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# 7: IIT level solved IV
We are dealing with the concept of SERIES here. It is quite simple actually. Instead of writing many equations in sequence, we write one i.e. x=r where r takes many values i.e. r= 1,2,3…. You write x=r simply but it means many straight lines.
As given each of these lines parallel to the y axis cuts a line, L passing through the Origin namely y=(sqrt 5)*x and thus the points where these numerous lines cuts this line L has been tabulated. The answer is in the form of a series too as we have shown int the diagram. Thereafter to find the sum of squares of the distances from Origin to these points we end up with a Sum of a Series and we use a simple formula to get the answer.
This is a very interesting question. Two random lines are taken, they cut the axes in 4 points as usual.
The additional fact is that these four points are concyclic. That is what we have to use.
So, I drew the two lines and drew the circle. We have to use the property of the circle. That is obvious.
Suddenly it flashes as you look at the circle and the two lines. Like AB & CD, AC & BD are also two chords of the circle and we have a rule that if two chords cut each other, the product of the segments of the chords are equal. THAT gives us an equation which helps us solve the problem.
A Problem on Locus (below)
Any line through the Origin is y=mx and that cuts the other two parallel lines given. So by equating the lines we obtain points A and B and thus we also find OA & OB.
As a technique to find the locus of a Point P we take tis coordinates temporarily as (h,k) then we establish the relationship that is given or sometimes, like in our previous example, from a geometrical property that emerges from the diagram given. Finally, once we use that property and do the algebra we end up with a relationship between h and k. It is then that we replace h by x and k by y and obtain the final locus.
Algebra is used to find unknown quantities. Look at the last sentence in the question. It says find the slope i.e. find m. So, we require an equation with m. So, we look at the information given:
a line passes through (1,2) and its slope is m. And the rest of the information we use to build up the equation containing m and thus find the same.
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### Course: Algebra 2>Unit 3
Lesson 2: Greatest common factor
# Greatest common factor of monomials
Follow along as Sal finds the greatest common factor of 10cd^2 and 25c^3d^2 and discover the secret to finding the greatest common factor of monomials! Dive into prime factorization and variable parts, and learn how to break down monomials into their simplest forms. Uncover the common factors and master the art of algebraic expressions. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• how is are the monomials used in life?
• We use monomials all the time, particularly when counting things. 5 gallons, 12 eggs, 20 dollars, 3 kids; all of these are monomials, or single term expressions where we are multiplying a constant by a variable. An example of a simple binomial equation could be for a classroom made up of 10 boys and 9 girls which we can express as something like:
c = 10b + 9g
The funny part is that we all use a fair bit of algebra on a regular basis without even knowing it because algebra is mostly just common sense calculations boiled down into pure logic.
• what dos monomial mean?
• it is a one-term polynomial (think of it as a constant multiplied by a variable (x) raised to some power)
• Just wondering, how do you calculate the greatest common factor of, say 56?
• The greatest common factor deals with two expressions, not one, so 56 doesn't have a GCF by itself because there's not another number to compare it to. If you're talking about breaking a constant expression down into its constituent factors, all you'd have to do is find all the numbers you can multiply by to get 56. Those would be 2(28), 4(14), and 8(7).
• what to do if a number is negative? for example -6t + 9t.
• To find the GCF you can ignore the sign. For -6t + 9t the GCF is 3.
• At , Sal Khan said "But I'll put that in quotes depending on whether c is negative or positive and d is greater than or less than 0". I think he made a mistake in this part, although I think it will also be helpful to others if I talk about why this is a mistake. The true greatest common factor does not depend on whether d is less than or equal to zero, as (-a)^2=(a)^2, as Sal Khan said, but rather on whether the absolute value of d is less than 1, in which case the absolute value of the entire monomial will decrease as x increases in d^x. For example, if d=1/3, then d^3 would be less than d^4, as d^3=1/27, and d^4=1/81. Now, if |d| is greater than 1, as x's value increases, it is true that the absolute value of the monomial 5cd^x's will increase, provided c and d are both non-zero numbers and |d| is not equal to 1. However, this does not translate to "If |d| is greater than 1 then as x's value increases, the value of 5cd^x will increase". This may or may not be true under certain circumstances. If c is positive, then yes, the value of 5cd^x will increase when x's value increases. However, if c is negative, the value of 5cd^x will only decrease when x's value increases. Now, using this knowledge, to know whether if 5cd^2 is truly greater than 5c by itself, requires knowledge of if c is negative or positive, and if |d| is less than 1. So, we have 4 different cases. I will use + for positive, - for negative, > for the absolute value is greater than 1, and < for the absolute value is less than 1. Increasing means that the value of the monomial increases from 5c to 5cd^2. (c+,d<)-Decreasing. (c-,d<)-Increasing. (c+,d>)-Increasing. (c+,d<)-Decreasing. Although Khan did say the part about c correctly, that the value of the monomial depended on whether c was negative or positive, I do believe that he meant to say "and d is greater than or less than 1" instead of "and d is greater than or less than 0".
• Well...What's the difference between a polynomial and a monomial?
• Also, polynomials can have any number of terms, as long as it is more than one.
• Is there any other way like Saxon's technique?
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# If Ben were to lose the championship (m07q12)
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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05 Nov 2013, 07:53
honchos wrote:
Bunuel wrote:
gurpreetsingh wrote:
I got it..
1/4 = 6/7 * m* (1-r) where m and r are prob of mike and rob winning.
1/3 = 6/7 * (1-m) * r
should be
1/4 = 1* m* (1-r) where m and r are prob of mike and rob winning.
1/3 = 1* (1-m) * r
Bunuel m I correct?
If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$ , and Rob - $$\frac{1}{3}$$ . If the probability of Ben being the winner is $$\frac{1}{7}$$ , what is the probability that either Mike or Rob will win the championship?
A. $$\frac{1}{12}$$
B. $$\frac{1}{7}$$
C. $$\frac{1}{2}$$
D. $$\frac{7}{12}$$
E. $$\frac{6}{7}$$
This is conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must loose --> the probability of Ben loosing is $$1-\frac{1}{7}=\frac{6}{7}$$.
Now out of these $$\frac{6}{7}$$ cases the probability of Mike winning is $$\frac{1}{4}$$ and the probability of Rob winning is $$\frac{1}{3}$$. So $$P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}$$.
Or consider the following:
Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).
Now, out of these 84 Ben will loose in $$\frac{6}{7}*84=72$$. Mike would be the winner in $$72*\frac{1}{4}=18$$ (1/4 th of the cases when Ben loose) and Rob would be the winner $$72*\frac{1}{3}=24$$ --> $$P=\frac{18+24}{84}=\frac{1}{2}$$.
Hope it's clear.
Out of 84 only 24 and 18 are won = 42, How come?
Sorry, I don't get your question at all... Please elaborate. Thank you.
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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16 Nov 2013, 00:43
voodoochild wrote:
Bunuel wrote:
gurpreetsingh wrote:
I got it..
1/4 = 6/7 * m* (1-r) where m and r are prob of mike and rob winning.
1/3 = 6/7 * (1-m) * r
should be
1/4 = 1* m* (1-r) where m and r are prob of mike and rob winning.
1/3 = 1* (1-m) * r
Bunuel m I correct?
If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$ , and Rob - $$\frac{1}{3}$$ . If the probability of Ben being the winner is $$\frac{1}{7}$$ , what is the probability that either Mike or Rob will win the championship?
A. $$\frac{1}{12}$$
B. $$\frac{1}{7}$$
C. $$\frac{1}{2}$$
D. $$\frac{7}{12}$$
E. $$\frac{6}{7}$$
This is conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must loose --> the probability of Ben loosing is $$1-\frac{1}{7}=\frac{6}{7}$$.
Now out of these $$\frac{6}{7}$$ cases the probability of Mike winning is $$\frac{1}{4}$$ and the probability of Rob winning is $$\frac{1}{3}$$. So $$P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}$$.
Or consider the following:
Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).
Now, out of these 84 Ben will loose in $$\frac{6}{7}*84=72$$. Mike would be the winner in $$72*\frac{1}{4}=18$$ (1/4 th of the cases when Ben loose) and Rob would be the winner $$72*\frac{1}{3}=24$$ --> $$P=\frac{18+24}{84}=\frac{1}{2}$$.
Hope it's clear.
Bunuel - I have a question -
In your explanation, Out of 84 people, Ben will lose in 12 cases. Therefore, we have 72 possibilities open for either Mike or Rob.
Now out of 72 cases, Mike would win in 18 and 24 cases respectively. My question is that what would happen to the remaining 84-12-18-24 = 30 cases? I am not sure whether I understood the break-up correctly.
Thanks
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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16 Nov 2013, 00:45
2
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Probability tree is very useful to solve this kind of question. C is the answer.
Here is my diagram.
Hope it helps
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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16 Nov 2013, 00:57
pqhai wrote:
Probability tree is very useful to solve this kind of question. C is the answer.
Here is my diagram.
Hope it helps
This is one of the very amazing way of solving, But i am still trying to grab the logic behinds Bunel's solution-
Bunuel's approach-
Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).
My Doubt:
Out of 84 only 24 and 18(1/4th and 1/3rd) are won = 42, How come?
Even if we add 1/7 X 84 = 12. it becomes 54, what about other matches, is this question correct?
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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16 Nov 2013, 01:23
honchos wrote:
pqhai wrote:
Probability tree is very useful to solve this kind of question. C is the answer.
Here is my diagram.
Hope it helps
This is one of the very amazing way of solving, But i am still trying to grab the logic behinds Bunel's solution-
Bunuel's approach-
Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).
My Doubt:
Out of 84 only 24 and 18(1/4th and 1/3rd) are won = 42, How come?
Even if we add 1/7 X 84 = 12. it becomes 54, what about other matches, is this question correct?
Hello honchos.
The logic behind Bunuel's approach is that if you have two variables having own probabilities
==> The probability of happening BOTH two variables = (Probability of variable 1) x (Probability of variable 2)
Let apply the logic above to the question:
Mike win: Mike wins ONLY IF Ben loses ==> The probability of Mike wins MUST be = (Probability of Ben loses) x(probability of Mike wins) = (84*6/7)*1/4 = 72*1/4 = 18
Rob win: The same pattern is true for Rob ==> Probability of Rob wins = (Probability of Ben loses) x(probability of Rob wins) = (84*6/7)*1/3 = 72*1/4 = 24
==> Total = 42 ==> Probability of either Mike or Rob win = 42/84 = 1/2
About your second question. Why 84*1/7 + 42 is only 54. As you thought, it SHOULD be 84?
No, 84 = [probability of Ben lose WITHOUT the impact of Mike and Rob] + Probability of Ben win = 84*6/7 + 84*1/7
Hope it's clear.
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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16 Nov 2013, 01:25
I d not agree with you-
"No, 84 = [probability of Ben lose WITHOUT the impact of Mike and Rob] + Probability of Ben win = 84*6/7 + 84*1/7 "
84 is the total #(Sample Space) of matches thats the reason we chose 6/7 and 1/7 to calculate.
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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16 Nov 2013, 01:34
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honchos wrote:
I d not agree with you-
"No, 84 = [probability of Ben lose WITHOUT the impact of Mike and Rob] + Probability of Ben win = 84*6/7 + 84*1/7 "
84 is the total #(Sample Space) of matches thats the reason we chose 6/7 and 1/7 to calculate.
I think you may misunderstand the question a bit. You assume there are only three people Ben, Mike and Rob. So, if Ben loses, either Mike or Rob is automatically a winner? If that's the case, why probability of Mike's win and that of Rob are 1/4 and 1/3 respectively?
Do not assume that Ben loses = Mike or Rob is automatically a winner. Because if that's the case ==> The probability of Mike or Rob win is automatically = 6/7 = probability of Ben loses. That's wrong.
Hope it makes sense.
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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16 Nov 2013, 02:48
I have noted, and don't want to do much research as my exam is too close on December 10 2013.
Thx.
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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17 Nov 2013, 02:29
Why the probability for Rob is -1/3?Is this a typo?
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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17 Nov 2013, 02:36
sunita123 wrote:
Why the probability for Rob is -1/3?Is this a typo?
This information is furnished in the question stem.
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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17 Nov 2013, 03:09
honchos wrote:
sunita123 wrote:
Why the probability for Rob is -1/3?Is this a typo?
This information is furnished in the question stem.
In question ,it says probability -1/3.Am i missing anything?
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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22 Nov 2013, 14:27
Does conditional probability figure out in GMAT syllabus?
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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22 Nov 2013, 14:36
Sunita123, it is dash 1/3 and not -1/3. The probability of happening or non happening of an event lies between 0 and 1 both inclusive. So, probability can NEVER be negative or greater than 1.
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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09 May 2014, 04:21
Bunuel wrote:
gurpreetsingh wrote:
I got it..
1/4 = 6/7 * m* (1-r) where m and r are prob of mike and rob winning.
1/3 = 6/7 * (1-m) * r
should be
1/4 = 1* m* (1-r) where m and r are prob of mike and rob winning.
1/3 = 1* (1-m) * r
Bunuel m I correct?
If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$ , and Rob - $$\frac{1}{3}$$ . If the probability of Ben being the winner is $$\frac{1}{7}$$ , what is the probability that either Mike or Rob will win the championship?
A. $$\frac{1}{12}$$
B. $$\frac{1}{7}$$
C. $$\frac{1}{2}$$
D. $$\frac{7}{12}$$
E. $$\frac{6}{7}$$
This is conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must loose --> the probability of Ben loosing is $$1-\frac{1}{7}=\frac{6}{7}$$.
Now out of these $$\frac{6}{7}$$ cases the probability of Mike winning is $$\frac{1}{4}$$ and the probability of Rob winning is $$\frac{1}{3}$$. So $$P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}$$.
Or consider the following:
Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).
Now, out of these 84 Ben will loose in $$\frac{6}{7}*84=72$$. Mike would be the winner in $$72*\frac{1}{4}=18$$ (1/4 th of the cases when Ben loose) and Rob would be the winner $$72*\frac{1}{3}=24$$ --> $$P=\frac{18+24}{84}=\frac{1}{2}$$.
Hope it's clear.
Bunuel It is no where mentioned that events are mutually exclusive or noe, whats wrong in this method:
6/7 X 1/4 X 2/3 + 6/7 X 3/4X1/3
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Re: If Ben were to lose the championship (m07q12) [#permalink]
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09 May 2014, 05:02
honchos wrote:
Bunuel wrote:
gurpreetsingh wrote:
I got it..
1/4 = 6/7 * m* (1-r) where m and r are prob of mike and rob winning.
1/3 = 6/7 * (1-m) * r
should be
1/4 = 1* m* (1-r) where m and r are prob of mike and rob winning.
1/3 = 1* (1-m) * r
Bunuel m I correct?
If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$ , and Rob - $$\frac{1}{3}$$ . If the probability of Ben being the winner is $$\frac{1}{7}$$ , what is the probability that either Mike or Rob will win the championship?
A. $$\frac{1}{12}$$
B. $$\frac{1}{7}$$
C. $$\frac{1}{2}$$
D. $$\frac{7}{12}$$
E. $$\frac{6}{7}$$
This is conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must loose --> the probability of Ben loosing is $$1-\frac{1}{7}=\frac{6}{7}$$.
Now out of these $$\frac{6}{7}$$ cases the probability of Mike winning is $$\frac{1}{4}$$ and the probability of Rob winning is $$\frac{1}{3}$$. So $$P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}$$.
Or consider the following:
Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).
Now, out of these 84 Ben will loose in $$\frac{6}{7}*84=72$$. Mike would be the winner in $$72*\frac{1}{4}=18$$ (1/4 th of the cases when Ben loose) and Rob would be the winner $$72*\frac{1}{3}=24$$ --> $$P=\frac{18+24}{84}=\frac{1}{2}$$.
Hope it's clear.
Bunuel It is no where mentioned that events are mutually exclusive or noe, whats wrong in this method:
6/7 X 1/4 X 2/3 + 6/7 X 3/4 X1/3
Mike winning automatically means Rob loosing. Similarly Rob winning automatically means Mike loosing. So, the assumption is that only one person can win the championship.
Mike would be the winner with a probability of $$\frac{1}{4}$$ IF Ben were to lose: 1/4*6/7.
Rob would be the winner with a probability of $$\frac{1}{3}$$ IF Ben were to lose: 1/3*6/7.
So, no need to multiply above by the probability of Rob/Mike losing too.
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Re: If Ben were to lose the championship (m07q12) [#permalink] 09 May 2014, 05:02
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# If Ben were to lose the championship (m07q12)
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# A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.
Question:
A point on the hypotenuse of a triangle is at distance and b from the sides of the triangle.
Show that the minimum length of the hypotenuse is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$
Solution:
Let ΔABC be right-angled at B. Let AB = x and BC = y.
Let P be a point on the hypotenuse of the triangle such that P is at a distance of and b from the sides AB and BC respectively.
Let C = θ.
We have,
$\mathrm{AC}=\sqrt{x^{2}+y^{2}}$
Now,
PC = cosec θ
And, AP = a sec θ
AC = AP + PC
$\Rightarrow \mathrm{AC}=b \operatorname{cosec} \theta+a \sec \theta \ldots$ (1)
$\therefore \frac{d(\mathrm{AC})}{d \theta}=-b \operatorname{cosec} \theta \cot \theta+a \sec \theta \tan \theta$
$\therefore \frac{d(\mathrm{AC})}{d \theta}=0$
$\Rightarrow a \sec \theta \tan \theta=b \operatorname{cosec} \theta \cot \theta$
$\Rightarrow \frac{a}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta}=\frac{b}{\sin \theta} \frac{\cos \theta}{\sin \theta}$
$\Rightarrow a \sin ^{3} \theta=b \cos ^{3} \theta$
$\Rightarrow(a)^{\frac{1}{3}} \sin \theta=(b)^{\frac{1}{3}} \cos \theta$
$\Rightarrow \tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}$
$\therefore \sin \theta=\frac{(b)^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}$ and $\cos \theta=\frac{(a)^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}$ ....(2)
It can be clearly shown that $\frac{d^{2}(\mathrm{AC})}{d \theta^{2}}<0$ when $\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}$.
Therefore, by second derivative test, the length of the hypotenuse is the maximum when $\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}$.
Now, when $\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}$, we have:
$\mathrm{AC}=\frac{b \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{1}{3}}}+\frac{a \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}} \quad[$ Using (1) and (2) $]$
$=\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}\left(b^{\frac{2}{3}}+a^{\frac{2}{3}}\right)$
$=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$
Hence, the maximum length of the hypotenuses is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$.
|
Free Mensuration 02 Practice Test - 6th grade
Perimeter is the distance covered along the boundary forming a/an________ when you go round the figure once.
A.
open figure
B. open curve
C.
closed figure
D.
line
SOLUTION
Solution : C
The length of the boundary of a closed figure is called its perimeter.
Hence, perimeter is the distance covered along the boundary forming a closed figure when you go round the figure once.
Perimeter is not valid for
A.
irregular and closed figures
B.
all closed figures
C.
closed regular figures
D.
open figures
SOLUTION
Solution : D
Perimeter is basically the total length of the boundary of a figure. This is valid for closed figures only (both regular and irregular).
Perimeter of any figure is
A.
Sum of lengths of sides
B.
Product of length of all sides
C.
Square of lengths of all sides
D.
None of these
SOLUTION
Solution : A
The perimeter of a figure is the sum of lengths of all sides. In other words, perimeter is the distance covered along the boundary of the figure when you go round the figure once.
A rectangle has a length of 10 inches and breadth of 5 inches. The perimeter of the rectangle will be
A.
30 square inches
B.
30 inches
C.
15 square inches
D.
15 inches
SOLUTION
Solution : B
Perimeter of a rectangle
= 2 × (Length + Breadth)
= 2 × ( 10 inches + 5 inches )
= 30 inches.
The perimeter of a regular octagon with each side measuring 15 cm is
A.
275 cm
B.
100 cm
C.
120 cm
D.
None of above
SOLUTION
Solution : C
Perimeter = Sum of all the sides
Since it is a regular octagon, all sides will be equal.
Perimeter = 8 × length of a side
= 8 × 15 cm
= 120 cm
The area of a square of side 14 cm will be:
A.
42 cm2
B.
28 cm2
C.
196 cm2
D.
56 cm2
SOLUTION
Solution : C
Area of square = Side × Side
= 14 × 14
= 196 cm2
The perimeter of a square of side 15 cm will be:
A.
90 cm
B.
60 cm
C.
225 cm
D.
34 cm
SOLUTION
Solution : B
Perimeter of square = 4 × length of side
= 4 × 15
= 60 cm
Karan wants to cover the floor of his room with lavish tiles. His room is square in shape with sides of 6 m. The shopkeeper told him that the area of each tile is 0.5 m2. Can you tell how many tiles Karan should buy?
A.
108
B.
36
C.
72
D.
144
SOLUTION
Solution : C
Given a square room of side = 6 m.
Area of each tile = 0.5 m2
We know that, area of a square = Side × Side
Area of a square room
= 6 × 6 = 36 m2.
So, number of tiles required
= Area of roomArea of a single tile
= 360.5 = 72
Hence, Karan should buy 72 tiles to cover the floor of his room.
Find the cost of ploughing a rectangular farm at the rate of ₹ 12/m2 , having length equal to 12 m and width equal to 5 m.
A.
₹ 120
B.
₹ 720
C.
₹ 360
D.
₹ 240
SOLUTION
Solution : B
Area of rectangular field
= 12 × 5
= 60 m2
Cost of ploughing
= Total area × cost per sq. metre
= 60×12
=
₹ 720
The area of given figure is:
A. 41 sq. units
B. 42 sq. units
C. 24 sq. units
D. 40 sq. units
SOLUTION
Solution : B
Area of the figure = Area of A + Area of B + Area of C
|
# Kindergarten Math 2nd grading period
```Kindergarten
Mathematics
Instructional Sequence
Second Nine Weeks
Number and Operations in Base Ten “Building Numbers”
Work with numbers 11–19 to gain foundations for place value.
MGSEK.NBT.1
Former: Compose and decompose numbers from 11 to 19 into ten ones and some further ones,
e.g., by using objects or drawings, and record each composition or decomposition by a drawing or
equation (e.g., 18 = 10 + 8); understand that these numbers are composed of ten ones and one,
two, three, four, five, six, seven, eight, or nine ones.
Revised for 2015-2016:
Compose and decompose numbers from 11 to 19 into ten ones and some further ones to
understand that these numbers are composed of ten ones and one, two, three, four, five, six ,
seven, eight, or nine ones, e.g., by using objects or drawings, and record each composition or
decomposition by a drawing or equation (e.g., 18 = 10 + 8)
Counting and Cardinality
Know number names and the count sequence.
MGSEK.CC.3 Write numbers from 0 to 20. Represent a number of objects with a written numeral
0-20 (with 0 representing a count of no objects).
Count to tell the number of objects.
MGSEK.CC.4 Understand the relationship between numbers and quantities; connect
counting to cardinality.
a. When counting objects, say the number names in the standard order, pairing each object
with one and only one number name and each number name with one and only one object.
MGSEK.CC.5
Former: Count to answer “how many?” questions about as many as 20 things arranged in a line, a
rectangular array, or a circle, or as many as 10 things in a scattered configuration; given a number
from 1–20, count out that many objects.
Revised for 2015-2016:
Count to answer ‘how many?” questions.
a. Count to answer “how many?” questions about as many as 20 things arranged in a variety of ways
(a line, a rectangular array, or a circle), or as many as 10 things in a scattered configuration.
b. Given a number from 1-20, count out that many objects.
c. Identify and be able to count pennies within 20. (Use pennies as manipulatives in multiple
mathematical contexts.)
Compare numbers.
MGSEK.CC.6 Identify whether the number of objects in one group is greater than, less than, or equal
to the number of objects in another group, e.g., by using matching and counting strategies.
MGSEK.CC.7 Compare two numbers between 1 and 10 presented as written numerals.
Measurement and Data “Measuring and Analyzing Data”
Classify objects and count the number of objects in each category.
MGSEK.MD.3 Classify objects into given categories; count the numbers of objects in each category
and sort the categories by count.
Measurement and Data
Describe and compare measurable attributes.
MGSEK.MD.1
Former: Describe measurable attributes of objects, such as length or weight. Describe several
measurable attributes of a single object.
Revised for 2015-2016:
Describe several measurable attributes of an object, such as length or weight. For
example, a student may describe a shoe as, “This shoe is heavy! It is also really long!”
MGSEK.MD.2 Directly compare two objects with a measurable attribute in common, to see
which object has “more of”/“less of” the attribute, and describe the difference. For example,
directly compare the heights of two children and describe one child as taller/shorter.
```
##### Related flashcards
Complex analysis
23 Cards
Special functions
14 Cards
Numeral systems
15 Cards
|
Powers and Roots Powers and Roots Operations with powers. Multiplication and division of powers. Power of product of some factors. Power of a quotient (fraction). Raising of power to a power. Operations with roots. Arithmetical root. Root of product of some factors. Root of quotient (fraction). Raising of root to a power. Proportional change of degrees of a root and its radicand. Negative, zero and fractional exponents of a power. About meaningless expressions. Operations with powers. 1. At multiplying of powers with the same base their exponents are added: a m · a n = a m + n . 2. At dividing of powers with the same base their exponents are subtracted: 3. A power of product of two or some factors is equal to a product of powers of these factors: ( abc… ) n = a n · b n · c n … 4. A power of a quotient (fraction) is equal to a quotient of powers of a dividend (numerator) and a divisor (denominator): ( a / b ) n = a n / b n . 5. At raising of a power to a power their exponents are multiplied: ( a m ) n = a m n . Know More About :- Adding Polynomials
Math.Edurite.com
Page : 1/3
All above mentioned formulas are read and executed in both directions – from the left to the right and back. E x a m p l e . ( 2 · 3 · 5 / 15 ) 2 = 2 2 · 3 2 · 5 2 / 15 2 = 900 / 225 = 4 . Operations with roots. In all below mentioned formulas a symbol means an arithmetical root ( all radicands are considered here only positive ). 1. A root of product of some factors is equal to a product of roots of these factors: 2. A root of a quotient is equal to a quotient of roots of a dividend and a divisor: 3. At raising a root to a power it is sufficient to raise a radicand to this power: 4. If to increase a degree of a root by n times and to raise simultaneously its power, the root value doesn’t change:
5. If to decrease a degree of a root by n times and to extract simultaneously the n-th degree root of the radicand, the root value doesn’t change: Widening of the power notion. Till now we considered only natural exponents of powers; but operations with powers and roots can result also to negative, zero and fractional exponents. All these exponents of powers require to be defined. Negative exponent of a power. A power of some number with a negative (integer) exponent is defined as unit divided by the power of the same number with the exponent equal to an absolute value of the negative exponent. Now the formula a m : a n = a m - n may be used not only if m is more than n , but also for a case if m is less than n . E x a m p l e . a4 : a7 = a4 - 7 = a-3 . If we want the formula a m : a n = a m - n to be valid at m = n we need the definition of zero exponent of a power. Zero exponent of a power. A power of any nonzero number with zero exponent is equal to 1. E x a m p l e s . 2 0 = 1, ( – 5 ) 0 = 1, ( – 3 / 5 ) 0 = 1. Fractional exponent of a power. To raise a real number a to a power with an exponent m / n it is necessary to extract the n-th degree root from the m-th power of this number a. Read More About :- Subtracting Polynomials
Math.Edurite.com
Page : 2/3
Thank You
Math.Edurite.Com
Powers and Roots
|
If you find any mistakes, please make a comment! Thank you.
## Compute the order of 5 in the integers mod a power of 2
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.22
Let $n$ be an integer with $n \geq 3$. Use the Binomial Theorem to show that $$(1+2^2)^{2^{n-2}} = 1 \pmod {2^n}$$ but that $$(1+2^2)^{2^{n-3}} \neq 1 \pmod {2^n}.$$ Deduce that 5 is an element of order $2^{n-2}$ in the group $(\mathbb{Z}/(2^n))^\times$.
Solution: We begin with some lemmas.
Lemma 1: Let $k$ be a positive integer and write $k = 2^m \cdot a$, where 2 does not divide $a$. Then $m+2 \leq 2k$.
Proof: If $k = 1$, we have $m=0$. Then $0 + 2 \leq 2 \cdot 1$, so the conclusion holds for $k=1$. If $k \geq 2$, we have $m < k$ by a lemma to a previous exercise; then $m+2 < k+k = 2k$. Thus the lemma holds for $k \geq 2$. $\square$
Lemma 2: Let $k$ be a positive integer with $k \geq 2$, and write $k = 2^m \cdot a$ where 2 does not divide $a$. Then $m+3 \leq 2k$.
Proof: If $k = 2$, we have $m = 1$. Then $1+3 \leq 2 \cdot 2$, so the conclusion holds. If $k \geq 3$, we have $m < k$ by a lemma to the previous exercise. Now $m+3 < k+k = 2k$, so the conclusion holds for all $k \geq 3$. $\square$
Now to the main result.
By the Binomial Theorem, we have $$(1+2^2)^{2^{n-2}} = \displaystyle\sum_{k=0}^{2^{n-2}} {2^{n-2} \choose k} \cdot 2^{2k}.$$Now by a lemma to the previous problem, if $2^{m_k}$ divides $k$, then $2^{n-m-2}$ divides $p^{n-2} \choose k$. Thus we have $$(1+2^2)^{2^{n-2}} = \displaystyle\sum_{k=0}^{2^{n-2}} 2^{n+2k-m_k-2} \cdot \beta_k,$$ for some integers $\beta_k$. Now by Lemma 1, $2k \geq m_k+2$ for all $k \geq 1$, so that $n+2k-m_k-2 \geq n$. Modulo $2^n$, all terms with $k \geq 1$ are zero, leaving only the $k=0$ term; thus we have $$(1+2^2)^{2^{n-2}} \equiv 1\ \mathsf{mod}\ 2^n.$$ We now consider $(1+2^2)^{2^{n-3}}$. Again by the Binomial Theorem, $$(1+2^2)^{2^{n-3}} = \displaystyle\sum_{k=0}^{2^{n-3}} {2^{n-3} \choose k} \cdot 2^{2k},$$ and using a lemma to the previous problem, if $2^{m_k}$ divides $k$ times, we have (for some integers $\beta_k$) $$(1+2^2)^{2^{n-3}} = \displaystyle\sum_{k=0}^{2^{n-3}} 2^{n+2k-m_k-3} \cdot \beta_k.$$ By Lemma 2, for all $k \geq 2$ we have $n+2k-m_k-3 \geq n$, so that, modulo $2^n$, these terms vanish. This leaves only the $k=0$ and $k=1$ terms. Thus, $$(1+2^2)^{2^{n-3}} = 1 + 2^{n-1},$$
which is not $1\pmod {2^n}$.
Thus, $|\bar 5|$ divides $2^{n-2}$ but does not divide $2^{n-3}$, so that we have $|\bar 5| = 2^{n-2}$ in $(\mathbb{Z}/(2^n))^\times$.
|
## Spinning Tops
• Lesson
• 1
• 2
6-8
1
Students measure distances using standard and nonstandard units and record their measurement in various tables. Then they are asked to use descriptive statistics to report the results. During a top-spinning contest, students measure the distance along a curve using indirect measurement. They record the data for their group in a chart, and compute their individual median and the group median.
Guide the students as they decorate tops made by pushing a writing pen through a plastic cover, such as those from yogurt and margarine containers.
Distribute a large sheet of white paper to each group, and demonstrate a spin on the top and how to "measure" the distance—with yarn or string, follow the path of the pen, and then measure the string with a centimeter ruler.
Distribute the Spinning Tops Activity Sheet to each student.
Guide the students as they record the data in their charts and compute the medians. Students should remember that the median is found by placing the data in order (from least to greatest, for example) and identifying the middle number. If there are two middle numbers, the median is the average of the two.
Encourage discussion of the tables created.
### Reference
Silverman, Helene. "IDEAS: Games, Measurement, and Statistics." The Arithmetic Teacher. April, 1990, pp. 27-32.
• Spinning Tops Activity Sheet
• Small Circular Plastic Tops (1 per student)
• Felt tip pens
• Large Sheets of White Paper
• Skeins of Yarn (1 per group)
• Scissors
Extension
Demonstrate how to make a stem-and-leaf plot with the data from all the groups. Guide the students as they display and analyze their position within the table, as well as the general tendency of their group.
none
### The Celebrated Jumping Frog
6-8
Using the story "The Celebrated Jumping Frog of Calveras County" by Mark Twain, students simulate a jumping-frog contest and determine the distances "jumped." The students record the distance of individual jumps in centimeters and determine the total distance jumped (the sum of the three separate jumps) and the official distance (the straight-line distance from the starting line to the end of the frog's third jump). The students compare the range and median of the total distances with those of the official distances of the group.
### Learning Objectives
Students will:
• Measure the distance along a curve using indirect measurement.
• Record data in chart form.
• Compute individual and group medians.
• Analyze data.
### NCTM Standards and Expectations
• Use common benchmarks to select appropriate methods for estimating measurements.
• Find, use, and interpret measures of center and spread, including mean and interquartile range.
|
# Lesson Worksheet: Double Line Graphs Mathematics
In this worksheet, we will practice analyzing and comparing two related sets of data presented in a double line graph.
Q1:
The following line graphs compare the average number of first graders in two different schools for 10 years. The orange line represents school 1, and the blue line represents school 2.
Which school had a higher capacity in the year 2010?
• ASchool 2
• BNeither
• CSchool 1
Q2:
In the diagram, the orange line represents the number of burgers sold in a restaurant in a particular week and the blue line represents the number of hot dogs sold by the restaurant in the same week.
What is the total number of sandwiches sold on Saturday ?
Q3:
The following line graphs show the number of members of each gender in a gym over several months. The orange line represents the males, and the blue line represents the females.
What is the total number of members in January?
Q4:
This graph shows the number of goals scored by two players in the last ten weekly games. The orange line represents player 1, and the blue line represents player 2.
Which player scored more goals in week 1 and by how many?
• APlayer 1, 1 goal
• BPlayer 2, 1 goal
• CPlayer 1, 3 goals
• DPlayer 1, 4 goals
• EPlayer 2, 2 goals
Q5:
This graph shows the number of students who got an A grade in two classes over 7 years. The orange line represents class 1, and the blue line represents class 2.
Which class has more students who achieved an A grade in the year 2013 and by how many?
• AClass 2, 8 students
• BClass 1, 5 students
• CClass 2, 5 students
• DClass 2, 3 students
• EClass 1, 3 students
Q6:
This graph shows the number of soccer games that two teams played in a year.
What is the difference between the number of games played by the two teams in February?
• A3 games
• B2 games
• C1 game
• D6 games
• E4 games
Q7:
The graph shows the Internet usage of two students over 5 days.
Which student used the Internet the most?
• AStudent A
• BStudent B
Q8:
Look at the double line graph to find the factory that made more sales in a period of 5 months.
• AFactory A
• BFactory B
Q9:
The following table shows the average monthly temperature in two different regions for a certain year.
MonthTemperature in Region ATemperature in Region B
January
February
March
April
May
June
July
August
September
October
November
December
Which of the following graphs shows the given data?
• A
• B
• C
• D
• E
Q10:
The double line graph shows the number of hours that Benjamin and Chloe studied over a week.
On which day did they study for the same number of hours?
• AWednesday
• BTuesday
• CSunday
• DMonday
• EThursday
### Practice Means Progress
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ISEE Middle Level Quantitative : How to multiply
Example Questions
1 2 8 9 10 11 12 13 14 16 Next →
Example Question #151 : Whole Numbers
Solve:
Explanation:
In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.
For this problem, 34 is the multiplier and 86 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.
First, we multiply 4 and 6
Then, we multiply 4 and 8 and add the 2 that was carried
Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.
Next, we multiply 3 and 6
Then, we multiply 3 and 8and add the 1 that was carried
Finally, we add the two products together to find our final answer
Example Question #152 : Whole Numbers
Solve:
Explanation:
In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.
For this problem, 33 is the multiplier and 67 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.
First, we multiply 3 and 7
Then, we multiply 3 and 6 and add the 2 that was carried
Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.
Next, we multiply 3 and 7
Then, we multiply 3 and 6and add the 2 that was carried
Finally, we add the two products together to find our final answer
Example Question #153 : Whole Numbers
Solve:
Explanation:
In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.
For this problem, 72 is the multiplier and 60 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.
First, we multiply 2 and 0
Then, we multiply 2 and 6
Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.
Next, we multiply 7 and 0
Then, we multiply 7 and 6
Finally, we add the two products together to find our final answer
Example Question #154 : Whole Numbers
Solve:
Explanation:
In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.
For this problem, 67 is the multiplier and 61 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.
First, we multiply 7 and 1
Then, we multiply 7 and 6
Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.
Next, we multiply 6 and 1
Then, we multiply 6 and 6
Finally, we add the two products together to find our final answer
Example Question #155 : Whole Numbers
Solve:
Explanation:
In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.
For this problem, 41 is the multiplier and 67 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.
First, we multiply 1 and 7
Then, we multiply 1 and 6
Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.
Next, we multiply 4 and 7
Then, we multiply 4 and 6and add the 2 that was carried
Finally, we add the two products together to find our final answer
Example Question #156 : Whole Numbers
Solve:
Explanation:
In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.
For this problem, 23 is the multiplier and 84 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.
First, we multiply 3 and 4
Then, we multiply 3 and 8 and add the 1 that was carried
Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.
Next, we multiply 2 and 4
Then, we multiply 2 and 8
Finally, we add the two products together to find our final answer
1 2 8 9 10 11 12 13 14 16 Next →
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maths > wholedivisors
Factors, Multiples, Prime Factorization
what you'll learn...
overview
This page provides a brief overview of factors of a number. multiplies of a number. prime factorization of numbers
Factors
When the remainder is 0$0$, a dividend is "divisible" by a divisor.
The divisors that divide, with remainder 0$0$, are called factors of the dividend.
eg: Factors of 6$6$ are 1$1$, 2$2$, 3$3$, and $6$. Because the remainder is $0$ for $6 \div 1$, $6 \div 2$, $6 \div 3$ and $6 \div 6$
eg: Factors of $20$ are $1$, $2$, $4$, $5$, $10$, and $20$
eg: Factors of $13$ are $1$ and $13$
The word "factor" means "a part or component of something bigger".
Factors of a number : The divisor that divides with remainder $0$, is a factor of the dividend.
the dividend, divisor are non-zero whole numbers.
The factors of $26$ are "$1 , 2 , 13 , 26$".
The factors of $28$? are "$1 , 2 , 4 , 7 , 14 , 28$".
Multiples
Consider the number $6$. The following shows result of multiplying $6$ by the sequence of numbers $1 , 2 , 3 , 4 , 5 , \cdots$
"$6$, $12$, $18$, $24$, $30$, $\cdots$ "
A dividend that has another number as factor is called multiple of the factor.
eg: $24$ is a multiple of $3$, as $24 \div 3 = 8$ with remainder $0$
eg: $24$ is also a multiple of $2$
eg: $24$ is also a multiple of $6$
eg: $24$ is also a multiple of $24$
The word "multiple" means "having several parts of something".
Multiples of a number : The dividend that is divided by a factor with remainder $0$, is a multiple of the factor.
The multiples of $6$ are "$6 , 12 , 18 , \cdots$".
The multiples of $34$ are "$34 , 68 , 102 , 136 , \cdots$".
Prime Factorization
The factors of $24$ are "$1 , 2 , 3 , 4 , 6 , 8 , 12 , 24$"
Consider the number $24$. Factors of $24$ are $1$, $2$, $3$, $4$, $6$, $8$, $12$, and $24$. The following ways, the number is expressed as a product of some of the factors.
$4 \times 6$
$2 \times 4 \times 3$
$3 \times 8$
and more like $2 \times 12$
The following is a product of prime numbers
"$2 \times 2 \times 2 \times 3$". The number $24$ is represented as product of prime factors. This is called prime factorization.
Prime Factorization : A number represented as product of prime numbers.
What is the prime factorization of $264$?
The answer is "product of $2 \times 2 \times 2 \times 3 \times 11$".
A procedure is illustrated in the figure.
What is the prime factorization of $17$?
Since $17$ is a prime number, the answer is "$17$".
summary
Factors of a number : The divisor that divides with remainder $0$, is a factor of the dividend.
the dividend, divisor are non-zero whole numbers.
Multiples of a number : The dividend that is divided by a factor with remainder $0$, is a multiple of the factor.
Prime Factorization : A number represented as product of prime numbers.
Outline
The outline of material to learn "Divisibility in Whole Numbers" is as follows.
Note: click here for detailed outline of Whole divisors
→ Classification as odd, even, prime, and composite
→ Factors, Multiples, Prime factorization
→ Highest Common Factor
→ Lowest Common Multiple
→ Introduction to divisibility tests
→ Simple Divisibility Tests
→ Simplification of Divisibility Tests
→ Simplification in Digits for Divisibility Tests
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# Question #4a68b
Oct 19, 2016
$x = \frac{15}{4}$
#### Explanation:
Note that as we have $4 - x$ under a radical, we must have $x \le 4$ to avoid taking the root of a negative number.
$4 + \sqrt{10 - x} = 6 + \sqrt{4 - x}$
$\implies \sqrt{10 - x} = 2 + \sqrt{4 - x}$
$\implies {\left(\sqrt{10 - x}\right)}^{2} = {\left(2 + \sqrt{4 - x}\right)}^{2}$
$\implies 10 - x = {2}^{2} + 2 \left(2\right) \sqrt{4 - x} + {\left(\sqrt{4 - x}\right)}^{2}$
$\implies 10 - x = 4 + 4 \sqrt{4 - x} + 4 - x$
$\implies 2 = 4 \sqrt{4 - x}$
$\implies \sqrt{4 - x} = \frac{1}{2}$
$\implies {\left(\sqrt{4 - x}\right)}^{2} = {\left(\frac{1}{2}\right)}^{2}$
$\implies 4 - x = \frac{1}{4}$
$\implies x = 4 - \frac{1}{4}$
$\therefore x = \frac{15}{4}$
Checking our result:
$4 + \sqrt{10 - \frac{15}{4}} = 4 + \sqrt{\frac{25}{4}}$
$= 4 + \frac{5}{2}$
$= \frac{13}{2}$
$= 6 + \frac{1}{2}$
$= 6 + \sqrt{\frac{1}{4}}$
$= 6 + \sqrt{4 - \frac{15}{4}}$
as desired
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Lesson plan
# Fraction division Word Problems Lesson Plan
## Overview
At this point, students have already learned about dividing fractions and how to solve by multiplying by the reciprocal. In this lesson, we’ll introduce different real-world scenarios and how to solve those problems using models, and being able to interpret what leftovers mean. You can expect this lesson to take one 45-minute class period.
You could teach fraction division including word problems after yo have taught ratios and rates. Students then have the opportunity to apply this thinking to solve fraction division word problems.
Number System
6.NS.A.1
Step-by-step help
ByteLearn gives students targeted feedback and hints based on their specific mistakes
Preview step-by-step-help
## Objective
Students will be able to divide fractions from word problems and interpret what leftovers represent.
## Materials
• Teacher Slideshow
• Partner Activity
• Online Practice
## How to Teach Dividing Fraction Word Problems
### Warm-up
Start the lesson with a warm-up problem that is accessible to students. Ask them to show their thinking using a drawing or words. After many or all of them have put down their thinking, ask them to share their work with their table partner.
Ask students “Is there anybody whose partner followed a different strategy? Can you explain their strategy? After students have shared strategies, make a note of their various drawings and thinking.
### Strategies and tools for solving fraction division word problems.
Typically, I teach fraction division and word problems after the ratio and rates units. Students are likely to use strategies from fraction division and using ratio tables to solve problems like these. Here are some sample drawings and thinking you are likely to see.
### Drawing circles
This comes naturally to students since this is how they would they have learned about fractions since elementary school.
### Number line
Students would have have been exposed to this kind of thinking when going over dividing a whole number by a fraction.
Students might keep adding 1/4 till they reach 2 wholes. Or they might add 1/4 till they reach 1 whole and then conclude that you can make 4 baggies from 1 pound, so you can make 8 baggies from 2 pounds.
### Ratio tables
Students who are comfortable with ratio tables might express their thinking this way. Encourage to explain why they set it up this way, what the 4 and 1 mean, and what does multiplying by 2 mean. This helps them to make connections between their ratio table operations and the actual problem.
### Number of groups and size of groups
You can also encourage students to think about what do they know in the problem:
• Total
• Number of groups
• Size of each group
This kind of thinking helps students to solve multiplication and division problems. You can give problems like to help them identify what they are given and what is missing.
Once you work through these problems, students might conclude that when the number of groups and size of groups is given, you would multiply. When the total is given and either number or size are given, you would divide. You do not want to enforce this as a rule… you really want to provide them with various tools to think through word problems.
### Fraction division word problems with mixed numbers
In this problem you are given the total and the size of a group and the number of groups is missing. We have given a tape diagram as a model to represent this problem; you could also use a number line.
Start with an open-ended question and see what relationships students can identify. Then ask more targeted questions around number, size, and total. Ask them what they are given and what is missing.
