Cheese11
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Ti-84 Silver Edition. Whats up.
Mathematics Thread
- Thread starter IanTheCuber
- Start date
Ti-84 Silver Edition. Whats up.
MTGjumper
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119, 637.
You generated the sequence with the following polynomial:
\( \frac{1}{120}x^7 - \frac{4}{15}x^6 + \frac{421}{120}x^5 - \frac{195}{8}x^4 + \frac{2861}{30}x^3 - \frac{24763}{120}x^2 + \frac{13447}{60}x - 85 \)
Christopher Mowla
Premium Member
@MTGjumper, the method of Lagrange Polynomials cannot be used to yield realistic values for extrapolation for such a high degree polynomial.
@brandbest1,
The sequence of numbers you listed can be represented with the following composition of step functions (and polynomial functions):
\( 7\left\lceil \frac{n}{7} \right\rceil +\left( n-1 \right)+\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \left( 1-n \right) \)
[Link]
[Link]
, which yields the values from the first value to the 23rd value (n = 1 to n = 23)
{7,7,7,10,11,12,13,21,22,23,24,25,26,27,35,36,37,38,39,40,41,49,50,...}
Derivation
So the answer is 22 and 23, assuming that the pattern {0,0,0,3,4,5,6,7} continues as {0,0,0,3,4,5,6,7,8,9,10,11,12,...,n+1} (from n = 1 to infinity).
EDIT:
{7,7,7,10,11,12,13,21,22,23,24,25,26,27,35,36,37,38,39,40,41,49,50,...}
Derivation
Given the sequence, \( \left\{ 7,7,7,10,11,12,13,21,\text{ }\!\!\_\!\!\text{ },\text{ }\!\!\_\!\!\text{ } \right\} \),
[Step 1]
Let's look at each term as a (multiple of 7)+ (some other integer).
That is, {7,7,7,10,11,12,13,21,...} = {7(1)+0,7(1)+0,7(1)+0,7(1)+3,7(1)+4,7(1)+5,7(1)+6,7(2)+7,...}
For all of the terms, the 7(integer) part can be represented as \( 7\left\lceil \frac{n}{7} \right\rceil \) (the brackets are the ceiling function, which means you round up to the nearest integer).
\( 7\left\lceil \frac{n}{7} \right\rceil \) gives the values {7,7,7,7,7,7,7,14,...} = {7(1),7(1),7(1),7(1),7(1),7(1),7(1),7(2),...} (for n = {1,2,3,4,5,6,7,8,...,n}).
[Step 2]
Now to handle the values for the integers which we add, {0,0,0,3,4,5,6,7...}, I made the assumption that the pattern continues as {0,0,0,3,4,5,6,7,8,9,10,11,12,...} (it's ambiguous really).
What we first need to do is construct a function which outputs non-zero integers values only for n = 1 through n = 3. The first function that came to my mind which is definitely indisputable is \( \left\lfloor \frac{3}{n} \right\rfloor \) (the brackets represent the floor function, which means to round down to the nearest integer).
And \( \left\lfloor \frac{3}{n} \right\rfloor \) = {3,1,1,0,0,0,0,0,...,0} for the entire set of positive integers n = {1,2,3,4,5,6,7,8,...,n}.
We need to make all three non-zero integer values (which is the first three) the same value. Why not make them 1? To make them one, we first multiply the entire sequence by (1/3) and then take the ceiling of it.
\( \left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \).
This gives us {1,1,1,0,0,0,...,0} for n = {1,2,3,4,5,6,...,n}.
Next, let's just pretend that we are just considering the sequence {n} = {1,2,3,4,5,6,7,8,...,n}. If we want to make only the first three terms zero, we can just add {n} to the additive inverse of the step functions we found before times n. That is,
\( n-n\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \)
,which generates the values {0,0,0,4,5,6,7,8,...,n}.
