) (2,9) (3,12) (4,48) then:
[MATH]p(x)=7 \frac{(x-2)(x-3)(x-4)}{(1-2)(1-3)(1-4)} + 9 \frac{(x-1)(x-3)(x-4)}{(2-1)(2-3)(2-4)}+12 \frac{(x-1)(x-2)(x-4)}{(3-1)(3-2)(3-4)}+48 \frac{(x-1)(x-2)(x-3)}{(4-1)(4-2)(4-3)}[/MATH]
works, for obvious reasons!
If we want p(x) to go through [MATH](i, a_i), i=1...n[/MATH], then
[MATH]p(x) = \sum_{i=1}^n a_i \prod_{j=1, i≠j}^n \frac{x-j}{i-j} \hspace1ex \text{ does it.}[/MATH]
D
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This is called:
It is a standard method for numerical analysis.
Legendre Polynomial fit
It is a standard method for numerical analysis.
Aha. I thought there might be a relationship to the fundamental theorem of algebra. But are there really an infinite number of quintics? (I know there are an infinite number of polynomials, but I thought I remembered that, with a quintic and six data points, there were no degrees of freedom and so the quintic was unique..
I was following on from your previous discussions, saying that the set of possible solutions to the problem is infinite and each of the solutions can be achieved using a unique quintic.
In other words if you wish to produce: 7, 9, 12, 48, 190, A....
the following quintic polynomial will produce this sequence, substituting your chosen value for A:
[MATH]\frac{1}{120}[\hspace1ex A (n^5 - 15 n^4 + 85 n^3 - 225 n^2 + 274 n - 120)-552n^5+8485n^4-48330n^3+127595n^2-154398n+68040)\hspace1ex][/MATH]Hence the set of possible solutions to the original problem is uncountably infinite and it's quite nice to be able to produce the whole sequence with any 6th value you like, just by changing one number in the quintic.
It would appear the answer is B 149;
the presumed, unlikely pattern being [MATH]a_n= 15 + (n-1)^3 +\text{ max}(0,2n-3)[/MATH]
@lex Got it. Sorry to be so stupid. I was dealing with medical people all day, and one gets infantalized pretty quickly. I actually was writing posts while hooked up to monitors. (No problem, just tests, acting as a test tube.)
Yes, given five pairs of arguments and resultants and a sixth argument, we can generate an infinite number of possibilities for the sixth resultant by varying a parameter in a quintic template. Very, very nice.
But from a given six pairs of arguments and resultants, the quintic generating function is unique.
I need to have a martini or something. My neurons need to start firing.
Although I still think the initial problem should not be given to students for the reasons that I have adumbrated previously, the resulting thread turned out to have a lot of merit. Thanks to pka, nasi, Subhotosh, Dr. Peterson, and lex (if I forgot jomo I do apologize), we touched on significant sequences, discriminating among answers, polynomials, numerical methods, puzzles versus proofs, etc. Saumyojit got a lot more than he bargained for with his initial post. I fear he is preparing for exams, a process that can interfere with learning some very material things. One of the bad things the Brits gave to India is crammers.
Sounds like a tough day. Very tedious but hopefully worth the inconvenience. I think a well-deserved break is in order!
You have summarised it all perfectly and I do agree with your assessment of the topic. It's funny how something apparently quite banal can produce such an interesting discussion.
You have summarised it all perfectly and I do agree with your assessment of the topic. It's funny how something apparently quite banal can produce such an interesting discussion.
D
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Correct - but cramming is not all that bad. Imagine how long it would have taken us (old >60 yrs old) to learn multiplication if we did not cram multiplication table.
I still cannot explain π * √2 to my grandchild. I just tell him - we will never know exactly.
I think these days cramming has been replaced by "button-pushing". Of course I blame Sumerians for that - blindly sliding those little balls!!!
Cubist
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I think this kind of question isn't always bad. It can teach thinking in different ways, and it can also reward perseverance.
And if the question is marked fairly (I accept that often it isn't) then it can illustrate to the class that some questions can have several, or even infinite, valid answers.
And if the question is marked fairly (I accept that often it isn't) then it can illustrate to the class that some questions can have several, or even infinite, valid answers.
Subhotosh
Some rote learning is necessary. But the essence of cramming is rote learning of formulas for specific types of question typical of standardized exams. No one learns anything permanent from that experience.
The Sumerians had computers with rollerballs rather than keys? Very ingenious. Of course, cuneiform did not adapt easily to a qwerty keyboard.