You would then identify the total and the size of the group. The total is 3\frac{1}{3} and the size of each group is 2/3.
Ask students if they can figure out the number of groups. Students are likely to show this kind of work.
### Write the division equation
The next step is to formalize this thinking. Tell students to set up an equation to represent this problem. Some might set up a multiplication equation and some might argue that this problem is represented by a division equation.
Once students have 3\frac{1}{3} \div 2/3 = 5, ask them if they can use the division algorithm to check whether the answer is correct. Hopefully students will show this work:
3\frac{1}{3} \div 2/3 = \frac{10}{3} \div 2/3 = 10/3 \times 3/2 = 30/6 = 5
Always encourage students to write an answer sentence so that they make a connection between their calculations and the problem.
### Fraction division word problems with dividing by a whole number
In this problem, the total and the number of groups is given but the size of groups is missing. Typically you would represent such a problem using an area model.
Students might struggle to see that 7/8 is the total since it is less than 1. Once they have convinced each other that the total and the number of groups is given, ask them these questions:
• What does the size of each group represent?
• Do you expect the size of each group to be less than 1 or more than 1? Less than 7/8 or more than 7/8?
This helps students to check for reasonableness of their final answer.
Ask students to now divide the total into 4 groups. This is a representation of what students might draw. Ideally, you should use their drawings to work through the problem. If students are stuck, you can use this slide as a backup.
The next question is how much pizza did each friend get? Ask students to work with their partner to highlight the row that shows how much one friend got.
Once they have highlighted one row, ask them what fraction of the whole pizza did each friend get? Students might have some trouble identifying the denominator. They might want to use 28 as the denominator since 28 cells have some color. This will inspire a good debate. Hopefully some students will point out that we are looking for a fraction of the whole pizza and that the whole pizza has 32 parts out of which each friend get 7 parts.
You should again ask students to set up the division equation and check if it is true.
### Building flexibility in using strategies
As students work through word problems, you should let them use the tools and strategies that they are comfortable with and make sense to them. You should ask students to share their strategies with the class so that students are exposed to different ways of thinking.
Typically, i give division problems mixed up with multiplication problems so that students don’t automatically know that it is a division problem. The key in word problems is to understand the situation and use any tools and strategies they know to solve the problem.
### Partner Activity
This partner activity allows students to work together to draw a model for a dividing fraction word problem. Give each student a copy of the partner activity.
Have students read the scenario, and work together to decode the problem. They can draw a model to help them. Tape diagrams or fraction models are great visual pictures to create for dividing fractions.
When it seems like students are finished, have a group come to the front of the room to share how they were able to solve the problem. If any groups drew their picture other ways, allow them to come up and explain too. It is interesting to see the similarities and differences between the way each student draws their picture, but how they are able to answer the same question.
The last question of the activity has to do with leftovers. This will be a great transition into dividing fraction word problems with leftovers. Being able to interpret what leftovers mean is a very important part of division of fractions.
## Fraction Division Word Problems Practice
After students have completed their lesson, it’s time for some independent practice! ByteLearn gives you access to tons of dividing fraction word problem activities. Check out the online practice and assign them to your students for classwork and/or homework!
Fraction Division Word Problems Practice
Problem 1 of 8
<p>Danielle made 2/3 of a quart of onion dip for a party. She divided the dip into small individual cups that each hold 1/12 of a quart of the dip. </p><p>How many individual cups will Danielle be able to make? <br><highlight data-color="#666" data-style="italic">Write your answer as a whole number, fraction, or exact decimal. </highlight> </p>
View this practice
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# Math Expressions Grade 2 Student Activity Book Unit 4 Lesson 20 Answer Key Mixed Word Problems
This handy Math Expressions Grade 2 Student Activity Book Pdf Answer Key Unit 4 Lesson 20 Mixed Word Problems provides detailed solutions for the textbook questions.
## Math Expressions Grade 2 Student Activity Book Unit 4 Lesson 20 Mixed Word Problems Answer Key
Solve and Discuss
Make a drawing. Write an equation. Solve.
Question 1.
Maxine cuts out 48 squares to make a quilt. She needs 16 more squares to complete the quilt. How many squares will be in the quilt altogether?
Given,
Maxine cuts out 48 squares to make a quilt.
She needs 16 more squares to complete the quilt
48 + 16 = 64 squares
Thus 64 squares will be in the quilt all together.
Question 2.
Mr. Adams buys 93 paper plates for a party. He buys 43 large plates. The rest are small. How many small plates does he buy?
Given,
93 – 43 = 50 small plates
So, he buys 50 small plates.
Question 3.
Chad collects stamps. He has 32 stamps. Loren gives him some more stamps. Now Chad has 51 stamps. How many stamps did Loren give Chad?
Given,
Chad collects stamps. He has 32 stamps.
Loren gives him some more stamps. Now Chad has 51 stamps.
51 – 32 = 19 stamps
Thus Loren give 19 stamps to Chad.
Question 4.
Trina’s team scores 56 points at the basketball game. This is 30 more points than the other team scores. How many points does the other team score?
Given,
Trina’s team scores 56 points at the basketball game.
This is 30 more points than the other team scores.
56 – 30 = 26 points
Thus other team score 26 points.
Question 5.
Maura gives 19 trading cards to Jim. Now she has 24 trading cards. How many trading cards did Maura have to start?
Given,
Maura gives 19 trading cards to Jim.
Now she has 24 trading cards.
24 – 19 = 5 trading cards
Question 6.
Jamal has 63 toy cars. Luis has 24 fewer toy cars than Jamal. How many toy cars does Luis have?
Given,
Jamal has 63 toy cars.
Luis has 24 fewer toy cars than Jamal.
63 – 24 = 39
Thus Luis has 39 toy cars.
Question 7.
Anna has some red balloons and some blue balloons. Altogether she has 46 balloons. How many balloons of each color could she have?
and
Given,
Anna has some red balloons and some blue balloons. Altogether
she has 46 balloons.
46 – 23 = 23
She could have 23 red balloons and 23 blue balloons
and
Question 8.
Jon has 71 stickers. Ken has 53 stickers. How many fewer stickers does Ken have than Jon?
Given,
Jon has 71 stickers.
Ken has 53 stickers.
71 – 53 = 18 stickers
Thus Ken have 18 stickers than Jon.
Question 9.
Amanda has 22 more color pencils than Troy. Troy has 38 color pencils. How many color pencils does Amanda have?
Given,
Amanda has 22 more color pencils than Troy.
Troy has 38 color pencils.
38 + 22 = 50 color pencils
Amanda have 50 color pencils.
Question 10.
Nicole is matching spoons and forks. She finds 36 spoons and 50 forks. How many more spoons does Nicole need to have the same number of spoons as forks?
Given,
Nicole is matching spoons and forks.
She finds 36 spoons and 50 forks.
50 – 36 = 14 spoons as forks.
Question 11.
Kristi has some shells. Then she finds 24 more shells at the beach. Now Kristi has 100 shells. How many shells did she start with?
Given,
Kristi has some shells. Then she finds 24 more shells at the beach.
Now Kristi has 100 shells.
100 – 24 = 76 shells.
Thus she starts with 76 shells.
Question 12.
Gabby has 84 beads. She uses some beads to make a necklace. Now she has 45 beads left. How many beads does Gabby use to make the necklace?
Given,
Gabby has 84 beads. She uses some beads to make a necklace.
Now she has 45 beads left.
84 – 45 = 39 beads
What’s the Error?
Sona has 63 balloons. That is 16 more balloons than Molly. How many balloons does Molly have?
Question 13.
Draw comparison bars to help Puzzled Penguin. Write an equation to solve the problem.
Molly has ballons.
Given,
Sona has 63 balloons.
That is 16 more balloons than Molly.
63 – 16 = 47 balloons
Molly has 47 balloons.
PATH TO FLUENCY Add and Subtract Within 100
Question 14.
By adding 34 and 84 we get 80.
Question 15.
By adding 13 and 78 we get 91.
Question 16.
By adding 49 and 26 we get 75.
Subtract.
Question 17.
By subtracting 38 from 95 we get 57.
Question 18.
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# Mock AIME 1 2007-2008 Problems/Problem 7
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Consider the following function $g(x)$ defined as $$(x^{2^{2008}-1}-1)g(x) = (x+1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - 1$$ Find $g(2)$.
## Solution
Multiply both sides by $x-1$; the right hand side collapses by the reverse of the difference of squares.
\begin{align*}(x-1)(x^{2^{2008}-1}-1)g(x) &= (x-1)(x+1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - (x-1)\\ &= (x^2-1) (x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - (x-1)\\ &= \cdots\\ &= \left(x^{2^{2008}}-1\right) - (x-1) = x^{2^{2008}} - x \end{align*} Substituting $x = 2$, we have $$\left(2^{2^{2008}-1}-1\right) \cdot g(2) = 2^{2^{2008}}-2 = 2\left(2^{2^{2008}-1}-1\right)$$ Dividing both sides by $2^{2^{2008}-1}$, we find $g(2) = \boxed{002}$.
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# Find the Volume of Prisms Using a Formula
4 teachers like this lesson
Print Lesson
## Objective
SWBAT find the volume of prisms using a formula
#### Big Idea
Move from the concrete to the abstract. Students turn the modeling of a prism's volume into an algebraic formula.
5 minutes
## Teacher Model
10 minutes
I will ask students to recall the two things that we considered in a previous activity for volume. I expect to hear answers that say we first found the number of unit cubes that would fit on the base and then we found out how many layers of unit cubes were used. We will then (again) refer to the formula. Example 1 is similar to what we have already done. Before we do example 2, I will ask students if the volume can be found by multiplying 5.5 * 4 * 2. It is a triangular prism so the answer is no, but students who have been taught l * w * h will sometimes apply this blindly to any prism. I ask the question to provide an opportunity for students to quickly put MP3 to use. A thoughtful response may include an explanation that the base shape is a triangle and only 0.5 * 5.5 * 2 unit cubes would fit on the base; and, we would use 4 layers.
As students use the generic equation to solve the problems today they are practicing MP4.
I will expect students to write out solutions in a manner similar to how I present. This helps me assess how students are doing.
## Guided Practice
10 minutes
Students will now answer 6 guided practice problems. They should refer to their notes and their neighbors as a resource. This allows me to spend time checking in on how students are doing. During these direct instruction lessons, I focus my check for understandings on students who I know generally struggle with math topics because I don’t usually have enough time to thoroughly check in with everyone. When I see that they are doing well, I know the majority are doing well.
I love GP5 (pentagonal prism) because, while I think it may be the easiest, it will cause the most confusion. The area of the base is given. GP6 is just another chance to make sure students distinguish surface area and volume (MP6). We will quickly go over the solutions to GP1 – GP6.
## Independent Practice
15 minutes
Students will work on the independent practice without the help of their partners. I will pull aside students that I’ve already identified are not ready for the independent practice. The problems (as they should) mirror the work already done. Problem 6 provides an opportunity for students to practice MP3 as they critique an error in work.
Most students should make it to the extension. Problem number 9 requires some proportional reasoning. Problem number 10 is a test in a student’s ability to be able to use MP1.
Solutions will be discussed as necessary.
## Exit Ticket
5 minutes
The exit ticket has 3 questions. The first two are pretty straight forward. Problem number 3 may cause problems for some students only because I have provided an image of a heptagonal prism. However, it will let me know if students understand how to find the volume of any prism.
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## Lesson 9: Areas (II)
May 9, 2011
### 4.Perimeter of a circumference and length of a an arc
The length of a circumference of radius and diameter d = 2r is L =2πr or L=πd.
The length of an arc of aº is Larc = 2πr · aº/360º.
You don’t need to learn this formula because it is easy to find out by applying a rule of three direct:
2πr ———- 360º
Larc ———– aº Larc = 2πr · aº/360º
The following table revises the formulae we studied in this lesson.
### 5.Area of Plane Shapes
Triangle AreaA = ½b · h b = base h = vertical height Rectangle Area Area = w · h w = width h = height Trapezoid (US) Area Trapezium (UK) Area A = ½(a+b) × h h = vertical height Regular polygon Area A = (p · a)/2 p = perimeter a = apothem
Square AreaA = a2 a = length of side Parallelogram Area A = b · h b = base h = vertical height Circle AreaA = πr2 Circumference=2πr r = radius Sector circle AreaA = (πr2 · θ) /360º r = radius θº = angle in degrees
### 6.Area of a Circular Ring or Annulus
A circular ring (annulus or crown) is plane figure bounded by the circumference of two concentric circles of two different radii. The area of a circular ring is found by subtraction the area of small circle from that of the large circle.
Ar = π(R2 – r2)
### 7.Composite Figures
Composite figures (or shapes) are figures (or shapes) that can be divided into more than one of the basic figures. Their area can be calculated by adding (or subtracting) the areas of the figures they are divided into.
One example would be the annulus.
Here you are some links to practice with areas and revise perimeters. The level is elemental.
2.Area
### 8. Sum of the interior angles of a polygon
The sum of the interior angles of a n-sided polygon is (n-2) · 180º.
In this case, we have an hexagon so the sum of the interior angles is (6-2) · 180º.
In a regular n-sided polygon all the interior angles are equal so each interior angle measures (n-2) · 180º/n.
Here you are a video on calculating the sum of the interior angles of 17-sided polygon.
### 9. Central angle of a regular n-sided polygon
The central angle of a regular n-sided polygon measures 360º/n
In this case we have a pentagon so its central angle is 360º/5 =72º
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# Class 10 NCERT Solutions- Chapter 13 Surface Areas And Volumes – Exercise 13.3
### Question 1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Radius of sphere (R)= 4.2 cm
Radius of cylinder (r)= 6 cm
In recasting process the volume will be same, so
Volume of cylinder = Volume of sphere
πr2h = πR3
π(6)2h = π(4.2)3 (cancel π from both side)
36h = (4.2 × 4.2 × 4.2)
h = (cm)
h = 1.4 × 1.4 × 1.4 (cm)
h = 2.74 cm
### Question 2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radius of sphere 1 (r1)= 6 cm
Radius of sphere 2 (r2)= 8 cm
Radius of sphere 3 (r3)= 10 cm
Let Radius of resulting sphere = R
In recasting process the volume will be same, so
Volume of Resulting sphere = Volume of sphere 1 + Volume of sphere 2 + Volume of sphere 3
π(R)3π(r1)3π(r2)3 +π(r3)3 (cancel π from both side)
R3 = (r1)3 + (r2)3 + (r3)3
R3 = (6)3 + (8)3 + (10)3
R3 = 216 + 512 + 1000
R3 = 1728
R = (1728) 1/3
R = 12 cm
### Question 3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
So basically here, the digging of cylindrical shape is changed to cuboidal shape
Given values,
Diameter of cylinder = 7 m
Radius of cylinder (r)= m
Height of cylinder (H)= 20 m
Length of Cuboid (l) = 22 m
Breadth of Cuboid (b) =14 m
Let Height of Cuboid = h
In this process the volume will be same, so
Volume of Cuboid = Volume of Cylinder
l × b × h = πr2H
22 × 14 × h = π × × 20
h = m (taking π =)
h = m
h = 2.5 m
### Question 4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
So basically here, the digging of cylindrical shape is changed to another Hollow cylindrical shape
Given values,
Diameter of cylinder = 3 m
Radius of cylinder (r) = m
Height of cylinder (h) = 14 m
Width of embankment = 4 m
Outer Radius of embankment R1= radius of cylinder + width = 3/2 + 4 = m
Inner Radius of embankment R2= radius of cylinder = m
Height of embankment = H
In this process the volume will be same, so
Volume of Embankment = Volume of Cylinder
(Ï€R12H) – (Ï€R22H) = Ï€r2h
(R12 R22)H = ()2× 14 (cancel π from both side)
H =
H =
H =
H =
H =
H = m
H = 1.125 m
### Question 5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
Given values,
Radius of cylinder (r) = 6 cm
Height of cylinder (h) = 15 cm
Radius of each cone (R) = 3 cm
Height of each cone (H) = 12 cm
Let n be the total number of ice creams
In this process the volume will be same, so
n × (Volume of each Cone + Volume of each hemisphere) = Volume of Cylinder
n × (πR2H + πR3) = πr2h
n = (cancel π from both side)
n =
n =
n =
n = 10
Hence, 10 numbers of cones filled with ice-cream.
### Question 6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Solution:
Given values,
Radius of cylindrical coin (r) =
Height of cylindrical coin (H) = 2 mm = cm
Length of Cuboid (l) = 5.5 cm
Breadth of Cuboid (b) =10 cm
Height of Cuboid (h)= 3.5 cm
Let n be the total number of coins
In this process the volume will be same, so
n × (Volume of each Coin) = Volume of Cylinder
n × (πR2H) = l × b × h
n × π × ()2× = 5.5 × 10 × 3.5
n = (taking π = )
n =
n = 400
Hence, 400 silver coins must be melted to form this cuboid.
### Question 7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
So basically here, the cylindrical shape is changed to conical shape
Given values,
Radius of cylinder (r) = 18 cm
Height of cylinder (h) = 32 cm
Height of cone (H) = 24 cm
Let Radius of cone = R
In this process the volume will be same, so
Volume of Cone = Volume of Cylinder
πR2H = πr2h
R2 × (24) = 182 × 32 (cancel π from both side)
R2
R = √(18 × 18 × 4)
R = 36 cm
Slant height (l) = √H2 + R2
l = √(242 + 362)
l = √(12×2)2 + (12×3)2
l =12 √(4+9)
l = 12 √13 cm
### Question 8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Given values,
Width of canal (w) = 6 m
Depth of canal (h) = 1.5 m
Speed of canal = 10 km/hr = (10,000 m/hr)
For 1hr (60 mins), we can take length (l) as = 10,000 m and
Volume in 1hr = (l × w × h)
= (10,000 × 6 × 1.5) m3
= 90,000 m3
So for 30 mins volume will be = m3
= 45,000 m3
Area irrigated in 30 minutes will be as:
Area × length of standing = Volume of canal in 30mins
Area =
Area = 562500 m2
### Question 9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Given values,
Radius of Pipe (R) = 10 cm = m
Radius of Cylindrical tank (r) = 5 m
Depth of Cylindrical tank (h) = 2 m
Speed of water flows in pipe = 3 km/hr = (3,000 m/hr)
Let Time to fill cylindrical tank = t
For 1hr (60 mins), we can take height of Pipe (H) as = 3,000 m and
Volume in 1hr = (Ï€R2H)
= (π × ()2 × 3,000) m3
= 30Ï€ m3
In this process the volume will be same, so
t (in hr) × (Volume of Pipe in 1hr) = Volume of Cylindrical Tank
t × (30π) = πr2h
t × (30π) = π × 52 × 2
t =
t = hr
t = × 60 mins
t = 100 minutes
Hence, 100 minutes will be taken to fill the tank.
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# Variance
## Variance Definition
In probability and statistics, variance is defined as the expected value of the squared deviation of a random variable from its mean value. It can also be described as the measurement of the spread between each number in a data set.
Variance indicates variability. A data set will have a mean value and variance indicates how far the elements are spread out around this mean value. If the elements are largely apart from each other, the variance will be high and if closer, the variance will be low.
If the elements of dataset A are 2, 3, 4, 5 and 3 and the elements of dataset B are 2, 7, 10, 3 and 17, the data set B will have a higher variance as the elements in data set B are more spread out and more apart from each other.
The lowest value for variance will be zero. A variance value of zero will be expected if the elements are identical. To calculate the variance of a given data set, it is necessary to know the mean value of the data set.
## Variance Formula
There are two types of variances. Population variance and sample variance. The formula for each is given below and what exactly each of the two are and what the differences between the two are will be studied later in the article
## Variance of a Random Variable
Variance is also used in binomial distribution where the probability of success and failure is determined. In binomial distribution. Variance is used for finding the variability of a random variable X.
The equation for the variability for the random variable X, Var(x) is given by:
Where P(x) is the probability mass function
## Variance Example
Let us illustrate the calculation of both population variance and sample with an example first and understand how the formulas are used and then take a look at the differences between population and sample variance
Question: It is given that the ages of the children in a housing colony are: 3, 8, 6, 10, 12, 9, 11, 10, 12, 7. Calculate the population and sample variance for this dataset.
Solution:
Let us calculate the variance of the ages of the children step by step:
Step 1: Calculating the mean value
The number of children = N = 10
Mean = [ 3+8+6+10+12+9+11+10+12+7] / 10
Mean = 8.8 years
Step 2: Calculate the deviation and squaring it
Step 3: Calculating the population variance
Substituting the values in the above formula, we get
σ2 = 73.6 / 10 = 7.36 years
Hence, the population variance = σ2 = 7.36 years
Step 4: Calculating the sample variance
Now let us say that the 10 children whose age are provided are from 10 families who are the oldest occupants of the colony. That is there are more families with children in the area who are recent occupants.
Now the data we are provided with is sample data i.e let us assume that our dataset above is a sub-set of a much larger dataset. In such a case, we are calculating the sample variance. Hence we have to use the sample variance equation to find the variance.
S2 = 73.6 / (10-1) = 8.17 years
Hence, the sample variance = S2 = 8.17 years
## Squaring of Differences
The squaring of each difference is done to prevent the numbers from canceling out each other. This is one of the significant advantages of the variance. It measures the variability of each element from the mean value without considering its direction. However, the squaring will lead to the unit of the variance to be different from the original unit of the data set. Hence the unit of the original data will get squared when the variance is calculated.
In the above example, it is observed clearly that the squaring of the deviations are necessary to get an accurate variance value. Sometimes the value of elements will be smaller than the mean. Hence the difference will give a negative number. When we take the average of this difference as such, the negative and positive numbers might cancel out each other. Squaring the difference will eliminate this problem.
If the squaring of differences is not done in the above example, the sum of the deviation will be equal to zero and thereby we get a result as the variance is equal to zero. This cannot be true as the ages of the children are not identical and there exists a degree of variation. Hence in order to avoid the direction of the deviation of the elements from the mean, the deviation value is squared, and then the variance is calculated.
Let us understand the same using an example:
Example: Suppose the differences are given for data set A that has 4 elements as -1, 2, -4, 9
If we take the absolute value and divide by 4, we get:
[ |-1| + |2| + |-4| + |9| ] / 4 = 4
Now if another data set B has the differences as -4, 4,4,4
We get [ |-4| + |4| + |4| +|4| ] / 4 = 4
Now, both have the same variability even though data set A has a greater spread than dataset B.
By squaring each difference, we ensure that this problem will not arise and when substituted in the formulas for variance, will ensure that dataset A reports greater variance than dataset B.
## Population Variance Vs Sample Variance
In the above example, we saw the terms population variance and sample variance. Let us understand each in detail.
If we are provided with the whole of the data in a dataset, then we calculate the population variance. The example we have taken in the previous section can be used to illustrate this. If the 10 children whose age we are provided with are the only number of children in the particular housing colony, then we are calculating the population variance. On the other hand, let us say that there are a total of 20 children in the housing colony and the ages of 10 children that are listed are the children who were selected under a specific criteria. In this case, we are asked to calculate the variance of the ages of the total children by providing only a portion of the data. This portion of the data can be called a sample data of the whole population. This is where the sample variance is calculated.
While calculating the sample variation, we are trying to calculate the population variance with only a sample of the data. Therefore the chances of error occurrence might be high as we are not provided with the entire data set. To avoid this, a change in the equation of the population variance is made known as Bessel’s correction.
The value of the variance calculated from the sample data might be slightly different from the actual variance calculated by taking the whole population. This difference will be higher especially if the number of elements in the sample is low or the corresponding population size is high. So we need to correct the error to get the actual variance.
As we are working with the sample mean (x̄), the elements will be much closer to the sample mean. Hence the squared differences will be smaller. This will lead to a smaller value in the numerator of the sample variance equation compared to the population variance equation. If we divide this squared difference with the sample size n, this will lead to a variance value that is smaller than the actual value.
But the population variance value should be accurate. For this, we subtract 1 from the sample size so as to get a smaller denominator in the sample equation and thereby producing a larger variance value from the equation. This correction of the sample size from n to n-1 is known as Bessel’s correction.
In the example, the sample variance is slightly higher than the population variance as we consider the ages of the 10 children as a sample from the total ages of 20 children. When the total data set of 20 elements is taken into consideration, the variance value will be obviously higher than the value we obtain when we take the 10 elements that are provided as the total population size.
In most of the real world scenarios, we calculate variance using sample data as the whole of the population data is rarely given or difficult to achieve.
Variance Vs Standard Deviation
Standard deviation can be defined as a statistic used to measure the dispersion of a given dataset in relation to it’s mean and can be calculated by taking the square root of the variance.
The equation for the population standard deviation and sample standard deviation is given as:
A standard deviation is always a positive number. The lowest value for a standard deviation is zero when the elements are identical. Both standard deviation and variance are two distinct parameters with few similarities.
Let us look at the similarities :
• Variance and standard deviation give the measure of the spread between the elements around the mean.
• The mean value should be determined to calculate both.
• Variance and standard deviation values are always positive. As variance is a squared value, it is always positive and the standard deviation is the positive root of the variance.
• Both variance and standard deviation will be equal to zero if the elements are identical.
• Both take the square of the differences of each element from the mean.
• Both variance and standard deviation do not consider the direction of the variability or deviation of the elements from the mean.
Nevertheless, these two parameters are distinct and different.
The standard deviation is introduced in statistics due to a disadvantage of the variance which is that the variance calculation give added weight to the elements that are far from the mean which makes it less accurate. The variance is calculated primarily to take the square root value of it which is the standard deviation.
Z TABLE
Z Table. Z Score Table. Normal Distribution Table. Standard Normal Table.
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Difference between revisions of "2015 AMC 10A Problems/Problem 13"
Problem 13
Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of his coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$
Solution 1
Let Ashraful have $x$ 5-cent coins and $\left( 12 - x \right)$ 10-cent coins. It is easily observed that any multiple of $5$ between $5$ and $5x + 10(12 - x) = 120 - 5x$ inclusive can be obtained by a combination of coins. Thus, $24 - x = 17$ combinations can be made, so $x = 7$. But the answer is not $7,$ because we are asked for the number of 10-cent coins, which is $12 - 7 = \boxed{\textbf{(C) } 5}$
Solution 2
Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of $5.$ To have exactly $17$ different multiples of $5,$ we will need to make up to $85$ cents. If all twelve coins were 5-cent coins, we will have $60$ cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain $5$ cents, and as we need to gain $25$ cents, the answer is $\boxed{\textbf{(C) } 5}$
For small scale substitutions of 5 cent for 10 cent, like in this problem, it is quite fast, however, if the problem is a little bit more complex, knowing that you need 85 cents, explained above, it is also possible to use a system of equations. It would be $$5x+10y=85$$ $$x+y=12$$ Solving this $x$ (5-cent coins) $= 7$ and $y$ (10-cent coins) $= 5$, so again the answer is $\boxed{\textbf{(C) } 5}$
~savannahsolver
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The 3n+1 Problem
The 3n+1 Problem
© January 2020, Darrell Cox (A link to "Fermat's Last Theorem and Related Problems" is at fermat and a link to "Farey Series and the Riemann Hypothesis" is at riemann. Other links are given at the end of this article.)
The 3n+1 problem appears to have been first posed by Collatz in 1937. In this problem, a sequence is generated starting with an initial natural number n. The rule for generating the next natural number in the sequence is; if n is even, the next natural number in the sequence is n/2, or if n is odd, the next natural number in the sequence is 3n+1. For example, if the initial value of n is 17, the sequence generated is {17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ...}. If the natural number 4 is encountered in the sequence, the sequence starts repeating ({4, 2, 1} is generated over and over). There are three possibilities; (1) the natural number 4 is encountered in the sequence, (2) the sequence starts repeating, but for a different cycle than {4, 2, 1}, or (3) the sequence doesn't repeat (in which case the natural numbers generated in the sequence have to keep getting larger and larger). The 3n+1 conjecture states that only the first possibility can occur. Let c be an odd integer greater than or equal to -1 that is not divisible by 3. The next number in the 3n+c sequence is defined to be 3n+c if n is odd, or n/2 if n is even (the next number in the sequence is always a natural number). The sequence repeats for {4c, 2c, c} if c>-1 or {2, 1} if c=-1. In this more general sequence, there are usually cycles other than {4c, 2c, c}. Although the 3n+1 and 3n-1 sequences have some unique properties, the 3n+c problem is essentially the same for all c values.
The 3n+1 problem is difficult and is not likely to be solvable by an amateur. Although it might seem that the problem should be classified as "recreational" mathematics, there is a considerable body of mainstream mathematical literature on the subject. (See Jeffrey C. Lagarias' "The 3x+1 problem: An annotated bibliography (1963-1999)" at http://arxiv.org/abs/math.NT/0309224 and "The 3x+1 Problem: An Annotated Bibliography, II (2000-2009)" at http://arxiv.org/abs/math.NT/0608208. Also, see Lagarias' 1985 article "The 3x+1 problem and its generalizations" at http://www.cecm.sfu.ca/organics/papers/lagarias/index.html.) This article reviews some of the highlights of the 3n+1 problem literature and presents some original research on the matter (in the form of empirically derived "results"). The level of the presentation is elementary and empirically derived propositions are specifically identified when practical. Sometimes entire sections are mostly empirically derived. In these cases, a statement to that effect is made at the beginning of the section.
Newcomers to the 3n+1 problem are frequently mystified by how the sequence increases and decreases in a seemingly random fashion, but always appears to arrive at 1. The fluctuations of the sequence are of little interest once the probability of the situation is taken into account (whether there are cycles other than {4, 2, 1} is far more interesting). See the section "A probabilistic heuristic" in the Wikipedia article at http://en.wikipedia.org/wiki/Collatz_conjecture for a brief introduction to the subject, the section "A heuristic argument" in Lagarias' 1985 article for more details, and the section "A Random-Walk Argument" in Richard E. Crandall's1 1978 article "On the "3x+1" Problem". The probabilistic argument to be given here avoids the "mixing" assumptions in other approaches and is amenable to empirical verification. If the 3n+1 sequence is bounded and there are no cycles other than {4, 2, 1}, then the sequence element 1 must eventually be reached. The argument to be given is then that the 3n+1 sequence is bounded.
A Probabilistic Argument that the 3n+1 Sequence is Bounded
The probabilistic argument entails a restructuring of the Collatz graph and an alternate definition of the 3n+1 sequence. The Collatz graph is a tree-like structure that shows how the sequence element 1 can be arrived at starting from different initial n values. (See the Wikipedia article for the standard depiction of the Collatz graph. Of course, the graph can only be constructed if the 3n+1 conjecture is assumed to be true.) When n is even and n-1 is divisible by 3, there is a node in the graph where the previous elements in the sequence are 2n and (n-1)/3. For example, in the Collatz graph, two limbs ending in 32 and 5 converge to form a limb starting at 16. In the restructured Collatz graph, 5 is considered to be a continuation of the limb segment starting with 16. In general, (n-1)/3 is considered to be a continuation of the limb segment at such nodes. (That is, "odd" paths are taken when tracing back through the nodes.) The limbs in the restructured Collatz graph are then;
{4, 2, 1}
{..., 24, 12, 6, 3, 10, 5, 16, 8}
{..., 72, 36, 18, 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20}
{..., 120, 60, 30, 15, 46, 23, 70, 35, 106, 53, 160, 80}
{..., 168, 84, 42, 21, 64, 32}
.
.
.
Each limb other than (4, 2, 1) contains exactly one odd element divisible by 3. Let j denote the odd natural number that is divisible by 3. The natural numbers to the right of j are not divisible by 3 (since 3 does not divide 3j+1, (3j+1)/2, ...). The natural numbers to the left of j must be of the form 2ij where i is a natural number (no natural number can attach to this part of the limb since 3 does not divide 2j-1, 4j-1, 8j-1, ...). (The limbs listed above [starting with the second limb] contain 3, 9, 15, and 21 respectively. The second limb attaches to the first limb, the third and fifth limbs attach to the second limb, and the fourth limb attaches to the third limb.) Note that other than the limb {4, 2, 1}, the next-to-last element in a limb is divisible by 8 and that no natural number to the right of the odd natural number divisible by 3 and before the next-to-last natural number is divisible by 8 (this is known as a 1-2 sequence vector since there are either one or two even sequence elements between successive odd sequence elements).
The 3n+c sequence can be defined by the recurrence operation [(3/2)h(n+c)-c]/2n, 2h divides n+c, 2h+1 does not divide n+c. Each iteration of the recurrence operation starting with an odd element will be referred to as a "jump". If n is odd and 22 does not divide n+c, then the sequence {n, 3n+c, (3n+c)/2=(3/2)(n+c)-c, ...} is generated where every other element of the sequence up to (3/2)(n+c)-c is odd. Similarly, if n is odd and 22 divides n+c, 23 does not divide n+c, then the sequence {n, 3n+c, (3n+c)/2, 3[(3n+c)/2]+c, [3[(3n+c)/2]+c]/2=(3/2)2(n+c)-c, ...} is generated where every other element of the sequence up to (3/2)2(n+c)-c is odd. In general, if 2h divides n+c, 2h+1 does not divide n+c, then the first element in the sequence that is divisible by 4 is the one just before (3/2)h(n+c)-c. If [(3/2)h(n+c)-c]/2 is even, then the first element in the sequence that is divisible by 8 has been found. Each limb of the restructured Collatz graph other than {4, 2, 1} consists of a series of jumps, the last jump ending in an even natural number (which connects the limb to the rest of the tree). The question of whether the 3n+1 sequence (or the 3n+c sequence) is bounded then reduces to the question of whether there are any jumps ending in an even natural number. The distribution of the number of jumps (denoted by i) before an even natural number is reached for the 1000000 sequences starting with 3, 9, 15, ..., 5999997 is;
i=1, 500002
i=2, 250004
i=3, 124998
i=4, 62498
i=5, 31211
i=6, 15683
i=7, 7782
i=8, 3897
i=9, 2000
i=10, 972
i=11, 497
i=12, 230
i=13, 109
i=14, 58
i=15, 23
i=16, 13
i=17, 15
i=18, 3
i=19, 3
i=20, 0
i=21, 1
i=22, 0
i=23, 0
i=24, 1
The probability that i iterations are required is about (1/2)i. Even though the probability that the sequence is unbounded is effectively 0, it's unlikely that this part of the 3n+1 conjecture is provable. (The data indicates that the process which forms 3n+1 sequences [having 1-2 sequence vectors] consisting of an arbitrarily large number of jumps is random. [In the mathematical literature, such processes are usually said to be "pseudo-random", but in the absence of a rigorous definition of "random", one could argue that saying the process is random is acceptable.]). The probability that h=1 is about 1/2, the probability that h=2 is about 1/4, the probability that h=3 is about 1/8, etc. These are the expected probabilities. Even if such a limb (one having a 1-2 sequence vector starting with an odd natural number divisible by 3) attaches to another such limb, and that limb attaches to another such limb, etc., there is no guarantee that the trunk of (4, 2, 1) would be reached. The 3n+1 sequence may then be unbounded in this way. About 90% of the time, the odd natural number divisible by 3 in a limb that attaches to another limb is larger than the odd natural number divisible by 3 in that limb. (The proportions for the first 10, 100, 1000, 10000, 100000, 10000000, and 10000000 odd natural numbers divisible by 3 are 0.9, 0.91, 0.905, 0.9022, 0.90333, 0.903506, and 0.903254 respectively. The limb (..., 24, 12, 6, 3, 10, 5, 16, 8) doesn't attach to another limb containing an odd natural number divisible by 3, but this is still counted as if 3 were greater than an odd natural number divisible by 3 in another limb.) There is then a strong tendency for the limbs to eventually be attached to the trunk.
If c=1 and negative n values are allowed, the absolute value of the integer divisible by 3 in a limb that attaches to another limb is larger than the absolute value of the odd integer divisible by 3 in that limb about 90% of the time. Also, the probabilities for the number of jumps required to reach an even integer starting with an odd integer divisible by 3 are the same as when only positive n values are allowed. If c>1 and negative n values are allowed, the same probabilities apply. Let t denote the odd integer divisible by 3. For example, for c=5, 5 does not divide t, -10k≤(t-3)/6≤10k, and k=1, 2, 3, ..., 7, the proportions are 0.941176, 0.913043, 0.906933, 0.903944, 0.903457, 0.903540, and 0.903280 respectively. For k=4, a histogram of the differences in absolute values of t values divided by 1536 (to scale the values) is;
For c=1 and k=4, a histogram of the differences in absolute values of t values divided by 1536 is;
(There are fewer values for c=5 due to excluding t values divisible by 5.) In general, 8/9 appears to be a lower bound of these proportions.
In 1972, John H. Conway2 showed that a more general function iteration problem similar in form to the 3n+1 problem is computationally undecidable. This lends some credence to the notion that this part of the 3n+1 problem is unprovable. However, some progress can be made in this area; in 1976, Riho Terras3 showed that almost all numbers have finite "stopping time". Let S0=N and Si=Si-1/2 if Si-1 is even or 3(Si-1+1)/2 otherwise. (This is the way the 3n+1 sequence is usually defined in the mathematical literature. This turns out to be the "natural" way to define the sequence.) The smallest value of i such that Si<S0 is defined to be the stopping time of N.
Necessary and Sufficient Conditions for Cycles in the 3n+c Sequence to Exist
An element s in the 3n+c sequence can be expressed as (Xa-cZ)/Y where a is a subsequent element in the sequence, X is a natural number and a power of 2 (X equals 2l where l is the length of the sequence [where the element after an odd element i is defined to be (3i+c)/2]), Y is a natural number and a power of 3 (Y equals 3m where m is the number of odd elements in the sequence), and Z is a natural number and a 3-adic number with power-of-two coefficients (see Bohm and Sontacchi's4 article and page 41 of Gunther Wirsching's5 book [for discussions of the c=1 case] ). (An example of Z is 2233+2332+2531+2630 =340. A formula for Z will be given later.) If s=a, then s=cZ/(X-Y). Other than setting c to X-Y, no method for finding X, Y, and Z where cZ/(X-Y) is an integer is known, so other requirements for cycles will be investigated.
The 3n+c Sequence in the Integer Domain
In the mathematical literature, the cycles {4, 2, 1} for c=1 and {2, 1} for c=-1 are usually said to be "trivial" (but there doesn't appear to be any rationale for this designation). For a given c value, there appear to be few (if any) non-trivial cycles. When c is composite, some of the cycles that occur are essentially no different from the cycles that occur for a factor of c (the elements of the cycle are just a common multiple of the elements of the cycle for the factor of c). (Cycles that are not multiples of other cycles are said to be primitive. Note that the cycle {4c, 2c, c} for c>1 is not primitive.) Also, some cycles for a given c value are interrelated; they have the same lengths (number of elements) and the same number of odd elements (and approximately the same dynamic range of elements). Counting these types of cycles as redundant, a histogram of the apparent number of cycles (including the "trivial" cycles) for c values less than or equal to 151 is;
number of cycles number of c values
0 0
1 20
2 21
3 10
4 1
5 0
6 0
total=52
A Poisson probability distribution with a parameter of 1 can be used to model the number of 3n+c cycles for a given c value (see the author's "The 3n+1 Problem: A Probabilistic Approach" at http://www.cs.uwaterloo.ca/journals/JIS/VOL15/Cox/cox10.pdf).
Associated cycles are defined in the Journal of Integer Sequences article. A Poisson probability distribution with a parameter of 1.082 superimposed on a histogram of the number of cycles for a given c value (allowing negative n values and counting only one of interrelated or associated cycles) for c values greater than 0 and less than or equal to 19999 is;
(The mean of the actual distribution is 1.082.) A Poisson probability distribution with a parameter of 1.070 superimposed on a histogram of the number of cycles for a given c value (allowing negative n values and counting only one of interrelated or associated cycles) for c values greater than 0 and less than or equal to 29999 is;
When c=-1, there are two known cycles other than (2, 1); the cycle (20, 10, 5, 14, 7) and the cycle (68, 34, 17, 50, 25, 74, 37, 110, 55, 164, 82, 41, 122, 61, 182, 91, 272, 136). A frequently asked question is why there are non-trivial cycles for the 3n-1 sequence, but apparently not for the 3n+1 sequence. There are many such examples when general negative c values are allowed. For example, there is at least one primitive cycle for the 3n+7 sequence, but apparently none for the 3n-7 sequence. However, there appears to be at least one primitive cycle in either the 3n+c sequence or the 3n-c sequence. This indicates that the distinction between the 3n+c and 3n-c sequences is artificial and that negative n values should be allowed (and that the c values should be required to be positive). (However, it is sometimes convenient to use the original definition of the 3n+c sequence.) More evidence that the 3n+1 and 3n-1 sequences (n>0) should be considered to be the same sequence will be presented later.
Interrelated 3n+c Cycles
In this section, negative n values are allowed, c is required to be positive, and the element after an odd element i in the 3n+c sequence is defined to be (3i+c)/2. A parity vector gives the order of even and odd elements in a 3n+c sequence, a "'1" for an odd element and a "0" for an even element. Let k0, k1, k2, ..., km-1 denote the positions of the 1's in a parity vector containing m 1's and having a length of l. Let Z denote 3m-12k(0)+3m-22k(1)+3m-32k(2)+...+302k(m-1) (due to typographical difficulties, k(0), k(1), k(2), ..., k(m-1) is used to denote k0, k1, k2, ..., km-1). A cycle exists if and only if cZ/(2l-3m) is an integer. (Note that every parity vector containing at least one 1 corresponds to a cycle for some c value [duplicated parity sub-vectors correspond to duplicated sub-cycles].) For example, for the c=17 cycle of (2, 1, 10, 5, 16, 8, 4), k0=1 and k1=3, Z=3121+3023=14, 119=27-32, and (17·14)/119 is an integer. There do not appear to be any factors of c left over after the division by 2l-3m. This has been confirmed experimentally (for the 73916 distinct (c, l, m) values of the 3n+c cycles for c<100000):
(1) A 3n+c cycle exists only if c divides 2l-3m.
If this proposition holds, Z need not be considered when searching for cycles; only the factors of 2l-3m need be considered. When c=|2l-3m|, an arbitrarily constructed parity vector with a length of l and containing m 1's corresponds to a cycle (or possibly duplicated sub-cycles if l and m are not relatively prime), but the cycle is not guaranteed to be primitive. When reduced, such a cycle corresponds to a cycle for a c value that is a proper divisor of 2l-3m. In this sense, there is no problem of finding cycles; they're all well-defined and determined by the parity vectors. Even the number of interrelated cycles is determined by the combinatorics of generating parity vectors that are distinct under rotation (see the above Journal of Integer Sequences article for a table of the numbers of distinct parity vectors for l less than or equal to 20 and m less than or equal to 10). There is, however, the problem of determining which c values the primitive cycles map to. For example, for l=1 and m=1, the parity vector is (1) (corresponding to the cycle {-1} for c=1), for l=2 and m=1, the parity vector is (0, 1) (corresponding to the cycle {2, 1} for c=1), and for l=3 and m=2 the parity vector is (1, 1, 0) (corresponding to the cycle {-5, -7, -10} for c=1). For l=11 and m=7, there are 30 distinct parity vectors, the first few of which are (0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0), (1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0), (1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0), ... (corresponding to cycles for c=139). The cycle corresponding to the first parity vector is not primitive (when c=139) and corresponds to the remaining known c=1 cycle of {-34, -17, -25, -37, -55, -82, -41, -61, -91, -136, -68}. Another example factorization will be given. For l=8 and m=4, the distinct parity vectors are (1, 1, 0, 0, 1, 1, 0, 0), (0, 1, 0, 1, 0, 1, 0, 1), (1, 0, 1, 1, 0, 1, 0, 0), (0, 0, 1, 1, 1, 1, 0, 0), (0, 1, 1, 0, 1, 1, 0, 0), (1, 0, 0, 1, 1, 1, 0, 0), (0, 1, 0, 1, 1, 1, 0, 0), (1, 0, 1, 0, 1, 1, 0, 0), (1, 1, 0, 1, 0, 1, 0, 0), and (0, 1, 1, 1, 0, 1, 0, 0) (corresponding to cycles for c=175). The first parity vector corresponds to two duplicated primitive cycles for c=7, the second parity vector corresponds to four duplicated primitive cycles for c=1, the third parity vector corresponds to a primitive cycle for c=25, the fourth and fifth parity vectors correspond to primitive cycles for c=35, and the remaining parity vectors correspond to primitive cycles for c=175. Although 5 also divides 175, there are no primitive cycles for this c value when l=8 and m=4 (the parity vectors have been used up by cycles for the larger c values and c=1).
Apparently, every c value is covered in this mapping. A more appropriate question to ask than why cycles don't exist for a given c value is why they do exist and what the expected number of cycles is (see the Journal of Integer Sequences article for more details).