Remember that all we have been aiming for in step 2 is to come up with the sequence {0,0,0,3,4,5,6,7,...,n+1} (from n = 1 to infinity)
All we have to do to the above (well it seems) is to subtract 1.
\( n-1-n\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \) = {-1,-1,-1,3,4,5,6,7,...,n+1}.
All we have to do now to finish this piece is to add the first step function we made, \( \left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \) = {1,1,1,0,0,0,...}.
\( -1+n-n\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil +\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \) = {0,0,0,3,4,5,6,7,...,n+1} (from n = 1 to infinity).
\( =n-1+\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \left( 1-n \right) \).
[Step 3]
Now all we do is add the terms developed in steps 1 and 2.
\( =\left( 7\left\lceil \frac{n}{7} \right\rceil \right)+\left( n-1+\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \left( 1-n \right) \right) \).
[Step 1]
Let's look at each term as a (multiple of 7)+ (some other integer).
That is, {7,7,7,10,11,12,13,21,...} = {7(1)+0,7(1)+0,7(1)+0,7(1)+3,7(1)+4,7(1)+5,7(1)+6,7(2)+7,...}
For all of the terms, the 7(integer) part can be represented as \( 7\left\lceil \frac{n}{7} \right\rceil \) (the brackets are the ceiling function, which means you round up to the nearest integer).
\( 7\left\lceil \frac{n}{7} \right\rceil \) gives the values {7,7,7,7,7,7,7,14,...} = {7(1),7(1),7(1),7(1),7(1),7(1),7(1),7(2),...} (for n = {1,2,3,4,5,6,7,8,...,n}).
[Step 2]
Now to handle the values for the integers which we add, {0,0,0,3,4,5,6,7...}, I made the assumption that the pattern continues as {0,0,0,3,4,5,6,7,8,9,10,11,12,...} (it's ambiguous really).
What we first need to do is construct a function which outputs non-zero integers values only for n = 1 through n = 3. The first function that came to my mind which is definitely indisputable is \( \left\lfloor \frac{3}{n} \right\rfloor \) (the brackets represent the floor function, which means to round down to the nearest integer).
And \( \left\lfloor \frac{3}{n} \right\rfloor \) = {3,1,1,0,0,0,0,0,...,0} for the entire set of positive integers n = {1,2,3,4,5,6,7,8,...,n}.
We need to make all three non-zero integer values (which is the first three) the same value. Why not make them 1? To make them one, we first multiply the entire sequence by (1/3) and then take the ceiling of it.
\( \left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \).
This gives us {1,1,1,0,0,0,...,0} for n = {1,2,3,4,5,6,...,n}.
Next, let's just pretend that we are just considering the sequence {n} = {1,2,3,4,5,6,7,8,...,n}. If we want to make only the first three terms zero, we can just add {n} to the additive inverse of the step functions we found before times n. That is,
\( n-n\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \)
,which generates the values {0,0,0,4,5,6,7,8,...,n}.
Remember that all we have been aiming for in step 2 is to come up with the sequence {0,0,0,3,4,5,6,7,...,n+1} (from n = 1 to infinity)
All we have to do to the above (well it seems) is to subtract 1.
\( n-1-n\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \) = {-1,-1,-1,3,4,5,6,7,...,n+1}.
All we have to do now to finish this piece is to add the first step function we made, \( \left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \) = {1,1,1,0,0,0,...}.
\( -1+n-n\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil +\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \) = {0,0,0,3,4,5,6,7,...,n+1} (from n = 1 to infinity).
\( =n-1+\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \left( 1-n \right) \).
[Step 3]
Now all we do is add the terms developed in steps 1 and 2.
\( =\left( 7\left\lceil \frac{n}{7} \right\rceil \right)+\left( n-1+\left\lceil \frac{1}{3}\left\lfloor \frac{3}{n} \right\rfloor \right\rceil \left( 1-n \right) \right) \).
EDIT:
TI-84 Plus Silver Present.