(For a given c value, two primitive cycles cannot have the same parity vector. For example, the 3n+5 sequence 83, 254, 127, 386, 193, 584, 292, 146, and 73 has the same parity vector as the 3n+5 cycle 19, 62, 31, 98, 49, 152, 76, 38, and 19. The ratios of the corresponding odd elements decrease monotonically and are 4.368, 4.097, 3.939, and 3.842. More generally, a primitive 3n+c1 cycle cannot have the same parity vector as a primitive 3n+c2 cycle. For example, the 3n+7 sequence 65, 202, 101, 310, 155, 472, 236, 118, and 59 has the same parity vector as the above 3n+5 cycle, but the ratios of the corresponding odd elements are 3.421, 3.258, 3.163, and 3.105. The 3n+7 sequence 1, 10, 5, 22, 11, 40, 20, 10, and 5 has the same parity vector as the above 3n+5 cycle, but the ratios of the corresponding odd elements are 0.053, 0.161, 0.224, and 0.263. For a primitive 3n+c cycle where c≠|2l-3m|, there is always another cycle [not necessarily primitive] having the same parity vector where c=|2l-3m|. This accounts for Proposition (1).)
The Minimum Element in a 3n+c Cycle
In 1997, Halbeisen and Hungerbühler6 proved optimal estimates for the length of a Collatz cycle in terms of its minimum using the formula Ml,m=∑(]jm/l[ - ](j-1)m/l[)2j-13m-]jm/l[ where the summation is from j=1 to l and the reversed brackets denote the ceiling function (the minimum is Ml,m/(2l-3m)). For a Collatz cycle, Ml,m/(2l-3m) is larger than the minimum element in other cycles having a length of l and containing m odd elements. Their Lemma 4 is;
Let s = (s1,..., sl) and t = (t1,..., tl) be two distinct elements of Sl,n. If ∑ki=1 si ≤ ∑ki=1 ti for all k an element of {1,..., l}, then φ(s) > φ(t).
Sl,n denotes the set of all 0-1 sequences of length l containing exactly n ones (the number of ones is denoted by m in this article). φ is defined recursively by φ({})=0, φ(s0)=φ(s), φ(s1)=3φ(s)+2l(s) where s denotes an arbitrary element of S (S denotes the union of all Sl,n sets from n=0 to l) and l(s) the length of s. (The explicitly computed form of φ is denoted by Z is this article.) For a given s, some left-shift permutation of s (denoted by t) gives a minimum φ(t) value. For all s an element of Sl,n, Ml,n is defined to be the maximum of these minimum φ(t) values. Their Lemma 5 (where the sequence s̃ for which φ attains the value Ml,n is determined) is;
Let n ≤ l be natural numbers. Let s̃ = ]in/l[ - ](i - 1)/l[ (for 1≤ il). Then φ(s̃) equals the minimum φ(t) value of s̃ (equal to Ml,n).
(Due to typographical difficulties, the exact form of this lemma is not duplicated here.) In the proof of this lemma, Lemma 4 is used and ki=1ti is represented by a staircase. For example, a staircase forwhere l=27 and n=17 along with a staircase representing the partial sums of [in/l]-[(i-1)/l] where the brackets denote the floor function is;
The staircase using the floor function (not used in the proof of Lemma 5) can be viewed as being an upside-down staircase where Halbeisen and Hungerbühler's logic can be used to find a lower bound of the maximum odd element in a Collatz cycle. Let tj=[jm/l] - [(j-1)m/l], j=1, 2, 3, ..., l. Let r denote gcd(l, m). The parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l, consists of r identical sub-vectors. Similarly, the parity vector tj consists of r identical sub-vectors and each of these sub-vectors is the same as the corresponding sub-vector of ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l, except for the first and last elements. First suppose that l and m are relatively prime. When the parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l, is right-rotated by one position (corresponding to a multiplication by 2), it matches tj except for the first two elements of each sub-vector. The first mismatch corresponds to a loss of 3m-1 and the second mismatch corresponds to a gain of 2∙3m-1. In general, the loss is ∑2i(l/r)3m-1-i(m/r) where the summation is from i=0 to r-1. Let Nl,m denote 2Ml,m- ∑2i(l/r)3m-1-i(m/r). An alternate way of looking at Halbeisen and Hungerbühler's result is;
(2) If c=|2l-3m| and the elements of the interrelated 3n+c cycles are positive, Ml,m is greater than or equal to the minimum elements in the interrelated 3n+c cycles (not necessarily primitive) and Nl,m is less than or equal to the maximum odd elements in the interrelated 3n+c cycles. Analogous results apply if the elements of the cycles are negative. (The sign of the elements of a cycle in the 3n+c sequence cannot change. The signs of the elements in interrelated 3n+c cycles are the same [the sign is determined by 2l-3m].) Also, both Ml,m and Nl,m are in one of these cycles.
All 3n+c cycles appear to be generated from cycles where c=|2l-3m|. For example, when l=11, and m=7, Ml,m=3767, Nl,m=6805 and |2l-3m|=139. By Proposition (2), 3767/139 (approximately equal to 27) must be greater than or equal to the minimum element in the c=-1 cycle (17) and 6805/139 (approximately equal to 49) must be less than or equal to the maximum odd element in the c=-1 cycle (91). For the c=-1 cycle of (5, 7, 10), M3,2=5 and N3,2=7 (-1 equals 23-32). Note the relevance of the Catalan conjecture (now Mihăilescu's theorem). For c=-17, the cycles are (85, 119, 170, 85, 119, 170), (103, 146, 73, 101, 143, 206), and (65, 89, 125, 179, 260, 130). -17 equals 26-34, M6,4=85, and N6,4=119. 85 is greater than 73 and 65. 119 is less than 143 and 179. The cycle (85, 119, 170) is not primitive and reduces to the c=-1 cycle. Note that 17 must divide 33+23·3 for this to be possible. For the c=1 cycle of (4, 1), 1 equals 22-31, and M2,1=N2,1=1. There can be no other such c=1 cycles. This leaves the possibility of interrelated 3n+c cycles where one of the cycles is not primitive and reduces to a c=1 cycle.
When l and m are not relatively prime and c=|2l-3m|, the cycles generated from Ml,m are not primitive (due to properties of binomial expansions). For example, when c=29-36, reducing the cycle generated by M9,6 effectively divides 29-36 by (29-36)/(23-32). For l=6, 9, 12, 15, 18, ..., and m=3, 6, 9, 12, 15, ..., there are 0, 0, 1, 2, 3, ..., odd cycle elements between Ml,m and Nl,m. In this case, c, Ml,m, and Nl,m have at least one prime factor in common. The respective prime factors are 37, 7·31, 13·109, 10177, 78697, .... The respective c values (after reduction) are 1, -1, -11, -49, -179, ..., and the respective cycles are (1, 2), (5, 7, 10), (19, 23, 29, 38), (65, 73, 85, 103, 130), and (211, 227, 251, 287, 341, 422), .... Similar results are obtained for other combinations of l and mWhen l and m are relatively prime and c=|2l-3m|, the cycles generated from Ml,m are primitive.
The largest upper bound of minima in potential 3n+c cycles having the same number of even elements can also be easily determined. An empirical result is;
(3) Ml,m/|2l-3m|, l=m+k, m=1, 2, 3, ...., increases until the sign of 2l-3m changes and then decreases. More precisely, the maximum Ml,m/|2l-3m| value occurs immediately before the sign change or at the sign change.
For example, M6,4/|26-34| is greater than M5,3/|25-33|, 25-33 is positive, and 26-34 is negative (the maximum occurs at the sign change). M8,5/|28-35| is greater than M9,6/|29-36|, 28-35 is positive, and 29-36 is negative (the maximum occurs before the sign change).
The Minimum Element in a 3n+c Cycle and the Continued-Fraction Convergents of Log(3)/Log(2)
In his 1978 article, Crandall proved a lower bound of 17985 for the number of odd elements in a 3n+1 cycle (n>0) using the minimum in a hypothetical cycle and the continued-fraction convergents of log(3)/log(2). Recent 3n+1 cycle research frequently involves use of the continued-fraction convergents of log(3)/log(2). For some purposes, this is too restrictive; valuable information is thrown away. This is the rationale for defining generalized continued-fraction convergents (allowing less accurate approximations). Let (c, d), (e, f), and (g, h) denote successive continued-fraction convergents of a real number and let j denote the ceiling of h/f. (ei, fi) and (ei+c, fi+d), i=1, 2, 3, ..., j, can be considered to be generalized continued-fraction convergents of the real number. (Sometimes it is convenient to consider just (ei+c, fi+d), i=1, 2, 3, ..., j, to be the generalized continued-fraction convergents of the real number; ei+c and fi+d are then relatively prime.) The generalized continued-fraction convergents of log(3)/log(2) are (2, 1), (3, 2), (4, 2), (5, 3), (6, 4), (8, 5), (9, 6), (11, 7), (16, 10), (19, 12), (24, 15), (27, 17), (38, 24), (46, 29), (57, 36), (65, 41), (76, 48), (84, 53), .... An empirical result is;
(4) The absolute values of 2l-3m increase monotonically for (l, m) values that are generalized continued-fraction convergents of log(3)/log(2) (excluding (2, 1), (4, 2), (6, 4), and (9, 6)).
The significance of this is that for a given c value, there can be at most one solution of c=|2l-3m| where (l, m) are generalized continued-fraction convergents of log(3)/log(2) (excluding (2, 1), (4, 2), (6, 4), and (9, 6)).
An algorithm for generating the lengths and numbers of odd elements of 3n+1 or 3n-1 (n>0) sequences associated with potential cycles having 1-2 sequence vectors almost always gives the generalized continued-fraction convergents of log(3)/log(2) (and only these values). The algorithm gives four (l, m) values that are not generalized continued-fraction convergents of log(3)/log(2) and four generalized continued-fraction convergents of log(3)/log(2) are not given. (This algorithm will be discussed in detail later. See the section "The Minimum in a Collatz Cycle".) This is more evidence that the generalized continued-fraction convergents of log(3)/log(2) are of significance to the 3n+1 problem. The continued-fraction convergents of log(3)/log(2) will be further generalized in the next section.
M-Cycles and an Algorithm for Determining When the Sign of 2l-3m for a Fixed l-m Value Changes
A cycle with no even elements can occur only if (3/2)i(n+c)-c=n where n is odd, i<h, and 2h is the largest power of 2 that divides n+c ((3/2)i(n+c)-c is an incomplete jump). The only primitive cycle with no even elements is the cycle {-1} for c=1 (where c=-n). An m-cycle is a hypothetical cycle in the 3n+1 sequence (n>0) having m local minima. (In this article, such cycles in the 3n+c sequence will be referred to as being M-cycles.) In 1977, R. P. Steiner7 proved that 1-cycles can't exist. A jump in the 3n+c sequence ending in the initial odd element is analogous to a 1-cycle (there would be exactly one even element in the cycle). Empirical results are;
(5) A cycle in the 3n+c sequence having only one even element can occur only if c=|2l-3l-1|.
(6) A cycle in the 3n+c sequence having only one odd element can occur only if c=|2l-3|.
A successor of a, a>0, in the 3n+c sequence is greater than a if 3ma>2la-cA (that is, 3m/2l>1-cA/(2la)) where the values of l, m, and the integer A depend on the parity vector between a and the successor of a. For a sufficiently large a, the successor of a is larger than a if 3m/2l>1. For a sufficiently large initial sequence value a, the following algorithm generates the parity vector (denoted by p) of a 3n+c sequence (with positive elements) where twice the minimum element in the sequence is larger than the maximum odd element in the sequence. Set x to 1, p1 to 1, and i to 2. Then repeat the following operations. Set pi to 1, x to (3/2)x, and increment i. If x>1, set pi to 0, x to (1/2)x, and increment i. The resulting parity vector is 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, .... The portion of the parity vector up to the Mth 0 gives an M-cycle. (The first 1 in the parity vector is taken to follow the 0 so that a cycle is formed.) The resulting (l, m) values (where l is the number of elements in the cycle and m is the number of odd elements) for the M-cycles are (3, 2), (5, 3), (8, 5), (11, 7), (13, 8), (16, 10), (19, 12), (21, 13), (24, 15), (27, 17), (30, 19), (32, 20), (35, 22), (38, 24), (40, 25), (43, 27), (46, 29), (49, 31), (51, 32), (54, 34), (57, 36), (59, 37), (62, 39), (65, 41), (68, 43), (70, 44), (73, 46), (76, 48), (78, 49), (81, 51), (84, 53), .... Empirical results are;
(7) These (l, m) values include the generalized continued-fraction convergents of log(3)/log(2) (except for (2, 1), (4, 2), (6, 4), and (9, 6)). Also, 2m>l.
(8) The difference in l values of a pair of successive M-cycles is 2 or 3 and the difference in m values is 1 or 2 respectively.
(9) There are either 3 or 4 successive M-cycles where the difference in l values is 3 (2l-3m is negative for the third or fourth M-cycle and positive otherwise).
(10) When there are 4 successive M-cycles where the difference in l values is 3, the values of Ml,m/|2l-3m| increase up until the third M-cycle and then decrease. The absolute values of 2l-3m usually increase, but may decrease for the third M-cycle. When there are 3 successive M-cycles where the difference in l values is 3, the values of Ml,m/|2l-3m| usually increase, but may decrease for the third M-cycle. The absolute values of 2l-3m usually increase, but may decrease for the third M-cycle (but only if the value of Ml,m/|2l-3m| increases). After the third or fourth successive M-cycle where the difference in l values is 3, the value of Ml,m/|2l-3m| decreases and the absolute value of 2l-3m increases (the difference in l values is 2).
(11) There are no successive M-cycles where the difference in l values is 2.
A table of the ]jm/l[ -](j-1)m/l[, j=1, 2, 3, ..., l, values (given in the columns) for (l, m)=(3, 2), (5, 3), (8, 5), ..., (103, 65) is;
``` 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 0 1 0 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 0 1 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 0 1 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 0 1 0 0 1 1 0 1 1
0 1 1 0 1 1 0 0 1 1 0 1 1 0 0 1 0 0 1 1 0 1 1 0 0 1 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 0 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1
0 1 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1
0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 1 0 0 1
0 1 1 0 1 1 0 0 1 1 0 1 1 0 1 1 1 0 1 1 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 0 1 1 0 1 1 0 0 1 0 0 1 1 0 1 1
0 1 1 0 0 1 0 0 1 1 0 1 1 0 0 1 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 0 1 1 0 1 1 0 1 1 1 0 1 1
0 0 1 0 0 1 0 0 1 0 0 0 1 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 0 1 0 0 1 0 0 0 1 0 0 1
0 1 1 0 1 1 0 1 1 1 0 1 1 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 0 1 1 0 1 1 1 1 1 1
0 0 1 0 0 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 0 1 0 0 1 1 0 0 1
0 1 1 0 1 1 0 0 1 1 0
1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1
0 1 1 0 1 1 1 1 1 1
0 0 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1
0 1 0 0 1 1 0 1 1
0 1 1 0 0 1 0 0
1 1 1 1 1 1 1 1
0 0 0 0 1 0 0 1
1 1 1 0 1 1 0
0 1 1 1 1 1 1
0 0 0 0 0 0
1 1 1 1 1 1
0 1 1 0 1 1
0 0 1 0 0
1 1 1 1 1
0 1 0 0 1
0 1 1 0
1 1 1 1
0 0 0 0
1 1 1
0 1 1
0 0
1 1
0 1
0
1
0
```
Note that there are rows of values consisting of all 0's, rows of values consisting of all 1's, rows of values consisting of all 1's except for an initial value of 0, and pairs of adjacent rows of values where the element-by-element sum consists of all 1's (disregarding the first element of the first row). The elements in one of these pairs of adjacent rows appear to become all 0's or 1's eventually. Let qj-1=]jm1/l1[ - ](j-1)m1/l1[, j=2, 3, 4, ..., l1, where (l1, m1) is a generalized continued-fraction convergent of log(3)/log(2), gcd(l1, m1)=1, and (l1, m1)≥(11, 7). Let l2, l2<l1, denote the number of elements in an M-cycle generated from the parity vector p and let rj=qj, j=1, 2, 3, ..., l2. If l2 is the last of a succession of l values having a difference of 3, switch the last two elements of r. An empirical result is;
(12) A rotation of r matches the first l2 elements of the parity vector p (usually, a right-rotation of 3 suffices).
The absolute values of 2l-3m where (l, m) are multiples of a continued-fraction convergent (as usually defined) of log(3)/log(2) increase monotonically since the values of Ml,m/|2l-3m| are the same. Truncated Ml,m/(2l-3m) values corresponding to the generalized continued-fraction convergents of log(3)/log(2) (excluding the multiples of convergents) are;
l=2, m=1, 1
l
=3, m=2, -5
l
=5, m=3, 4
l=8, m=5, 24
l=11, m=7, -27
l=19, m=12, -219
l=27, m=17, 108
l=46, m=29, 281
l=65, m=41, 867
l=84, m=53, -6143
l=149, m=94, 2419
l=233, m=147, 4862
l=317, m=200, 9266
l=401, m=253, 19584
l=485, m=306, 75028
l=569, m=359, -81063
l=1054, m=665, -3664765
l=1539, m=971, 72058
.
.
.
The initial difference between successive (l, m) values is (1, 1). After the 2l-3m values change signs, the difference between successive (l, m) values becomes (2, 1) (the (l, m) value immediately before the sign change). After the 2l-3m values change signs again, the difference between successive (l, m) values becomes (3, 2) (the (l, m) value immediately before the sign change), etc. Note that the Ml,m/|2l-3m| values increase until the sign changes.
Empirical results (showing why the parity vector p is of importance) are;
(13) If the maximum Ml,m/|2l-3m|, l=m+k, m=1, 2, 3, ..., value occurs immediately before the sign change of 2l-3m, then the (l, m) value is that of an M-cycle generated from the parity vector p.
(14) If the maximum Ml,m/|2l-3m|, l=m+k, m=1, 2, 3, ..., value occurs at the sign change of 2l-3m and the (l, m) value is not that of an M-cycle generated from the parity vector p, then the (l-1, m-1) value is that of an M-cycle generated from p.
A graph of Ml,m/|2l-3m|, l=m+k, m=1, 2, 3, ..., 10 and k=1, 2, 3, ..., 10 is;
Let k1 denote the difference between the l and m values of a generalized continued-fraction convergent of log(3)/log(2) where l and m are relatively prime. Let k2 denote the difference between the l and m values of the next generalized continued-fraction convergent of log(3)/log(2) where gcd(l, m)=1 and denote Ml,m/|2l-3m| for this (l, m) value by min2. An empirical result is;
(15) The Ml,m/|2l-3m| values where k1<l-m<k2 are less than min2 except possibly when (l, m) is a generalized continued-fraction convergent of log(3)/log(2) where gcd(l, m)≠1.
For example, (149, 94) is a generalized continued-fraction convergent of log(3)/log(2) where gcd(l, m)=1 and in this case k1=55 and Ml,m/|2l-3m|=2419. For the next generalized continued-fraction convergent of log(3)/log(2) where gcd(l, m)=1 ((233, 147)), k2=86, and Ml,m/|2l-3m|=4862. For k=55, 56, 57, ..., 86, the largest Ml,m/|2l-3m| values occur for (l, m)=(149, 94), (152, 96), (154, 97), (157, 99), (160, 101), (163, 103), (165, 104), (168, 106), (171, 108), (173, 109), (176, 111), (179, 113), (182, 115), (184, 116), (187, 118), (190, 120), (192, 121), (195, 123), (198, 125), (201, 127), (203, 128), (206, 130), (209, 132), (211, 133), (214, 135), (217, 137), (219, 138), (222, 140), (225, 142), (228, 144), (230, 145), and (233, 147) respectively and the Ml,m/|2l-3m| values (truncated) are 2419, 219, 130, 388, 433, 142, 220, 6143, 219, 158, 558, 390, 147, 281, 1605, 219, 191, 867, 361, 152, 356, 1004, 219, 231, 1561, 340, 175, 463, 763, 219, 281, and 4862 respectively. The only Ml,m/|2l-3m| value larger than 4862 (6143) occurs when (l, m)=(168, 106), a generalized continued-fraction convergent of log(3)/log(2) where gcd(l, m)≠1.
Upper bounds of minima in 3n+c cycles can then be determined using the generalized continued-fraction convergents of log(3)/log(2).
Miscellaneous Properties of Minima in 3n+c Cycles
When the element following an odd element i in the 3n+c sequence is defined to be (3i+c)/2 instead of 3i+c, 1-2 sequence vectors become 0-1 sequence vectors. Note that 2m must be greater than l for there to be any cycles having 0-1 sequence vectors. For example, when l=11 and m=7, the parity vectors of cycles having 0-1 sequence vectors are (1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1), (1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0), (1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1), (1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1), and (1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1) (corresponding to primitive cycles for c=139). The 3n+139 cycle corresponding to the second parity vector is {-4151, -6157, -9166, -4583, -6805, -10138, -5069, -7534, -3767, -5581, -8302} and twice the largest element in the cycle (-3767) is greater than the smallest odd element (-6805). In the other interrelated 3n+139 cycles having 0-1 sequence vectors, twice the largest element in a cycle is not greater than the smallest odd element in the cycle. Let min denote the element of a cycle in the 3n+c sequence having the smallest absolute value and let max denote the odd element having the largest absolute value. More empirical results are;
(16) If c=|2l-3m| and 2m>l, there is at least one primitive cycle having a 0-1 sequence vector and there are usually primitive cycles not having 0-1 sequence vectors. If c=|2l-3m|, 2m>l, and the greatest common divisor of l and m is 1, there is at least one cycle among the interrelated primitive cycles having 0-1 sequence vectors where 2|min|>|max| (there is no cycle among the interrelated primitive cycles not having 0-1 sequence vectors where 2|min|>|max|). If c=|2l-3m|, 2m<l, and gcd(l, m)=1, there is at least one cycle among the interrelated primitive cycles where 2|min|>|max|.
(17) 2|min|>|max| for a primitive 3n+c cycle where gcd(l, m)=1 only if c=|2l-3m|.
Usually, if c=|2l-3m|, 2m>l, and gcd(l, m)=1, there is exactly one cycle among the interrelated primitive cycles having 0-1 sequence vectors where 2|min|>|max|. When c=1675, l=9, and m=7, there are two primitive cycles where 2|min|>|max|. For one of these cycles, min=-2219, max=-4429, and 2|min| is barely greater than |max| (min=-2363 and max=-3997 for the other cycle). Usually, if c=|2l-3m|, 2m>l, and gcd(l, m)≠1, there is at least one cycle among the interrelated primitive cycles having 0-1 sequence vectors where 2|min|>|max|. When c=9823, l=14, and m=8, there doesn't appear to be a primitive cycle where 2|min|>|max|. If c=|2l-3m|, 2m<l, and gcd(l, m)=1, there appears to be exactly one cycle among the interrelated primitive cycles where 2|min|>|max|. If c=|2l-3m|, 2m<l, and gcd(l, m)≠1, there doesn't appear to be a primitive cycle where 2|min|>|max|. Usually, if a primitive 3n+c cycle where c properly divides 2l-3m exists and gcd(l, m)≠1, then 2|min| is not greater than |max|. When c=145, l=12, and m=8, min=-617, max=-1231, and 2|min| is barely greater than |max|. Also, when c=493, l=12, and m=8, min=-1829, max=-3635, and 2|min| is barely greater than |max|.
A graph of the proportions of 3n+c cycles having 0-1 sequence vectors for c less than or equal to 499, 997, 1499, 1999, ..., 9997 is;
More on m-Cycles
Let K denote the number of odd elements in a 3n+1 cycle (n>0) and L the number of even elements. In 2004, Simons8 proved that if a nontrivial 2-cycle exists, then (K+L, K) must be a convergent in the continued-fraction expansion of log2 3. Let δ denote log(3)/log(2). Using an empirically derived lower bound for the minimum element in an m-cycle, Simons and de Weger9 proved that δK<K+L<1.000001δK. As shown above, 3n+c cycles with large elements and having a constrained dynamic range behave similarly, that is, the (K+L, K) values include the generalized continued-fraction convergents of log(3)/log(2). This will be discussed in more detail in following sections.
Characterization of Cycles in the 3n+c Sequence
Attachment points in a cycle are defined to be even integers i such that i-c is divisible by 3 (and (i-c)/3 is not already in the cycle). An attachment point i will be said to be primary if 4i is not an attachment point. If i is a primary attachment point and 4 divides i, then i/4 will be said to be the secondary attachment point, etc. In the cycle for c=1, -68 is the only attachment point. (There are no attachment points in the other known 3n+1 cycles.) In the following, 3n+c sequences are extended backwards from attachment points. "Odd" paths are taken, that is, if i is an even integer in the extended sequence and 3 divides i-c, then the path (i-c)/3 (and not 2i) is taken. Presumably, an odd integer divisible by 3 (denoted by t) will be reached. The t value of a primary attachment point will be said to be the "proxy" of the first odd element in the cycle after the preceding primary attachment point (in the case where there is only one primary attachment point, the preceding primary attachment point is the primary attachment point). (Note that the sequence vector of the cycle elements starting with the first odd element after the preceding primary attachment point and ending with the primary attachment point is a 0-1 sequence vector; this is the motivation for taking "odd" paths when extending sequences backwards from attachment points. For the c=1 cycle, t equals -21 and is the proxy of the cycle element -17.) As will be shown, the proxy concept allows a simplified description of 3n+c cycles with attachment points. The extended sequences have many properties and, unlike Bohm and Sontacchi's formula, are amenable to Diophantine analysis. Cycles with at least one attachment point will be considered in this section and the next seventeen sections (cycles with no attachment points will be discussed at the end of the article). An empirical result is;
(18) There exists at least one primitive 3n+c cycle having an attachment point for every c value.
Let u denote the first odd element in a 3n+c cycle after a primary attachment point. In this article, the average of the |u| values and the absolute values of their proxies is used to characterize a 3n+c cycle. Let a denote the number of primary attachment points in a cycle. When a>0, the average of the |u| values and the absolute values of their proxies will be taken to be the maximum likelihood estimator for the parameter λ of an exponential probability distribution (f(x; λ)=λe-λx, x≥0). The function g(x)=λe-λx where positive and negative x values are allowed will be used to compute the domain of the absolute values of the u values and the absolute values of their proxies. The domain of the absolute values of the u values and the absolute values of their proxies is determined by -log(|u1|/λ)/λ, -log(|t1|/λ)/λ, -log(|u2|/λ)/λ, -log(|t2|/λ)/λ, -log(|u3|/λ)/λ, -log(|t3|/λ)/λ, ..., -log(|ua|/λ)/λ, -log(|ta|/λ)/λ. For example, the (u, t) values for a cycle for c=121 are (19, -9), (395, 51), (65, 39), (119, 183), (335, 147), (281, 27), (101, -237), (53, -63), (35, 21), and (23, -159). A plot of the |u| and |t| values versus their domain is;
(The x values have been scaled up by a factor of 104 and all but two data points are shown. No shape-preserving interpolation of the data points is done.) Note that all the u and t values in a cycle are distinct (although two t values can have the same absolute value). The domain of the absolute values of the u values and the absolute values of their proxies appears to be a small interval about zero, the largest x value usually being larger than the absolute value of the smallest x value (so that the maximum |u| or |t| value times the minimum |u| or |t| value is usually less than λ2). Let utmax denote the maximum |u| or |t| value in a cycle and utmin the minimum |u| or |t| value. An empirical result (based on the 663743 cycles for the c values less than or equal to 199999) is;
(19) utmaxutmin2 is almost always less than δ.
A histogram of the utmaxutmin2 values for the cycles where c≤199999 is;
Fifty bins are used. A histogram of the 65 values greater than δ is;
Ten bins are used. The maximum value is 2.1673 (less than δ2). The a values of these cycles are small (less than or equal to 28) and the |L-K| values are relatively small. In fourteen of these cycles, L=K. Also, 2KL if L>K and 2L>K if L<K. The rationale for including the absolute values of the proxies in the average is that they appear to have the same exponential curve as the |u| values. Of course, |u| values are local minima and their average is related to the minimum in a cycle.
Belaga10 proved that if 2l-3k>0 where l is the length of and k is the number of odd elements in a 3x+d cycle (x>0), then 1 ≤ n < d/(2l/k-3) where n is the smallest odd element in the cycle. Let umax denote the largest |u| value in a cycle and umin the smallest |u| value. Let tmax denote the largest |t| value in a cycle and tmin the smallest |t| value. Empirical results (based on the cycles where c≤199999) are;
(20) If L<K, 2c/utmin>|2(K+L)/K-3|. If LK, c/utmin>|2(K+L)/K-3|. If L<K, 3c/umin>|2(K+L)/K-3|. If L≥K, c/umin>|2(K+L)/K-3|.
(21) |1-c/utmin|, |1-c/umin|, and |1-c/tmin| are usually greater than |2(K+L)/K-3|.
A histogram of the a values for the 4458 exceptions for utmin is;
The largest
a value is 15. A histogram of the a values for the 13811 exceptions for umin
is;
The largest
a value is 30. A histogram of the a values for the 13698 exceptions for tmin
is;
The largest
a value is 17. The respective percentages of exceptions for utmin
, umin, and tmin are 0.67%, 2.08%, and 2.06%. The expressions "1+c/X0" and "|1-c/X0|" where X0 is a lower bound of the minimum in an m-cycle are relevant to Simons and de Weger's work. An empirical result (based on the 27041 cycles having the (K+L, K) value of an M-cycle generated from the parity vector p for c less than or equal to 199999) is;
(22)
If the (K+L, K) value of a cycle equals that of an M-cycle generated from the parity vector p, |1-c/utmin| and |1-c/tmin| are almost always greater than |2(K+L)/K-3|. If the (K+L, K) value of a cycle equals that of an M-cycle generated from the parity vector p, |1-c/umin| is greater than |2(K+L)/K-3|.
The (c, L, K, a) values for the exceptions (for utmin and tmin) are (1631, 5, 8, 2), (36791, 15, 25, 3), (36791, 15, 25, 4), and (186793, 9, 15, 2). Empirical results (based on the cycles having the (K+L, K) value of an M-cycle generated from the parity vector p for c less than or equal to 199999) are;
(23) If the (K+L, K) value of a cycle equals that of an M-cycle generated from the parity vector p, utmax is greater than cNK+L,K/|2K+L-3K|/6 and utmin is less than cMK+L,K/|2K+L-3K|.
(24) If the (K+L, K) value of a cycle equals that of an M-cycle generated from the parity vector p, δ2a+14>utmax/utmin.
(The latter proposition may fail for very large upper bounds of c values.)
Let i denote the smallest power of δ such that δi>utmax/utmin. A histogram of the i-2a values of the cycles having (K+L, K) values equal to those of M-cycles generated from the parity vector p for c less than or equal to 997 is;
This distribution has a mean, standard deviation, and sample size of -1.1163, 2.3629, and 301 respectively. A histogram of the i-2a values of the cycles having (K+L, K) values equal to those of M-cycles generated from the parity vector p for c=7153 is;
This distribution has a mean, standard deviation, and sample size of -0.6905, 1.3898, and 2843 respectively. A histogram of the i-2a values of the cycles having (K+L, K) values equal to those of M-cycles generated from the parity vector p for c less than or equal to 49999 is;
This distribution has a mean, standard deviation, and sample size of -1.3532, 2.4842, and 9332 respectively. A histogram of the i-2a values of the cycles having (K+L, K) values equal to those of M-cycles generated from the parity vector p for c less than or equal to 99997 is;
This distribution has a mean, standard deviation, and sample size of -1.4960, 2.6018, and 14663 respectively. For c upper bounds of 19999, 39997, 59999, 79999, 99997, 119999, 139999, 159997, 179999, and 199999, the distributions have respective means of -1.1470, -1.3009, -1.3818, -1.4450, -1.4960, -1.5158, -1.4805, -1.5319, -1.6246, and -1.4388, respective standard deviations of 2.3157, 2.4634, 2.5125, 2.5437, 2.6018, 2.6466, 2.6293, 2.6861, 2.7751, and 2.5766, and respective sample sizes of 5811, 7604, 9629, 14365, 14663, 14704, 15672, 15919, 16443, and 27041. The standard deviation generally increases as the c upper bound increases, but in an irregular fashion. When c=185357, the distribution has a mean, standard deviation, and sample size of -1.4773, 2.3212, and 6321 respectively. When c=186793, the distribution has a mean, standard deviation, and sample size of -0.5107, 1.5784, and 4161 respectively. For a c value having a large sample size, the standard deviation of the distribution is generally smaller than that of the composite distribution.
Let i denote the smallest power of δ such that δi>utmax/utmin. A table of the number of i values for given a values of the 3n+c cycles for c less than or equal to 99997 is;
``` a=1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
i=1 3423 253 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 1845 1858 182 29 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3 1003 4035 1254 254 45 8 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4 428 3412 2103 791 213 59 28 4 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 267 2542 3113 1688 689 256 101 29 13 6 0 1 0 0 0 0 0 0 0 0 0 0 0 0
6 145 1829 3082 2322 1389 707 332 153 74 44 22 10 5 2 1 0 0 0 0 0 0 0 0 0
7 103 1245 2546 2579 1919 1242 726 452 294 161 86 70 37 13 10 5 6 4 1 0 0 0 0 0
8 66 886 2149 2488 2066 1539 1183 851 592 363 306 166 125 76 47 39 24 14 13 8 7 3 4 2
9 45 602 1590 2012 1968 1770 1461 1089 913 703 516 410 252 212 161 132 99 80 49 35 34 18 12 13
10 25 397 1161 1592 1662 1625 1526 1335 1051 914 775 590 482 389 332 257 210 173 146 118 81 88 80 61
11 31 297 726 1116 1340 1345 1310 1250 1206 1048 893 735 641 543 485 393 331 311 267 217 215 156 134 121
12 13 193 483 782 943 1081 1140 1117 1041 923 932 755 704 634 551 518 494 423 411 283 308 294 257 220
13 11 139 326 505 661 750 869 856 866 815 802 727 696 609 603 534 552 521 442 455 379 381 321 287
14 8 96 193 324 448 596 616 700 697 696 664 626 609 602 552 547 523 494 501 397 402 423 355 363
15 6 49 135 222 318 368 462 501 506 503 519 474 473 491 489 524 485 479 413 412 435 409 334 369
16 7 43 78 127 197 261 320 343 363 372 419 397 410 411 388 417 417 386 400 362 370 355 349 305
17 4 18 67 95 123 151 205 229 232 283 260 264 276 322 309 313 318 316 312 290 315 309 298 255
18 3 10 34 58 66 117 136 164 178 185 200 216 214 217 216 227 264 257 260 261 229 234 234 259
19 1 6 17 36 59 62 82 90 124 141 131 134 153 138 155 158 169 172 178 208 193 185 185 167
20 2 8 15 23 34 54 54 92 65 64 88 93 103 115 129 134 138 138 152 160 127 146 131 129
21 1 1 10 22 26 29 38 43 63 49 65 98 76 74 106 96 92 110 111 114 101 110 105 115
22 2 4 5 5 10 26 25 34 42 44 44 58 55 65 64 62 70 65 74 72 75 68 75 70
23 0 2 1 7 14 6 17 19 18 21 33 32 36 29 40 42 41 53 48 56 57 55 51 65
24 0 0 1 5 4 6 8 9 12 18 16 24 21 28 21 30 25 40 26 42 36 38 22 37
25 1 0 0 2 2 3 5 9 8 13 13 17 15 19 20 21 28 25 24 30 20 27 20 25
26 0 0 0 1 1 4 5 2 4 8 11 9 7 11 8 15 12 11 17 21 17 21 22 18
27 0 0 3 0 2 2 4 5 7 6 8 3 9 10 4 9 15 11 18 10 14 15 8 7
28 0 0 0 0 1 0 2 4 4 3 0 3 7 4 5 9 9 7 8 7 4 6 9 16
29 0 0 0 1 2 2 0 1 0 1 1 4 2 5 4 5 5 7 8 6 5 7 2 8
30 0 0 0 0 0 0 1 1 1 2 1 3 1 2 3 3 2 2 3 2 1 3 2 5
31 0 0 0 0 0 0 0 0 1 1 0 1 1 1 0 1 3 4 2 0 0 0 3 0
32 0 0 0 0 0 0 0 0 0 0 0 2 1 0 0 1 0 0 2 3 1 1 0 3
33 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 2 0 0 2 3
34 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 0 1 0 0 0
35 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1
36 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 2
37 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0
38 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
39 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
```
For a=1, 2, 3, ..., 24, the means of these distributions are 2.3075, 5.0025, 6.9688, 8.2655, 9.3172, 10.2088, 10.9385, 11.5711, 12.0145, 12.4725, 12.8472, 13.3144, 13.6613, 14.0181, 14.2853, 14.6265, 14.8641, 15.0889, 15.3236, 15.6876, 15.6275, 15.8181, 15.9344, and 16.2146 respectively, the standard deviations are 2.0704, 2.7025, 2.8688, 2.9650, 3.0535, 3.1151, 3.1678, 3.2250, 3.2571, 3.3044, 3.3179, 3.5003, 3.4540, 3.4775, 3.4843, 3.5874, 3.5740, 3.6121, 3.6397, 3.6950, 3.5507, 3.5912, 3.5758, and 3.7751 respectively, and the sample sizes are 7440, 17925, 19279, 17086, 14210, 12069, 10658, 9383, 8375, 7388, 6805, 5922, 5412, 5023, 4704, 4495, 4334, 4105, 3887, 3572, 3428, 3353, 3016, and 2926 respectively. A plot of the means and standard deviations is;
A linear least-squares fit of the means plotted against log(a) is;
For c less than or equal to 199999 and a=1, 2, 3, ..., 24, the means of the distributions are 2.3960, 5.0181, 6.9037, 8.2448, 9.3514, 10.2428, 10.9874, 11.5606, 12.0565, 12.4427, 12.8846, 13.3813, 13.7249, 14.0554, 14.2985, 14.5896, 14.8762, 15.0961, 15.3512, 15.6148, 15.6866, 15.8611, 15.9868, and 16.2639 respectively, the standard deviations are 2.1291, 2.7093, 2.8870, 2.9846, 3.0753, 3.1413, 3.1873, 3.2188, 3.2653, 3.3025, 3.3770, 3.5027, 3.4663, 3.5132, 3.4893, 3.5289, 3.5896, 3.5850, 3.6588, 3.7029, 3.6014, 3.6314, 3.6232, and 3.7260 respectively, and the sample sizes are 10995, 31363, 38306, 33299, 27300, 23335, 20718, 18800, 16910, 15307, 13815, 12411, 11597, 10821, 10147, 9817, 9199, 8784, 8420, 8048, 7673, 7422, 6682, and 6371 respectively. The means and standard deviations of the distributions do not change much for different upper bounds of c values.
An empirical result (based on the 663743 cycles with attachment points for c less than or equal to 199999) is;
(25) δa+1>utmax/λ.
Let i denote the smallest power of δ such that δi>utmax/λ. A table of the number of i values for given a values of the 3n+c cycles for c less than or equal to 99997 is;
``` a=1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
i=1 6226 4689 984 224 43 11 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 1214 10462 9699 5654 2684 1338 697 332 154 92 49 21 10 12 1 2 1 0 0 0 0 0 0 0
3 0 2774 7635 9026 8045 6440 5074 3883 2929 2135 1716 1270 913 701 571 466 366 316 224 169 142 115 101 73
4 0 0 961 2121 3161 3788 4081 4052 3908 3562 3303 2779 2494 2237 2015 1784 1730 1471 1353 1132 1067 1001 822 782
5 0 0 0 61 277 487 774 1043 1249 1428 1495 1530 1583 1572 1555 1555 1487 1517 1457 1373 1265 1278 1161 1135
6 0 0 0 0 0 5 30 73 134 169 235 298 383 462 499 607 644 664 704 727 742 762 710 686
7 0 0 0 0 0 0 0 0 1 2 7 24 29 38 62 79 102 129 139 157 198 187 205 227
8 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 4 8 10 14 14 10 16 21
9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2
```
The means and standard deviations of these distributions for a=1, 2, 3, ..., and 24 are 1.1632, 1.8932, 2.4447, 2.7741, 3.0665, 3.2831, 3.4708, 3.6421, 3.7950, 3.9260, 4.0253, 4.1496, 4.2777, 4.3761, 4.4613, 5.5660, 4.6297, 4.7181, 4.7970, 4.8917, 4.9501, 4.9806, 5.0491, and 5.0967 respectively and 0.3695, 0.6364, 0.6699, 0.6861, 0.7050, 0.7165, 0.7371, 0.7547, 0.7717, 0.7878, 0.8047, 0.8381, 0.8502, 0.8718, 0.8869, 0.9147, 0.9237, 0.9445, 0.9407, 0.9536, 0.9793, 0.9554, 0.9822, and 0.9891 respectively. A plot of these values is;
The means and standard deviations of these distributions (and the distributions themselves) do not change much for different upper bounds of c values. A linear least-squares fit of the means plotted against log(a) is;
For c less than or equal to 199999 and a=1, 2, 3, ..., 24, the means of the distributions are 1.1843, 1.9007, 2.4258, 2.7702, 3.0672, 3.2924, 3.4819, 3.6423, 3.8029, 3.9201, 4.0369, 4.1636, 4.2806, 4.3784, 4.4652, 4.5543, 4.6415, 4.7115, 4.7964, 4.8831, 4.9571, 4.9980, 5.0503, and 5.1163 respectively and the standard deviations are 0.3877, 0.6345, 0.6626, 0.6896, 0.7049, 0.7218, 0.7356, 0.7564, 0.7710, 0.7909, 0.8143, 0.8350, 0.8608, 0.8690, 0.8859, 0.9013, 0.9226, 0.9423, 0.9380, 0.9441, 0.9708, 0.9649, 0.9948, and 0.9876 respectively.
Usually, a |u| or |t| value greater than λ can be "chained" to the next larger |u| or |t| value by multiplying it by δ.
The Chain Equation
The chain equation for an m-cycle is given by 3k(i)ai-1=2k(i+1)+l(i)ai+1-2l(i), i=0, 1, 2, ..., m-1, where k(i) is the number of successive odd numbers leading up to a local maximum (due to typographical difficulties, k(i) is used to denote ki), l(i) is the number of successive even numbers going down to the next local minimum (due to typographical difficulties, l(i) is used to denote li), and ai are natural numbers (see Simons and de Weger's article). The chain equation is the starting point of m-cycle theory and is used to derive Simons and de Weger's Lemma 4;
0 < Λ < ∑1/xi
where xi is a local minimum, the summation is from i=1 to m, and Λ denotes (K+L) log(2)-K log(3). The above characterization of a 3n+c cycle is related to this lemma. For most 3n+c cycles, c times the sum of the reciprocals of |u| and |t| is greater than the absolute value of Λ. When (K+L, K) is equal to (or approximately equal to) a generalized continued-fraction convergent of log(3)/log(2), c times the sum of reciprocals is sometimes smaller than the absolute value of Λ. An empirical result (based on the cycles with attachment points for c≤199999) is;
(26) 10c∑(1/|ui| + 1/|ti|) where the summation is from i=1 to a is greater than |Λ|. 16c∑1/|ui| where the summation is from i=1 to a is greater than |Λ|. 128c∑1/|ti| where the summation is from i=1 to a is greater than |Λ|.
Simons and de Weger's Corollary 5 (of Lemma 4) is;
0 < Λ < m/xminm/X0
where xmin is the global minimum of an m-cycle and X0 is an empirically derived lower bound for xmin. (As will be shown, the chain equation is not entirely applicable to the above characterization of a 3n+c cycle, so even if Proposition (26) is accepted, a proposition analogous to Corollary 5 can't be derived without some knowledge of the relative sizes of the |u| and |t| values.) Simons and de Weger show that Λ is exponentially small in terms of its coefficients and use a result from transcendence theory due to Rhin11 to derive Lemma 12;
Λ > e-13.3(0.46057+log K).
Using Corollary 5 gives Simons and de Weger's Corollary 13 (where the global minimum of an m-cycle is estimated in terms of K);
xmin < me13.3(0.46057+log K)
Brox12 proved that for any m there are only finitely many m-cycles.
The Number of M-Cycles
A histogram of the [3·log(c)]-a values for the 27014 cycles for c less than or equal to 199999 having (K+L, K) values equal to those of M-cycles generated by the parity vector p is;
This distribution has a mean of 29.2260 and a standard deviation of 4.3270. The only negative value (-1) occurs for the c=1 cycle. Also, small values only occur for small c values. Denote the (L, K) value of a cycle generated from the parity vector p by (L', K'). If the (L, K) value of a cycle doesn't equal that of an M-cycle generated from the parity vector p, but K=K' for some K', set d to |L-L'|, otherwise set d to zero. If KK' for any K' then K+1=K' for some K', and if K≠1, K-1=K' for some K'. In this case, set d to the absolute value of L minus the average of the corresponding L' values. If K=1, set d to |L-L'| where K+1=K'. An empirical result (based on the 663743 cycles with attachment points for c less than or equal to 199999) is;
(27) [3·log(c)+d/δ] is almost always greater than a.
This proposition is true for 99.9235% of the cycles. (Current m-cycle theory lacks such a relationship between (L, K) and m.) A histogram of the [3·log(c)+d/δ]-a values for the 23132 cycles with attachment points for c less than or equal to 9997 superimposed on a normal probability distribution having the same mean and standard deviation is;
When c=32389, the difference is -14 for a cycle where (L, K)=(876, 984) and a=234. When c=33989, the difference is -13 for a cycle where (L, K)=(1116, 1224) and a=296. When c=22429, the difference is -13 for a cycle where (L, K)=(553, 689) and a=137. When c=45679, the difference is -12 for a cycle where (L, K)=(527, 646) and a=138. When c=43925, the difference is -10 for a cycle where (L, K)=(680, 800) and a=175. When c=45613, the difference is -10 for a cycle where (L, K)=(1181, 1345) and a=290. When c=21809, the difference is -9 for a cycle where (L, K)=(480, 576) and a=129. When c=6515, the difference is -8 for a cycle where (L, K)=(436, 506) and a=122. When c=11707, the difference is -7 for a cycle where (L, K)=(624, 713) and a=165. When c=43739, the difference is -7 for a cycle where (L, K)=(969, 1083) and a=250. When c=46823, the difference is -7 for a cycle where (L, K)=(1105, 1192) and a=296. When c=5827, the difference is -6 for a cycle where (L, K)=(426, 504) and a=114. When c=27547, the difference is -5 for a cycle where (L, K)=(588, 696) and a=149. When c=39605, the difference is -5 for a cycle where (L, K)=(912, 1016) and a=237. When c=28771, the difference is -5 for a cycle where (L, K)=(738, 843) and a=190. When c=42641, the difference is -5 for a cycle where (L, K)=(1555, 1722) and a=382. When c=46841, the difference is -5 for a cycle where (L, K)=(585, 690) and a=151. When c=48091, the difference is -5 for a cycle where (L, K)=(596, 720) and a=147. When c=35191, the difference is -4 for a cycle where (L, K)=(522, 672) and a=135. When c=8243, the difference is -4 for a cycle where (L, K)=(805, 929) and a=196. When c=16001, the difference is -4 for a cycle where (L, K)=(613, 726) and a=151. When c=14957, the difference is -3 for a cycle where (L, K)=(500, 591) and a=129. When c=16487, the difference is -3 for a cycle where (L, K)=(529, 625) and a=134. When c=17315, the difference is -3 for a cycle where (L, K)=(696, 750) and a=194. When c=26213, the difference is -3 for a cycle where (L, K)=(428, 514) and a=113. When c=46801, the difference is -3 for a cycle where (L, K)=(564, 664) and a=145. When c=17449, the difference is -2 for a cycle where (L, K)=(336, 454) and a=94. When c=38689, the difference is -2 for a cycle where (L, K)=(633, 726) and a=164. When c=20807, the difference is -2 for a cycle where (L, K)=(484, 590) and a=119. When c=22343, the difference is -2 for a cycle where (L, K)=(1185, 1282) and a=306. When c=41893, the difference is -2 for a cycle where (L, K)=(471, 549) and a=128. (There is only one cycle each for c equal to 6515, 14957, 17315, 28771, 32389, 33989, 39605, 42641, and 48091. There are only two cycles each for c equal to 5827, 8243, 11707, 16001, 17449, 20807, 21809, 22429, 27547, 35191, 41893, 43925, 45613, and 45679. The largest number of cycles occurs for c=46841; there are ten cycles having four distinct (L, K) values. Except for c=22343, the (L, K) values of all the cycles for the above c values are large. When c=22343, there are four cycles having (L, K) values of (1185, 1282), (1080, 1027), (7, 17), and (7, 17).) Also, the difference is -1 for 9 cycles when c is less than or equal to 49999. Similar results apply for c greater than 50000. Let cmax denote the upper bound of the c values. Other than being discrete-valued, the distribution of differences resembles a normal probability distribution where the mean is approximately equal to the variance for relatively small cmax values (the mean of the above distribution is 24.579 and the standard deviation is 5.061). Assuming the d/δ-a values cancel each other out, the expected mean is (1/n)∑(3·log(c)) where n is the number of c values less than or equal to cmax. The expected mean of the above distribution is 24.631. A normal probability plot of the differences for the 102 cycles for the c values less than or equal to 97 is;
The mean of this distribution is 9.716 with a 95% confidence interval of (9.066, 10.365) and the standard deviation is 3.308 with a 95% confidence interval of (2.907, 3.836). The expected mean is 10.785. When cmax=997, the mean of the distribution of differences is 16.949 and the standard deviation is 3.918 (there are 1556 cycles with attachment points for this cmax value). A plot of the distribution of differences superimposed on a normal probability distribution having the same parameters is;
For larger cmax values, a normal probability distribution doesn't fit the data very well. When cmax=19999, the mean of the distribution of differences is 26.879 and the standard deviation is 5.957 (there are 48449 cycles with attachment points for this cmax value). A plot of the distribution of differences superimposed on a normal probability distribution having the same parameters is;
When cmax=29999, the mean of the distribution of differences is 28.332 and the standard deviation is 6.612. A plot of the distribution of differences superimposed on a normal probability distribution having the same parameters is;
A detailed plot showing the left-hand tails of the above distributions is;
A plot of the means, standard deviations, and expected means of the distributions of differences versus log(cmax) for cmax values of 997, 1999, 2999, 3997, 4999, 5999, 6997, ..., 49999 is;
(The curves for the means and expected means intersect.) Although the mean starts out being smaller than the expected mean and eventually becomes larger than the expected mean, the curve of means still appears to be mostly linear (when plotted against log(cmax)). A quadratic (f(x)=p1x2+p2x+p3) least-squares fit of the means plotted against log(cmax) for cmax=997, 1999, 2999, ..., 199999 is;
p1=0.1109 with a 95% confidence interval of (0.1067, 0.1151), p2=1.346 with a 95% confidence interval of (1.258, 1.434), and p3=2.688 with a 95% confidence interval of (2.23, 3.147). SSE=0.3537, R-square=0.9999, adjusted R-square=0.9999, and RMSE=0.04237.
The non-linearity of the standard deviation curve is due to a large number of cycles with attachment points (2849) for c=7153. A cubic (f(x)=p1x3+p2x2+p3x+p4) least-squares fit of the standard deviations plotted against log(cmax) for cmax=997, 1999, 2999, ..., 199999 is;
p
1=0.02579 with a 95% confidence interval of (0.0208, 0.03079), p2=-0.5298 with a 95% confidence interval of (-0.6803, -0.3792), p3=4.046 with a 95% confidence interval of (2.55, 5.542), and p4=-7.176 with a 95% confidence interval of (-12.07, -2.28). SSE=1.053, R-square=0.9976, adjusted R-square=0.9976, and RMSE=0.07328. When c=40741, there are 1429 cycles with attachment points. The total number of cycles is large enough for cmax=40999 that there is only a minor drop in the standard deviation due to these cycles. There are also 3782 cycles with attachment points for c=71515, 1238 cycles with attachment points for c=72023, and 1658 cycles with attachment points for c=84095 (and larger numbers of cycles for larger c values).
The expected number of cycles is cmax/log(cmax) times the expected mean. A plot of the expected number of cycles and the number of cycles having attachment points for cmax values of 997, 1999, 2999, ..., 49999 is ;
For these cmax values, the expected number of cycles is larger than the actual number of cycles. The expected number of cycles becomes smaller than the actual number of cycles for larger cmax values.
A quadratic least-squares fit of the number of cycles plotted against (cmax+1)/1000 or (cmax+3)/1000 (when 1000 does not divide cmax+1) for cmax=997, 1999, 2999, ..., 199999 is;
(The coefficients are poorly conditioned when the number of cycles is plotted against cmax.) p1=3.081 with a 95% confidence interval of (2.951, 3.211), p2=2750 with a 95% confidence interval of (2723, 2777), and p3=-8472 with a 95% confidence interval of (-9645, -7299). SSE=1.517e+09, R-square=0.9998, adjusted R-square=0.9998, and RMSE=2775.
A linear least-squares fit of the proportions of cycles in the left-hand tail of the distribution of differences (up to and including cycles for which the difference is 0) plotted against (cmax+1)/1000 or (cmax+3)/1000 for cmax=19999, 39997, 59999, ..., 199999 is;
p1=5.014e-5 with a 95% confidence interval of (4.652e-5, 5.376e-5) and p2=0.000286 with a 95% confidence interval of (0.0002635, 0.0003084). SSE=1.627e-9, R-square=0.9922, adjusted R-square=0.9912, and RMSE=1.426e-5. Why these proportions appear to be increasing linearly is unknown.
Modifying the scaling factor (1/δ) slightly has a significant effect on the distribution of differences. A plot of [3·log(c)+d/1.5]-a superimposed on a normal probability distribution having the same parameters for cmax=29999 is;
Even this small of a change in the scaling factor (from 1/1.584962501 to 1/1.5) skews the distribution significantly (there are fewer negative differences and more large positive differences). When c=6515, (L, K)=(436, 506), and a=122, the difference is -3 for a cycle. When c=21809, (L, K)=(480, 576), and a=129, the difference is -4 for a cycle. When c=22429, (L, K)=(553, 689), and a=137, the difference is -7 for a cycle. Also, the difference is -1 for two cycles when cmax=39997. A plot of the means, standard deviations, and expected means of the distributions of differences versus log(cmax) for cmax values of 997, 1999, 2999, ..., 29999 is;
The mean of the distribution of differences still appears to increase linearly (mostly) with log(cmax). A linear least-squares fit of the means plotted against log(cmax) for cmax=997, 1999, 2999, ..., 29999 is;
A linear least-squares fit of the means plotted against log(cmax) for cmax=997, 1999, 2999, ..., 29999 when the scaling factor of 1/δ is used is;
A plot of [3·log(c)+d/1.67]-a superimposed on a normal probability distribution having the same parameters for cmax=29999 is;
A linear least-square fit of the means plotted against log(cmax) for cmax=997, 1999, 2999, ..., 29999 is;
The slight deviation in the linearity of the means when plotted against log(cmax) is due to the large number of cycles for c=7153.
A plot of [3·log(c)+d/1.75]-a superimposed on a normal probability distribution having the same parameters for cmax=9997 is;
For this scaling factor, the distribution is skewed to the left. The "best" scaling factor is then somewhere between 1/1.75 and 1/1.5. The best least-squares fit of the means plotted against log(cmax) appears to occur for the scaling factor of 1/δ.
The [3·log(cmax)]-[3·log(c)]+a values have a probability distribution resembling that of a chi-square distribution with 6 degrees of freedom. A histogram of the [3·log(cmax)]-[3·log(c)]+a values and a plot of the expected numbers of values for a chi-square distribution for cmax=997 is;
For larger cmax values, a chi-square probability distribution doesn't fit the data very well. A cubic (f(x)=p1x3+p2x2+p3x+p4) least-squares fit of the means of the [3·log(cmax)]-[3·log(c)]+a distributions plotted against log(cmax) for cmax=997, 1999, 2999, ..., and 199999 is;
p1=0.3299 with a 95% confidence interval of (0.2874, 0.3584), p2=-7.309 with a 95% confidence interval of (-8.379, -6.239), p3=58.28 with a 95% confidence interval of (47.64, 68.91), and p4=-152.6 with a 95% confidence interval of (-187.4, -117.8). SSE=53.2, R-square=0.9982, adjusted R-square=0.9982, and RMSE=0.521. Much of the non-linearity of the curve is due to the large number of cycles with attachment points for c=7153 (and for a few other c values). A quadratic (f(x)=p1x2+p2x+p3) least-squares fit of the corresponding variances of the distributions plotted against (cmax+1)/1000 or (cmax+3)/1000 (when cmax+1 is not divisible by 1000) is;
(The coefficients are poorly conditioned when the variances are plotted against cmax.) p1=0.02988 with a 95% confidence interval of (0.02814, 0.03163), p2=39.04 with a 95% confidence interval of (38.68, 39.41), and p3=-44.09 with a 95% confidence interval of (-59.83, -28.36). SSE=2.732e+05, R-square=0.9998, adjusted R-square=0.9998, and RMSE=37.24.
For a fixed c value, the [3·log(c)]-a values may be distributed over several (L, K) values. For a cycle having the (K+L, K) value of an M-cycle generated from the parity vector p, 3·log(c)+1≥a. For every c value, there should be only finitely many cycles having these (K+L, K) values.
More on the Characterization of 3n+c Cycles
Let λu denote the average of the |u| values for a cycle and λt the average of the |t| values. A plot of the |u| values versus their domain for a cycle for c=85 is;
(The x values have been scaled up by a factor of 103 and all but one data point is shown. No shape-preserving interpolation of the data points is done. λu equals 368.0.) A plot of the |t| values versus their domain for the same cycle is;
(The x values have been scaled up by a factor of 103 and all but one data point is shown. No shape-preserving interpolation of the data points is done. λt equals 324.0.) An empirical result (based on the 663743 cycles for c less than or equal to 199999) is;
(28) umaxuminu2 is almost always less than δ and tmaxtmint2 is almost always less than δ.
A histogram of the 83 exceptions for u is;
Ten bins are used. The maximum value is 2.1759 (less than δ2). The a values of these cycles are small (less than or equal to 77) and the |L-K| values are relatively small. A histogram of the L-K values is;
The |L-K|/L values are less than 3/8. A histogram of the 51 exceptions for t is;
Ten bins are used. The maximum value is 2.7221 (greater than δ2). The a values of these cycles are small (less than or equal to 59) and the |L-K| values are relatively small. A histogram of the L-K values is;
An empirical result (based on the cycles for c≤199999) is;
(29) δ (a+2)/2>umaxu and δ(a+2)/2>tmaxt.
Let i denote the smallest power of δ such that δi>umaxu. A table of i values versus a values for the cycles for c less than or equal to 9997 is;
a
=2 3 4 5 6 7 8 9 10
i
=1 2979 912 284 101 31 7 6 2 0
2 870 1315 1048 671 445 302 167 104 93
3 0 139 362 468 439 413 337 304 296
4 0 0 0 35 100 157 161 176 172
5 0 0 0 0 0 2 16 19 46
The means and standard deviations of these distributions for a=2, 3, 4, ..., and 10 are 1.2260, 1.6733, 2.0460, 2.3427, 2.5990, 2.8241, 3.0204, 3.1752, and 3.2817 respectively and 0.4183, 0.5810, 0.6160, 0.6625, 0.7063, 0.7290, 0.7785, 0.7562, and 0.8133 respectively. A plot of these values is;
A least-squares fit of the means plotted against log(a) is;
Let i denote the smallest power of δ such that δi>tmaxt. A table of i values versus a values for the cycles for c less than or equal to 9997 is;
a
=2 3 4 5 6 7 8 9 10
i
=1 2857 803 261 75 31 19 4 4 1
2 992 1331 962 613 376 240 151 83 78
3 0 232 471 535 495 462 354 309 294
4 0 0 0 52 113 160 164 190 196
5 0 0 0 0 0 0 14 19 38
The means and standard deviations of these distributions for a=2, 3, 4, ..., and 10 are 1.2578, 1.7587, 2.1240, 2.4424, 2.6798, 2.8661, 3.0480, 3.2264, and 3.3163 respectively and 0.4374, 0.6159, 0.6457, 0.6680, 0.7085, 0.7232, 0.7496, 0.7436, and 0.7806 respectively. A plot of these values is;
A least-squares fit of the means plotted against log(a) is;
An advantage to working with the average of the |u| and |t| values (that is, λ) is that the |u| and |t| values are more evenly distributed along the exponential curve (also, there are distinct maximum and minimum |u| or |t| values when a=1). Let λ1 denote the average of the |u| and |t| values less than or equal to λ. If λ1utmin, let λ2 denote the average of the |u| and |t| values less than or equal to λ1, if λ2utmin, let λ3 denote the average of the |u| and |t| values less than or equal to λ2, etc. An empirical result is;
(30) δa+1 is usually greater than λ/λ1, λ12 (when λ2 is defined), λ23 (when λ2 and λ3 are defined), etc.
When c is less than or equal to 9997, the proportions of λ/λ1, λ12, λ23, ..., λ56 values greater than δa+1 are 0.02892 (669/23132), 0.02224 (453/20369), 0.01443 (241/16700), 0.00528 (67/12678), 0.00065 (6/9164), and 0.00000 (0/6051) respectively. When c is less than or equal to 19999, the proportions of λ/λ1, λ12, λ23, ..., λ56 values greater than δa+1 are 0.02334 (1131/48449), 0.01827 (818/44776), 0.01219 (475/38967), 0.00423 (133/ 31452), 0.00063 (15/23967), and 0.00006 (1/16878) respectively. When c is less than or equal to 29999, the proportions of λ/λ1, λ12, λ23, ..., λ56 values greater than δa+1 are 0.01920 (1411/73486), 0.01616 (1122/69438), 0.01105 (691/62531), 0.00391 (204/ 52145), 0.00066 (27/40996), and 0.00007 (2/29809) respectively. When a≥3, λ2 and λ3 are usually defined. This is the basis of the following empirical result;
(31) δ4a+4 is usually greater than utmax/utmin. If a>3, δ4a+5>utmax/utmin.
Of the 23132 cycles for c less than or equal to 9997, a=1 for 2562 cycles and δ4a+4<utmax/utmin for 46 cycles (about 1.80% of the cycles), a=2 for 3849 cycles and δ4a+4<utmax/utmin for 56 cycles (about 1.45% of the cycles), a=3 for 2366 cycles and δ4a+4<utmax/utmin for 20 cycles, and a≥4 for 14355 cycles and δ4a+4<utmax/utmin for 4 cycles (in every instance, a=4). If a=1, δ20>utmax/utmin. If a=2, δ20>utmax/utmin. If a=3, δ22>utmax/utmin. For very large cmax values, larger multiples of a+1 would be required for Proposition (31) to still be valid.
Let i denote the smallest power of δ such that δi>λ/λ1. For c≤99997 and a=1, 2, 3, ..., 12 , the means of the distributions of numbers of i values are 1.7859, 2.1929, 2.3974, 2.5008, 2.6018, 2.6874, 2.7393, 2.8119, 2.8246, 2.8736, 2.8835, and 2.9667 respectively, the standard deviations are 1.7134, 1.2101, 0.9375, 0.8372, 0.8025, 0.7870, 0.7788, 0.8094, 0.7969, 0.8103, 0.8091, and 0.8625 respectively, and the sample sizes are 7440, 17925, 19279, 17086, 14210, 12069, 10658, 9383, 8375, 7388, 6805, and 5922 respectively. A linear least-squares fit of the means plotted against log(a) is;
p1=0.4509 with at 95% confidence interval of (0.4186, 0.4832) and p2=1.854 with a 95% confidence interval of (1.796, 1.913). SSE=0.01321, R-square=0.9898, adjusted R-square=0.9887, and RMSE=0.03634. For c≤199999 and a=1, 2, 3, ..., 12, the means of the distributions are 1.8542, 2.2050, 2.3763, 2.4966, 2.6051, 2.6894, 2.7532, 2.8125, 2.8349, 2.8644, 2.9013, and 2.9741 respectively, the standard deviations are 1.7636, 1.2116, 0.9308, 0.8453, 0.8076, 0.7930, 0.7929, 0.8023, 0.7961, 0.8096, 0.8341, and 0.8607 respectively, and the sample sizes are 10995, 31363, 38306, 33299, 27300, 23335, 20718, 18800, 16910, 15307, 13815, and 12411 respectively. The means and standard deviations of the distributions do not change much for different c upper bounds.
For the 48449 cycles for c less than or equal to 19999, there are 3374, 5899, 4419, 3297, 2567, and 2073 cycles where a equals 1, 2, 3, 4, 5, and 6 respectively. A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=1 superimposed on the corresponding chi-square probability distribution with 2 degrees of freedom is;
A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=2 superimposed on the corresponding chi-square distribution with 3 degrees of freedom is;
A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=3 superimposed on the corresponding chi-square distribution with 4 degrees of freedom is;
A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=4 superimposed on the corresponding chi-square distribution with 5 degrees of freedom is;
A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=5 superimposed on the corresponding chi-square distribution with 6 degrees of freedom is;
A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=6 superimposed on the corresponding chi-square distribution with 7 degrees of freedom is;
For a given a value, the distribution of the minimum power of δ greater than utmax/utmin doesn't depend on the upper bound of c very muchPlots of the means of the distributions for a=1, 2, 3, ..., 10 and cmax=1999, 3997, 5999, ..., 29999 are;
From bottom to top, the curves are for a equal 1 to 10. Plots of the standard deviations of the distributions for a=1, 2, 3, 4, and 5 and cmax=1999, 3997, 5999, ..., 29999 are;
From bottom to top, the curves are for a equal 1 to 5. Plots of the standard deviations of the distributions for a=6, 7, 8, 9, and 10 for cmax=1999, 3997, 5999, ..., 29999 are;
Plots of the sample sizes of the distributions for a=1, 2, 3, ..., 10 and cmax=997, 1999, 2999, ..., 99997 are;
The aberrant curve is for a=1. From top to bottom, the other curves are for a equal 2 to 10 (the curves for a=2 and a=3 cross). The number of cycles having a given a value appears to increase linearly with cmax (although the curve eventually begins to resemble a step function).
The sample sizes for a=1, 2, 3, ..., 24 plotted against (cmax+1)/1000 or (cmax+3)/1000 for cmax=9997, 19999, 29999, ..., 199999 are;
A linear least-squares fit of the curve for a=1 and cmax=997, 1999, 2999, ..., 99997 is;
p1=0.06089 with a 95% confidence interval of (0.05805, 0.06373) and p2=1640 with a 95% confidence interval of (1474, 1805). SSE=1.676e+007, R-square=0.9485, adjusted R-square=0.948, and RMSE=413.6. A linear least-squares fit of the curve for a=2 and cmax=997, 1999, 2999, ..., 99997 is;
p
1=0.1691 with a 95% confidence interval of (0.1647, 0.1734) and p2=1670 with a 95% confidence interval of (1418, 1922). SSE=3.89e+007, R-square=0.9839, adjusted R-square=0.9838, and RMSE=630. A linear least-squares fit of the curve for a=3 and cmax=997, 1999, 2999, ..., 99997 is;
p
1=0.1959 with a 95% confidence interval of (0.1935, 0.1983) and p2=336.9 with a 95% confidence interval of (196.5, 477.2). SSE=1.208e+007, R-square=0.9962, adjusted R-square=0.9962, and RMSE=351.
Although a chi-square probability distribution doesn't model the distribution very well for larger a values, the number of values in the right-hand tails of the distributions is still about the same. For the 194664 cycles for c less than or equal to 69997, there are 5463, 12533, 13505, 12072, 9941, 8240, 7103, 6118, 5521, and 4913 cycles where a equals 1, 2, 3, ..., and 10 respectively. A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=7 superimposed on the corresponding chi-square probability distribution with 8 degrees of freedom is;
A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=8 superimposed on the corresponding chi-square probability distribution with 9 degrees of freedom is;
When the curve for the actual distribution rises more steeply than the curve for the chi-square distribution (when a>7), there are fewer values in the right-hand tail of the distribution than for the chi-square distribution. A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=9 superimposed on the corresponding chi-square probability distribution with 10 degrees of freedom is;
A plot of the distribution of the minimum power of δ greater than utmax/utmin for the cycles where a=10 superimposed on the corresponding chi-square probability distribution with 11 degrees of freedom is;
Given the number of cycles for an a value, the number of values of the smallest power of δ greater than utmax/utmin in the right-hand tail of the distribution can be estimated using the chi-square distribution. Proposition (27) then gives a relationship between the L-K value of a cycle and an upper bound of utmax/utmin.
The Minimum |L-K| Value of the 3n+c Cycles for a Given c Value
(c, L, K) values of cycles with no attachment points where there are also no interrelated cycles with attachment points for c less than or equal to 99997 are (1, 0, 1), (1, 1, 1), (1, 1, 2), (11, 1, 3), (49, 1, 4), (179, 1, 5), (601, 1, 6), (1931, 1, 7), (6049, 1, 8), (18659, 1, 9), (57001, 1, 10), (791, 2, 8), (85, 4, 8), (145, 4, 8), (57001, 1, 10), (2167, 4, 12), (66469, 3, 13), (8497, 4, 15), (29267, 4, 16), (53095, 4, 16), (16133, 12, 24), and (78313, 5, 19). These values are included in the following. For the c values less than or equal to 19999, there is only 1 cycle (counting only one of interrelated cycles) for 2260 c values and the mean and standard deviation of the L-K values are 1.013 and 21.245 respectively. A histogram of the L-K values superimposed on a normal probability distribution having this mean and standard deviation is;
(There is a tendency for the L-K values to be a multiple of 6.) For the c values less than or equal to 19999, there are exactly 2 cycles (counting only one of interrelated cycles) for 2351 c values and the mean and standard deviation of the L-K values of the smaller |L-K| values are 0.185 and 11.385 respectively. A histogram of the L-K values superimposed on the corresponding normal probability distribution is;
For the c values less than or equal to 19999, there are exactly 3 cycles (counting only one of interrelated cycles) for 1132 c values and the mean and standard deviation of the L-K values of the smallest |L-K| values are 0.281 and 7.669 respectively. A histogram of the L-K values superimposed on the corresponding normal probability distribution is;
For the c values less than or equal to 19999, there are exactly 4 cycles (counting only one of interrelated cycles) for 452 c values and the mean and standard deviation of the L-K values of the smallest |L-K| values are -0.215 and 5.457 respectively. A histogram of the L-K values superimposed on the corresponding normal probability distribution is;
The corresponding standard deviations when there are 5, 6, 7, and 8 cycles (counting only one of interrelated cycles) are 4.195, 3.277, 3.109, and 2.515 respectively (when there are j cycles, the standard deviation is about 1/j times the standard deviation when there is 1 cycle). Plots of j times the standard deviation for j=1, 2, and 3 versus log(cmax) for cmax values of 997, 1999, 2999, ..., 99997 are;
Plots of j times the standard deviation for j=4, 5, 6, 7, and 8 versus log(cmax) for cmax values of 997, 1999, 2999, ..., 39997 are;
Plots of the proportions of c values having 1, 2, 3, 4, 5, and 6 cycles (counting only one of interrelated cycles) versus cmax for cmax values of 997, 1999, 2999, ..., 49999 are;
For cmax=49999, an upper bound of the proportion of c values having only one cycle is about 0.354 (the upper bound of the proportion may increase further as cmax increases, but appears to be leveling off). For relatively small cmax values, an upper bound of |L-K| can be computed beforehand using the slope (about 4.74) and intercept (about -25.67) of the line giving the standard deviation of the distribution of L-K values versus log(cmax). A linear least-squares fit for cmax values up to 24997 when there is only one cycle is;
For example, the standard deviation for cmax=19999 should be 21.272 (compared to the actual standard deviation of 21.245). There are 6667 c values less than or equal to 19999, so there should be less than 2360 (6667·0.354) c values with only one cycle (counting only one of interrelated cycles). For a normal probability distribution with a mean of 0, a standard deviation of 21.272, and a sample size of 2360, the expected number of L-K values becomes less than 1/2 for |L-K|>62. In the actual distribution, the L-K values range from -112 to 108. Out of the 2260 L-K values, 14 values are less than -62 and 13 values are greater than 62. About 99% of the L-K values are then in the expected range.
The standard deviation of the L-K values of the smallest |L-K| values doesn't increase linearly with log(cmax)A cubic least-squares fit for cmax values up to 99997 when there is only one cycle (counting only one of interrelated cycles) is;
p1=0.1906 with a 95% confidence interval of (0.1675, 0.2136), p2=-4.622 with a 95% confidence interval of (-5.28, -3.963), p3=41.51 with a 95% confidence interval of (35.29, 47.73), and p4=-121.5 with a 95% confidence interval of (-140.9, -102.2) . SSE=3.046, R-square=.9992, adjusted R-square=.9992, and RMSE=0.1782. ( The adjusted R-square value for a quadratic least-squares fit is .9968.) A quadratic least-squares fit of the standard deviation of the L-K values when there is only one cycle (counting only one of interrelated cycles) plotted against log(cmax) for cmax=9997, 19999, 29999, ..., 199999 is;
p1=1.57 with a 95% confidence interval of (1.476, 1.665), p2=-25.25 with a 95% confidence interval of (-27.32, -23.19), and p3=117.2 with a 95% confidence interval of (106.1, 128.4). SSE=0.3439, R-square=0.9997, adjusted R-square=0.9996, and RMSE=0.1422.
Let (L1, K1), (L2, K2), (L3, K3), ..., (Ln, Kn) denote the (L, K) values of the cycles for a given c value sorted by increasing L values. For the c values less than or equal to 39997, the standard deviations of the L-K values of the smallest |L-K| values when (L1, K1), (L2, K2), (L3, K3), ..., (Li, Ki), i=1, 2, 3, ..., 7, are not included and there are 2, 3, 4, ..., 8 cycles are as follows;
# cycles 2 3 4 5 6 7 8
i=1 28.617 14.938 9.807 7.066 5.640 4.343 3.372
2 n/a 26.498 14.273 8.897 7.540 5.456 3.876
3 n/a n/a 25.309 14.022 10.020 6.825 5.020
4 n/a n/a n/a 24.508 14.577 9.484 7.686
5 n/a n/a n/a n/a 23.872 14.445 9.315
6 n/a n/a n/a n/a n/a 24.387 13.634
7 n/a n/a n/a n/a n/a n/a 18.637
When none of the (L, K) values are excluded, the standard deviations of the L-K values of the smallest |L-K| values for 1, 2, 3, ..., and 8 cycles are 26.214, 13.923, 9.465, 6.918, 5.316, 4.213, 3.336, and 3.007 respectively. For the c values less than or equal to 99997, the standard deviations of the L-K values of the smallest |L-K| values when (L1, K1), (L2, K2), (L3, K3), ..., (Li, Ki), i=1, 2, 3, ..., 7, are not included and there are 2, 3, 4, ..., 8 cycles are as follows;
# cycles 2 3 4 5 6 7 8
i=1 37.735 18.932 12.238 8.602 6.502 5.382 4.545
2 n/a 34.344 18.520 11.294 8.939 6.320 5.110
3 n/a n/a 31.834 16.662 12.025 8.160 6.014
4 n/a n/a n/a 30.582 17.415 11.584 7.892
5 n/a n/a n/a n/a 29.094 16.554 11.606
6 n/a n/a n/a n/a n/a 32.073 18.093
7 n/a n/a n/a n/a n/a n/a 32.363
When none of the (L, K) values are excluded, the standard deviations of the L-K values of the smallest |L-K| values for 1, 2, 3, ..., and 8 cycles are 34.620, 18.719, 12.196, 8.565, 6.482, 5.307, 4.256, and 3.966 respectively. For the c values less than or equal to 199999, the standard deviations of the L-K values of the smallest |L-K| values when (L1, K1), (L2, K2), (L3, K3), ..., (Li, Ki), i=1, 2, 3, ..., 7, are not included and there are 2, 3, 4, ..., 8 cycles are as follows;
# cycles 2 3 4 5 6 7 8
i=1 46.682 23.462 14.347 10.278 7.922 6.452 4.885
2 n/a 41.609 21.938 13.610 10.057 7.595 5.576
3 n/a n/a 38.775 20.521 13.873 9.730 6.622
4 n/a n/a n/a 35.499 20.758 13.006 8.901
5 n/a n/a n/a n/a 35.544 18.489 12.346
6 n/a n/a n/a n/a n/a 34.944 18.806
7 n/a n/a n/a n/a n/a n/a 34.597
When none of the (L, K) values are excluded, the standard deviations of the L-K values of the smallest |L-K| values for 1, 2, 3, ..., and 8 cycles are 43.014, 23.079, 15.069, 10.349, 7.923, 6.528, 5.402, and 4.150 respectively. Not including the sorted (L, K) values does not change the standard deviations much. A cubic least-squares fit of the standard deviations of the L-K values when there are exactly two cycles and (L1, K1) is excluded for cmax values up to 99997 is;
p
1=0.1765 with a 95% confidence interval of (0.1543, 0.1987), p2=-4.005 with a 95% confidence interval of (-4.64, -3.37), p3=33.87 with a 95% confidence interval of (27.88, 39.87), and p4=-90.65 with a 95% confidence interval of (-109.3, -71.98). SSE=2.833, R-square=0.993, adjusted R-square=0.993, and RMSE=0.1718.
(L, K) Trees
Cycles for a given c value with (L, K) values that are in arithmetic progression with an increment of (L1, K1) (and aren't multiples of (L1, K1)) are said to be associated with each other. For the 3333 c values less than or equal to 9997, the numbers of c values having 1, 2, 3, ..., and 12 cycles (counting only one of interrelated or associated cycles) are 1115, 1319, 595, 186, 76, 31, 6, 2, 1, 0, 1, and 1 respectively. This distribution has a mean of 1.0816. For a Poisson probability distribution having this parameter, the expected numbers of c values having 1, 2, 3, ..., and 12 cycles are 1130, 1222, 611, 238, 64, 14, 3, 0, 0, 0, 0, and 0 respectively. For the 6667 c values less than or equal to 19999, the numbers of c values having 1, 2, 3, ..., and 12 cycles (counting only one of interrelated or associated cycles) are 2260, 2618, 1152, 393, 161, 53, 17, 9, 2, 0, 1, and 1 respectively. This distribution has a mean of 1.0817. For a Poisson probability distribution having this parameter, the expected numbers of c values having 1, 2, 3, ..., and 12 cycles are 2260, 2445, 1322, 477, 129, 28, 5, 1, 0, 0, 0, and 0 respectively. For the 10000 c values less than or equal to 29999, the numbers of c values having 1, 2, 3, ..., and 12 cycles (counting only one of interrelated or associated cycles) are 3450, 3896, 1707, 597, 232, 72, 28, 11, 4, 0, 2, and 1 respectively. This distribution has a mean of 1.0697. For a Poisson probability distribution having this parameter, the expected numbers of c values having 1, 2, 3, ..., and 12 cycles are 3431, 3670, 1963, 700, 187, 40, 7, 1, 0, 0, 0, and 0 respectively. There are too many cycles for the larger x values for the Poisson probability distribution to model the actual distribution very well. (The number of c values having only 1 cycle is modeled fairly well though. For c less than or equal to 39997, there is only 1 cycle for 4681 c values, the distribution has a mean of 1.0601, and the expected number of c values having only 1 cycle is 4619. For c less than or equal to 49999, there is only 1 cycle for 5909 cycles, the distribution has a mean of 1.0538, and the expected number of c values having only 1 cycle is 5811.) (L, K) trees (generalized associated cycles) model the number of cycles better and are defined in the Journal of Integer Sequences article (cycles with (L, K) values that are in arithmetic progression with an increment of (L2, K2) are also said to be associated with each other, etc.) . For the c values less than or equal to 9997, the numbers of c values having 1, 2, 3, ..., and 8 cycles (counting only one of interrelated cycles or one of cycles in (L, K) trees) are 1115, 1338, 636, 178, 58, 6, 2, and 0 respectively. This distribution has a mean of 1.0255. For a Poisson probability distribution having this parameter, the expected numbers of c values having 1, 2, 3, ..., and 8 cycles are 1195, 1226, 629, 215, 55, 11, and 2 respectively. For the c values less than or equal to 1999, the numbers of c values having 1, 2, 3, ..., and 8 cycles (counting only one of interrelated cycles or one of cycles in (L, K) trees) are 2260, 2664, 1228, 381, 117, 14, 3, and 0 respectively. This distribution has a mean of 1.0228. For a Poisson probability distribution having this parameter, the expected numbers of c values having 1, 2, 3, ..., and 8 cycles are 2397, 2454, 1254, 428, 109, 22, 4, and 1 respectively. For the c values less than or equal to 29999, the numbers of c values having 1, 2, 3, ..., and 8 cycles (counting only one of interrelated cycles or one of cycles in (L, K) trees) are 3450, 3980, 1805, 582, 158, 22, 3, and 0 respectively. This distribution has a mean of 1.0096. For a Poisson probability distribution having this parameter, the expected numbers of c values having 1, 2, 3, ..., and 8 cycles are 3644, 3679, 1857, 625, 158, 32, 5, and 1 respectively. For the c values less than or equal to 39997, the numbers of c values having 1, 2, 3, ..., and 8 cycles (counting only one of interrelated cycles or one of cycles in (L, K) trees) are 4681, 5273, 2369, 773, 205, 27, 5, and 0 respectively. This distribution has a mean of 0.9987. For a Poisson probability distribution having this parameter, the expected numbers of c values having 1, 2, 3, ..., and 8 cycles are 4912, 4905, 2449, 815, 204, 41, 7, and 1 respectively. For the c values less than or equal to 49999, the numbers of c values having 1, 2, 3, ..., and 8 cycles (counting only one of interrelated cycles or one of cycles in (L, K) trees) are 5909, 6579, 2907, 970, 262, 35, 5, and 0 respectively. This distribution has a mean of 0.9933. For a Poisson probability distribution having this parameter, the expected numbers of c values having 1, 2, 3, ..., and 8 cycles are 6172, 6131, 3045, 1008, 250, 50, 8, and 1 respectively. A plot of the distributions for c less than or equal to 49999 is;
In this case, there are too few cycles for the larger x values but the Poisson probability distribution still models the distribution fairly well. For the c values less than or equal to 199999, the numbers of c values having 1, 2, 3, ..., and 10 cycles (counting only one of interrelated cycles or one of cycles in (L, K) trees) are 25074, 26248, 10797, 3486, 893, 137, 28, 2, 2, and 0 respectively. This distribution has a mean of 0.9413. For a Poisson probability distribution having this parameter, the expected numbers of c values having 1, 2, 3, ..., and 10 cycles are 26008, 24482, 11522, 3615, 851, 160, 25, 3, 0, and 0 respectively. The Poisson probability distribution still models the distribution fairly well. The means of the distributions plotted against (cmax+1)/1000 or (cmax+3)/1000 for cmax=9997, 19999, 29999, ..., 199999 are;
Another way to model the number of cycles is to count only one of interrelated cycles, only one of associated cycles, and only one of cycles where the (L, K) values are a multiple of another common (L, K) value. Counting the number of cycles this way, the numbers of c values for c less than or equal to 9997 having 1, 2, 3, ..., and 7 cycles are 1721, 1221, 318, 61, 11, 1, and 0 respectively. This distribution has a mean of 0.6268. For a Poisson probability distribution having this parameter, the expected numbers of cycles having 1, 2, 3, ..., and 7 cycles are 1781, 1116, 350, 73, 11, 1, and 0 respectively. Counting the number of cycles this way, the numbers of c values for c less than or equal to 19999 having 1, 2, 3, ..., and 7 cycles are 3501, 2408, 624, 108, 22, 3, and 1 respectively. This distribution has a mean of 0.6133. For a Poisson probability distribution having this parameter, the expected numbers of cycles having 1, 2, 3, ..., and 7 cycles are 3611, 2214, 679, 139, 21, 3, and 0 respectively. Counting the number of cycles this way, the numbers of c values for c less than or equal to 29999 having 1, 2, 3, ..., and 7 cycles are 5294, 3581, 937, 155, 28, 3, and 2 respectively. This distribution has a mean of 0.6059. For a Poisson probability distribution having this parameter, the expected numbers of cycles having 1, 2, 3, ..., and 7 cycles are 5456, 3306, 1001, 202, 31, 4, and 0 respectively. Counting the number of cycles this way, the numbers of c values for c less than or equal to 39997 having 1, 2, 3, ..., and 7 cycles are 7124, 4765, 1196, 201, 40, 5, and 2 respectively. This distribution has a mean of 0.5968. For a Poisson probability distribution having this parameter, the expected numbers of cycles having 1, 2, 3, ..., and 7 cycles are 7341, 4381, 1307, 260, 39, 5, and 0 respectively. Counting the number of cycles this way, the numbers of c values for c less than or equal to 49999 having 1, 2, 3, ..., and 7 cycles are 8925, 5963, 1461, 259, 49, 8, and 2 respectively. This distribution has a mean of 0.5946. For a Poisson probability distribution having this parameter, the expected numbers of cycles having 1, 2, 3, ..., and 7 cycles are 9197, 5468, 1626, 322, 48, 6, and 1 respectively. (The numbers of cycles for the larger x values are modeled better.) A plot of the distributions for c less than or equal to 49999 is;
For large upper bounds of c values (such as 199999), this approach doesn't model the number of cycles very well; there are too few cycles for the larger x values. For cmax=199999, the numbers of cycles are 36517, 23653, 5268, 958, 193, 51, 17, 3, 2, and 0 respectively and the expected numbers of cycles are 37568, 21544, 6178, 1181, 169, 19, 2, 0, 0, and 0 respectively.
The Number of Prime Factors of 2K+L-3K
The number of prime factors of a natural number n where the prime factors are not necessarily distinct is denoted by Ω(n). For 1≤L, K≤29, the numbers of values of 2K+L-3K where Ω(|2K+L-3K|) equals 1, 2, 3, 4, 5, 6, 7, and 8 are 117, 267, 229, 119, 63, 30, 8, and 6 respectively (there are no prime factors of 2K+L-3K in two instances, so the number of samples is 839 [292-2]). A Poisson probability distribution where λ=1.8760 can be used to model this data. A plot of the data superimposed on a Poisson probability distribution having the same mean (where the numbers of values are counted as being 0, 1, 2, 3, 4, 5, 6, and 7) is;
Corresponding data for smaller L, K upper bounds is;
# prime factors= 1 2 3 4 5 6 7 8
L, K upper bound=2 2
3 7
4 11 2 1
5 17 5 1
6 19 11 4
7 26 16 5
8 29 21 11 1
9 36 27 15 1
10 37 36 22 3
11 46 46 23 4
12 48 54 28 8 3 1
13 54 65 36 8 3 1
14 56 75 41 17 4 1
15 58 90 48 21 5 1
16 61 100 54 28 7 4
17 67 113 66 29 8 4
18 67 121 76 36 14 8
19 75 133 87 40 15 9
20 76 144 97 47 22 11 0 1
21 83 157 108 55 24 11 0 1
22 83 166 124 68 27 13 0 1
23 90 185 133 75 30 13 0 1
24 91 192 148 79 38 18 3 5
25 98 209 164 84 42 18 3 5
26 101 219 179 95 50 22 3 5
27 107 229 201 104 55 22 4 5
28 113 240 210 113 62 30 8 6
A plot of the means of these distributions is;
Corresponding Poisson probability distributions model the number of prime factors fairly well. A linear least-squares fit of the means plotted against the logarithm of the maximum L, K value is;
p1=0.8852 with a 95% confidence interval of (0.838, 0.9324) and p2=-1.111 with a 95% confidence interval of (-1.24, -0.9823). SSE=0.09954, R-square=0.9842, adjusted R-square=0.9836, and RMSE=0.0644. The number of prime factors of 2K+L-3K affects the number of c values covered by the parity vectors corresponding to (K+L, K).
More on the Domain of 3n+c Cycles
In the following, all cycles are assumed to have been found for c values up to certain limits. For the c values less than 1000, at least one primitive cycle where a>1 occurs for every c value except 1, 7, and 37. The largest domain of the absolute values of the u values and the absolute values of their proxies occurs for a cycle for c=5; the domain is (-2.027326e-001, 3.465736e-001). The domains of the cycles for most c values are one, two, or three orders of magnitude smaller. For the c values less than 1000, the smallest domain occurs for a cycle for c=467; the domain is (-1.051072e-007, 1.944498e-007). Small domains also occur for cycles for c=311 and c=343 (and other c values). The smallest domains for cycles for larger c values appear to become increasingly smaller. For example, for c=4501, nine primitive cycles where a>1 occur and the domains are as large as (-3.956808e-004, 1.654824e-003) and as small as (-7.278549e-008, 2.117179e-007).
When the domain of the absolute values of the u values and the absolute values of their proxies is small, the (K+L, K) value of the cycle is likely to be a generalized continued-fraction convergent of log(3)/log(2). (Even for the c=1 cycle, the (K+L, K) value [(11, 7)] is a generalized continued-fraction convergent of log(3)/log(2).) For example, when c=467, the (K+L, K) value is (84, 53), a continued-fraction convergent of log(3)/log(2). When c=311, the (K+L, K) value is (19, 12), a continued-fraction convergent of log(3)/log(2). When c=343, the (K+L, K) value is (149, 94), a generalized continued-fraction convergent of log(3)/log(2). When the domain is small, (K+L, K) values that are small multiples (usually 2) of generalized continued-fraction convergents of log(3)/log(2) occur. For example, when c=4501, the (K+L, K) value is (92, 58), twice a generalized continued-fraction convergent of log(3)/log(2). When the domain is small, (K+L, K) values that are approximately equal to generalized continued-fraction convergents of log(3)/log(2) occur. For example, for a cycle for c=407, the (K+L, K) value is (44, 28), almost equal to a generalized continued-fraction convergent ((46, 29)) of log(3)/log(2). However, cycles occur for some c values where the domain appears to be small but there is no apparent relationship between the (K+L, K) values and the generalized continued-fraction convergents of log(3)/log(2). This occurs for a cycle for c=1813 where the domain is (-7.915834e-007, 3.418000e-006) and the (K+L, K) value is (228, 132) and for a cycle for c=2009 where the domain is (-4.947314e-006, 1.590942e-005) and the (K+L, K) value is (175, 98). The (K+L, K) values are fairly large multiples of approximations of generalized continued-fraction convergents of log(3)/log(2). The domains are no longer "small" for all of the cycles of these large c values.