Last edited:
IanTheCuber
Regular
I'm not that advanced, I'm in the 6th grade. But I can graph some cubic/quadratic equations.
aaronb
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I'm a Freshman, and since I'm pretty good at math, I got put in Geometry Honors; so that would be the highest math class I've taken. But I'd take that with a grain of salt, because my city (Lewiston, Maine) isn't exactly known for its quality education. 
So despite being one of the best at math in my class, in any good education system, I would probably still be put in the higher math classes, but not be anywhere near the best.
(After like 5 minutes, I am still trying to figure out if this post makes me sound conceited or humble)
So despite being one of the best at math in my class, in any good education system, I would probably still be put in the higher math classes, but not be anywhere near the best. (After like 5 minutes, I am still trying to figure out if this post makes me sound conceited or humble)
brandbest1
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Actually, Sarah's answer was the one I was thinking of. Except that the first term is in base 10.
I'm only in 8th grade, and I'm WAY ahead of my class.
I never really knew what the difference between prealgrebra, algebra 1 and 2, until now.
I'm only in 8th grade, and I'm WAY ahead of my class.
I never really knew what the difference between prealgrebra, algebra 1 and 2, until now.
Last edited:
MTGjumper
Veteran
hyunchoi98
Regular
I have a TI-89 Titanium that my dad gave me.
It seems live everyone else at school either has a TI-84, or a TI Nspire
It seems live everyone else at school either has a TI-84, or a TI Nspire
I'm currently in my first year of IB Math (High Level), which covers what I guess you would call trig, pre-calc and calc. I sort of did a pre-calc class last year, but it was more of a pre-IB class. We haven't done any calculus this year so far, but I think we should be starting it next month or so.
Jaycee
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8th Grade, in Algebra 1, but it turns out we're using the book that the highschoolers use for Algebra 2. So I guess I'm in Algebra 2? xD I have no idea what math class I'll be in next year :/
jeff081692
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I might have to take my answer of calc 2 back. I forgot I was taking a discrete structures class which is basically math for computer science.
brandbest1
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Since I'm a math nerd, I want to learn Trig and Calc so badly, even though I'm in 8th.
jeff081692
Veteran
khan academy my friend.
asportking
Veteran
I think pre-calc was probably the hardest "class" for me, since I decided to take it over the summer using an old textbook I picked up at the library. It was totally worth it though; calc this year is the best and most fun class I've ever gone into.
Uberzj
Member
For the 8th graders who don't know the math progression, I will lay it out how I took it.
I was in the Gifted program(many of you will call it enriched).
Algebra I in the 8th grade.
Algebra II in the 9th.
Geometry in the 10th.
Pre-calc in the 11th.
Calculus in the 12th.
I am about to be a sophomore in college.
The honors program at my college has math 240 241 and 242, which are the equivalent of Pre-Cal, Calc I, and Calc II. Which I had in highschool.
Next year I will be taking Calc III, Calc IV, and Differential Equations.
This is in an honors engineering environment at a fairly prestigious school.
I was in the Gifted program(many of you will call it enriched).
Algebra I in the 8th grade.
Algebra II in the 9th.
Geometry in the 10th.
Pre-calc in the 11th.
Calculus in the 12th.
I am about to be a sophomore in college.
The honors program at my college has math 240 241 and 242, which are the equivalent of Pre-Cal, Calc I, and Calc II. Which I had in highschool.
Next year I will be taking Calc III, Calc IV, and Differential Equations.
This is in an honors engineering environment at a fairly prestigious school.
brandbest1
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Somebody tell me what a differential equation is in an easy-to-understand-not-nerdy explanation?
Sa967St
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Well, you need to know some basic calculus to easily understand what they are.
A differential equation is an equation containing independent and dependent variables, as well as derivatives of the dependent variable.
e.g. dy/dx =2xy
Last edited:
vcuber13
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isn't that partial derivatives?
Sa967St
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MTGjumper
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In the case Sarah gave, I can only assume y := y(x), just to make it more clear.