Extended Sequences of 3n+c Cycles
In this section, the element after an odd element i is defined to be 3i+c. The shortest possible jump from an odd element in the 3n+c sequence to another odd element will be referred to as a "hop" (when the element after an odd element i is defined to be 3i+c, there are two even elements between the initial odd element and the destination of the jump). An example of a "multiple-jump" attachment point (for c=11) is -21→-13→-7→-1, 8. The attachment point is several jumps away from the odd integer divisible by 3, that is, -21 jumps to -13, -13 jumps to -7, -7 jumps to -1, and 8 immediately follows -1. The jumps in multiple-jump attachment points usually consist of a jump followed by one or more hops or one or more hops followed by a jump. On average, there are 1.2 jumps and 2.0 hops in a multiple-jump (based on the 3767 multiple-jump attachment points for the c values less than or equal to 1999). One-jump attachment points (such as -21→-23, -68 [for c=1]) have some special properties.
An example of a "jumped-over" attachment point (for c=13) is 51→358. The destination of the jump is not the attachment point associated with the odd integer divisible by 3; the attachment point (262 in this case) has been "jumped over". Empirical results are;
(32) The destination of a jumped-over attachment point is another attachment point if the destination is even.
(33) The sign in an extended sequence can change only once.
If the sign in the extended sequence of a jumped-over attachment point changes, it usually does so at the attachment point. For example, -63→74 for a cycle for c=95 where the attachment point 26 is jumped over. The extended sequence up to 26 is {-63, -94, -47, -46, -23, 26}. Sometimes there are several jumps before the attachment point is jumped over. For example, -249→-287→-173→-89→118 for a cycle for c=169 where the attachment point 22 is jumped over. The sequence up to 22 (starting with -89) is {-89, -98, -49, 22}. Jumped-over attachment points become rare when the domain of the absolute values of the u values and the absolute values of their proxies becomes small. As will be shown (by way of empirical evidence), the largest possible li value (in the chain equation) in a 3n+1 cycle (n>0) without a jumped-over attachment point is 5. (Of course, determining whether jumped-over attachment points cease to exist for a sufficiently small domain would be difficult.)
An example of a "no-jump" attachment point (for c=11) is -3, 2. The odd integer divisible by 3 is immediately before the attachment point. Empirical results are;
(34) In a cycle having an attachment point, there is at least one no-jump or one-jump attachment point.
(35) In primary, secondary, tertiary, etc., attachment points, the types of attachment points that can occur have a specific order; a no-jump attachment point (not necessarily the primary attachment point) is followed by a jumped-over attachment point, the jumped-over attachment point is followed by a multiple-jump (or one-jump) attachment point, the multiple-jump (or one-jump) attachment point is followed by a no-jump attachment point, etc.
(36) Except when the last attachment point in primary, secondary, tertiary, etc., attachment points is a jumped-over attachment point, the destination of a jumped-over attachment point is the next attachment point.
(37) A jumped-over attachment point cannot be a primary attachment point.
In the chain equation, the element after an odd element i in the 3n+1 sequence (n>0) is defined to be (3i+1)/2. Counting the even element before a primary attachment point and the even elements up to the tertiary attachment point gives 6 even elements. If there are primary, secondary, and tertiary attachment points in a cycle, one of them must be a jumped-over attachment point (by Proposition (35)).
Solving for a one-jump attachment point followed by a no-jump attachment point (in primary, secondary, tertiary, etc., attachment points) gives the Diophantine equation (3/2)h=(8t2+3c)/(t1+c) where 2h is the largest power of 2 that divides t1+c. Solving for a no-jump attachment point followed by a jumped-over attachment point gives (3/2)i=(3/8)(t2+3c)/(t3+c) (usually, 2i is not the largest power of 2 that divides t3+c). Solving for a jumped-over attachment point followed by a one-jump attachment point gives (3/2)i-j=3(t4+c)/(t3+c), etc.
Hardy and Littlewood13 proved that the sequence {frac(xn)} where frac(x) is the fractional part of x is equidistributed for almost all real numbers x>1 (the exceptional set has Lesbeque measure zero). (A sequence of real numbers {xn} is equidistributed on an interval [a, b] if the probability of finding xn in any subinterval is proportional to the subinterval length. The points of an equidistributed sequence form a dense set on the interval [a, b].) The properties of {frac(3/2)n} have been extensively studied. Since {frac(3/2)n} appears to be equidistributed (and thus dense), the equation (3/2)h=(8t2+3c)/(t1+c) (or one of the other equations) is likely to have many solutions (as can be easily verified).
The exponents of 3/2 in the above equations are small. For example, for c=41, the attachment points for primary, secondary, tertiary, and quatenary attachment points for a cycle are 3→29, 128 (a one-jump attachment point), -3, 32 (a no-jump attachment point), -21→2 (an attachment point where 8 is jumped over [the sequence up to 8 is {-21, -22, -11, 8}]), and -87→-55→-31→-13, 2 (a multiple-jump attachment point). The last attachment point is not a one-jump attachment point, but -31 can be substituted for t4 in the equation above. t1=3, t2=-3, and t3=-21 and the values of h, i, and j are 2, 2, and 1 respectively. For c=137, there are primary, secondary, tertiary, and quatenary attachment points for a cycle (starting with a one-jump attachment point) and the successive exponents of 3/2 are 4, 2, and 1. For c=107, there are primary, secondary, tertiary, etc., attachment points for a cycle (starting with a no-jump attachment point) and the successive exponents of 3/2 are 2, 1, 1, and 3. For c=4159, there are primary, secondary, tertiary, etc., attachment points for a cycle (starting with a one-jump attachment point) and the successive exponents of 3/2 are 1, 3, 2, 2, 2, and 1. For c=4519, there are primary, secondary, tertiary, etc. attachment points for a cycle (starting with a one-jump attachment point) and the successive exponents of 3/2 are 2, 2, 1, 1, 2, and 1. When there are more than three attachment points in primary, secondary, tertiary, etc., attachments points, the attachment points are frequently powers of 2 (for c≤151, the only exception occurs for the above cycle for c=137). Another empirical result is;
(38) The last attachment point in the group of primary, secondary, tertiary, etc., attachment points preceding a primary multiple-jump attachment point is a no-jump or jumped-over attachment point.
A Diophantine equation (similar in form to those derived for primary, secondary, tertiary, etc., attachment points) involving the t value of a multiple-jump attachment point (or at least the odd element at the beginning of the last jump of the multiple-jump) and the t value of the preceding no-jump or jumped-over attachment point can be derived.
The Simplified Structure of a 3n+c Cycle Having Attachment Points
Another empirical result is;
(39) A primary attachment point is a multiple-jump attachment point only if the attachment point is one jump away from the first odd cycle element after the preceding primary attachment point.
A consequence of the above proposition (and Proposition (37)) is that a u value jumps (in one jump) to the next primary attachment point, or if its proxy isn't the odd integer immediately before the next primary attachment point, the proxy jumps (in one jump) to the odd integer immediately before the next primary attachment point, etc. This is the aforementioned simplified structure of a cycle.
The Largest Power of 2 That Divides the Difference Between a u Value and Its Proxy
Let 2j be the largest power of 2 that divides the difference between a u value and its proxy. Empirical results are;
(40) For a one-jump attachment point that is a primary attachment point, the expected value of j+1 is the number of odd elements in the jump (counting the destination of the jump).
(41) For a multiple-jump attachment point that is a primary attachment point, j is small (the numbers of j values equal to 1, 2, 3, 4, 5, and 6 for the c values less than or equal to 997 are 555, 447, 82, 9, 1, and 1 respectively).
There is usually a one-jump attachment point in a cycle where j or j+1 equals the number of odd elements in the jump. (For the c=1 cycle, -21≡-17(mod 22) and there are 3 odd elements in the jump.) For example, when c=3013, there are 93 primitive cycles (with a total of 338 attachment points) having an (K+L, K) value of (38, 24) (a generalized continued-fraction convergent of log(3)/log(2)). For 72 of the 202 primary one-jump attachment points, j+1 equals the number of odd elements in the jump and for 48 of the primary one-jump attachment points, j equals the number of odd elements in the jump. A histogram of the differences between the number of odd elements in the jump and j is;
More on the Chain Equation
Note that the modulo operations (ut(mod 2j)) introduce variables corresponding to the ai values in the chain equation (in the chain equation, xi≡-1(mod 2k(i)) where xi is a local minimum). Similarly, the odd integers before the attachment points in no-jump attachments points correspond to ai values so that there are the same number of variables as in the chain equation. From this perspective, the c=1 cycle is a 1-cycle. [The variables of the chain equation are k0=4, l0=1, a0=-1, k1=3, l1=3, and a1=-5.]) However, these equations don't have the same form as the chain equation (except for multiple-jump attachment points where the usual "path" is taken) since, for a one-jump attachment point, the attachment point equals 3i+c where i is the destination of the jump. Similarly, these equations don't have the same form as the chain equation for no-jump attachment points. Simons and de Weger show that all xi are about the same size by "chaining" them. Their Lemma 6 is;
For all i
= 0,1,..., m-1 we have xi+1 < bδxiδ
where b = (1 + X0-1)/21/δ and X0 is a lower bound of xmin. For a one-jump attachment point, the next u value equals {[(3/2)f+1(t+c)-(c/2)]/2}/2g where 2f is the largest power of 2 that divides t+c. This expression is less than (1/2)(3/2)f+1(t+c)=(1/2)(2f+1)δ-1(t+c). If u and t are positive, t=2ja+u where a≥1, so the next u value is less than (1/2)[2f+1-j(2ja+u+c)]δ-1(t+c)=(2f+1-j)δ-1(1/2)(t+c)δ. So if u and t are positive and f+1 (the number of odd elements in the jump [counting the destination of the jump]) is less than or equal to j, a result analogous to Lemma 6 can be derived (where b=(1+cX0-1)/21/δ and X0 is a lower bound for tmin). For a no-jump attachment point, the next u value equals [(3/2)(t+c)-c]/2g. This expression is less than (1/2)(3/2)(t+c), so if u and t are positive, a result analogous to Lemma 6 can be derived. For a multiple-jump attachment point consisting of a jump followed by a hop, the odd element immediately before the attachment point is [(3/2)f+1(t+c)-(c/2)]/22. The next u value then equals {[(3/2)f+2(t+c)+(5/4)c]/22}/2g. This expression is not necessarily less than (1/2)(3/2)f+2(t+c). For a multiple-jump attachment point consisting of a jump followed by two hops, the next u value equals {[(3/2)f+3(t+c)+(47/8)c]/23}/2g.
Simons and de Weger's Lemma 7 (derived from Lemma 6 and Corollary 5) is;
0 < Λ < mcm2-((δ -1)/ζ)K
where cm= 2(m/δ)(δ -1)/ζbδ/(δ-1)-m and ζ denotes δm-1. (The symbol ζ is used due to typographical difficulties.) Simons and de Weger's Lemma 14 (their main result) is;
Let x=K1(m) be the largest solution of e-13.3(0.46057+log x)=mcm2-((δ -1)/ζ)x. Then K<K1(m).
Complementary Variables
The expected number of odd elements in the jump (or jumps in the case of a multiple-jump attachment point) from t to the odd integer before a primary attachment point (including the destination of the jump or t in the case of a no-jump attachment point) is the number of even elements after the even element immediately before the primary attachment point (including the even element) plus the number of jumps (none for a no-jump attachment point, 1 for a one-jump attachment point, and at least 2 for a multiple-jump attachment point) minus the number of hops between u and the primary attachment point. For example, for the c=1 cycle, there are 3 odds elements in the jump from -21 to -23, the even elements are -136, -68, and -34, and the hop between -17 and -68 is {-55, -164, -82, -41} (3=3+1-1). A histogram of the number of odd elements in the jump (or jumps) from t to the odd integer before the primary attachment point minus the number of jumps plus the number of hops between u and the primary attachment point minus the number of even elements after the even element before the primary attachment point for the 22851 primary attachment points for the c values less than or equal to 1999 superimposed on a normal probability distribution having the same parameters is;
The mean of the distribution is -0.004901 and the standard deviation is 2.3789. Other than being discrete-valued, the distribution resembles a normal probability distribution for small cmax values. For c values less than or equal to 49, the mean is 0.5319 with a 95% confidence interval of (0.1571, 0.9067) and the standard deviation is 1.8299 with a 95% confidence interval of (1.6005, 2.1367). A normal probability plot of the data is;
For c values less than or equal to 199, the mean is 0.3458 with a 95% confidence interval of (0.1694, 0.5221) and the standard deviation is 2.4152 with a 95% confidence interval of (2.2968, 2.5466). A normal probability plot of the data is;
A histogram of the number of odd elements in the jump (or jumps) from t to the odd integer before the primary attachment point minus the number of jumps plus the number of hops between u and the primary attachment point minus the number of even elements after the even element before the primary attachment point for the 36112285 primary attachment points for the c values less than or equal to 199999 superimposed on a normal probability distribution having the same parameters is;
This distribution has a mean of -0.1381 and a standard deviation of 2.3462. A plot of the standard deviations of the distributions versus cmax for cmax=9997, 1999, 2999, ..., 199999 is;
The distributions appear to be tending towards a fixed probability distribution.
A histogram of the sum of the number of odd elements in the jumps from t values to the odd elements before the primary attachment points minus the sum of the number of jumps plus the number of hops in the cycle minus the sum of the number of even elements after the even element before the primary attachment points for the 3213 cycles less than or equal to 1999 is;
The above distribution has a mean of -0.03486 and a standard deviation of 6.1492. The number of odd elements in the jumps from t to the odd element before a primary attachment point and the number of even elements after the even element before the primary attachment point are "complementary" variables. A histogram of the sum of the number of odd elements in the jumps from the t values to odd elements before the primary attachment points minus the sum of the number of even elements after even elements before the primary attachment points for the 3213 cycles less than or equal to 1999 is;
The above distribution has a mean of -0.4591 and a standard deviation of 6.6471. A quadratic least-squares fit of the standard deviations of these distributions plotted against log(cmax) for cmax=9997, 19999, 29999, ..., 199999 is;
p1=0.8128 with a 95% confidence interval of (0.6758, 0.9497), p2=-12.93 with a 95% confidence interval of (-15.92, -9.95), and p3=59.63 with a 95% confidence interval of (43.45, 75.8). SSE=0.7203, R-square-0.9975, adjusted R-square=0.9972, and RMSE=0.2058. A quadratic least-squares fit of the means of these distributions plotted against log(cmax) for cmax=9997, 19999, 29999, ..., 199999 is;
p1=0.01189 with a 95% confidence interval of (0.008548, 0.1524), p2=-0.5579 with a 95% confidence interval of (-0.6302, -0.4855), and p3=-1.275 with a 95% confidence interval of (-1.605, -0.9451). SSE=0.7506, R-square=0.9887, adjusted R-square=0.9874, and RMSE=0.2101.
The number of jumps from t to the odd element before a primary attachment point and the number of hops between the corresponding u value and the primary attachment point are complementary variables. A histogram of the number of hops in the cycle minus the sum of the number of jumps from t values to the odd elements before the primary attachment points for the 3213 cycles less than or equal to 1999 is;
The above distribution has a mean of 0.4242 and a standard deviation of 5.2375. A quadratic least-squares fit of the standard deviations of the distributions plotted against log(cmax) for cmax =9997, 19999, 29999, ..., 199999 is;
p1=0.2415 with a 95% confidence interval of (0.2038, 0.2793), p2=-2.73 with a 95% confidence interval of (-3.552, -1.908), and p3=11.83 with a 95% confidence interval of (7.368, 16.28). SSE=0.0547, R-square=0.9993, adjusted R-square=0.9992, and RMSE=0.05672. The means of the distributions are approximately equal to 0.
Chaining u Values Together and Chaining t Values Together
Let n denote the number of jumps in the multiple-jump of a primary multiple-jump attachment point. Also, let n equal 1 for a primary one-jump attachment point and let n equal 0 for a primary no-jump attachment point. Let h denote the number of hops between u and the primary attachment point, let f denote the number of odd elements in the jump (or jumps) from t to the odd element immediately before the attachment point (let f equal 0 for a no-jump attachment point), and let g denote the number of even elements after the even element before the primary attachment point. As previously shown, {[(3/2)f+1(t+c)]/2n}/2g is less than the next u value for no-jump and one-jump attachment points and greater than the next u value for multiple-jump attachment points. Assuming these deficiencies and excesses cancel out, ∏{[(3/2)f(i)+1(ti+c)]/2n(i)}/2g(i) is approximately equal to ∏ui where the product is from i=1 to a. [2∑(f(i)+1)-∑j(i)]δ-1/2g(i) is usually less than 1, so if the u and t values are positive, ∏(ti+c )δ/2n(i) should be greater than ∏ui. (Of the 3212 cycles for the c values less than or equal to 1999, [2∑(f(i)+1)-∑j(i)]δ-1/2g(i) is greater than 1 in only 37 instances and its largest value is 17.31905.) In proving Lemma 6, Simons and de Weger show that (1/2)(xi+1)δ>xi+1. Similarly, if the u values are positive, (1/2)(ui+c)δ is greater than the next local minimum (possibly the destination of a hop). An empirical result is;
(42) The u values can be chained together, that is, |ui+c|δ is almost always greater than |ui+1|. The corresponding t values can be chained together, that is, 2n(i)|ti+c|δ is almost always greater than |ti-1|. Also, 2n(i)|ti+c|δ is almost always greater than |ui+1|. (The u and t values are indexed circularly.)
Note that there are no restrictions on the signs of the u and t values in the above proposition. Of the 22843 primary attachment points of the c values less than or equal to 1999, the chain of u values is broken only 23 times (in every instance, the next primary attachment point doesn't have an associated secondary attachment point). Of the 22843 primary attachment points of the c values less than or equal to 1999, the chain of proxies is broken only 47 times. The t value of a primary multiple-jump attachment point where the multiple-jump consists of two jumps, two hops, or a jump and a hop can be chained (without scaling by a factor of 22) with the t value of the preceding attachment point (not necessarily primary). (If |t+c|=2h, then the absolute value of the preceding t value equals 3h [an empirical result] so that |t+c|δ equals the absolute value of the preceding t value.) Chaining one of these t values to the t value of the preceding primary attachment point is then apt to fail (even after scaling by a factor of 22) when the preceding primary attachment point has associated secondary, tertiary, quatenary, etc. attachment points. This accounts for 38 out of the 47 instances where the chain of proxies is broken. In 3 other instances, the t value of a primary one-jump attachment point can't be chained to the t value of the preceding primary attachment point. This occurs when the preceding primary attachment point has associated secondary and tertiary (and quatenary) attachment points. Of the 22843 primary attachment points of the c values less than or equal to 1999, 2n(i)|ti+c|δ is less than |ui+1| in 99 instances and 96 of these instances occur when the multiple jump of a primary multiple-jump attachment point consists of two jumps, two hops, or a jump and a hop. The corresponding definition of b in Lemma 6 is b=|1+cX0-1| where X0 is multiple-valued, that is, if u or t is positive, then u or t is greater than X0, or if u or t is negative, then u or t is less than -X0. (Note that b=1+cX0-1 or |1-cX0-1|.) m-cycles then become a-cycles. ca is then defined so that |1+cX0-1|ab(δ/(δ-1))(ζ/(δ-1)-a)=caζ/(δ-1), that is, ca=bδ/(δ-1)-a/ζ where ζ denotes δa-1. ∑(1/|ui|+1/|ti|) where the summation is from i=1 to a is less than a/|u|min+a/|t|min, so the corresponding version of Lemma 7 (empirically derived) is;
(43) |Λ|<5caca2-((δ-1)/ζ)K
X0 can be defined to be the smallest |u| or |t| value in the cycle or an empirically derived lower bound of the minimum. For the c=121 cycle, L=42, K=46, a=10, the largest negative proxy value is -9, the smallest positive proxy value is 21, and the smallest u value is 19. Λ equals 10.46078661. Assuming this is the only c=121 10-cycle in existence, X0 can be set to 9. b then equals 14.44444444 or 12.44444444. ca then equals 1059.577577 or 718.2799973. 5caca2-((δ-1)/ζ)K then equals 5310117 or 359969. This test (where X0 is set to the smallest |u| or |t| value in the cycle) fails for 489 of the 663743 cycles with attachment points for the c values less than or equal to 199999. In almost every case, the (K+L, K) value of the cycle is approximately equal to the (K+L, K) value of an M-cycle generated from the parity vector p and a is small. In every case, L<K and a≤10. Usually (in all but 45 cases), the cycle is a 1-cycle where the inequality simplifies to |Λ|<5cb2-K. Setting the X0 value to 17 for the c=1 cycle gives 5cb2-K equals 0.04136029 or 0.03676471 (Λ=-0.06566703). Denote the (L, K) value of a cycle generated from the parity vector p by (L', K'). If the (L, K) value of a cycle doesn't equal that of an M-cycle generated from the parity vector p, but K=K' for some K', set d to L'-L, otherwise set d to zero. If KK' for any K' then set d to L'-L where L' corresponds to the first K' >K. A histogram of the d values for the cycles where a≠1 is;
In two instances, the d value is negative. A histogram of the d values for the cycles where a=1 is;
The (c, L, K, a, utmin) values for the exceptions when a≠1 are (295, 12, 18, 2, 281), (679, 24, 36, 5, 669), (2381, 20, 30, 5, 2349), (2647, 23, 32, 4, 2589), (3221, 14, 21, 2, 3187), (3623, 15, 23, 3, 3609), (4741, 13, 19, 2, 4743), (6511, 15, 22, 2, 6441), (6661, 23, 33, 5, 6621), (9145, 20, 30, 3, 9171), (12769, 21, 28, 3, 12699), (13085, 12, 18, 3, 13123), (16105, 14, 21, 4, 16083), (17753, 16, 24, 4, 17751), (22885, 24, 38, 6, 22859), (24817, 15, 21, 2, 25083), (27713, 38, 62, 10, 27753), (31727, 14, 23, 4, 31713), (33313, 9, 12, 3, 33303), (37465, 12, 18, 2, 37479), (42665, 16, 40, 3, 43689), (50857, 9, 30, 2, 52393), (58045, 12, 16, 3, 58117), (71873, 16, 14, 2, 72041), (71873, 16, 24, 3, 71899), (72023, 14, 20, 4, 72073), (73811, 20, 30, 6, 74013), (73811, 20, 30, 5, 73901), (82955, 12, 16, 2, 82827), (86233, 24, 36, 4, 86515), (103427, 8, 12, 2, 103395), (104551, 39, 60, 9, 104845), (108521, 22, 34, 4, 108487), (116449, 20, 28, 5, 1166871), (127735, 16, 24, 5, 127593), (130955, 20, 30, 4, 130827), (130955, 20, 30, 5, 130969), (138503, 14, 21, 2, 138173), (138503, 14, 21, 3, 138609), (145201, 23, 38, 5, 145359), (154403, 12, 18, 2, 154365), (154403, 12, 18, 4, 154395), (159523, 24, 36, 5, 159387), (172555, 14, 21, 3, 172713), and (192685, 16, 24, 192191). In every instance, the test passes when b is set to 1+cX0-1 and fails when b is set to |1-cX0-1| (the respective values of 5caca2-((δ-1)/ζ)K in the latter case are 0.3691, 0.3922, 1.4575, 2.7252, 0.2127, 0.1860, 0.0061, 0.3373, 0.6257, 0.1063, 0.6044, 0.8059, 0.1618, 0.0011, 0.0397, 1.6251, 0.0417, 0.0299, 0.0977, 0.0529, 4.6475, 1.2214, 1.1050, 0.2577, 0.0566, 0.2069, 1.1974, 0.2520, 1.4361, 1.5065, 0.5626, 0.9057, 0.0226, 1.2505, 0.4705, 0.3435, 0.0024, 1.1474, 0.5873, 0.2706, 0.1224, 0.0032, 0.1931, 0.9929, and 0.7907 and the respective values of |Λ| are 1.0194, 2.0388, 1.6990, 2.9675, 1.1893, 1.0715, 1.3071, 1.4770, 2.5620, 1.6990, 3.2031, 1.0194, 1.1893, 1.3592, 1.2279, 1.8824, 1.2008, 0.3784, 1.3727, 1.0194, 5.1282, 5.9256, 1.8303, 1.3592, 1.3592, 1.5948, 1.6990, 1.6990, 1.8303, 2.0388, 0.6796, 2.7048, 1.4634, 2.5099, 1.3592, 1.6990, 1.6990, 1.1893, 1.1893, 0.5347, 1.0194, 1.0194, 2.0388, 1.1893, and 1.3592). In every instance, the smallest |u| or |t| value in the cycle is approximately equal to c. A plot of c versus utmin is;
A plot of the relative errors is;
The two large negative spikes in the middle of the graph correspond to the two negative d values in a previous graph. A similar relationship between c and utmin is not valid for exceptions when a=1.
When only umin is considered (and a factor of 16.0 is used instead of 5.0), the inequality fails for 359 of the cycles for c≤199999. In all but 72 instances, the cycle is a 1-cycle. In every instance, L<K and a≤13. A histogram of the d values for a≠1 is;
Only one value is negative. A histogram of the d values when a=1 is;
A plot of c versus umin is;
A plot of the relative errors is;
The large negative spike in the middle of the graph corresponds to the negative d value in a previous graph.
When only tmin is considered (and a factor or 128.0 is used), the inequality fails for 140 of the cycles where c≤199999. In all but 12 cycles, the cycle is a 1-cycle. In every instance L≤K and a≤13. A histogram of the d values for a≠1 is;
A histogram of the d values for a=1 is;
A plot of c versus tmin is;
For a linear least-squares fit of the curve, the slope is 0.9996 with a 95% confidence interval of (0.9993, 1), the intercept is 10.85 with a 95% confidence interval of (-14.19, 35.9), and R-squared=1. A plot of the relative errors is;
Local Maximum and Minimum Odd Elements in a 3n+c Cycle
Let locmin denote the odd element between two successive primary attachment points having the smallest absolute value. Usually, for a one-jump attachment point, locmin equals u or is a few successive hops away from u. For the 4234 primary one-jump attachment points for the c values less than or equal to 997, the numbers of attachment points where 0, 1, 2, 3, 4, 5, and 6 hops are required to reach locmin are 3677, 358, 102, 19, 5, 1, and 0 respectively. For 72 of the attachment points, locmin can't be reached by successive hops away from u. These proportions don't change much for different c upper bounds; a table of proportions where i denotes the number of hops and n denotes the number of primary one-jump attachment points is;
``` cmax= 49999 99997 149999 199999
i=0 9.087466e-001 9.103450e-001 9.108970e-001 9.111167e-001
1 6.377740e-002 6.288359e-002 6.253354e-002 6.238482e-002
2 1.576506e-002 1.560997e-002 1.560444e-002 1.560868e-002
3 4.056629e-003 3.968739e-003 3.938980e-003 3.916318e-003
4 9.995667e-004 9.866914e-004 9.902790e-004 9.858214e-004
5 2.581761e-004 2.566830e-004 2.553726e-004 2.519073e-004
6 6.479131e-005 6.738539e-005 6.547580e-005 6.579589e-005
7 1.088098e-005 1.285856e-005 1.365493e-005 1.372299e-005
8 0.000000e+000 9.765999e-007 2.544396e-006 2.819064e-006
9 4.945902e-007 4.882999e-007 5.936925e-007 6.382787e-007
10 0.000000e+000 0.000000e+000 8.481321e-008 5.318989e-008
- 6.320368e-003 5.867575e-003 5.698006e-003 5.652702e-003
n= 2021876 6143765 11790616 18800566
```
Let locmax denote the odd element between two successive primary attachment points having the largest absolute value. Let i denote the last odd element before the second primary attachment point. If locmax≠i, the attachment point (that is, the second primary attachment point) is usually a no-jump attachment point, but can occasionally be a one-jump attachment point (but not a jumped-over or multiple-jump attachment point). For the c values less than or equal to 997, there are 1622 primary attachment points where locmax≠i and 1489 of these are no-jump attachment points and 133 are one-jump attachment points.
A histogram of the differences between the number of odd elements in primary one-jump attachment points and the j values for the c=467 cycles where the (K+L, K) values are (84, 53) is;
A histogram of the differences between the number of odd elements in primary one-jump attachments points and the j values for the c=311 cycles where the (K+L, K) values are (19, 12) is;
As expected, the shoulders on the peak for the c=311 cycles are broader than those on the peak for the c=467 cycles (taking into account the different numbers of primary one-jump attachment points). When j is greater than the number of odd elements in the jump, locmin is usually not equal to u. An empirical result is;
(44) When j equals the number of odd elements in a primary one-jump attachment point, locmin equals u.
When locmin doesn't equal u, the difference between the number of odd elements in the jump and the j value ranges from -11 to 10 for the c values less than or equal to 997. The corresponding numbers of attachment points are 1, 0, 0, 1, 2, 8, 15, 7, 30, 69, 113, 0, 25, 118, 89, 45, 18, 8, 2, 2, 3, and 1. The corresponding numbers of attachment points where locmin can't be reached by successive hops away from u are 0, 0, 0, 0, 0, 1, 0, 0, 4, 5, 6, 0, 25, 12, 11, 5, 2, 0, 0, 0, 1, and 0. Note that these attachment points account for all of the attachment points where j+1 equals the number of odd elements in the jump and that most of these attachment points occur when j is less than the number of odd elements in the jump. A histogram of the number of odd elements in the jump minus the j value for the primary one-jump attachment points where locmin doesn't equal u and locmin can't be reached by successive hops away from u for the c values less than or equal to 3997 is;
This distribution has a mean of 1.056657 and a standard deviation of 2.295541. A histogram of the number of odd elements in the jump minus the j value for the primary one-jump attachment points where locmin doesn't equal u but locmin can be reached by successive hops away from u for the c values less than or equal to 3997 is;
This distribution has a mean of .6050493 and a standard deviation of 2.969974. When j is greater than the number of odd elements in the jump, locmax is usually not equal to the last odd element before the attachment point. An empirical result is;
(45) When j equals the number of odd elements in a primary one-jump attachment point, locmax equals the last odd element before the attachment point.
When locmax doesn't equal the last odd element before the attachment point, the difference between the number of odd elements in the jump and the j value ranges from -6 to 2 for the c values less than or equal to 997. The corresponding numbers of attachment points are 1, 5, 0, 5, 10, 24, 0, 87, and 1. For the c=467 cycles where (K+L, K) equals (84, 53), the difference between the number of odd elements in the jump (1) and the j value ranges from -4 to 0. The corresponding numbers of attachment points are 1, 4, 7, 6, and 14.
For primary multiple-jump attachment points, locmin equals u and locmax equals the last odd element before the attachment point. For the c values less than or equal to 997, the numbers of primary multiple-jump attachments points where j equals 1, 2, 3, ..., 7 are 556, 447, 82, 9, 1, 1, and 0 respectively (there are 1096 primary multiple-jump attachment points). The proportions don't change much for different c upper bounds; a table of proportions where n denotes the number of primary multiple-jump attachment points is;
``` cmax= 49999 99997 149999 199999
j=1 5.018008e-001 5.008240e-001 5.004788e-001 5.004156e-001
2 4.099361e-001 4.108239e-001 4.110778e-001 4.110728e-001
3 7.693787e-002 7.689667e-002 7.703393e-002 7.706442e-002
4 9.129327e-003 9.315844e-003 9.303581e-003 9.330295e-003
5 1.855497e-003 1.810723e-003 1.794086e-003 1.802707e-003
6 2.786773e-004 2.807546e-004 2.665826e-004 2.699150e-004
7 5.467719e-005 4.341567e-005 3.855657e-005 3.664347e-005
8 7.055122e-006 3.473253e-006 5.723242e-006 6.233167e-006
9 0.000000e+000 1.157751e-006 9.036697e-007 1.133303e-006
10 0.000000e+000 0.000000e+000 0.000000e+000 1.888838e-007
11 0.000000e+000 0.000000e+000 0.000000e+000 0.000000e+000
n= 566964 1727487 3319797 5294259
```
For a primary no-jump attachment point, the probability that j equals 1 is about 1/2, the probability that j equals 2 is about 1/4, the probability that j equals 3 is about 1/8, etc. For the c values less than or equal to 997, the numbers of primary no-jump attachment points where j equals 1, 2, 3, ..., 12 are 1265, 627, 302, 152, 83, 37, 18, 18, 3, 1, 2, and 0 respectively. For the c values less than or equal to 199999, the numbers of primary no-jump attachment points where j equals 1, 2, 3, ..., 21 are 6008288, 2998990, 1505485, 750484, 377569, 188532, 94326, 46823, 23613, 11593, 5914, 3000, 1368, 729, 423, 230, 68, 21, 2, 2, and 0 respectively (there are 12017460 primary no-jump attachment points).
The Largest Power of 2 That Divides the Difference Between a u Value and Its Proxy and the Generalized Continued-Fraction Convergents of log(3)/log(2)
A histogram of j minus the number of jumps from t to the odd integer before the primary attachment point plus the number of hops between u and the primary attachment point minus the number of even elements after the even element before the primary attachment point for the 22851 primary attachment points for the c values less than or equal to 1999 is;
j and the number of even elements after the even element before a primary attachment point are complementary variables. A histogram of j minus the number of even elements after the even element before the primary attachment point for the 22851 primary attachments points for the c values less than or equal to 1999 is;
A histogram of j minus the number of even elements after the even element before the primary attachment point for the 36112285 primary attachment points for the c values less than or equal to 199999 is;
This distribution has a mean of -0.7955 and a standard deviation of 2.0347. For cmax=9997, 1999, 2999, ..., 199999, the standard deviations of the distributions are 2.0402, 2.0386, 2.0378, 2.0377, 2.0359, 2.0356, 2.0346, 2.0348, 2.0349, 2.0346, 2.0348, 2.0347, 2.0346, 2.0346, 2.0347, 2.0345, 2.0346, 2.0345, 2.0346, and 2.0347 respectively. The means of the distributions are approximately equal to -0.8. The distribution doesn't change much for different c upper bounds.
For c values less than or equal to 499, this distribution has a mean of -0.5522 and a standard deviation of 2.0706 (there are 3200 primary attachment points). For c=467 (and (K+L, K)=(84, 53)), the distribution has a mean of 0.1892 and a standard deviation of 2.1805 (there are 111 primary attachment points). Since there are no jumped-over attachment points, the differences between the j values and the numbers of even elements after the even element before the primary attachment point are skewed to the positive side. (The differences range from -3 to 9 and the histogram values are 1, 15, 39, 21, 15, 8, 4, 1, 3, 1, 0, 2, and 1 respectively.) Apparently, there is a limit to how skewed the distribution can be for a given c value. For c=107, there is apparently only one cycle and the (K+L, K) value is (106, 53). Since the number of even and odd elements in the cycle are the same, the differences between the j values and the numbers of even elements after the even element before the primary attachment point are skewed to the negative side. The distribution has a mean of -1.4 and a standard deviation of 3.0984 (there are 10 primary attachment points). (The differences range from -9 to 2 and the histogram values are 1, 0, 0, 0, 0, 0, 1, 2, 2, 1, 2, and 1 respectively.) In cycles having the same number of even and odd elements and where the number of odd elements equals the denominator of a generalized continued-fraction convergent of log(3)/log(2), j minus the number of even elements after the even element before the primary attachment point is no more negative (approximately) than the difference for cycles having the (K+L, K) value of a generalized continued-fraction convergent of log(3)/log(2) is positive (an empirical result).
A histogram of the sum of the j values minus the sum of the number of even elements after the even element before the primary attachment point for the 3213 cycles with attachment points for the c values less than or equal to 1999 is;
This distribution has a mean of -5.2770 and a standard deviation of 8.9742. The difference between the sum of the j values and the sum of the numbers of even elements after the even element before the primary attachment point is likely to be the most negative in cycles having more even elements than odd elements. The largest difference appears to very slowly increase as c increases. A quadratic least-squares fit of the means of the distributions plotted against (cmax+1)/1000 or (cmax+3)/1000 for cmax=9997, 19999, 29999, ..., 199999 is;
(The coefficients are poorly conditioned when the means are plotted against cmax.) p1=0.0561 with a 95% confidence interval of (0.04223, 0.06996), p2=-2.778 with a 95% confidence interval of (-3.078, -2.478), and p3=-9.464 with a 95% confidence interval of (-10.83, -8.097). SSE=12.89, R-square=0.9927, adjusted R-square=0.9919, and RMSE=0.8709.
A histogram of j minus the number of even elements after the even element before the primary attachment point where the (K+L, K) value of the cycle equals a generalized continued-fraction convergent of log(3)/log(2) for the c values less than or equal to 9997 is;
This distribution has a mean of 0.03171 and a standard deviation of 2.0258 (there are 9462 samples). A histogram of the sum of the j values minus the sum of the numbers of even elements after the even element before the primary attachment point where the (K+L, K) value of the cycle equals a generalized continued-fraction convergent of log(3)/log(2) for the c values less than or equal to 9997 is;
This distribution has a mean of 0.07266 and a standard deviation of 3.0704 (there are 4129 samples).
A histogram of j minus the number of even elements after the even element before the primary attachment point where the number of even and odd elements in the cycle are the same and the K value equals the denominator of a generalized continued-fraction convergent of log(3)/log(2) for the c values less than or equal to 9997 is;
This distribution has a mean of -0.6945 and a standard deviation of 1.9292 (there are 2236 samples). A histogram of the sum of the j values minus the sum of the numbers of even elements after the even element before the primary attachment point where the number of even and odd elements in the cycle are the same and the K value equals the denominator of a generalized continued-fraction convergent of log(3)/log(2) for the c values less than or equal to 9997 is;
This distribution has a mean of -4.0262 and a standard deviation of 4.8975 (there are 764 samples).
A histogram of the sum of the j values plus d/δ minus the sum of the numbers of even elements after the even element before the primary attachment point where the number of even and odd elements in the cycle are the same and the K value equals the denominator of a generalized continued-fraction convergent of log(3)/log(2) for the c values less than or equal to 9997 is;
This distribution has a mean of 1.0524 and a standard deviation of 4.1845 (there are 764 samples). This distribution resembles the corresponding distribution where the (K+L, K) values equal those of a generalized continued-fraction convergent of log(3)/log(2). A histogram of these values scaled by 4129/764 (the ratio of the sample sizes) superimposed on the corresponding distribution where the (K+L, K) values equal those of a generalized continued-fraction convergent of log(3)/log(2) is;
For c≤19999, there are 5288 cycles having the (K+L, K) value of a generalized continued-fraction convergent of log(3)/log(2) and there are 1903 cycles having an equal number of even and odd elements and a K value equal to the denominator of a generalized continued-fraction convergent of log(3)/log(2). The corresponding histograms are;
In general, adding d/δ "normalizes" values so that they can be compared to those of cycles having the (K+L, K) values generated by the parity vector p.
The Distribution of [3·log(c)]-a Values
A histogram of [3·log(cmax)]-[3∙log(c)]+a for the 4897 cycles for cmax=2999 superimposed on a chi-square distribution with 6 degrees of freedom is;
This distribution has a mean of 12.1873 and a standard deviation of 10.0945. [3·log(cmax)]=24, so about 91% of the [3·log(c)]-a values are non-negative. A histogram of [3·log(cmax)]-[3·log(c)]+a for the 23132 cycles for cmax=9997 is;
This distribution has a mean of 15.2292 and a standard deviation of 19.0883. [3·log(cmax)]=27, so about 85% of the [3·log(c)]-a values are non-negative. For cmax equal to 997, 1999, 2999, ..., and 19999, the means of these distributions are 7.6247, 9.5976, 12.1873, 12.4800, 13.6209, 15.1857, 14.8543, 13.2578, 14.7518, 15.2292, 15.8069, 17.2515, 17.6940, 18.1386, 18.1259, 19.6618, 19.7332, 20.1047, 20.4672, and 20.9466 respectively and the standard deviations are 5.6051, 8.5752, 10.0945, 12.1485, 13.3792, 14.7405, 15.9844, 16.5977, 18.0075, 19.0883, 20.1579, 21.1702, 21.8124, 22.9374, 23.5984, 24.4908, 25.1141, 25.8074, 26.3705, and 27.1986 respectively. (The values of [3·log(cmax)] minus the means are 12.3753, 12.4024, 11.8127, 11.5200, 11.3791, 10.8143, 11.1457, 12.7422, 12.2482, 11.7708, 11.1931, 10.7485, 10.3060, 9.8614, 9.8741, 9.3382, 9.2667, 8.8953, 8.5328, and 8.0534 respectively.) The distributions resemble a chi-square probability distribution (or an exponential probability distribution), but since the means and standard deviations increase, the probability that [3·log(c)]-a takes on a specific value is not fixed. For example, proportions of some [3·log(c)]-a values (denoted by i) for the different distributions are;
cmax= 997 1999 2999 3997 4999 5999 6997 7999 8999 9997
i=0 .0045 .0056 .0078 .0081 .0082 .0086 .0080 .0070 .0072 .0072
1 .0071 .0084 .0102 .0113 .0102 .0101 .0096 .0082 .0083 .0083
2 .0109 .0115 .0127 .0127 .0120 .0113 .0106 .0085 .0083 .0081
3 .0084 .0100 .0098 .0102 .0113 .0118 .0108 .0089 .0089 .0091
4 .0161 .0153 .0159 .0143 .0133 .0131 .0122 .0101 .0098 .0095
5 .0141 .0140 .0153 .0146 .0147 .0144 .0135 .0112 .0114 .0114
6 .0206 .0174 .0169 .0169 .0158 .0151 .0140 .0117 .0117 .0117
7 .0308 .0240 .0216 .0184 .0168 .0168 .0151 .0127 .0122 .0124
8 .0219 .0215 .0196 .0177 .0186 .0179 .0166 .0133 .0129 .0131
9 .0379 .0321 .0296 .0265 .0239 .0232 .0204 .0166 .0165 .0162
10 .0463 .0352 .0333 .0290 .0259 .0243 .0219 .0177 .0171 .0171
11 .0668 .0464 .0392 .0335 .0312 .0282 .0258 .0209 .0198 .0188
12 .0521 .0445 .0392 .0327 .0304 .0279 .0251 .0201 .0197 .0195
13 .0758 .0542 .0506 .0437 .0396 .0373 .0329 .0262 .0248 .0243
As expected, the proportions for the larger [3·log(c)]-a values decrease more rapidly than those for the smaller values. The large number of cycles (2849) for c=7153 doesn't appear to affect the proportions for cmax=7999 substantially. The cmaxδmax/log(cmax) values for cmax equal to 997, 1999, 2999, ..., 9997 are 890.3, 2255.4, 3781.3, 5855.0, 7852.8, 10164.9, 12633.0, 14772.9, 17798.1, and 20719.3 respectively and the numbers of cycles are 1556, 3213, 4897, 6796, 8870, 10807, 13477, 18448, 20820, and 22132 respectively (δmax denotes the standard deviation of the distribution for cmax).
This composite curve indicates that the a values of the cycles for a given c value should have a distribution resembling that of an exponential probability distribution. (Also, the exponential distribution applies to the first success in a Poisson process and the "waiting time" between successes and would explain the relationship to Poisson probability distributions.) For example, for K+L=16 and K=10, there are 504 parity vectors that are distinct under rotation. 216-310=6487=13·499 and there are 456 primitive cycles for c=6487 having a (K+L, K) value of (16, 10) (438 of which have attachment points), 41 primitive cycles for c=499 having a (K+L, K) value of (16, 10) (39 of which have attachment points), and 7 primitive cycles for c=13 having a (K+L, K) value of (8, 5). Note that 504=456+41+7 (since 16 and 10 are not relatively prime, there are 7 parity vectors consisting of duplicated sub-vectors). For the 438 c=6487 cycles with attachment points and a (K+L, K) value of (16, 10), there are 11 [3·log(c)]-a values equal to 23, 191 [3·log(c)]-a values equal to 24, and 236 [3·log(c)]-a values equal to 25. These counts would more closely resemble an exponential probability distribution if there were other c=6487 cycles having different (K+L, K) values. For c=6487, there are cycles having (K+L, K) values of (16, 10), (32, 20), (48, 30), (64, 40), (80, 50), and (96, 60). There are 516 cycles and there are 1, 2, 3, 7, 19, 40, 207, and 237 [3·log(c)]-a values of 18, 19, 20, 21, 22, 23, 24, and 25 respectively. These counts more closely resemble an exponential probability distribution. The counts for the individual (L, K) values are;
(L, K)=(6, 10) (12, 20) (18, 30) (24, 40) (30, 50) (36, 60)
[3·log(c)]-a=18 0 0 0 0 0 1
19 0 0 0 1 1 0
20 0 0 1 2 0 0
21 0 2 5 0 0 0
22 0 12 5 2 0 0
23 11 28 1 0 0 0
24 191 16 0 0 0 0
25 236 1 0 0 0 0
As discussed in the Journal of Integer Sequences article, there are frequently cycles for a given c value with (L, K) values that are multiples of other (L, K) values. In this example, the function of the cycles with (L, K) values that are a multiple of (6, 10) is apparent; the modal class (or the average of the modal classes if the distribution is multimodal) of the distributions of [3·log(c)]-a values for the different (L, K) values (that is, the values in the columns of the above matrix) decreases linearly (roughly). (In this example, the modal classes are 25, 23, 21.5, 21, 19, and 18 respectively.)
For c=7153 (equal to |219-312|), there are cycles having (L, K) values of (7, 12), (14, 24), (21, 36), (28, 48), (35, 60), (42, 72), (56, 96), (63, 108), (271, 221), (292, 257), (341, 341), and (362, 377). There are 2849 cycles and there are 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 1, 0, 0, 2, 4, 4, 11, 25, 55, 125, 314, 1333, and 968 [3·log(c)]-a values of -60, -59, -58, ..., 25 respectively. Including these cycles (say for cmax in the range from 6499 to 7499) doesn't change the smoothness of the composite curve substantially.
There are also frequently associated cycles (defined and discussed in the Journal of Integer Sequences article) for a given c value. Let (L1, K1) denote the number of even and odd elements in the cycles with the smallest number of even elements. In associated cycles, the (L, K) values of cycles are in arithmetic progression with an increment of (L1, K1) and aren't multiples of (L1, K1). For example, for c=1843, the (L, K) values of the cycles (sorted by increasing L values) are (6, 11), (12, 22), (18, 9), (24, 20), (30, 31), (36, 42), (42, 53), (48, 64), (54, 51), and (90, 117) (the cycles with (L, K) values of (18, 9), (24, 20), (30, 31), (36, 42), (42, 53), and (48, 64) are the associated cycles). There are 44 cycles and there are 1, 0, 0, 0, 0, 0, 1, 1, 0, 3, 1, 2, 0, 2, 2, 1, 2, 3, 11, and 14 [3·log(c)]-a values of 1, 2, 3, ..., 21 respectively. The counts for the individual (L, K) values are;
(L, K)= (6, 11) (12, 22) (18, 9) (24, 20) (30, 31) (36, 42) (42, 53) (48, 64) (54, 51) (90, 117)
[3·log(c)]-a= 1 0 0 0 0 0 0 0 0 0 1
2 0 0 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 0 0 0
4 0 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 1 0
9 0 0 0 0 0 0 0 1 0 0
10 0 0 0 0 0 0 0 0 0 0
11 0 0 0 0 0 1 2 0 0 0
12 0 0 0 0 0 0 0 1 0 0
13 0 0 0 0 1 1 0 0 0 0
14 0 0 0 0 0 0 0 0 0 0
15 0 0 0 1 1 0 0 0 0 0
16 0 0 1 1 0 0 0 0 0 0
17 0 0 0 1 0 0 0 0 0 0
18 0 1 0 1 0 0 0 0 0 0
19 0 3 0 0 0 0 0 0 0 0
20 7 3 1 0 0 0 0 0 0 0
21 14 0 0 0 0 0 0 0 0 0
The associated cycles have the same function as the multiples of cycles; the modal classes of the distributions of [3·log(c)]-a values decrease linearly (roughly). The modal classes in this example are 21, 19.5, 18, 16.5, 14, 12, 11, 10.5, 8, and 1. Note that there is a big gap between the modal classes for the cycles having the last two (L, K) values (which aren't associated cycles). When the (L, K) values are sorted by increasing L values, the modal classes almost always decrease. As previously shown, a Poisson probability distribution models the number of cycles fairly well if only one of the interrelated cycles is counted, only one of the associated cycles is counted, and only one of cycles with (L, K) values that are a multiple of another (L, K) value is counted.
For c=2425, the (L, K) values of the cycles (sorted by increasing L values) are (6, 11), (12, 22), (18, 33), (24, 44), (36, 26), (42, 37), (48, 48), and (54, 59) (the cycles with (L, K) values of (36, 26), (42, 37), (48, 48), and (54, 59) are the associated cycles). (Note that the L value of the first associated cycle is frequently a multiple of L1.) There are 40 cycles and there are 1, 0, 0, 2, 1, 0, 3, 1, 0, 0, 0, 2, 1, 0, 19 and 10 [3·log(c)]-a values of 7, 8, 9, ..., 22 respectively. The counts for the individual (L, K) values are;
(L, K)= (6, 11) (12, 22) (18, 33) (24, 44) (36, 26) (42, 37) (48, 48) (54, 59)
[3·log(c)]-a= 7 0 0 0 0 0 0 0 1
8 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 2
11 0 0 0 0 0 0 1 0
12 0 0 0 0 0 0 0 0
13 0 0 0 0 1 1 1 0
14 0 0 0 0 0 1 0 0
15 0 0 0 0 0 0 0 0
16 0 0 0 0 0 0 0 0
17 0 0 0 0 0 0 0 0
18 0 0 1 1 0 0 0 0
19 0 0 1 0 0 0 0 0
20 0 0 0 0 0 0 0 0
21 18 1 0 0 0 0 0 0
22 10 0 0 0 0 0 0 0
The modal classes are 21, 21, 18.5, 18, 13, 13.5, 12 and 10.
For c=3689, the (L, K) values of the cycles are (6, 12), (12, 24), (111, 102), (123, 126), and (147, 174) (there are no associated cycles, but the difference between (111, 102) and (123, 126) is (12, 24))There are 15 cycles and there are 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 3, and 7 [3·log(c)]-a values of -11, -10, -9, ..., 23 respectively. The counts for the individual (L, K) values are;
(L, K)= (6, 12) (12, 24) (111, 102) (123, 126) (147, 174)
[3·log(c)]-a=-11 0 0 0 0 1
-10 0 0 0 0 0
-9 0 0 0 0 0
-8 0 0 0 0 0
-7 0 0 0 0 0
-6 0 0 0 1 0
-5 0 0 1 0 0
-4 0 0 0 0 0
-3 0 0 0 0 0
-2 0 0 0 0 0
-1 0 0 1 0 0
0 0 0 0 0 0
1 0 0 0 0 0
2 0 0 0 0 0
3 0 0 0 0 0
4 0 0 0 0 0
5 0 0 0 0 0
6 0 0 0 0 0
7 0 0 0 0 0
8 0 0 0 0 0
9 0 0 0 0 0
10 0 0 0 0 0
11 0 0 0 0 0
12 0 0 0 0 0
13 0 0 0 0 0
14 0 0 0 0 0
15 0 0 0 0 0
16 0 0 0 0 0
17 0 0 0 0 0
18 0 0 0 0 0
19 0 0 0 0 0
20 0 0 0 0 0
21 0 1 0 0 0
22 3 0 0 0 0
23 7 0 0 0 0
In this case, there are two separate populations of (L, K) values, one where 2K+L-3K is negative and the other where it is positive (the same as for c=7153 and other values).
Another Way for 3n+c Cycles to be Interrelated
An empirical result is;
(46) For a given c value, if there are two 3n+c cycles having (L, K) values of (L1, K1) and (L2, K2) where K1=K2 and L1L2, then either L1K1 and L2K2 or L1K1 and L2K2. If there are several such pairs of cycles, say K1=K2, L1L2, K3=K4 (K3K1), L3L4, and K5=K6 (K5K1, K5K3), L5L6, then one of L1-L2, L3-L4, L5-L6 divides the other differences in L values.
Of the 23132 3n+c cycles for c less than or equal to 9997, there are 24 such pairs of cycles (not counting interrelated cycles). Note that for a given c value and number of odd elements, there can be at most three cycles (not counting interrelated cycles) having different L values. Belaga14 calls numbers of the form Bk,l = 2l - 3k > 0 "Collatz numbers" and gives a rephrasing of the Diophantine interpretation of the Collatz problem: no non-trivial Collatz number can be a divisor of numbers from a certain finite set of natural numbers called the "Collatz (k, l)–corona". A Collatz number such that 2l-1 - 3k < 0 is called "narrow". When 2K+L-3K is positive and (K+L, K) is a generalized continued-fraction convergent of log(3)/log(2), the Collatz number is narrow. In all of the above 24 pairs of cycles, neither of the corresponding Collatz numbers is narrow. An empirical result is;
(47) For a given c value, if there are two 3n+c cycles have (L, K) values of (L1, K1) and (L2, K2) where L1=L2 and K1K2, then either L1K1 and L2K2 or L1K1 and L2K2 (usually). If there are several such pairs of cycles, then one of the differences in K values divides the other differences in K values (apparently always).
For the 3n+c cycles for c less than or equal to 9997, there are 107 such pairs of cycles (not counting interrelated cycles). An exception occurs for a pair of cycles for c=145 having (L, K) values of (16, 18) and (16, 32) (16 is approximately equal to 18). An exception occurs for a pair of cycles for c=1009 having (L, K) values of (56, 37) and (56, 46) (these are the only cycles for this c value, so conditions are less stringent). A similar exception occurs for a pair of cycles for c=7063 having (L, K) values of (168, 192) and (168, 174). A third type of exception occurs when there are many such pairs of cycles for a given c value. For example, when c=6305, the (L, K) values that can be paired are [(24, 12), (24, 20), (24, 28), (24, 36)], [(48, 40), (48, 48), (48, 56)], and [(72, 60), (72, 68), (72, 84)]. Another empirical result is;
(48) For a given c value, if there are two 3n+c cycles having (K+L, K) values of (K1+L1, K1) and (K2+L2, K2) where K1+L1=K2+L2, K1K2, then either L1K1 and L2K2 or L1K1 and L2K2. If there are several such pairs of cycles, then one of the differences in K values divides the other differences in K values and one of the differences in L values divides the other differences in L values.
Note that for a given c value and length, there can be at most three cycles (not counting interrelated cycles) having different numbers of odd elements. Of the 23132 3n+c cycles for c less than or equal to 9997, there are 11 such pairs (not counting interrelated cycles) and in every case, neither of the corresponding Collatz numbers is narrow. Proposition (48) is usually valid (with the same kinds of exceptions as for Proposition (47)) for pairs of cycles where fK1+gL1=fK2+gL2 (this would account for there being few cycles that aren't interrelated). For the 333 c values less than or equal to 997, the number of c values having 1, 2, 3, 4, 5, 6, 7, and 8 cycles (not counting interrelated cycles) are 109, 121, 70, 20, 8, 1, 4, and 0 respectively. (As previously mentioned, the number of interrelated cycles having a given (K+L, K) value is determined by the number of parity vectors having this length and number of 1's that are distinct under rotation.) For a given c value, pairs of cycles that satisfy the equation fK1+gL1=fK2+gL2 (or have equal L values and unequal K values or equal K values and unequal L values) can also be considered to be interrelated. For example, when c=9215, the pairs of cycles having [(L1, K1) (L2, K2)] values of [(24, 20), (12, 22)], [(18, 9), (6, 11)], [(36, 42), (24, 44)], [(30, 31), (18, 33)], [(42, 29), (18, 33)], [(42, 29), (30, 31)], [(48, 40), (24, 44)], [(48, 40), (36, 42)], [(42, 53), (54, 51)], and [(48, 64), (60, 62)] can be considered to be interrelated. For all combinations of relatively prime f and g values less than 13 (which usually accounts for most solutions), the adjusted number of c values having 1, 2, 3, 4, 5, 6, 7, and 8 cycles are 127, 125, 61, 12, 7, 1, 0, and 0 respectively. This appears to be a Poisson probability distribution where λ is approximately equal to 1. For the 667 c values less than or equal to 1999, the number of c values having 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 cycles (not counting interrelated cycles) are 220, 246, 134, 34, 21, 6, 5, 0, 0, and 1 respectively. Accounting for the second type of interrelated cycles, the number of c values having 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 cycles are 247, 254, 121, 25, 15, 5, 0, 0, 0, and 0 respectively. For the 1000 c values less than or equal to 2999, the number of c values having 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 cycles (not counting interrelated cycles) are 321, 361, 198, 66, 34, 9, 8, 1, 1, and 1 respectively. Accounting for the second type of interrelated cycles, the number of c values having 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 cycles are 359, 368, 188, 54, 23, 6, 2, 0, 0, and 0 respectively. The mean of this distribution (where x=0, 1, 2, ...) is 1.04. A difficulty with this approach is determining an upper bound of f and g. This approach hasn't been thoroughly investigated.
The Order of a 3n+c Cycle
The order of a cycle is defined to be the smallest natural number of the form 3∙2k greater than or equal to the absolute value of every element of the cycle. In the following, if all the absolute values of the elements of an extended sequence of a cycle are less than the cycle order, then the extended sequence is further extended backwards until the cycle order has been exceeded (the elements of this portion of the extended sequence will be integers of the form 2it where t is the odd integer divisible by 3). The extension order is then taken to be the cycle order. If the cycle order has already been exceeded between the odd integer divisible by 3 and the attachment point, the order of the extension is defined to be the smallest natural number of the form 3∙2k greater than or equal to the absolute value of every element of the extended sequence. The extended sequence is then further extended backwards until this order has been exceeded. (For the remainder of this article, this will be referred to as being the extended sequence of an attachment point.) For example, the extended sequence for the c=1 cycle is {-336, -168, -84, -42, -21, -62, -31, -92, -46, -23, -68, -34, -17, -50, -25, -74, -37, -110, -55, -164, -82, -41, -122, -61, -182, -91, -272, -136, -68} and the cycle order and extension order are 384. An empirical result is;
(49) For a one-jump attachment point, the cycle order equals the extension order.
Let P denote the primary attachment point having the largest absolute value. In most 3n+c cycles, 4 times the absolute value of P determines the orders (all the same) of all the extended sequences of the cycle. Many exceptions are associated with multiple-jump attachments points. When there is a multiple-jump to P and there is a secondary attachment point, the order of the extended sequence of the primary attachment point may be different from the order of the extended sequence of the secondary attachment point.
Let m be the number of odd elements in an extended sequence up to the attachment point (where the element following an odd element i is temporarily defined to be (3i+c)/2) and e the number of even elements. For a no-jump attachment point, e is greater than or equal to m. For a one-jump attachment point where c≤151, e+1 is greater than or equal to m. For a multiple-jump or jumped-over attachment point where c≤151 and the absolute value of the odd element immediately before the attachment point is greater than order/6, e+2 is greater than or equal to m. Of the 649 attachment points of the c values less than or equal to 151, e is less than m for only 32 attachment points. The only c values less than or equal to 151 where e is less than m for a one-jump attachment point are 19, 31, 97, 115, and 139. In every case, the attachment point equals P and 4|P| determines the order of the extended sequence. A histogram of the e-m values for the 1807 one-jump attachment points for c values less than or equal to 499 is;
As c increases, the upper bound of m-e appears to gradually increase. For example, for c=259 and 271, m-e=2 for a one-jump attachment point. In both cases, the attachment point equals P and 4|P| determines the order of the extended sequence. A histogram of the e-m values for the 19223351 one-jump attachment points for cmax=199999 superimposed on a normal probability distribution having the same parameters (a mean of 8.5393 and a standard deviation of 3.4633) is;
A linear least-squares fit of the means of these distributions plotted against log((cmax+1)/1000) or log((cmax+3)/1000) for cmax=997, 1999, 2999, ..., 199999 is;
p1=0.7478 with a 95% confidence interval of (0.7447, 0.7509) and p2=4.567 with a 95% confidence interval of (4.553, 4.581). SSE=0.08784, R-square=0.9991, adjusted R-square=0.9991, and RMSE=0.02106. A cubic least-squares fit of the standard deviations of these distributions plotted against log((cmax+1)/1000) or log((cmax+3)/1000) for cmax=997, 1999, 2999, ..., 199999 is;
p1=0.006052 with a 95% confidence interval of (0.005319, 0.006784), p2=-0.07311 with a 95% confidence interval of (-0.08006, -0.06616), p3=0.3758 with a 95% confidence interval of (0.3558, 0.3958), and p4=2.628 with a 95% confidence interval of (2.61, 2.646). SSE=0.02255, R-square=0.9901, adjusted R-square=0.99, and RMSE=0.01073.
For c=187, m-e=3 for a multiple-jump attachment point where the absolute value of the odd element immediately before the attachment point is greater than order/6. In general, m-e is relatively large for multiple-jump or jumped-over attachment points when there are associated primary, secondary, tertiary, etc., attachment points.
Let max denote the odd integer in the cycle having the largest absolute value and let i denote the odd integer immediately before the primary attachment point following max ((3i+c)/4 equals the primary attachment point). For example, for a cycle for c=5, max=3397 and the sequence between max and the primary attachment point is {3397, 10196, 5098, 2549, 7652, 3826, 1913, 5744, 2872, 1436} (i=1913 and 3397→2549→1913). Empirical results are;
(50) If i≠max and max jumps to i (either by one or multiple jumps), then the primary attachment point is a no-jump attachment point and the corresponding t value is less than max.
(51) If i≠max and max jumps over i, then the primary attachment point is a one-jump attachment point and the corresponding t value is less than max.
If the primary attachment point is P (that is, i=max) and 4|P| determines the extension order, then the primary attachment point is usually a one-jump attachment point (but may be a multiple-jump attachment point or even a no-jump attachment point).
3n+c Cycles With Only One Attachment Point
In this paragraph, cycles containing only one attachment point are discussed. Also, the order of the extended sequence is required to have been determined by 4|P|. The sequence vector of such an extended sequence mostly has bilateral symmetry when the number of 2's in the sequence vector (not counting the last element of the sequence vector) is even (usually, it is necessary to disregard the first and last elements of the sequence vector when considering the symmetry, but sometimes it is also necessary to disregard the second or next-to-last elements of the sequence vector). Since there is only one attachment point, the sequence vector in the interior of the cycle is a 1-2 sequence vector. Then the sequence vector of the extended sequence is a 1-2 sequence vector except for the first and last elements of the sequence vector. (The first element of the sequence vector does not indicate that there is an odd sequence element immediately before the extended sequence; it just counts the number of even sequence elements before the first odd sequence element. A similar situation applies for the last element of the sequence vector.) The sequence vectors of the extended sequences of the interrelated cycles for c=13 are (4, 1, 1, 1, 1, 2, 2), (3, 2, 1, 1, 1, 2, 1, 2), (4, 1, 2, 1, 1, 2, 1, 1, 2), (4, 1, 1, 2, 1, 2, 1, 1, 1, 2), and (5, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2). The sequence vectors of the extended sequences of the interrelated cycles for c=37 and a cycle length of 9 are (6, 2, 1, 2, 1, 2) and (5, 1, 2, 2, 1, 1, 2). The sequence vectors of the extended sequences of the interrelated cycles for c=47 and a cycle length of 11 are (4, 1, 1, 1, 2, 2, ), (4, 2, 1, 1, 2, 1, 2), (6, 1, 2, 1, 2, 1, 1, 2), and (7, 1, 1, 2, 2, 1, 1, 1, 2). The sequence vectors of the extended sequences of the interrelated cycles for c=101 and a cycle length of 10 are (6, 2, 1, 2, 2) and (5, 1, 2, 2, 2). In all of these cases, it is possible to list the interrelated cycles so that the number of 1's between the pair of 2's (if there is a pair) decreases by one and the extension order remains the same or increases for each successive interrelated cycle.
Consider a 3n+1 sequence where the fourth-to-last element of the sequence (denoted by i) is odd, the third-to-last element of the sequence is divisible by 8, and the sequence has been extended backwards (without any restrictions on the path taken), until an odd integer divisible by 3 has been encountered. The order of this sequence is defined to be the smallest natural number of the form 3∙2k greater than or equal to the element in the sequence with the largest absolute value. The sequence is then further extended backwards until the order is exceeded. Note that if the absolute value of i is greater than the order divided by 6, then the absolute value of the third-to-last element of the sequence (3i+1) is greater than the order divided by 2, that is, the third-to-last element of the sequence (along with other elements of the sequence) determines the order. When the absolute value of i is greater than the order divided by 6, these sequences will be referred to as "generalized dead limbs" (ordinary dead limbs will be defined later). In the following, the element after an odd element j is temporarily defined to be (3j+1)/2 (for the purpose of determining lengths). Let e denote the number of even elements in a generalized dead limb, m the number of odd elements, and l the number of elements. Also, generalized dead limbs containing duplicated cycles are excluded. An empirical result is;
(52) For a given number of odd elements in a generalized dead limb, the number of even elements is mostly fixed; the number of even elements can deviate by at most 1.
For example, for m=1, e can equal 4 or 3. (The (e, m) value will be denoted by ((4, 3), 1).) (e, m) values for generalized dead limbs are ((4, 3), 1), ((5, 4), 2), ((5, 4), 3), ((6, 5), 4), ((6, 5), 5), ((7, 6), 6), ((8, 7), 7), ((8, 7), 8), ((9, 8), 9), ((9, 8), 10), ((10, 9), 11), ((11, 10), 12), ((11, 10), 13), ((12, 11), 14), ((12, 11), 15), ((13, 12), 16), ((13, 12), 17), ((14, 13), 18), ((15, 14), 19), ((15, 14), 20), ((16, 15), 21), (16, 15), 22), ((17, 16), 23), ((18, 17), 24), .... An empirical result is;
(53) The adjusted (l, m) values (where l is decremented by 4 or 3) of generalized dead limbs include the generalized continued-fraction convergents of log(3)/log(2).
The (l, m) value of a 3n+1 cycle is a generalized continued-fraction convergent of log(3)/log(2) (based on empirical evidence). (Assuming the factor of 1.000001 in Simons and de Weger's inequality can be reduced to 1.0, the (l, m) value of a 3n+1 cycle [n>0] is a continued-fraction convergent of log(3)/log(2).) For example, the (l, m) value of the 3n+1 cycle {-34, -17, -25, ..., -68} is (11, 7) and the (l, m) value of the extended sequence {-336, -168, -84, ..., -68} (a generalized dead limb containing the cycle) is (19, 10). An empirical result is;
(54) The (l, m) value of a generalized dead limb minus the (l, m) value of a generalized continued-fraction convergent of log(3)/log(2) is the (l, m) value of a shorter generalized dead limb.
Let f denote the mapping of the number of odd elements in a generalized dead limb to the smaller of the two possible numbers of even elements in the limb. Attachment points were previously defined for cycles. The definition of attachment points for generalized dead limbs is the same; the attachment points are just not necessarily in cycles. An empirical result is;
(55) The number of even elements in a generalized dead limb containing x primary attachment points is optimal in that the number of even elements up to the yth primary attachment point, y<x, is greater than or equal to f(m) where m is the number of odd elements up to the yth primary attachment point.
Weibull probability distributions are used to model birth and decay processes and have a scaling and shaping parameter. For the generalized dead limbs where the odd elements divisible by 3 are less than a fixed amount and the orders are less than a fixed amount, the number of even elements (for a given number of odd elements) up to the yth primary attachment point has a Weibull probability distribution. For example, for generalized dead limbs (in the 3n-1 sequence, n>0) where the odd elements divisible by 3 are less than or equal to 99999999, x=5, and y=3, the numbers of generalized dead limbs where m=10 and e-8 equals 0, 1, 2, ..., 19 are 62, 1246, 2939, 3626, 3213, 2097, 1307, 651, 338, 168, 99, 51, 15, 8, 0, 0, 1, 1, 0, and 1 respectively (f(10)=8). The parameters of the Weibull probability distribution for this data (excluding the 62 values equal to 0) are 4.2974 (with a 95% confidence interval of (4.2640, 4.3310) and 2.1167 (with a 95% confidence interval of (2.0921, 2.1415). A Weibull plot of the data is;
A histogram of the data and the corresponding Weibull probability distribution is;
A plot of the Weibull scaling parameters for m=3, 4, 5, ..., 20 and where the odd elements divisible by 3 are less than or equal to 99999999 is;
The curve of smaller values is for when the odd elements divisible by 3 are less than or equal to 999999999. The scaling parameters don't change much for this larger sample size. A plot of the Weibull shaping parameters for m=3, 4, 5, ..., 20 and where the odd elements divisible by 3 are less than or equal to 99999999 is;
The curve of smaller values is for when the odd elements divisible by 3 are less than or equal to 999999999. The shaping parameters don't change much for this larger sample size.
When c>1, the behavior of such sequences approaches that of c=1 in a limiting sense, that is, for a fixed c value and a sufficiently large order, the sequences have the same (l, m) values as for c=1 (an empirical result).
Hypothetical 3n+1 Cycles
Properties of 3n+1 cycles can be inferred from those of general 3n+c cycles that are in generalized dead limbs and have (l, m) values of generalized continued-fraction convergents of log(3)/log(2). That a 3n+1 cycle having an attachment point should be in a generalized dead limb is to be expected since most 3n+c cycles having attachment points are in generalized dead limbs and the 3n+1 sequence is comparatively "well-behaved". Catalan's conjecture (proved by Mihăilescu15) states that the only natural number solutions of xa-yb=1 are x=3, a=2, y=2, and b=3. The only solution of 2l-3m=-1 is then (3, 2). Also, |2l-3m| increases monotonically (apparently) for (l, m) values that are generalized continued-fraction convergents of log(3)/log(2) (excluding (2, 1), (4, 2), (6, 4), and (9, 6)), so there shouldn't be any more 3n+1 cycles where |2l-3m|=1. Study of 3n+c cycles where c properly divides |2l-3m| and the (l, m) values are relatively large should yield expected properties of any other 3n+1 cycles (the number of odd elements in a hypothetical 3n+1 cycle [n>0] has been confirmed to be very large). A c value of 467 (and several others) fulfills these and the above requirements. When c=467, there are fifteen primitive cycles having (l, m) values of (84, 53) (a continued-fraction convergent of log(3)/log(2)) and these cycles have 120 attachment points. Empirical results are;
(56) When the domain of the absolute values of the u values and the absolute values of their proxies is sufficiently small, the (l, m) value of a 3n+c cycle is a generalized continued-fraction convergent of log(3)/log(2), the cycle is in a generalized dead limb, and the (l, m) value of the extended sequence of P minus the (l, m) value of the cycle is the (l, m) value of a shorter generalized dead limb.
(57) When the domain of the absolute values of the u values and the absolute values of their proxies is sufficiently small, all the extended sequences of a 3n+c cycle have the same order, the sign does not change in an extended sequence, and there are more even elements in an extended sequence than odd elements (the only possible (e, m) values of the extended sequence of P are then ((4, 3), 1), ((5, 4), 2), ((5, 4), 3), ((6, 5), 4), ((6, 5), 5), ((7, 6), 6), ((8, 7), 7), ((8, 7), 8), and ((9, 8), 9)).
(58) When the domain of the absolute values of the u values and the absolute values of their proxies is sufficiently small, there are no jumped-over attachment points (so that the attachment points are either primary or secondary attachment points), P is a one-jump attachment point, one-jump or multiple-jump attachment points are primary attachment points, secondary attachment points are no-jump attachment points, multiple-jump attachment points do not have associated secondary attachment points, and a multiple-jump attachment point is preceded by a no-jump attachment point (possibly a primary attachment point).
Here, "sufficiently small" means being on the verge of becoming too small. For a c=467 cycle, the plot of the |u| values and the absolute values of the corresponding proxy values versus their domain is;
(The x values have been scaled up by a factor of 108 and all but three data points are shown. No shape-preserving interpolation of data points is done.) The domain is (-2.390748e-007, 4.432486e-007).
As shown previously, the number of even elements in the extended sequence is sometimes smaller than the number of odd elements for a one-jump attachment point (especially when the attachment point is P). For two of the cycles for c=467, the number of even elements in an extended sequence is one less than the number of odd elements and this occurs for one-jump attachment points where the attachment point is P. For four of the 71 one-jump attachment points for the c=467 cycles, tu(mod 2j) where j is greater than the number of odd elements in the jump (in every case, locmin is not equal to u). For three of the 15 cycles, there is not a one-jump attachment point where j equals the number of odd elements in the jump.
When c=311, there are two primitive cycles having an (l, m) value of (57, 36) (a generalized continued-fraction convergent of log(3)/log(2)). There are 11 attachment points and these attachment points have the same properties as the c=467 cycles. For one of the cycles, the number of even elements in an extended sequence equals the number of odd elements and this occurs for a one-jump attachment point where the attachment point is P. For one of these cycles, there is not a one-jump attachment point where j equals the number of odd elements in the jump. When c=311, there are seventeen primitive cycles having an (l, m) value of (38, 24) (a generalized continued-fraction convergent of log(3)/log(2)). There are 66 attachment points and these attachment points have the same properties as the c=467 cycles except that there are two jumped-over attachment points (in different cycles). For two of the cycles the number of even elements in an extended sequence is one less than the number of odd elements and this occurs for one-jump attachment points where the attachment point is P. For fourteen of the 41 one-jump attachment points, tu(mod 2j) where j is greater than the number of odd elements in the jump (usually locmin is not equal to u). For seven of the cycles, there is not a one-jump attachment point where j equals the number of odd elements in the jump. The domains of the absolute values of the u values and the absolute values of their proxies are somewhat larger than the domains of the c=467 cycles having (l, m) values of (84, 53).
When c=343, there is a primitive cycle having an (l, m) value of (149, 94) (a generalized continued-fraction convergent of log(3)/log(2)). The domain of the absolute values of the u values and the absolute values of their proxies is (-6.216047e-007, 1.015725e-006), somewhat larger than the domains of the c=467 cycles having (l, m) values of (84, 53). There are 15 attachment points and these attachment points have the same properties as the c=467 cycles having (l, m) values of (84, 53) except that P is a no-jump attachment point. For all eight one-jump attachment points, tu(mod 2j) where j is less than or equal to the number of odd elements in the jump. For three of these attachment points, j equals the number of odd elements in the jump.
When c=1091, there is a primitive cycle having an (l, m) value of (130, 82) (a generalized continued-fraction convergent of log(3)/log(2)). The domain of the absolute values of the u values and the absolute values of their proxies is (-3.446479e-007, 8.889605e-007), somewhat larger than the domains of the c=467 cycles having (l, m) values of (84, 53). There are 12 attachment points and these attachment points have essentially the same properties as the c=467 cycles having (l, m) values of (84, 53) except that there is a jumped-over attachment point and the number of even elements in an extended sequence for a multiple-jump attachment point is equal to the number of odd elements. For one of the 7 one-jump attachment points, t≡u(mod 2j) where j is greater than the number of odd elements in the jump (locmin is not equal to u). For two of these attachment points, j equals the number of odd elements in the jump.
When c=2507, there are three primitive cycles having an (l, m) value of (76, 48) (a generalized continued-fraction convergent of log(3)/log(2)) and the domains of the absolute values of the u values and the absolute values of their proxies are (-1.018299e-006, 2.239814e-006), (-2.209624e-007, 7.065484e-007), and (-1.465981e-006, 2.967221e-006). In one of the cycles, P is a no-jump attachment point. In another cycle, there is a jumped-over attachment point and the number of even elements in an extended sequence for a one-jump attachment point (that of P) is equal to the number of odd elements. In the remaining cycle, there is a jumped-over attachment point preceding a multiple-jump attachment point. For one of the 11 one-jump attachment points for these cycles, t≡u(mod 2j) where j is greater than the number of odd elements in the jump (locmin is not equal to u). For one of these cycles, j equals the number of odd elements in the jump for three of the 5 one-jump attachment points. Although the domains appear to be small, they are not small enough for this large of a c value to exhibit all of the above properties; many anomalies occur. For comparison, when c=29, there are two primitive cycles having an (l, m) value of (65, 41) (a continued-fraction convergent of log(3)/log(2)) and the domains of the absolute values of the u values and the absolute values of their proxies are (-1.136015e-005, 2.425506e-005) and (-2.555711e-005, 3.583320e-005). There are 15 attachment points and these attachment points have the same properties as the c=467 cycles having (l, m) values of (84, 53) (except there are no anomalies).
When c=5, 13, 23, 71, 139, and 355, there are cycles having (l, m) values of generalized continued-fraction convergents of log(3)/log(2) that also have some properties of the c=467 cycles having (l, m) values of (84, 53) . Also, when c=407, cycles having an (l, m) value of (44, 28) (almost equal to (46, 29) [a generalized continued-fraction convergent of log(3)/log(2)]) have some properties of the c=467 cycles.
This is currently the extent to which 3n+c cycles with attachment points have been investigated. In the remainder of the article, 3n+c cycles not having attachment points are discussed.
Least-Residue Trees
In this section and the remaining sections, the original definition of 3n+c sequences (where n>0 and c≥-1) is used. Consider the following "least-residue" tree (where "{}" denotes a limb) consisting of the natural numbers less than 32k, k=4;
{4, 2, 1}
{24, 12, 6, 3, 10, 5, 16, 8}
{36, 18, 9, 28, 14, 7, 22, 11, 34, 17}
{30, 15, 46, 23}
{26, 13, 40, 20}
{38, 19}
{42, 21}
{32}
{44}
{25}
{27}
{29}
{31}
{33}
{35}
{37}
{39}
{41}
{43}
{45}
{47}
Each limb consists of a snippet from the 3n+c sequence (in this case, c=1). Note that "odd" paths are taken at nodes when tracing back through the sub-sequences (that is, if i is an even element of the sub-sequence and 3 divides i-c, then the element prior to i is (i-c)/3). A limb containing an odd natural number that is divisible by 3 will be referred to as a "dead" limb (and limbs that aren't dead will be referred to as being "alive"). A limb ending in an odd natural number greater than 2k attaches to the beginning of another limb (or possibly itself) when going from one size of the tree to the next. That is, if i is the last natural number in the limb, then 3i+c "cements" the limb ending in i to the beginning of another limb to form a longer limb in the next larger least-residue tree. A dead limb that starts with an even natural number and ends with an odd natural number cannot attach to itself at its beginning (or any point up to and including the odd natural number divisible by 3) since (3i+c)/2 (where i is the last natural number in the limb) is not divisible by 3. Similarly, a limb ending in an odd natural number greater than 2k cannot attach to the interior of another limb (or itself) since the even natural numbers in the interior of a limb (and, if the limb is dead, to the right of the odd natural number divisible by 3) are either of the form 3j+c where j is a natural number or are less than half the order, and the odd natural numbers in the interior of a limb are less than half the order. (These limbs have properties similar to those of interrelated 3n+c cycles having 1-2 sequence vectors. Also, as will be shown, there is a connection between these limbs and "admissible" parity vectors and Terras' stopping time.)
Note that there is no overall "gain" in a cycle. The feature of least-residue trees that makes them relevant to cycle formation is that there is no gain (at least in principle) from the end of a limb ending in an odd natural number greater than 2k to the beginning of the limb, that is, the last element of the limb is greater than order/3 and the first element of the limb is greater than order/2 (unproven as yet). Also, note that the definition of least-residue trees (where "odd" paths are taken) is not broad enough to encompass all possible 3n+c cycle formation. (All cycles will appear in the dead limbs of a least-residue tree of sufficiently large order, however, active limb formation is easier to analyze. More general least-residue trees will be introduced later.) Another feature of least-residue trees is that they provide a means to define "trivial" cycles. (What trivial cycles are is intuitively obvious, but they haven't been defined.) The 3n-1 cycles of {2, 1} and {20, 10, 5, 14, 7} will be classified as trivial cycles (the other known 3n-1 cycle contains an element divisible by 8, so this paves the way for a unified treatment of the 3n+1 and 3n-1 cycles). The objective in the following (not attained) is to use mathematical induction to prove the existence of a 3n+c "process". (In this process, a limb ending in an odd natural number greater than 2k cannot attach to itself at its beginning [apparently] when c=1 or -1. For a sufficiently large k, an "odd" path is in a limb of a least-residue tree [assuming that there is a jump in the "odd" path ending in an even natural number and that there is a backward jump in the "odd" path ending in an odd natural number divisible by 3]. By default, there would not be any cycles for 3n+1 or 3n-1 sequences having 1-2 sequence vectors.) A limb ending in an odd natural number greater than 2k must attach to the beginning of some limb when going to the next larger size of the tree; remaining questions are how many limbs end in an odd natural number greater than 2k and which limbs they attach to. When c>1, there are sometimes limbs in the least-residue trees starting with an even natural number and ending in an odd natural number i where 3i+c>32k and i<2k (this cannot occur for c=1 or -1, one of the unique properties of the 3n+1 and 3n-1 sequences). In the following, this is not considered to be a different kind of limb since the properties of these limbs haven't been fully determined yet. Nine different kinds of limbs are required to define the 3n+c process (this is an empirical result).
For convenience, the following sets will be defined for c=1 (when c≠1, certain sets are permuted). Let E denote the set of limbs that are not dead, have more than one element, and end in an even natural number. Let F denote the set of limbs that are dead and end in an even natural number. (Limbs in E or F are already attached to other limbs.) Let G denote the set of limbs that end in an odd natural number less than 2k (other than the previously mentioned limbs for c>1, these limbs always contain cycles [this is an empirical result]). Let A denote the set of limbs ending in an element of {2k+1, 2k+9, 2k+17, ..., 2k+1-7}, let B denote the set of limbs ending in an element of {2k+3, 2k+11, 2k+19, ..., 2k+1-5}, let C denote the set of limbs ending in an element of {2k+5, 2k+13, 2k+21, ..., 2k+1-3}, and let D denote the set of limbs ending in an element of {2k+7, 2k+15, 2k+23, ..., 2k+1-1}. (When c≡1(mod 8), the sets A, B, C, and D are not permuted. When c≡3(mod 8), the sets B, A, D, and C become the sets A, B, C, and D respectively. When c≡5(mod 8), the sets C, D, A, and B become the sets A, B, C, and D respectively. When c≡7(mod 8), the sets D, C, B, and A become the sets A, B, C, and D respectively. For example, when c=-1, the limbs in A end in odd natural numbers that are congruent to 7 modulo 8, the limbs in B end in odd natural numbers that are congruent to 5 modulo 8, the limbs in C end in odd natural numbers that are congruent to 3 modulo 8, and the limbs in D end in odd natural numbers that are congruent to 1 modulo 8.) Let T denote the set of limbs ending in an element {2k+1+1, 2k+1+3, 2k+1+5, ..., 3∙2k-1} (these are one-element limbs). Let U denote the set of limbs ending in an element of {[3(2k+5)+1]/2, [3(2k+13)+1]/2, [3(2k+21)+1]/2, ..., [3(2k+1-3)+1]/2} (these are one-element limbs). (When c is congruent to 1 modulo 6, the U set is the same. When c is congruent to 5 modulo 6, U is the set of limbs ending in an element of {[3(2k+1)+5]/2, [3(2k+9)+5]/2, [3(2k+17)+5]/2, ..., [3(2k+1-7)+5]/2}.) Also, let S denote the set of limbs ending in an element of {2k+1, 2k+3, 2k+5, ..., 2k+1-1}. The 3n+c process will be illustrated for k=5, 6, and 7.
The limb comprising E for k=6 and c=-1 is (118, 88) where the first element in the parentheses denotes the beginning element in the 3n+c sequence and the second element in the parentheses denotes the ending element. The limbs comprising F are (96, 4), (144, 28), (120, 16), (108, 40), (156, 64), (102, 76), and (126, 52), the limbs comprising G are (2, 1) and (20, 7), the limbs comprising T are (129), (131), (133), ..., (191), and the limbs comprising U are (100), (112), (124), (136), (148), (160), (172), and (184). In the following table, "a 2 (142, 71)→106(A)", for example, denotes that the limb is not dead (denoted by "a"), has 2 elements, and attaches to 106 (the first element of a limb in A). "d " denotes that the limb is dead. The limbs in A, B, C, and D (in the four columns respectively) are;
a 2 (142, 71)→106(A) d 2 (138, 69)→103(A) d 5 (180, 67)→100(U) a 2 (130, 65)→97(D)
a 4 (106, 79)→118(E) a 2 (154, 77)→115(C) d 2 (150, 75)→112(U) d 7 (132, 73)→109(B)
d 2 (174, 87)130(D) d 4 (114, 85)127(A) a 2 (166, 83)124(U) d 2 (162, 81)121(D)
a 2 (190, 95)142(A) d 2 (186, 93)139(T) d 25 (168, 91)136(U) a 2 (178, 89)133(T)
a 1 (103)154(B) a 1 (101)151(T) d 1 (99)148(U) a 1 (97)145(T)
d 1 (111)166(C) a 1 (109)163(T) a 1 (107)160(U) d 1 (105)157(T)
a 1 (119)178(D) d 1 (117)175(T) a 1 (115)172(U) a 1 (113)169(T)
a 1 (127)190(A) a 1 (125)187(T) d 1 (123)184(U) a 1 (121)181(T)
For k=7 and c=1, the limbs comprising E are (194, 188) and (218,164), the limbs comprising F are (192, 8), (288, 20), (240, 80), (336, 32), (264, 44), (312, 152), (360, 68), (204, 116), (228, 56), (276, 104), (300, 28), (324, 92), (372, 140), and (234, 176), the limbs comprising U are (200), (212), (224), ..., (380), and the limbs comprising T are (257), (259), (261), ..., (383). The limbs in A, B, C, and D consisting of more than one element are;
d 2 (258, 129)→194(E) d 5 (348, 131)→197(C) a 2 (266, 133)→200(U) d 2 (270,135)→203(B)
a 7 (242, 137)→206(B) a 2 (278, 139)→209(A) d 2 (282, 141)→212(U) d 7 (252, 143)→215(D)
a 2 (290, 145)→218(E) d 2 (294, 147)→221(C) d 4 (198, 149)→224(U) a 2 (302, 151)→227(B)
d 2 (306, 153)→230(C) a 4 (206, 155)→233(A) a 2 (314, 157)→236(U) d 2 (318, 159)→239(D)
d 17 (216, 161)→242(A) a 2 (326, 163)→245(C) d 2 (330, 165)→248(U) d 4 (222, 167)→251(B)
a 2 (338, 169)→254(D) d 2 (342, 171)→257(T) a 4 (230, 173)→260(U) a 2 (350, 175)→263(T)
d 2 (354, 177)→266(C) d 9 (210, 179)→269(T) a 2 (362, 181)→272(U) d 2 (366, 183)→275(T)
d 4 (246, 185)→278(B) a 2 (374, 187)→281(T) d 2 (378, 189)→284(U) a 4 (254, 191)→287(T)
When c=5 and k=5, the limbs in A, B, C, and D consisting of more than one element are;
a 7 (58, 37)→58(A), d 2 (78, 39)→61(A) d 2 (66, 33)→52(U) a 2 (70, 35)→55(D)
d 2 (90, 45)→70(D), a 2 (94, 47)→73(T) a 2 (82, 41)→64(U) d 4 (54, 43)→67(T)
When c=11 and k=5, the limbs in A, B, C, and D consisting of more than one element are;
a 2 (70, 35)→58(G), d 2 (66, 33)→55(C) d 2 (78, 39)→64(U) d 5 (84, 37)→61(D)
d 14 (48, 43)→70(A), a 2 (84, 41)→67(T) a 2 (94, 47)→76(U) d 2 (90, 45)→73(T)
The tables above show which types of limbs attach to other types of limbs and that only a limb in A can attach to itself at its beginning (of course, this hasn't been proven). Suppose i is the last element of a two-element limb in S. A two element limb in S cannot attach to itself at its beginning (except possibly for one order per c value) since (3i+c)/2 equals 2i only if i=c. An empirical result is;
(59) If a limb in S is not dead and has more than one element, then the second element is odd.
Suppose i is the last element of a four-element limb in S. In a four-element limb in S, the first and third elements are even and the second and fourth elements are odd. A four-element limb in S cannot attach to itself at its beginning (except possibly for one order when c=-1) since (3i+c)/2 equals 2(2i-c)/3 only if i=-7c. Usually, a limb in A cannot attach to itself at its beginning since the limb is either dead, a one-element limb, a two element limb, or a four-element limb. Usually, limbs in S do not attach to other limbs of equal length. When c=1 or -1, two-element, four-element, seven-element, twelve-element, seventeen-element, twenty-element, twenty-five element, and other longer length limbs in A sometimes attach to other limbs of equal length in A, B, C, or D. When c=1 or -1, a seven-element limb in A that attaches to a limb of equal length in A, B, C, or D is dead (this is an empirical result). For c=1 and k=18, the only limb in A that is not dead, does not consist of two or four elements, and that attaches to another limb in A with the same length is (484034, 306305). This limb has a length of 12 and attaches to the limb (459458, 290753). For c=1 and k=20, the only limb in A that is not dead, does not consist of two or four elements, and that attaches to another limb in A with the same length is (2056898, 1301633). This limb also has a length of 12. If c=-1, 11, 13, 25, 29, 31, 41, 43, 47, 59, 61, 73, 77, 79, 89, 91, or 95 (or larger unspecified values), let Ao denote the limbs in A where the integer portion of the last element of the limb divided by 8 is even (and let Ae denote the remaining limbs in A), or if c=1, 5, 7, 17, 19, 23, 25, 35, 37, 49, 53, 55, 65, 67, 71, 83, 85, or 97 (or larger unspecified values), let Ao denote the limbs in A where the integer portion of the last element of the limb divided by 8 is odd (and let Ae denote the remaining limbs in A). Another empirical result is;
(60) A limb in Ao attaches to a two-element or four-element limb in A, B, C, or D.
This property further reduces the probability of a limb in S attaching to itself at its beginning.
More Properties of Least-Residue Trees
In this section, c is restricted to being 1 or -1. (This is for the purpose of comparing 3n+1 and 3n-1 sequences; many of the properties to be given usually still apply for c>1.) Some empirical results are;
(61) If c=1, no limb in a least residue tree other than {4, 2, 1} ends in an odd natural number less than 2k (in the case of {4, 2, 1}, 1 is less than 2k for the order of 6). If c=-1, no limbs in a least residue tree other than {2, 1} and {20, 10, 5, 14, 7} end in an odd natural number less than 2k (in the case of {2, 1}, 1 is less than 2k for the order of 6 and in the case of {20, 10, 5, 14, 7}, 7 is less than 2k for the order of 24). (When c=-1 and k=2, the limb (10, 5) [in B] attaches to the limb (7) [in A], and the limb (7) attaches to the limb (10, 5). [There are no limbs in C, D, or U in this case.] When k=3, the limb in F is (12, 4), the limbs in G are (2, 1) and (20, 7), the limbs in S are (18, 9), (22, 11), (13), and (15), the limbs in T are (17), (19), (21), and (23), and the limb in U is (16). In essence, trivial cycles occur before the 3n+c process is fully in effect. Note that the dead limb {..., 54, 27, 80, 40} attaches to the limb {20, 10, 5, 14, 7} for k>3, so that the limbs in G could be viewed as being absorbed by dead limbs.)
(62) If k>2, there are 2k-3 limbs in A, B, C, or D.
(63) If k>2, there are 2k-3 limbs in E and F.
(64) If k>4, a fourth of the limbs in A attach to four-element limbs.
(65) If a limb in A ends in the natural number i where (3i+c)/2>2k+1, then the limb attaches to a two-element limb.
(66) The fourth element of a limb in E is even (there are at least four elements in a limb in E).
(67) A four-element limb in Ae cannot attach to a four-element limb in A, B, C, or D.
(68) A two-element limb in Ao cannot attach to a two-element limb in A or C.
(69) There are no limbs in E with lengths of 2, 3, 5, 6, 8, or 11.
The many common properties of the least-residue trees of the 3n+1 and 3n-1 sequences indicate that it would be more logical to allow negative n values and consider the two sequences to be the same sequence.
Other Cycles in Least-Residue Trees
Besides trivial cycles and limbs in A attaching to themselves at their beginning (as for the previously shown limb (58, 37) for c=5 and k=5), cycles can also be formed from limbs in F. When c=-1 and k=6, the limb (168, 91) (in C) attaches to 136 (in U) and when k=7, the limb {336, 168, ..., 91, 272, 136} contains 68 (136/2) and is in F. When c=5, the sequence 187, 566, 283, ..., 187 (consisting of 44 distinct elements) is not an "odd" path. When c=5 and k=10, the limb (2076, 1327) (in B) attaches to 1993 (in C) and when k=11, the limb {4152, 2076, ..., 1327, 3986, 1993, 5984, 2992} is in F. When k=11, the limb (3936, 2461) in (A) attaches to 3694 (in A). When k=12, the two dead limbs {8304, 4152, 2076, ..., 519, 1562, ..., 374} and {7872, 3936, 1968, ..., 123, 374, ..., 1562} form a cycle (containing all of 187, 566, 283, ..., 187) where the attachment points are immediately after the odd natural numbers divisible by 3.
When c=5, the sequence 347, 1046, 523, ..., 347 (consisting of 44 distinct elements) is not an "odd path". When k=12, the two dead limbs {7632, 3816, 1908, ..., 477, 1436, ..., 1334} and {7088, 3544, 1772, ..., 443, 1334, ..., 1436} form a cycle (containing all of 347, 1046, 523, ..., 347) where the attachment points are immediately after the odd natural numbers divisible by 3.
Limit Relationships in Least-Residue Trees and a Diophantine Equation
In this section, c is restricted to being 1 or -1. Most of the results in this section are empirically derived. Let y denote the first element of a limb in S and let z denote the first element of the limb that the limb in S attaches to. Let x denote y-zx/y is largely fixed for a particular limb length; these ratios approach limits as k approaches infinity. The x/y values for limbs that aren't dead approach their limits monotonically, that is, if c>0, then the maximum x/y value of live limbs of a given length for a given order is less than the maximum x/y value of live limbs of the given length for a larger order, and the minimum x/y value of live limbs of a given length for a given order is less than the minimum x/y value of live limbs of the given length for a larger order. Similarly, if c<0, then the maximum x/y value of live limbs of a given length for a given order is greater than the maximum x/y value of live limbs of the given length for a larger order, and the minimum x/y value of live limbs of a given length for a given order is greater than the minimum x/y value of live limbs of the given length for a larger order. The limits for limb lengths of 1, 2, 4, 5, 7, 9, 10, 12, 14, 15, 17, 18, 20, 22, 23, and 25 are -1/2, 1/4, -1/8, 7/16, 5/32, -17/64, 47/128, 13/256, -217/512, 295/1024, -139/2048, 1909/4096, 1631/8192, -3299/16384, 13085/32768, and 6487/65536 respectively. When c=1, the x/y values approach their limits from below, so x values for limb lengths where the limits are negative (1, 4, 9, 14, 17, 22, ...) can never reach 0. Similarly, when c=-1, the x/y values approach their limits from above, so x values for limb lengths where the limits are positive (2, 5, 7, 10, 12, 15, 18, 20, 23, 25, ...) can never reach 0. (Note that not all limb lengths are permissible. Lengths less than or equal to 101 that aren't permissible are 3, 6, 8, 11, 13, 16, 19, 21, 24, 26, 29, 32, 34, 37, 39, 42, 44, 47, 50, 52, 55, 57, 60, 63, 65, 68, 70, 73, 75, 78, 81, 83, 86, 88, 91, 94, 96, 99, and 101.) Additionally, x and y have to be solutions of the Diophantine equation ny-dx=ce for some n, d, and e values that are fixed for a particular limb length. The e values for limb lengths up to 18 are;
length=1, e=1
length=2, e=2
length=4, e=10
length=5, e=20
length=7, e=76, 58
length=9, e=260, 206
length=10, e=520, 412, 340
length=12, e=1688, 1364, 1148, 1004, 986, 842
length=14, e=5320, 4348, 3700, 3268, 3214, 2782
length=15, e=10640, 8696, 7400, 6536, 6428, 5960, 5564, 4988, 4916, 4340
length=17, e=32944, 27112, 23224, 20632, 20308, 18904, 17752, 17716, 15988, 15772, 14836, 14314, 14044, 12892, 12748, 12586, 11596, 11434, 11290, 10138
length=18, e=65888, 54224, 46448, 41264, 40616, 37808, 35504, 35432, 31976, 31544, 29672, 28628, 28088, 25784, 25496, 25172, 23192, 22868, 22580, 20276
Note that there are "consecutive length" pairs (1 and 2, 4 and 5, 9 and 10, 14 and 15, 17 and 18, ...) and that two times an e value for the smaller length of the pair is an e value for the larger length of the pair. Other than a length of 2, an e value for the larger length of the pair is divisible by 4. (Generalized dead limbs were previously discussed in connection with the properties of 3n+c cycles. When the element after an odd element i in a generalized dead limb is defined to be 3i+1, the lengths of generalized dead limbs that aren't permissible are 3, 6, 11, 16, 19, 24, 29, 32, 37, 42, 47, 50, 55, 60, 63, 68, 73, 78, 81, 86, 91, 94, 99, 104, .... The permissible lengths of generalized dead limbs include the permissible lengths of the limbs in S. New permissible lengths [in the generalized dead limbs] are three larger than the second element of a consecutive-length pair in S.) The origin of the Diophantine equation ny-dx=ce can be understood by using the previously mentioned formula s=(Xa-cZ)/Y. Let i denote the last element of a limb in S and let a=(3i+c)/2. Then x=(Xa-cZ)/Y-a, y=(Xa-cZ)/Y, n=X-Y, and d=X. Substituting into the equation ny-dx=ce and simplifying gives e=Z. Limbs in S having more than one element satisfy the Diophantine equation n(y/2)-(d/2)x=c(e/2) where d/2 is even. By Proposition (59), y/2 is odd when the limb is not dead, so e/2 must be odd. Limbs with lengths of 5, 10, 15, 18, 23, ... (where the e values are divisible by 4) must then be dead.
x>0 for limbs with positive x/y limits and x<0 for limbs with negative x/y limits (an empirical result), so a limb in S cannot attach to itself at its beginning. Also, y>order/2, therefore x=0 implies n(order/2)<ce, c>0, n>0, or (-n)(order/2)<(-c)e, c<0, n<0, a contradiction (based on empirical evidence). (Apparently, it is not necessary to solve the Diophantine equation ny=ce to arrive at a contradiction.) Proving that n(order/2)>ce, c>0, n>0, or (-n)(order/2)>(-c)e, c<0, n<0, would require finding the smallest order for which a given e value of a given limb length occurs. Let order1 denote the smallest order for which a given limb length occurs and let e11, e21, e31, ..., ei1 denote the e values that occur for this order. Let e12, e22, e32, ..., ej2 denote the remaining e values that occur for the given limb length and 2order1, let e13, e23, e33, ..., ek3 denote the remaining e values that occur for the given limb length and 4∙order1, etc. If c>0 and n>0, n(order1/2) is so much greater than ce11 for example, that n(order1/2)-ce11, n(order1/2)-ce21, n(order1/2)-ce31, ..., n(order1/2)-cei1 are all approximately equal. Similarly, n(2∙order1/2)-ce12, n(2∙order1/2)-ce22, n(2∙order1/2)-ce32, ..., n(2∙order1/2)-cej2 are all approximately equal and approximately twice as large as the differences for order1. Similar empirical results apply for c<0 and n<0.
The first step in proving that x cannot equal 0 is to find a formula for n. This paragraph applies for limbs in S. For a limb length of 12, X=28 and Y=35, for a limb length of 25, X=216 and Y=310, for a limb length of 38, X=224 and Y=315, ..., and for a limb length of 181, X=2112 and Y=370. (Note that 28-35>1, so that 28i-35i>1 where i is a natural number and hence the limits for limb lengths of the form 13m-1 where m is a natural number less than or equal to 14 are positive.) Blocks of eight adjacent limb lengths starting with a length of 1 and ending with a length of 181 consist of five types; an example of the first type is {1, 2, 4, 5, 7, 9, 10, 12}, an example of the second type is {40, 41, 43, 45, 46, 48, 49, 51}, an example of the third type is {66, 67, 69, 71, 72, 74, 76, 77}, an example of the fourth type is {105, 107, 108, 110, 111, 113, 115, 116}, and an example of the fifth type is {144, 146, 147, 149, 151, 152, 154, 155}. There are exactly three pairs of numerically consecutive limb lengths in each type. Denote the types by I, J, K, L, and M. The types of blocks of eight adjacent limb lengths starting with a length of 1 and ending with a length of 181 are I, I, I, J, J, K, K, K, L, L, L, M, M, and M. (In these blocks of eight adjacent limb lengths, the first limb length of a block is of the form 13m+1 where m is a non-negative integer.) The n values for the first type are 28i-7-35i-4, 28i-6-35i-4, 28i-5-35i-3, 28i-4-35i-3, 28i-3-35i-2, 28i-2-35i-1, 28i-1-35i-1, and 28i-35i respectively, i=1, 2, and 3. The n values for the second type are 28i-7-35i-4, 28i-6-35i-4, 28i-5-35i-3, 28i-4-35i-2, 28i-3-35i-2, 28i-2-35i-1, 28i-1-35i-1, and 28i-35i respectively, i=4 and 5. The n values for the third type are 28i-7-35i-4, 28i-6-35i-4, 28i-5-35i-3, 28i-4-35i-2, 28i-3-35i-2, 28i-2-35i-1, 28i-1-35i, and 28i-35i respectively, i=6, 7, and 8. The n values for the fourth type are 28i-7-35i-4, 28i-6-35i-3, 28i-5-35i-3, 28i-4-35i-2, 28i-3-35i-2, 28i-2-35i-1, 28i-1-35i, and 28i-35i respectively, i=9, 10, and 11. The n values for the fifth type are 28i-7-35i-4, 28i-6-35i-3, 28i-5-35i-3, 28i-4-35i-2, 28i-3-35i-1, 28i-2-35i-1, 28i-1-35i, and 28i-35i respectively, i=12, 13, and 14. Blocks of eight adjacent limb lengths starting with a length of 182 and ending with a length of 336 consist of five types. An example of the first type is {182, 183, 185, 186, 188, 190, 191, 193}, an example of the second type is {208, 209, 211, 213, 214, 216, 217, 219}, an example of the third type is {234, 235, 237, 239, 240, 242, 244, 245}, an example of the fourth type is {286, 288, 289, 291, 292, 294, 296, 297}, and an example of the fifth type is {312, 314, 315, 317, 319, 320, 322, 323}. Denote these types by I', J', K', L', and M'. The types of blocks of eight adjacent limb lengths starting with a length of 182 and ending with a length of 336 are I', I', J', J', K', K', K', K', L', L', M', and M'. There are exactly three pairs of numerically consecutive limb lengths in each type. In these blocks of eight adjacent limb lengths, the first limb length of a block is of the form 13m where m is a natural number. The n values for the first type are 28i-8-35i-4, 28i-7-35i-4, 28i-6-35i-3, 28i-5-35i-3, 28i-4-35i-2, 28i-3-35i-1, 28i-2-35i-1, and 28i-1-35i respectively, i=15 and 16. The n values for the second type are 28i-8-35i-4, 28i-7-35i-4, 28i-6-35i-3, 28i-5-35i-2, 28i-4-35i-2, 28i-3-35i-1, 28i-2-35i-1, and 28i-1-35i respectively, i=17 and 18. The n values for the third type are 28i-8-35i-4, 28i-7-35i-4, 28i-6-35i-3, 28i-5-35i-2, 28i-4-35i-2, 28i-3-35i-1, 28i-2-35i, and 28i-1-35i respectively, i=19, 20, 21, and 22. The n values for the fourth type are 28i-8-35i-4, 28i-7-35i-3, 28i-6-35i-3, 28i-5-35i-2, 28i-4-35i-2, 28i-3-35i-1, 28i-2-35i, and 28i-1-35i respectively, i=23 and 24. The n values for the fifth type are 28i-8-35i-4, 28i-7-35i-3, 28i-6-35i-3, 28i-5-35i-2, 28i-4-35i-1, 28i-3-35i-1, 28i-2-35i, and 28i-1-35i respectively, i=25 and 26. Blocks of eight adjacent limb lengths starting with a length of 338 and ending with a length of 454 consist of three types; an example of the first type is {338, 340, 341, 343, 345, 346, 348, 350}, an example of the second type is {377, 379, 381, 382, 384, 385, 387, 389}, and an example of the third type is {416, 418, 420, 421, 423, 425, 426, 428}. There are exactly two pairs of numerically consecutive limb lengths in each type (the last limb length in a block and the first limb length in the next block are numerically consecutive) and the first limb length of a block is of the form 13m where m is a natural number. Denote these types by N, O, and P. The types of blocks of eight adjacent limb lengths starting with a length of 338 and ending with a length of 454 are N, N, N, O, O, O, P, P, and P. The n values for the first type are 28i-8-35i-4, 28i-7-35i-3, 28i-6-35i-3, 28i-5-35i-2, 28i-4-35i-1, 28i-3-35i-1, 28i-2-35i, and 28i-1-35i+1 respectively, i=27, 28, and 29. The n values for the second type are 28i-8-35i-4, 28i-7-35i-3, 28i-6-35i-2, 28i-5-35i-2, 28i-4-35i-1, 28i-3-35i-1, 28i-2-35i, and 28i-1-35i+1 respectively, i=30, 31, and 32. The n values for the third type are 28i-8-35i-4, 28i-7-35i-3, 28i-6-35i-2, 28i-5-35i-2, 28i-4-35i-1, 28i-3-35i, 28i-2-35i, and 28i-1-35i+1 respectively, i=33, 34, and 35. An example of the type of blocks starting with a limb length of 456 is (456, 457, 459, 460, 462, 464, 465, 467). (There is a limb length of 455, but this limb length is considered to be associated with the previous block.) There are exactly three pairs of numerically consecutive limb lengths in each type and the first limb length of a block is of the form 13m+1 where m is a natural number.
The formula for the n values is valid for small limb lengths but gradually requires adjustments. Let a denote the number of 2's in a sequence vector of a live limb in S and b the number of 1's in the sequence vector (a and b are fixed for a particular limb length). The formula comes from the recursive algorithm for generating the a and b values corresponding to limb lengths of 2, 4, 7, 9, 12, 14, 17, 20, 22, 25, 27, 30, 33, 35, 38, 40, ..., (that is, limb lengths where not all of the limbs are dead). This algorithm is as follows. Set x to 1/2 and a and b to 0. If x is less than 1, set x to 3/2 times x and increment b, otherwise set x to 3/4 times x and increment a. (See "Power Fractional Parts" at Wolfram Mathworld [http://www.mathworld.wolfram.com/PowerFractionalParts.html] for a discussion of the inequality frac[(3/2)N]≤1-(3/4)N and its relationship to Waring's problem and Collatz 1-cycles.) The a values for limb lengths of 2, 4, 7, 9, 12, 14, 17, 20, 22, 25, 27, 30, 33, 35, 38, and 40 are 0, 0, 1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, and 8 respectively and the b values are 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, and 8 respectively. A steady-state where 5a is approximately equal to 7b is attained since 35 is approximately equal to 28. Note that the recursive algorithm has been defined for 1-2 sequence vectors; the 3/2 factor corresponds to the "gain" due to a 1 in the sequence vector and the 3/4 factor corresponds to the "loss" due to a 2 in the sequence vector. When the element after an odd element i is defined to be (3i+1)/2, the number of even elements in a live limb is a and the number of odd elements is a+b. If a limb attached to itself at its beginning, the length of the cycle would be 2a+b+1. Empirical evidence indicates that (2a+b)/(a+b) is approximately equal to log(3)/log(2). As previously seen, this irrational number (equal to log2(3)) frequently arises in 3n+1 cycle theory. Crandall showed (in his 1978 article) that if the main conjecture is true (that the cardinality of the "trajectory" of an odd natural number is finite), then powers of two and three tend to be poor approximations of each other. In imprecise terms, log2(3) must then be difficult to approximate with rational numbers. A parity vector is defined to be "admissible" when the sum of the elements in the vector equals the greatest integer in the length times θ where θ equals log(2)/log(3). For limbs having a length of 2a+b and containing a+b odd elements, the parity vectors are almost admissible. If another 0 is included in the parity vector (corresponding to a cycle), limbs with lengths of 2, 7, 12, 20, 25, 33, 38, ... (in the old way of defining the length and before the length is incremented by 1) are admissible. Quoting from Lagarias' 1985 article, Terras' "Theorem C" is;
(a) The set of integers with coefficient stopping time k are exactly the set of integers in those congruence classes n (mod 2k) for which there is an admissible vector v of length k with n=n0(v).
(b) Let n=n0(v) for some vector v of length k. If v is admissible, then all sufficiently large integers congruent to n (mod 2k) have stopping time k. If v is not admissible, then only finitely many integers congruent to n (mod 2k) have stopping time k.
By part (b) of Terras' theorem, if y were congruent to n=n0(v) modulo 2k (for an admissible parity vector v having a length of k=2a+b+1) for a large order of least-residue tree, then the stopping time would be k, a contradiction (since y/2 is less than y). As will be shown, there is a simpler way to arrive at the conclusion that a live limb in S of a given length (and with an arbitrary starting value y) cannot attach to itself at its beginning for a sufficiently large order of least-residue tree. Limbs with lengths of 4, 9, 14, 17, 22, 27, 30, 35, ... (in the old way of defining the length and before the length is incremented by 1) are still not admissible, but n (defined to equal X-Y) is negative for these lengths.
For limb lengths of 2, 4, 9, 14, 22, 27, 35, 40, ..., the maximum z value divided by the minimum y value for live limbs associated with a given sequence vector is less than 3/2 and for limb lengths of 7, 12, 17, 20, 25, 30, 33, 38, ...., the minimum z value divided by the maximum y value for live limbs associated with a given sequence vector is greater than 3/4. For live limbs with lengths of 4, 9, 14, 22, 27, ... (where the b values are 2, 3, 4, 5, 6, ... respectively and the b values for the next smaller limb lengths are 1, 2, 3, 4, 5, ..., respectively), the upper bound of the maximum y value to order ratio associated with the sequence vector giving the largest e value is (3/4)X/Y. (For large orders, the maximum y value to order ratio is approximately equal to the upper bound, and as the order decreases, the maximum y value to order ratio gradually decreases [but not monotonically].) For some limb lengths, the upper bound of the maximum y value to order ratio associated with a group of sequence vectors is different from the upper bounds associated with other groups of sequence vectors. (As will be shown, these groups of sequence vectors are the same as those giving different frequencies of occurrence of live limbs.) For example, for a live limb with a length of 12, the upper bound of the maximum y value to order ratio associated with the sequence vector giving the largest e value is 2/3((3/4)X/Y for a live with a length of 4) and the upper bound of the maximum y value to order ratio associated with the sequence vector giving the smallest e value is 16/27 ((3/4)X/Y for a live limb with a length of 9). For a live limb with a length of 17, the upper bound of the maximum y value to order ratio associated with two sequence vectors (one of which is the sequence vector giving the largest e value) is 2/3 ((3/4)X/Y for a live limb with a length of 4), the upper bound of the maximum y value to order ratio associated with two other sequence vectors (one of which is the sequence vector giving the smallest e value) is 128/243 ((3/4)X/Y for a live limb with a length of 14), and the upper bound of the maximum y value to order ratio associated with the remaining sequence vector is 16/27 ((3/4)X/Y for a live limb with a length of 9). The upper bound of the maximum y value to order ratio associated with any sequence vector of a live limb of a given length is one of 2/3, 16/27, 128/243, 4096/6561, 32768/50949, ....
These n values can then be substituted into the inequality n(order/2)>ce, c>0, n>0, or (-n)(order/2)>(-c)e, c<0, n<0. The next step is to find the maximum e value for a given limb length. Each sequence vector corresponds to exactly one e value (unproven as yet). Denote the sequence vector array by v (where the array is indexed starting with 0). Denote the number of elements in a sequence vector by l and the sum of the elements in the sequence vector by s. The length of the limb is then l+s, the Y value of the limb is 3l, and the X value of the limb is 2s+1. The e value of the limb is ∑3i∙2m where the summation is from i=0 to l-1 and where m=∑vj and the summation is from j=0 to l-1-i. The maximum e value for a given limb length then occurs when the 2's are at the beginning of the sequence vector (for a limb that is not dead, there is a single 1 at the beginning of the sequence vector and the 2's must follow this 1). The first odd element o in the limb after the 2's in the sequence vector is [(y/2)(3/4)a]+1 if c=1 or [(y/2)(3/4)a] if c=-1 where the brackets denote the greatest integer function. The 1's in the sequence vector (not counting the initial 1) correspond to a jump (which goes beyond the end of the limb). The first element of the limb that the limb attaches to is then z=(3/2)b(o+c)-c. The e value is then (zX-yY)/c. (The 2's in the sequence vector decrease the limb elements and the 1's in the sequence vector compensate for this by increasing the limb elements until the last element of the limb is greater than order/3.) For a live limb in S, (y/2)(3/4)a(3/2)b+δ=z where δ is relatively small and positive if c≥1 or negative if c=-1. x=0 then implies 1=(1/2)(3/4)a(3/2)b+δ/y. 3a+b/22a+b+1=Y/X, so whether x can equal 0 depends on how close n gets to 0 (x=0 implies n=X(δ/y)). (Approximating powers of 2 with powers of 3 becomes relevant at this point.) For a given order, the largest |X(δ/y)| values correspond to the sequence vector giving the largest e value, so the largest |X(δ/y)| values are easily quantifiable (the X(δ/y) values only depend on the y values due to the greatest integer function). As the order increases, the |X(δ/y)| values decrease. For example, for c=1 and a limb length of 25, the first live limbs occur when the order equals 393216. Live limbs occur that correspond to the ninth largest e value (for live limbs) and the fourteenth largest e value (for live limbs) and the respective X(δ/y) values are 2.624163 and 2.292368. For an order of 786432, live limbs occur that correspond to the largest e value, the fourth largest e value, the sixth largest e value, the seventh largest e value, the eleventh largest e value, the twelveth largest e value, the seventeenth largest e value, and the eighteenth largest e value and the respective X(δ/y) values are 1.693732, 1.415807, 1.285766, 1.281626, 1.156335, 1.135663, 1.079423, and 1.011811. The n value for this limb length is 6487. Since y>order/2, an upper bound for |X(δ/y)| (valid for both c=1 and c=-1) is 22a+b+2-k((2(3/2)b-1)/3) (when a>0). This can be used to find small orders for which n>X(δ/y) if c=1 or -n>X(-δ/y) if c=-1. For live limbs with a length of 25, a=5, b=5, and 6487 is greater than the upper bound until k=6. The smallest k values such that n is greater than the upper bound of X(δ/y), c>0, n>0, for live limbs with lengths of 7, 12, and 20 are 4, 7, and 5 respectively.
In his 1978 article, Crandall used continued fractions to show that if m>1 and Ck(m)=m, then k>17985 (C(x) is defined to be (3x+1)/2e(x) where 2e(x) is the highest power of two dividing 3x+1). The following lemma is used. Let 1<m=inf Tm and let k be the period of the trajectory Tm. Then m<k(3+1/m)k-1/(2A(k)-3k) (where A(k) [due to typographical difficulties, A(k) is used to denote Ak] is a sum of positive integer sequence values). For a live limb in S, this gives m<(a+b)(3+1/m)a+b-1/(22a+b+1-3a+b) where a and b are generated from the recursive algorithm above. The elements of a live limb in S with a length less than or equal to 20 are greater than order/12. (In general, the lower bound of the elements of the limb depends on the sequence vector giving the maximum e value. For example, for a limb length of 22, (1/4)(3/4)4<1/2.) The largest k values for which 2k-2<(a+b)(3+1/2k-2)a+b-1/(22a+b+1-3a+b) for live limbs with lengths of 7, 12, and 20 are 4, 7, and 5 respectively. (a+b)3a+b-1/(22a+b+1-3a+b) does not grow very fast as a and b increase, so the maximum order for which a cycle might occur can be approximated fairly accurately. The difficulty with this approach (and the above approach) is that small orders have to be checked for cycles for every limb length.
More miscellaneous results pertaining to allowable (not necessarily admissible) sequence vectors and the maximum e values are discussed in this section. The sequence vectors of limbs in S that aren't dead for lengths of two, four, seven, and nine are (1), (1, 1), (1, 2, 1), and (1, 2, 1, 1) respectively. The only possible sequence vectors for a live limb in S with a length of twelve are (1, 2, 2, 1, 1) and (1, 2, 1, 2, 1). The only possible sequence vectors for a live limb in S with a length of fourteen are (1, 2, 2, 1, 1, 1) and (1, 2, 1, 2, 1, 1). For a live limb in S with a length of 17 or 20, there are five different sequence vectors. For a live limb in S with a length of 22, there are nine different sequence vectors and for a live limb in S with a length of 25, there are nineteen different sequence vectors. A subset of the sequence vectors for a limb length of 22 is;
(1, 2, 1, 2, 1, 2, 2, 1, 1)
(1, 2, 2, 1, 1, 2, 2, 1, 1)
(1, 2, 2, 1, 2, 1, 2, 1, 1)
(1, 2, 2, 1, 2, 2, 1, 1, 1)
(1, 2, 2, 2, 1, 2, 1, 1, 1)
(1, 2, 2, 2, 2, 1, 1, 1, 1)
The first sequence vector corresponds to the smallest e value for a live limb and the last sequence vector corresponds to the largest e value for a live limb. The second sequence vector is obtained from the first sequence vector by transposing an adjacent 1 and 2, the third sequence vector is obtained from the second sequence vector by transposing an adjacent 1 and 2, the fourth sequence vector is obtained from the third sequence vector by transposing an adjacent 1 and 2, etc. until the last sequence vector is obtained. The number of elements in a sequence vector is l (9 in this case) and the number of sequence vectors (6) is approximately equal to l. The 1's and 2's that are transposed gradually move from left to right in a somewhat regular fashion, wrap around at the third-to-last sequence vector, and then resume moving from left to right. The minimum e value divided by the maximum e value is 0.62636. The corresponding subset of sequence vectors for a limb length of 25 is;
(1, 2, 1, 2, 2, 1, 2, 1, 2, 1)
(1, 2, 1, 2, 2, 2, 1, 1, 2, 1)
(1, 2, 1, 2, 2, 2, 1, 2, 1, 1)
(1, 2, 2, 1, 2, 2, 1, 2, 1, 1)
(1, 2, 2, 2, 1, 2, 1, 2, 1, 1)
(1, 2, 2, 2, 1, 2, 2, 1, 1, 1)
(1, 2, 2, 2, 2, 1, 2, 1, 1, 1)
(1, 2, 2, 2, 2, 2, 1, 1, 1, 1)
In this case, l=10 and the number of sequence vectors is 8. The minimum e value divided by the maximum e value is 0.55595. The corresponding subset of sequence vectors for a limb length of 27 is;
(1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 1)
(1, 2, 1, 2, 1, 2, 2, 2, 1, 1, 1)
(1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 1)
(1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 1)
(1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 1)
(1, 2, 2, 2, 1, 2, 1, 2, 1, 1, 1)
(1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 1)
(1, 2, 2, 2, 2, 1, 2, 1, 1, 1, 1)
(1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1)
In this case, l=11 and the number of sequence vectors is 9. The minimum e value divided by the maximum e value is 0.53027. The corresponding subset of sequence vectors for a limb length of 30 is;
(1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1)
(1, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1)
(1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1)
(1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1)
(1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 1)
(1, 2, 1, 2, 2, 2, 2, 1, 1, 2, 1, 1)
(1, 2, 1, 2, 2, 2, 2, 1, 2, 1, 1, 1)
(1, 2, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1)
(1, 2, 2, 1, 2, 2, 2, 2, 1, 1, 1, 1)
(1, 2, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1)
(1, 2, 2, 2, 2, 1, 2, 2, 1, 1, 1, 1)
(1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 1, 1)
(1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1)
In this case, l=12 and the number of sequence vectors is 13. The 1's and 2's to be transposed move from left to right in a more regular fashion and wrap around twice. The minimum e value divided by the maximum e value is 0.42839. The corresponding subset of sequence vectors for a limb length of 33 is;
(1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1)
(1, 2, 2, 2, 1, 1, 2, 2, 1, 2, 1, 2, 1)
(1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1)
(1, 2, 2, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1)
(1, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 1)
(1, 2, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1)
(1, 2, 2, 2, 2, 2, 1, 1, 2, 1, 1, 2, 1)
(1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 1, 2, 1)
(1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 1)
(1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 1, 1)
(1, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1)
(1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 1, 1, 1)
(1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1)
In this case, l=13 and the number of sequence vectors is 13. The 1's and 2's to be transposed wrap around once. Since the length of a sequence vector approximately equals the number of sequence vectors, the 1's and 2's to be transposed move on average one position to the right from one sequence vector to the next (disregarding wrap-around). The minimum e value divided by the maximum e value is 0.44269 (the minimum e value divided by the maximum e value approaches 1/2).
k=12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
L=2 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
7 0 0 0 0 1 -1 0 0 0 0 0 0 0 -1 1 -1 0 0 0 0 0 0
9 0 0 1 0 0 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0
12 0 0 0 0 0 -1 1 -1 1 -1 -1 0 0 0 -1 -1 0 0 0 0 0 0
14 -1 0 0 -1 0 0 -1 0 0 0 0 0 0 0 0 0 -1 0 -1 0 -1 0
17 0 1 1 1 0 1 0 1 -1 1 0 1 0 2 1 1 0 1 0 0 1 0
20 0 -1 0 -1 1 -1 1 0 1 -1 1 -1 1 0 1 -1 0 -1 1 -1 1 -1
22 0 -1 1 0 -3 -1 -2 0 -1 -1 -1 -2 -3 -2 -1 -3 -2 0 1 1 1 2
25 0 0 0 1 2 1 -2 3 0 5 4 4 1 2 0 5 0 -1 -4 1 1 6
27 0 0 0 0 1 -1 1 2 3 -1 0 4 -1 1 2 1 5 2 1 1 0 0
30 0 0 0 0 0 1 3 -3 1 0 -2 -2 -1 1 -2 -2 -2 1 0 1 1 4
33 0 0 0 0 0 0 0 0 -1 1 -1 1 0 -4 1 0 2 -3 2 -5 0 -7
For every order and limb length, the difference between the number of live limbs in S associated with one sequence vector in a group and the number of live limbs in S associated with another sequence vector in the group is at most 1. (This appears to still be true for live limbs in A, B, C, or D or any other similarly constructed subset of S where the last elements in the limbs are congruent to an odd natural number j modulo 2i, j<2i , i<k).
The above property can be used to estimate the smallest order for which a live limb in S with a given length occurs. For example, assuming that the upper bound of the difference between the estimated and actual number of live limbs with a length of 25 (where the difference is given by the formula [2k-1(3791/(212∙39))+0.5]-m') is 6 (a somewhat dubious proposition) gives m'≥1 when k=19 (when c=1, the smallest order for which a live limb with a length of 25 occurs is 3∙217). The n value for a limb length of 25 is 6487 and the largest e value is 811694, so n(order/2) is much greater than ce, c>0, n>0, when order=3∙219. The difficulty with this approach is determining the upper bound of the difference between the estimated and actual number of live limbs.
When c>1, there are relatively few limbs in S having lengths other than 1, 2, 4, 5, 7, 9, 10, 12, 14, 15, 17, 18, 20, 22, 23, 25, .... When k is sufficiently large for a given limb length and c value greater than 1, the limbs in S have many of the above properties.
Another Kind of Least-Residue Tree
In this section, c is restricted to being 1 or -1. Consider a tree where the order is of the form 3∙2k and no element of a limb is greater than or equal to the order or divisible by 3. The first element of a limb must be even and greater than the order divided by 2, the last element of a limb must be odd and greater than the order divided by 3, and every limb must contain at least one element that is divisible by 8. (As for a limb in S of a least-residue tree, a limb cannot attach to its interior or the interior of another limb.) The possible lengths of these limbs are 12, 14, 15, 17, 18, 20, 22, 23, .... The lengths of the limbs are the same as those for a limb in S of a least-residue tree (except there are no lengths of 1, 2, 4, 5, 7, or 9) and the X and Y values are the same. Denote the first element of a limb by y, the last element of a limb by i, (3i+c)/2 by z, and y-z by x. As before, x and y satisfy the Diophantine equation ny-dx=ce where n=X-Y, d=X, and the e values are fixed for a given limb length. Some e values are;
length=12, e=1202
length=14, e=3862
length=15, e=7724, 5780
length=17, e=24196, 22738, 18850, 18364, 16906, 16258, 14530, 14476, 13378, 13018
length=18, e=48392, 45476, 37700, 36728, 33812, 32516, 29060, 28952, 26756, 26036
For a given limb length, the largest e value is larger than the largest e value for a live limb in S of a least-residue tree (when not all of the limbs are dead for that length) and the smallest e value (when there is more than one e value) is smaller than the largest e value for a live limb in S of a least-residue tree (when not all the limbs are dead for that length). The lengths of the sequence vectors for limb lengths of 12, 14, 15, 17, 18, 20, 22, and 23 are 5, 6, 6, 7, 7, 8, 9, and 9 respectively. The largest elements in the sequence vectors for limb lengths of 12, 14, 15, 17, 18, 20, 22, and 23 are 3, 3, 3, 4, 4, 5, 5, and 5 respectively. For some limb lengths, limbs associated with certain sequence vectors occur more frequently than limbs associated with other sequence vectors. For example, for a limb length of 17, the number of limbs associated with two sequence vectors is about [2k-1(295/(29∙36))+0.5], the number of limbs associated with five other sequence vectors is about [2k-1(5/(210∙31))+0.5], the number of limbs associated with two other sequence vectors is about [2k-1(5/(29∙33))+0.5], and the number of limbs associated with the remaining sequence vector is about [2k-1(13/(210∙35))+0.5]. For every order and limb length, the difference between the number of limbs associated with one sequence vector in a group and the number of limbs associated with another sequence vector in the group is at most 2. For some limb lengths, the upper bound of the maximum y value to order ratio associated with a group of sequence vectors is different from the upper bounds associated with other groups of sequence vectors. (These groups of sequence vectors are the same as those giving different frequencies of occurrence of limbs.) For example, for a limb length of 17, the upper bound of the maximum y value to order ratio associated with two sequence vectors is 512/729, the upper bound of the maximum y value to order ratio associated with five other sequence vectors is 2/3, the upper bound of the maximum y value to order ratio associated with two other sequence vectors is 16/27, and the upper bound of the maximum y value to order ratio associated with the remaining sequence vector is 128/243. For limb lengths of 12, 14, 15, 17, 18, 20, 22, and 23, the number of limbs can be approximated by [2k-1f+0.5] where f equals 17/(25∙35), 13/(28∙35), 353/(28∙36), 1057/(29∙36), 695/(29∙37), 6733/(210∙37), 9817/(213∙36), and 56915/(214∙37) respectively. The number of limbs in one of these trees is approximately equal to the order divided by 144.
m-Cycles
In their 2005 article, Simons and de Weger gave bounds for Λ=(K+L) log 2 - K log 3 (where K is the total number of odd elements and L is the total number of even elements in the m-cycle). A corollary proved is 0<Λ<m/xmin. For a limb in S of a least-residue tree (with a length of 3a+2b) to attach to itself at its beginning, this gives xmin<(a+1)/((2a+b+1) log 2 - (a+b) log 3). This upper bound for the minimum is about twice as large as the minimum given by Halbeisen and Hungerbühler's formula. In the next two sections, the upper bound of the minimum will be computed when it's not computationally feasible to compute the minimum.
The Minimum in a Collatz Cycle
In this section, c is restricted to being 1 or -1. The results in this section are mostly empirically derived. The 1-2 sequence vector (of a limb in S having a length of l and containing m odd elements) corresponding to a small e value (possibly the smallest) can be constructed from ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l, by rotating (if necessary) the vector so that a (1, 1) is at the beginning of the vector and converting (1, 0)'s to 2's. The last element of the sequence vector will be a 2 and must be changed to a 1 to match the sequence vector corresponding to the smallest e value. (Note that this rotation of the parity vector does not correspond to the parity vector of a limb in S [since the first element must be even] and that the first element of a sequence vector just indicates that the first element of the limb is even [and not that there is an odd natural number in the 3n+c sequence immediately preceding it]. If a cycle were formed by a limb attaching to itself at its beginning, the first element of the sequence vector would be changed to a 2 [and the 2 would indicate that there were 2 even elements between odd elements in the sequence].) For example, for a limb length of 20, the parity vector must be rotated right by 5 positions for the resulting sequence vector to match the sequence vector corresponding to the smallest e value. (The possible right-rotations for a limb length of 20 are 0, 5, and 10. In general, the difference between successive right-rotations is 3 or 5.) The required right-rotations of the parity vector for limb lengths of 2, 4, 7, 9, 12, 14, 17, 20, 22, 25, 27, 30, 33, 38, and 43 are 0, 0, 0, 0, 0, 0, 0, 5, 11, 0, 11, 0, 5, 0, and 0 respectively. If the un-rotated parity vector gives an e value for a limb in S, then the e value is the smallest for a limb in S and all the possible right-rotations give e values for limbs in S. For limb lengths of 20, 22, 27, 33, 35, 40, 45, 48, 51, 53, and 56, the un-rotated parity vector gives e values too small to be a limb in S. In these cases, a right-rotation of 5 gives an e value (usually not the smallest) for a limb in S.
A rotation of the parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l, so that a (1, 0) is at the beginning of the vector does not correspond to a live limb in S since the second element of a live limb in S is odd. A rotation of the parity vector so that a (0, 1) is at the beginning of the vector corresponds to a live limb in S, but except for limb lengths of 7, 12, 17, 30, 43, ..., the e values given are too small. For a given limb length, live limbs in S occur for only one right-rotation of the parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l. (The resulting sequence vectors for limb lengths of 7, 12, 17, 30, and 43 are (1, 2, 1), (1, 2, 1, 2, 1), (1, 2, 1, 2, 1, 2, 1), (1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1), and (1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1) respectively. [For example, for a limb length of 43, the parity vector is 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, and 1. Changing the last 1 and the first 0 (a (1, 0)) to a 1 and the other (1, 0)'s to 2's gives the sequence vector.] These sequence vectors have bilateral symmetry and are the sequence vectors corresponding to the smallest e values.)
For limb lengths less than or equal to 51, no limb in S can have the parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l, if l and m are not relatively prime (51 is the largest limb length where it is computationally feasible to determine this [at least by an exhaustive search up to relatively large orders of least-residue trees]). For limb lengths of 7, 12, 17, 30, and 43, l and m are relatively prime and there are limbs in S having the parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l, (the (l, m) values for limb lengths of 12 and 30 [(8, 5) and (19, 12) respectively] are continued-fraction convergents of log(3)/log(2)). For limb lengths of 20, 22, 27, 33, 40, and 48, l and m are relatively prime, but there are no limbs in S having the parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l.
For a live limb in S, the second sequence vector value is a 2 if the limb length is greater than 4 (so that the parity vector must begin with a (0, 1, 0)). Let v1, v2, v3, ..., vl be a right-rotation of the parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l, such that v1=0, v2=1, and v3=0. Let s1=1.0 and si+1=(3/2)si if vi=1 or (1/2)si otherwise, i=1, 2, 3, ..., l-1. Let r denote gcd(l, m). Requirements for a live limb in S (with a sufficiently large beginning element y) to have this right-rotated parity vector are that (1) twice the minimum s value be greater than the maximum "odd" s value for all possible right-rotations of the parity vector where v1=0, v2=1, and v3=0, (2) the difference in successive right-rotations of the parity vector and the difference in successive indices (modulo l/r) of the minimum s value must be the same for all possible right-rotations of the parity vector where v1=0, v2=1, and v3=0, and (3) the difference in the indices (modulo l/r) of the minimum and maximum "odd" s values must be the same for all possible right-rotations of the parity vector where v1=0, v2=1, and v3=0. Limb lengths satisfying the first condition for m values less than or equal to 100 are 7, 12, 17, 20, 25, 30, 38, 43, 51, 56, 61, 74, 87, 92, 105, 123, 136, 211, and 242. Of the first 8192 limb lengths, 20 is the only limb length that satisfies the first and third conditions, but not the second condition. Of the first 8192 limb lengths, 56 is the only limb length that satisfies the first condition, but not the second and third conditions. (l, m) values satisfying the first condition are (5, 3), (8, 5), (11, 7), (13, 8), (16, 10), (19, 12), (24, 15), (27, 17), (32, 20), (35, 22), (38, 24), (46, 29), (54, 34), (57, 36), (65, 41), (76, 48), (84, 53), (130, 82), (149, 94), (168, 106), (233, 147), (252, 159), (317, 200), (336, 212), (401, 253), (420, 265), (485, 306), (504, 318), (569, 359), (970, 612), (1054, 665), (1455, 918), (1539, 971), (2108, 1330), (2593, 1636), (3162, 1995), (3647, 2301), (4216, 2660), (4701, 2966), (5270, 3325), (5755, 3631), (6324, 3990), (6809, 4296), 7378, 4655), (7863, 4961), (8432, 5320), (8917, 5626), (9486, 5985), (9971, 6291), (10540, 6650), (11025, 6956), (11594, 7315), (12079, 7621), (12648, 7980), .... Except for (13, 8), (32, 20), (35, 22), and (54, 34), these are generalized continued-fraction convergents of log(3)/log(2) (all the convergents except (1, 1), (2, 1), (3, 2), (4, 2), (6, 4), and (9, 6) are included).
The Minima in Collatz Cycles and Prime Gaps
In this section, c is restricted to being 1 or -1 and Halbeisen and Hungerbühler's formula is used to compute the minimum in a cycle formed by a limb in S of a least-residue tree attaching to itself at its beginning (this assumes the cycle has the above parity vector). l denotes the number of elements in the cycle and m denotes the number of odd elements. The absolute value of 2l-3m is not used when computing the "minimum". The minima where m is a multiple of a natural number d increase and decrease in a regular fashion. For example, the truncated minima for m=5, 10, 15, ..., 65 are 24, 24, 24, 24, 24, -19, -27, -35, -48, -70 -111, -219, and -1004 respectively, the truncated minima for m=70, 75, 80, ..., 130 are 463, 204, 137, 108, 88, -58, -70, -86, -111, -147, -219, -361, and -1004 respectively, and the truncated minima for m=135, 140, 145, ..., 195 are 1561, 463, 281, 204, 162, 137, -111, -133, -165, -219, -295, -462, and -1004 respectively. In general, the minima for d =5 have a roughly saw-tooth shaped curve with a period of 13 or 14. Some long periods (and large minima) occur for d=7, 12, 17, 19, 22, and 29; the periods are 10 or 11, 51 or 52, 17 or 18, 8 or 9, 7 or 8, and 27 or 28 respectively. The "superposition" of these quasi-periodic "waves" is of importance in determining the largest minima. For example, a large truncated minimum (for c=-1) of -3664765 occurs when m=665=5∙7∙19.
The periods for prime m values less than 41346 and of the form 12k-1, k=1, 2, 3, ..., are 17, 24, 24, 18, 18, 22, 18, 15, 19, 19, ..., 17, 14, 16, 13, 13, 13, 16, 17, 15, and 13. A period is considered to start with a negative minima value and end with a positive minima value. A "peak" occurs when there is a positive spike in minima values (at the end of a period) immediately followed by a negative spike. The difference in m values between successive peaks in minima values appears to be fairly constant (usually 612, but as small as 528 or as large as 696). Similar results apply for prime m values of the form 12k-5, 12k-7, or 12k+1. Let x denote one-twelveth of the difference in m values between successive peaks in minima values. For prime m values less than 2000000, the x values range from 31 to 107, the distribution of x values for 31, 32, 33, ..., 71 is 2, 5, 3, 11, 14, 22, 29, 29, 46, 98, 110, 156, 171, 228, 394, 418, 563, 697, 1047, 1578, 1561, 1367, 953, 819, 823, 477, 333, 270, 190, 175, 102, 70, 70, 40, 46, 24, 12, 10, 6, 9, and 3 respectively (a population centered around x=51), and the distribution of x values for 91, 92, 93, ..., 107 is 1, 1, 0, 0, 0, 0, 1, 0, 3, 3, 3, 1, 3, 4, 3, 0, and 1 respectively (a second population centered around x=102). Each of these populations has a Weibull probability distribution. A Weibull probability plot of the first population is;
For the first population and prime m values less than 4000000, the shaping parameter is .6692 (with a 95% confidence interval of (.5263, .8510)) and the scaling parameter is 477.4297 (with a 95% confidence interval of (294.3538, 774.3714)). A Weibull probability plot of the second population of x values for prime m values less than 4000000 is;
For the second population and prime m values less than 6000000, the shaping parameter is 1.1256 (with a 95% confidence interval of (0.8173, 1.5501)) and the scaling parameter is 6.777 (with a 95% confidence interval of (4.6156, 9.9505)). There should be infinitely many such populations. The minima appear to be bounded for all m values in a given population. For example, 240∙1.5704 is an upper bound of the minima (as computed using the m-cycle inequality) in the first population when m is less than 1000000 and is still an upper bound of the minima in the first population when m is less than 6000000. When c=-1, 239∙1.2002 is an upper bound of the minima in the first population when m is less than 3000000 and is still an upper bound of the minima in the first population when m is less than 6000000.
The minima in the second population do not increase monotonically within a period; there is a characteristic dip about half-way through the period. For example, a graph of truncated upper bounds of minima (scaled by 32768.0) for m=609757, 609781, 609877, ..., 610993 (primes of the form 12k+1) is;
Similarly, there is a dip about a third of the way through the period and another dip about two-thirds of the way through the period for the minima in the third population. For example, a graph of truncated upper bounds of minima (scaled by 32768.0) for m=5905433, 5905553, 5905673, ..., 5907353 (primes of the form 12k+5) is;
A graph of the maximum upper bounds of minima (scaled by 65536.0) for x values from 31 to 72 (in the first population) for prime m values less than 6000000 is;
(The scaled maximum upper bounds of minima for x=31, 32, 33, ..., 52 are 174, 188, 199, 226, 234, 248, 274, 295, 320, 344, 379, 418, 480, 532, 634, 746, 934, 1202, 1760, 3356, 24559, and 26346062 respectively. This appears to be an exponential probability distribution where μ=1.1993e+06 with a 95% confidence interval of (.8219e+06, 1.9136e+06). The scaled maximum upper bounds of minima for x=53, 54, 55, ..., 72 appear to be random samples from the same exponential probability distribution.) A graph of the maximum upper bounds of minima (scaled by 65536.0) for x values from 89 to 114 (in the second population) for prime m values less than 6000000 is;
A graph of the sorted maximum upper bounds of minima for x values from 89 to 144 for prime m values less than 6000000 is;
The periods for prime l values of the form 54k+1 are approximately equal. Similar results apply for prime l values of the form 54k+5, 54k+7, 54k+11, 54k+13, 54k+17, 54k+19, 54k+23, 54k+25, 54k+29, 54k+31, 54k+35, 54k+37, 54k+41, 54k+43, 54k+47, 54k+49, or 54k+53. Let x denote one fifty-fourth of the difference in l values between successive peaks in minima values. For prime l values less than or equal to 3169919, the distribution of x values for 11, 12, 13, ..., 17 is 504, 1488, 2848, 4712, 4054, 1907, and 1009 respectively (a population centered around x=14), the distribution of x values for 25, 26, 27, ..., 32 is 183, 937, 1510, 2826, 2409, 2020, 811, and 165 respectively (a population centered around x=28), the distribution of x values for 40, 41, 42, ..., 46 is 433, 664,1312, 1258, 899, 705, and 184 respectively (a population centered around x=42), etc. These are Weibull probability distributions (with different parameters). (The shaping parameters range from about 1.3 to 2.2, so these are "aging" processes.) For prime l values less than or equal to 6339847, the scaled (by 65536.0) maximum upper bounds of minima (as computed using the m-cycle inequality) for x values of 11, 12, 13, ..., 17 are 6220270, 436632, 744405, 514414, 597, 256, and 166 respectively. For prime l values less than or equal to 6339847, the scaled maximum upper bounds of minima for x values of 25, 26, 27, ..., 32 are 332686, 1366582, 93783, 526748, 831, 299, 179, and 130 respectively. For prime l values less than or equal to 6339847, the scaled maximum upper bounds of minima for x values of 40, 41, 42, ..., 46 are 301174, 540827, 101914, 1552, 362, 196, and 139 respectively. These appear to be exponential probability distributions. For prime l values less than or equal to 9509761, the scaled maximum upper bounds of minima for x values of 11, 12, 13, ..., 17 are 6220270, 436632, 744405, 916817, 900, 387, and 246 respectively. For prime l values less than or equal to 9509761, the scaled maximum upper bounds of minima for x values of 25, 26, 27, ..., 32 are 9037278, 1366582, 229310, 526748, 1320, 439, 269, and 195 respectively. For prime l values less than or equal to 9509761, the scaled maximum upper bounds of minima for x values of 40, 41, 42, ..., 46 are 301174, 540827, 149513, 2343, 521, 294, and 204 respectively. In general, the scaled maximum upper bounds of minima for the larger x values of each population are still growing. (As the prime gaps become larger, the scaled maximum upper bounds of minima for even the larger x values of say the first population should stop growing.)
A graph of minima (along the y axis) for m=11, 59, 71, ..., 5099 (primes of the form 12k-1 where m+2 is also a prime) is;
A graph of minima for m=13, 61, 73, ..., 5101 (primes of the form 12k+1 where m-2 is also a prime) is;
A graph of minima for m=5, 17, 29, ..., 5021 (primes of the form 12k-7 where m+2 is also a prime) is;
A graph of minima for m=7, 19, 31, ..., 5023 (primes of the form 12k-5 where m-2 is also a prime) is;
A graph of the upper bounds of the minima for m=11, 59, 71, ..., 9767 (primes of the form 12k-1 where m+2 is also a prime) is;
A graph of the upper bounds of the minima for m=5, 17, 29, ..., 9929 (primes of the form 12k-7 where m+2 is also a prime) is;
The x values corresponding to these m values also have a Weibull probability distribution (but with different shaping and scaling parameters).
Tumbles and Jumps
In this section, c is restricted to being 1. Let y be an even natural number such that y/2 is odd and denote [(y/2)(3/4)f]+1 where f is a natural number and the brackets denote the greatest integer function by v1. (The 3/4 factor corresponds to the "loss" due to a 2 in a sequence vector.) v1 will be referred to as a "tumble" if it is odd (tumbles are the counterparts of jumps). Usually, v1 is not in the 3n+1 sequence starting with y/2. Under the same order constraints as least-residue trees, a succession of tumbles and jumps (where the jumps start from the tumbles and the tumbles start from the jumps) have properties similar to a live limb in S of a least-residue tree. Denote the peaks and valleys in a succession of tumbles and jumps by y/2, v1, p1, v2, p2, v3, p3, ..., vi, pi and denote the exponents of the tumbles by f1, f2, f3, ..., fi and the exponents of the jumps by g1, g2, g3, ..., gi. The conditions to be imposed on y and the jumps are that y be less than the order (of the form 3∙2k) and greater than order/2, each of p1, p2, p3, ..., pi be less than the order, and pi be greater than order/3. These conditions do not always guarantee that v1 is in the 3n+1 sequence starting with y/2, v2 is in the 3n+1 sequence starting with p1, v3 is in the 3n+1 sequence starting with p2, etc., but they come close to doing so. For example, for i=2, f1=5, g1=4, f2=5, g2=3, y=282825586, and an order of 402653184, v1=33557919, p1=169886969, v2=40314975, and p2=136063043 where y, p1, and p2 are within the specified bounds. The 3n+1 sequence starting with 141412793 (y/2) and having a length of 3f1+1 ends in 33557920, which is one larger than v1 (the 3 factor of the tumble exponent corresponds to the number of elements in the 3n+1 sequence due to a 2 in a sequence vector) . The 3n+1 sequence starting with 169886969 (p1) and having a length of 3f2+1 ends in 241889848, where [241889848/6]+1 equals v2. Denote the natural numbers these 3n+1 sequences end in by j and k. Some possibilities for different y values giving tumbles and jumps with exponents of 5, 4, 5, and 3 respectively are; (1) v1=j and v2=k, (2) v1=j-1 and v2=[k/6]+1, (3) v1=j-1 and v2=[k/6], (4) v1=j and v2=[k/6], (5) v1=[j/6]+1 and v2=6k-5, (6) v1=j-1 and v2=k-1, (7) v1=6j-5 and v2=k, (8) v1=[j/6]+1 and 3v2+1=[k/12], (9) v1=[j/6]+1 and 3v2+1=[k/12]+1, and (10) v1=j-1 and v2=36k-9 (there are other possibilities). In general, if v1 is in the 3n+1 sequence starting with y/2, there are f1+1 odd natural numbers and 2f1 even natural numbers in this sub-sequence (the same as for a sequence vector consisting of f1 2's), if v2 is in the 3n+1 sequence starting with p1, there are f2+1 odd natural numbers and 2f2 even natural numbers in this sub-sequence (the same as for a sequence vector consisting of f2 2's), if v3 is in the 3n+1 sequence starting with p2, there are f3+1 odd natural numbers and 2f3 even natural numbers in this sub-sequence (the same as for a sequence vector consisting of f3 2's), etc. These limbs (not necessarily 1-2 sequence vectors) then have the same X and Y values as a live limb in S of a least-residue tree.
References
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Software
MSVC++™ C programs were used to confirm the above propositions. Readers may copy and modify the software in this section. No guarantee is made that it is error-free.
Use test0a to find cycles in the 3n+c sequence. The "iters" variable specifies the number of jumps from the initial n value in the sequence (an odd natural number divisible by 3). Usually, "iters" is set to 1. The list of cycles for c in the range from -151 to 151 was generated using this setting. The list of cycles was confirmed to be the same using an "iters" setting of 2 or 3. The cycles were verified to be the same as those given by Keith Matthews' cycle-finding program at "http://www.numbertheory.org/keith.html". An improved cycle-finding program is test0azf (double-words are used). A much faster cycle-finding program is test0azi. Example output of this program (after some minor editing to fill in array sizes) is c1. Another "include" file that is needed for subsequent processing is ln. A program that does this processing is test0cyu. Links to the subroutines "euclid" and "halbhung" (and other required subroutines) are given below. An output of this program is the (L, K) values of the cycles with attachment points. A table of the (L, K) values of the cycles with attachment points for c≤69997 is newlk. A table of the (L, K) values of the cycles with attachment points for 70001≤c≤99997 is newlk1. A table of the (L, K) values of the cycles with attachment points for 100001≤c≤149999 is newlk2A table of the (L, K) values of the cycles with attachment points for 150001≤c≤199999 is newlk3. These tables are "include" files for the program test23r (which finds associated cycles). The program test23s finds (L, K) trees. A still faster cycle-finding program is test0azk. TMS320C64™ assembly language subroutines used are jumps, cycle, regen and check. These subroutines are usually at least an order of magnitude faster than corresponding C code (depending on the compiler and microprocessor used for the C code). A program for finding cycles without attachment points is test0azj. Other input to "test0cyu" is c203, c401, c601, c803, c1001, c1201, c1403, c1601, c1801, c2003, c2201, c2401, c2603, c2801, c3001, c3203, c3401, c3601, c3803, c4001, c4201, c4403, c4601, c4801, c5003, c5201, c5401, c5603, c5801, c6001, c6203, c6401, c6601, c6803, c7001, c7201, c7403, c7601, c7801, c8003, c8201, c8401, c8603, c8801, c9001, c9203, c9401, c9601, and c9803. Attachment points for 3n+c cycles where c≤199999 are given in att1, att2, att3, att4, att5, att6, att7, att8, att9, att10, att11, att12, att13, att14, and att15. Only one attachment point per cycle is given and only the least significant word of the attachment point is given. The program test0cyw supplies the most significant word of the attachment point and regenerates the other attachment points. Otherwise, this program does the same processing as "test0cyu". A version of "test0cyw" where all the c values up to 99997 can be processed at once is test0cyx. "Include" files are att1t, att2t, att3t, att4t, and att5t.
Use test1c to histogram the |t2|-|t1| values of 3n+c sequences starting with an odd integer divisible by 3 and having a 1-2 sequence vector that attach to other such sequences.
Use test0b to regenerate a cycle in the 3n+c sequence given an entry point (must be even) into the cycle. The order of the cycle is computed. The output of "test0a" is input to this program.
Use test0c to regenerate cycles in the 3n+c sequence given a list of entry points into the cycles. (The list of cycles is confirmed.)
Use test0d to regenerate cycles in the 3n+c sequence given a list of entry points into the cycles. A check for 1-2 sequence vectors at entry points into the cycles is made (for "multiple jump" connection points).
Use test0e to regenerate cycles in the 3n+c sequence given a list of entry points into the cycles. A check for 1-2 sequence vectors at entry points into the cycles is made (for "jumps over" connection points).
list consists of the cycles found for c in the range from -151 to 151.
list1 consists of one-attachment-point cycles found for c in the range from -151 to 151.
Use test1 to find the number of jumps in the 3n+1 sequence before an even natural number is reached (starting with an odd natural number divisible by 3).
Use test2 to generate a histogram of limb lengths for a 3n+c sequence that is not an "odd" path. The fourth-to-last element of the sequence must be an odd natural number greater than order/6 and the third-to-last element of the sequence must be an even natural number divisible by 8. Use test2a to compute the (l, m) values of generalized dead limbs. The "iters" variable is the number of primary attachment points (usually, the "iters1" variable is set to the same value). An updated version of this program is at test2an.
Use test3a to generate limbs of least-residue trees where at least one element of the limb is divisible by 8.
Use test3b to generate limbs of least-residue trees where at least one element of the limb is divisible by 8. Multiple-word arithmetic is used.
Use test3c to generate limbs of least-residue trees where at least one element of the limb is divisible by 8. Multiple-word arithmetic is used. This C program is for use on the TMS320C64™ digital signal processing chip (with hand-optimized assembly language subroutines).
Use test4a12, test4a13, test4a14, test4a15, test4a16, test4a17, test4a18, test4a19, test4a20, test4a21, test4a22, test4a23, test4a24, test4a25, test4a26, test4a27, test4a28, test4a29, test4a30, test4a31, test4a32, and test4a33 to compare the number of limbs in S of a given length to the estimated number of limbs. Only live limbs are considered.
Use test5a to generate tumbles and jumps having three peaks and two troughs.
Use test5b to generate tumbles and jumps having four peaks and three troughs.
Use test6a to compute the largest y values, the smallest y values, the largest z values, and the smallest z values for live limbs in S associated with a given sequence vector. The maximum |X(δ/y)| value is computed.
Use test6b to compute the largest y values for live limbs in S associated with a given sequence vector. The maximum |X(δ/y)| value is computed. This C program is for use on the TMS320C64™ digital signal processing chip (with hand-optimized assembly language subroutines).
Use test7 to check that n(order/2)>ce, c>0, n>0, or (-n)(order/2)>(-c)e, n<0, c<0. Floating point arithmetic is used.
Use test8 to check that x>0 for limb lengths with positive limits and x<0 for limb lengths with negative limits.
Use test9 to generate least-residue trees. The permutation of A, B, C, and D for different c values is checked.
Use test10 to generate least-residue trees for the 3n-1 sequence, k≤15. Numerous properties of least-residue trees are tested. Limbs in S having lengths of 5, 10, 15, 18, 23, and 28 are verified to be dead.
Use test11 to generate long limbs in S. Sequence vectors are generated.
Use test12 to find sequence vectors corresponding to different e values. All the orders up to 3∙224 are computed. The maximum and minimum x/y values for live limbs are computed for every order and are verified to be either ascending of descending.
Use test13 to compute a and b values.
Use test14a to compute the minimum element in a cycle using the formula Ml,m=∑(]jm/l[-](j-1)m/l[)2j-13m-]jm/l[ where the summation is from j=1 to l. Use test14b to compute the minimum element in a cycle in a limb of S of a least-residue tree. Use test14c to compute the minimum element in a cycle in a limb of S of a least-residue tree (using the TMS320C64™ DSP.) Use test14d to compute the minimum element in a cycle in a limb of S of a least-residue tree when m and m+2 are primes. Use test14e to output the computed minima for a cycle in a limb of S of a least-residue tree. Use test14f to compute an upper bound of the minimum element in a cycle in a limb of S of a least-residue tree where m and m+2 are prime. Use test14g to compute an upper bound of the minimum element in a cycle in a limb of S of a least-residue tree when m is prime. Use test14h to compute an upper bound of the minimum element in a cycle in a limb of S of a least-residue tree when m is prime (using the TMS320C64™ DSP). Use test14i to generate the sequence vector corresponding to the smallest e from the parity vector. Use test14j to compute the largest odd element in a cycle using the floor function. Use test14k to compute an upper bound of the minimum element in a cycle in a limb of S of a least-residue tree when l is prime (using the TMS320C64™ DSP).
Use test15 to generate parity vectors (distinct under rotation) for given l and m values (using the TMS320C64™ DSP). Subroutines called are bitcnt and pack.
Use test16 to determine that the (l, m) value of a generalized dead limb minus the (l, m) value of a generalized continued-fraction convergent of log(3)/log(2) is the (l, m) value of a shorter generalized dead limb.
Use test17 to show that |2l-3m| increases monotonically when (l, m) are generalized continued-fraction convergents of log(3)/log(2) (excluding (2, 1), (4, 2), (6, 4), and (9, 6)). |2l-3m| is shown to increase monotonically for 105 generalized continued-fraction convergents of log(3)/log(2). This program is for use on the TMS320C64™ DSP.
Use test18 to generate the parity vector of a 3n+c cycle where twice the minimum element in the cycle is larger than the maximum odd element in the cycle (and the elements are large). This program is for use on the TMS320C64™ DSP. (A corresponding C program is test14uf.) sv gives a portion of this parity vector. Use test14da to compute Ml,m/(2l-3m) values for the corresponding M-cycles. Use test14db to determine if rotations of certain parity vectors match the parity vector p.
Use test19 to find rotations of the parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l, where (0, 1, 0) is at the beginning of the vector and twice the smallest s value is larger than the largest "odd" s value. This program is for use on the TMS320C64™ DSP. Use test19a to find rotations of the vector satisfying the two additional conditions.
Use test20 to find limbs in S almost having the parity vector ]jm/l[ - ](j-1)m/l[, j=1, 2, 3, ..., l (the parity vector is rotated so that a (1, 1) is at the beginning of the vector and the first 1 in the vector is changed to a 0 and the last element of the vector [a 0] is changed to a 1). A subroutine used is limb. This program is for use on the TMS320C64™ DSP. test20a doesn't use the "limb" subroutine.
Use test21ca to compute the number of odd elements in a one-jump (or multiple-jump) attachment point minus j where t≡u(mod 2j). test21cb computes these values for c values between 601 and 997. test21cc computes these values for c values between 1001 and 1499. test21cd computes these values for c values between 1501 and 1999. test21ce computes these values for c values between 2003 and 2497. Subroutines used are halbhung and euclid. The properties of no-jump, one-jump, multiple-jump, and jumped-over attachment points are tested.
Use log2 to compute the logarithm of 2. Use log3o2 to compute the logarithm of 3/2. Use expand to compute the generalized continued-fraction convergents of log(3)/log(2). Use test22 to compute the upper bound of the minimum in a 3n+c cycle using Proposition (43). Use test23b to factor Collatz numbers.
Multiple-word arithmetic C subroutines are add64, add128, add256, add512, add1024, carry, copy256, copy512, copy1024, div6432, div12864, div25632div102432, div204832, divn, lshift256, lshift512, lshift1024, lmbd, mul6432, mul6464, mul12832, set256, set512, set1024, shift64, shift128, shift256, shift512, shift1024, sub64, sub128, sub256, sub512, sub1024, subr2, n-word arithmetic, table, table1, table2, and table4.
Multiple-word arithmetic TMS320C64™ assembly language subroutines are add64, div12864, div6432, mul6432, mul6464, mul3232, shift64, sub64, subr2, and n-word arithmetic.
Other Topics
It's commonly known that there is a formula for Pythagorean triples (integer solutions of a2+b2=c2). It's less commonly known that there are "almost" integer solutions of a3+b3=c3. When 3 does not divide a+b and a+b is factored out of a3+b3, the equation a2-ab+b2=T3 is obtained and there are integer solutions of this equation. Solutions of this equation are of interest since classical results from algebraic number theory (Furtwängler's and Vandiver's theorems) are still applicable. This topic is discussed in the link fermat.
In his book Elements of Algebraic Coding Theory (Chapman & Hall, 1996), Lekh Vermani gives a proof of the quadratic reciprocity law using Perron's theorem. Perron's theorem concerns the number of consecutive quadratic residues and consecutive quadratic non-residues. Many textbooks on algebraic number theory have sections on cubic, biquadratic, and octic reciprocity. Whether a generalized version of Perron's theorem is still applicable to higher-order reciprocity is discussed in the link congr.
A relationship between the cross-ratio function of geometry and complex analysis, the Poisson probability distribution, and Fermat's congruence modulo a prime power is discussed in the link psquare.
Generalized Fibonacci and Lucas series are discussed in the link nroots.
The cross-ratio function is discussed in the link cross.
The Farey series is discussed in the link farey.
The relationship between the Farey series and the Riemann hypothesis is discussed in the link riemann.
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Paul's Online Notes
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### Section 3-5 : Graphing Functions
Now we need to discuss graphing functions. If we recall from the previous section we said that $$f\left( x \right)$$ is nothing more than a fancy way of writing $$y$$. This means that we already know how to graph functions. We graph functions in exactly the same way that we graph equations. If we know ahead of time what the function is a graph of we can use that information to help us with the graph and if we don’t know what the function is ahead of time then all we need to do is plug in some $$x$$’s compute the value of the function (which is really a $$y$$ value) and then plot the points.
Example 1 Sketch the graph of $$f\left( x \right) = {\left( {x - 1} \right)^3} + 1$$.
Show Solution
Now, as we talked about when we first looked at graphing earlier in this chapter we’ll need to pick values of $$x$$ to plug in and knowing the values to pick really only comes with experience. Therefore, don’t worry so much about the values of $$x$$ that we’re using here. By the end of this chapter you’ll also be able to correctly pick these values.
Here are the function evaluations.
$$x$$ $$f(x)$$ $$\left( {x,y} \right)$$
-1 -7 $$\left( { - 1, - 7} \right)$$
0 0 $$\left( {0,0} \right)$$
1 1 $$\left( {1,1} \right)$$
2 2 $$\left( {2,2} \right)$$
3 9 $$\left( {3,9} \right)$$
Here is the sketch of the graph.
So, graphing functions is pretty much the same as graphing equations.
There is one function that we’ve seen to this point that we didn’t really see anything like when we were graphing equations in the first part of this chapter. That is piecewise functions. So, we should graph a couple of these to make sure that we can graph them as well.
Example 2 Sketch the graph of the following piecewise function. $g\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{ - {x^2} + 4}&{{\mbox{if }}x < 1}\\{2x - 1}&{{\mbox{if }}x \ge 1}\end{array}} \right.$
Show Solution
Okay, now when we are graphing piecewise functions we are really graphing several functions at once, except we are only going to graph them on very specific intervals. In this case we will be graphing the following two functions,
\begin{align*} - {x^2} + 4\hspace{0.25in} & {\mbox{on}}\hspace{0.25in}x < 1\\ 2x - 1\hspace{0.25in} & {\mbox{on}}\hspace{0.25in}x \ge 1\end{align*}
We’ll need to be a little careful with what is going on right at $$x = 1$$ since technically that will only be valid for the bottom function. However, we’ll deal with that at the very end when we actually do the graph. For now, we will use $$x = 1$$ in both functions.
The first thing to do here is to get a table of values for each function on the specified range and again we will use $$x = 1$$ in both even though technically it only should be used with the bottom function.
$$x$$ $$- {x^2} + 4$$ $$\left( {x,y} \right)$$
-2 0 $$\left( { - 2,0} \right)$$
-1 3 $$\left( { - 1,3} \right)$$
0 4 $$\left( {0,4} \right)$$
1 3 $$\left( {1,3} \right)$$
$$x$$ $$2x - 1$$ $$\left( {x,y} \right)$$
1 1 $$\left( {1,1} \right)$$
2 3 $$\left( {2,3} \right)$$
3 5 $$\left( {3,5} \right)$$
Here is a sketch of the graph and notice how we denoted the points at $$x = 1$$. For the top function we used an open dot for the point at $$x = 1$$ and for the bottom function we used a closed dot at $$x = 1$$. In this way we make it clear on the graph that only the bottom function really has a point at $$x = 1$$.
Notice that since the two graphs didn’t meet at $$x = 1$$ we left a blank space in the graph. Do NOT connect these two points with a line. There really does need to be a break there to signify that the two portions do not meet at $$x = 1$$.
Sometimes the two portions will meet at these points and at other times they won’t. We shouldn’t ever expect them to meet or not to meet until we’ve actually sketched the graph.
Let’s take a look at another example of a piecewise function.
Example 3 Sketch the graph of the following piecewise function. $h\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{x + 3}&{{\mbox{if }}\,x < - 2}\\{{x^2}}&{{\mbox{if }} - 2 \le x < 1}\\{ - x + 2}&{{\mbox{if }}\,x \ge 1}\end{array}} \right.$
Show Solution
In this case we will be graphing three functions on the ranges given above. So, as with the previous example we will get function values for each function in its specified range and we will include the endpoints of each range in each computation. When we graph we will acknowledge which function the endpoint actually belongs with by using a closed dot as we did previously. Also, the top and bottom functions are lines and so we don’t really need more than two points for these two. We’ll get a couple more points for the middle function.
$$x$$ $$x + 3$$ $$\left( {x,y} \right)$$
-3 0 $$\left( { - 3,0} \right)$$
-2 1 $$\left( { - 2,1} \right)$$
$$x$$ $${x^2}$$ $$\left( {x,y} \right)$$
-2 4 $$\left( { - 2,4} \right)$$
-1 1 $$\left( { - 1,1} \right)$$
0 0 $$\left( {0,0} \right)$$
1 1 $$\left( {1,1} \right)$$
$$x$$ $$- x + 2$$ $$\left( {x,y} \right)$$
1 1 $$\left( {1,1} \right)$$
2 0 $$\left( {2,0} \right)$$
Here is the sketch of the graph.
Note that in this case two of the portions met at the breaking point $$x = 1$$ and at the other breaking point, $$x = - 2$$, they didn’t meet up. As noted in the previous example sometimes they meet up and sometimes they won’t.
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# How do you find the discriminant of -x^2-x=4 and use it to determine if the equation has one, two real or two imaginary roots?
May 2, 2018
The equation has no real roots and two imaginary roots.
#### Explanation:
The "discriminant" that the question is referring to is the discriminant of the quadratic formula, which is ${b}^{2} - 4 a c$. I've highlighted it in the quadratic formula here in red:
$x = \frac{- b \pm \sqrt{\textcolor{red}{{b}^{2} - 4 a c}}}{2 a}$
We know that by the fundamental theorem of algebra that any quadratic has exactly $2$ roots which can be real or imaginary.
If the value of the discriminant is less than $0$, the quadratic has no real roots. This is because the negative number would make the square root "unsolvable." According to the theorem above, there must be two roots, so that means that they are both imaginary (since there are no real ones).
If the value of the discriminant is more than $0$, the quadratic has two real roots. This is because of the "plus-or-minus" sign before the square root. Since there are only two roots according to the theorem, that means that there are no imaginary roots in this case.
If the value of the discriminant is exactly $0$, the quadratic has one real root. This is because the square root would be zero, eliminating the use of the "plus-or-minus", leaving just one solution. However, in this case, there isn't also one imaginary solution, because imaginary solutions only ever come in pairs (conjugate pairs).
In our case, first, we would have to get the quadratic to equal zero on one side:
$- {x}^{2} - x = 4$
$- {x}^{2} = 4 + x$
$0 = 4 + x + {x}^{2}$
$0 = {x}^{2} + x + 4$
Now, we plug our $a$, $b$, and $c$ values (which are $1$, $1$, and $4$, respectively) into the discriminant and see what we get:
$\textcolor{w h i t e}{=} {b}^{2} - 4 a c$
$= {\left(1\right)}^{2} - 4 \left(1\right) \left(4\right)$
$= 1 - 16$
$= - 15$
Since this number is less than $0$, the quadratic has no real roots. Therefore, it has two imaginary roots. We can verify that it has no real roots by looking at its graph:
graph{-x^2-x-4 [-18.3, 17.73, -13.97, 4.04]}
Hope this helped!
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# 12.15: Combination Problems
Difficulty Level: At Grade Created by: CK-12
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Remember the decorations in the last Concept? Take a look once again.
The decorating committee is getting the stage ready for the Talent Show. There was a bunch of different decorating supplies ordered, and the students on the committee are working on figuring out the best way to decorate the stage.
They have four different colors of streamers to use to decorate.
Red
Blue
Green
Yellow
“I think four is too many colors. How about if we choose three of the four colors to decorate with?” Keith asks the group.
“I like that idea,” Sara chimes in. “How many ways can we decorate the stage if we do that?”
The group begins to figure this out on a piece of paper.
Combinations are arrangements where order does not make a difference. The decorating committee is selected three colors from the possible four options. Therefore, the order of the colors doesn’t matter.
Combinations are the way to solve this problem. Look at the information in this Concept to learn how to figure out the possible combinations.
### Guidance
Once you figure out if you are going to be using permutations or combinations, it is necessary to count the combinations.
There are several different ways to count combinations. When counting, try to keep the following in mind:
• Go one by one through the items. Don’t stop your list until you’ve covered every possible link of one item to all other items.
• Keep in mind that order doesn’t matter. For combinations, there no difference between \begin{align*}AB\end{align*} and \begin{align*}BA\end{align*}. So if both \begin{align*}AB\end{align*} and \begin{align*}BA\end{align*} are on your list, cross one of the choices off your list.
• Check your list for repeats. If you accidentally listed a combination more than once, cross the extra listings off your list.
James needs to choose a 2-color combination for his intramural team t-shirts. How many different 2-color combinations can James make out of red, blue, and yellow?
One way to find the number of combinations is to make a tree diagram. Here, if red is chosen as one color, that leaves only blue and yellow for the second color.
The diagram shows all 6 permutations of the 3 colors. But wait–since we are counting COMBINATIONS here order doesn’t matter.
So in this tree diagram we will cross out all outcomes that are repeats. For example, the first red-blue is no different from blue-red, so we’ll cross out blue-red.
In all, there are 3 combinations that are not repeats.
This method of making a tree diagram and crossing out repeats is reliable, but it is not the only way to find combinations.
Let's look at a situation like this.
James has added a fourth color, green, to choose from in selecting a 2-color combination for his intramural team. How many different 2-color combinations can James make out of red, blue, yellow, and green?
Step 1: Write the choices. Match the first choice, red, with the second, blue. Add the combination, red-blue, to your list. Match the other choices in turn. Add the combinations to your list.
Step 2: Now move to the second choice, blue. Match blue up with every possible partner. Add the combinations to your list.
Step 3: Now move to the third choice, yellow. There is only one combination left to match it with. Add the combination to your list.
Your list is now complete. There are 6 combinations.
Sometimes, you won’t want to use all of the possible options in the combination. Think about it as if you have 16 flavors of ice cream, but you only want to use three flavors at a time. This is an example where there are 16 flavors to work with, but you can only use three at a time. With an example like this one, you are looking for combinations of object where only a certain number of them are used in any one combination.
This happens a lot with teams.
Here is a situation with teams.
How many different 2-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form?
\begin{align*}&\underline{\text{Combination}} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \underline{\text{List}}\\ &\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean}\\ &\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine}\\ &\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Lurleen}\\ &\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Doreen}\end{align*}
Step 2: You’ve covered all combinations that begin with Jean. Now go through all combinations that begin with Dean, Francine, and Lurleen.
\begin{align*}&\underline{\text{Combination}} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \underline{\text{List}}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Francine}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Francine-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Francine-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Lurleen-Doreen}\end{align*}
Your list is now complete. There are 10 combinations.
Look at this situation.
How many different 3-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form?
Use the process above to go through all of the combinations.
\begin{align*}&\underline{\text{Combination}} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \underline{\text{List}}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean-Francine}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Lurleen-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean, Francine-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Francine-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Lurleen-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Francine-Lurleen-Doreen}\end{align*}
Your list is now complete. There are 10 combinations.
We can use a formula to help us to calculate combinations. This is very similar to the work that you did in the last section with factorials and permutations.
Suppose you have 5 marbles in a bag–red, blue, yellow, green, and white. You want to know how many combinations there are if you take 3 marbles out of the bag all at the same time. In combination notation you write this as:
\begin{align*}{_5}C_3 \Longleftarrow 5 \ \text{items taken 3 at a time}\end{align*}
In general, combinations are written as:
\begin{align*}{_n}C_r \Longleftarrow n \ \text{items taken} \ r \ \text{at a time}\end{align*}
To compute \begin{align*}{_n}C_r\end{align*} use the formula:
\begin{align*}{_n}C_r = \frac{n!}{r!(n - r)!}\end{align*}
This may seem a bit confusing, but it isn’t. Notice that the factorial symbol is used with the number of object \begin{align*}(n)\end{align*} and the number taken at any one time \begin{align*}(r)\end{align*}. This helps us to understand which value goes where in the formula.
Now let’s look at applying the formula to the example.
For \begin{align*}{_5}C_3\end{align*}:
\begin{align*}{_5}C_3 = \frac{5!}{3!(5 - 3)!} = \frac{5!}{3! 2!}\end{align*}
Simplify.
\begin{align*}{_5}C_3 = \frac{5 (4)(3)(2)(1)}{(3 \cdot 2 \cdot 1)(2 \cdot 1)} = \frac{120}{12} = 10\end{align*}
There are 10 possible combinations.
Here is another one.
Find \begin{align*}{_6}C_2\end{align*}
Step 1: Understand what \begin{align*}{_6}C_2\end{align*} means.
\begin{align*}{_6}C_2 \Longleftarrow 6 \ \text{items taken 2 at a time}\end{align*}
Step 2: Set up the problem.
\begin{align*}{_6}C_2 = \frac{6!}{2!(6 -2)!}\end{align*}
Step 3: Fill in the numbers and simplify.
\begin{align*}{_6}C_2 = \frac{6(5)(4)(3)(2)(1)}{(2 \cdot 1)(4 \cdot 3 \cdot 2 \cdot 1)} = \frac{720}{48} = 15\end{align*}
There are 15 possible combinations.
Find the number of combinations in each example.
#### Example A
\begin{align*}{_5}C_2\end{align*}
Solution: \begin{align*}30\end{align*}
#### Example B
\begin{align*}{_4}C_3\end{align*}
Solution: \begin{align*}1\end{align*}
#### Example C
\begin{align*}{_6}C_4\end{align*}
Solution: \begin{align*}15\end{align*}
Here is the original problem once again.
The decorating committee is getting the stage ready for the Talent Show. There was a bunch of different decorating supplies ordered, and the students on the committee are working on figuring out the best way to decorate the stage.
They have four different colors of streamers to use to decorate.
Red
Blue
Green
Yellow
“I think four is too many colors. How about if we choose three of the four colors to decorate with?” Keith asks the group.
“I like that idea,” Sara chimes in. “How many ways can we decorate the stage if we do that?”
The group begins to figure this out on a piece of paper.
Combinations are arrangements where order does not make a difference. The decorating committee is selected three colors from the possible four options. Therefore, the order of the colors doesn’t matter.
We can use combination notation to figure out this problem.
\begin{align*}{_4}C_3 = \frac{4!}{3!(4 - 3)!} = \frac{4(3)(2)(1)}{(3 \cdot 2 \cdot 1)(1)}= \frac{24}{6} = 4\end{align*}
There are four possible ways to decorate the stage.
Now that the students have this information, they can look at their color choices and vote on which combination they like best.
### Vocabulary
Here are the vocabulary words used in this Concept.
Combination
an arrangement of objects or events where order does not matter.
Permutations
an arrangement of objects or events where the order does matter.
### Guided Practice
Here is one for you to try on your own.
\begin{align*}{_5} C_4\end{align*}
\begin{align*}{_5}C_4 = \frac{5!}{4!(5 - 4)!} = \frac{5(4)(3)(2)(1)}{(4 \times 3 \times 2 \times 1)1}= \frac{120}{24} = 5\end{align*}
### Video Review
Here is a video for review.
### Practice
Directions: Evaluate each factorial.
1. 5!
2. 4!
3. 3!
4. 8!
5. 9!
6. 6!
Directions: Evaluate each combination using combination notation.
7. \begin{align*}{_7} C_2\end{align*}
8. \begin{align*}{_7} C_6\end{align*}
9. \begin{align*}{_8} C_4\end{align*}
10. \begin{align*}{_9} C_6\end{align*}
11. \begin{align*}{_8} C_3\end{align*}
12. \begin{align*}{_{10}}C_7 \end{align*}
13. \begin{align*}{_{12}}C_9\end{align*}
14. \begin{align*}{_{11}}C_9\end{align*}
15. \begin{align*}{_{16}}C_{14}\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
combination Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.
combination notation Combination notation has the forms nCr and c(n, r) where n is the number of different units to choose from and r is the number of units in each group.
combinatorics Combinatorics is the study of permutations and combinations.
n value When calculating permutations with the TI calculator, the n value is the number of objects from which you are choosing.
Permutation A permutation is an arrangement of objects where order is important.
permutation notation Permutation notation is the form nPr or P(n, r), and indicates the number of ways that n objects can be ordered into groups of r items each.
TI-84 The TI-84 calculator is a graphing calculator produced by Texas Instruments and is considered an “industry standard” for more advanced calculations.
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# 5.6: Chapter 5 Review
Difficulty Level: At Grade Created by: CK-12
## Keywords, Theorems and Postulates
Midsegments in Triangles
• Midsegment
• Midsegment Theorem
Perpendicular Bisectors and Angle Bisectors in Triangles
• Perpendicular Bisector Theorem
• Perpendicular Bisector Theorem Converse
• Inscribe
• Circumscribe
• Angle Bisector Theorem
• Angle Bisector Theorem Converse
Medians and Altitudes in Triangles
• Median
• Centroid
• Median Theorem
• Altitude
Inequalities in Triangles
• Theorem 5-9
• Converse of Theorem 5-9
• Triangle Inequality Theorem
• SAS Inequality Theorem
• SSS Inequality Theorem
Extension: Indirect Proof
## Review
If C\begin{align*}C\end{align*} and E\begin{align*}E\end{align*} are the midpoints of the sides they lie on, find:
1. The perpendicular bisector of FD¯¯¯¯¯¯¯¯\begin{align*}\overline{FD}\end{align*}.
2. The median of FD¯¯¯¯¯¯¯¯\begin{align*}\overline{FD}\end{align*}.
3. The angle bisector of FAD\begin{align*}\angle FAD\end{align*}.
4. A midsegment.
5. An altitude.
6. A triangle has sides with length x+6\begin{align*}x + 6\end{align*} and 2x1\begin{align*}2x - 1\end{align*}. Find the range of the third side.
Fill in the blanks.
1. A midsegment connects the __________ of two sides of a triangle.
2. The height of a triangle is also called the __________.
3. The point of intersection for all the medians of a triangle is the __________.
4. The longest side is opposite the __________ angle in a triangle.
5. A point on the __________ bisector is __________ to the endpoints.
6. A point on the __________ bisector is __________ to the sides.
7. A circle is __________ when it touches all the sides of a triangle.
8. An __________ proof is also called a proof by contradiction.
9. For ABC\begin{align*}\triangle ABC\end{align*} and DEF: AB=DE, BC=EF\begin{align*}\triangle DEF: \ AB = DE, \ BC = EF\end{align*}, and mB>mE\begin{align*}m \angle B > m \angle E\end{align*}, then __________.
## Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9690.
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# Solve the following pairs of equations by reducing them to a pair of linear equations :
Question.
Solve the following pairs of equations by reducing them to a pair of linear equations :
(i) $\frac{1}{2 x}+\frac{1}{3 y}=2, \frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}$
(ii) $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2, \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$
(iii) $\frac{4}{x}+3 y=14, \frac{3}{x}-4 y=23$
(iv) $\frac{5}{(x-1)}+\frac{1}{(y-2)}=2, \frac{6}{(x-1)}-\frac{3}{(y-2)}=1$
(v) $\frac{7 x-2 y}{x y}=5, \frac{8 x+7 y}{x y}=15$
(vi) $6 x+3 y=6 x y, 2 x+4 y=5 x y$
(viii) $\frac{1}{(3 x+y)}+\frac{1}{(3 x-y)}=\frac{3}{4}, \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}$
Solution:
(i) $\frac{1}{2 x}+\frac{1}{3 y}=2, \frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}$
Substituting $\quad \frac{\mathbf{1}}{\mathbf{x}}=\mathbf{u a n d} \frac{\mathbf{1}}{\mathbf{v}}=\mathbf{v}$
We get $\frac{1}{2} \mathbf{u}+\frac{1}{3} \mathbf{v}=\mathbf{2}, \frac{1}{3} \mathbf{u}+\frac{1}{8} \mathbf{v}=\frac{13}{6}$
Multiplying by 6 on both sides, we get
$\Rightarrow 3 u+2 v=12$...(ii)
$2 \mathrm{u}+3 \mathrm{v}=13$..(ii)
Multiplying (i) by 3 and (ii) by 2, then subtracting later from first, we get
$3(3 u+2 v)-2(2 u+3 v)=3 \times 12-2 \times 13$
$\Rightarrow 9 \mathrm{u}-4 \mathrm{u}=36-26 \quad \Rightarrow \mathrm{u}=2$
Then substituting $u=2$ in (i), we get
$6+2 v=12$
$\Rightarrow v=3$
Now, $u=2$ and $v=3$
$\Rightarrow \frac{\mathbf{1}}{\mathbf{x}}=2$ and $\frac{\mathbf{1}}{\mathbf{y}}=3 \quad \Rightarrow x=\frac{\mathbf{1}}{\mathbf{2}}$ and $y=\frac{\mathbf{1}}{\mathbf{3}}$
(ii) $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$ ..(i)
$\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$...(ii)
Take $\frac{\mathbf{1}}{\sqrt{\mathbf{x}}}=\mathrm{a}, \frac{\mathbf{1}}{\sqrt{\mathbf{y}}}=\mathrm{b}$, we get
$2 a+3 b=2$ ...(iii)
$4 a-9 b=-1$ ...(iv)
Multiplying (iii) by 3 , we get
$6 a+9 b=6$ ...(v)
Adding (iv) and (v), we get
$10 a=5$
$\Rightarrow a=\frac{1}{2}$ Substituting $a=\frac{1}{2}$ in (iii), we get
$3 b=1$
$\Rightarrow \mathrm{b}=\frac{\mathbf{1}}{\mathbf{3}}$
Now, $\frac{1}{\sqrt{x}}=a=\frac{1}{2}$ and $\frac{1}{\sqrt{y}}=b=\frac{1}{3}$
$\Rightarrow \sqrt{\mathbf{x}}=2, \sqrt{\mathbf{y}}=3$
Squaring, we get
$x=4, y=9$
(iii) $\frac{4}{x}+3 y=14$ ..(i)
$\frac{3}{x}-4 y=23$ ..(ii)
Take, $\frac{\mathbf{1}}{\mathbf{x}}=\mathrm{a}$
$4 a+3 y=14$ ...(iii)
$3 a-4 y=23$ ...(iv)
Multiplying (iii) by 4 and (iv) by 3
$16 a+12 y=56$
$9 a-12 y=69$
Adding both, we get
$25 a=125$
$\Rightarrow a=5$
Substituting a in (iii), we get
$20+3 y=14$
$\Rightarrow 3 y=-6$
$\Rightarrow y=-2$
As, $\frac{1}{x}=a=5$
$\Rightarrow x=\frac{1}{5}$
Hence, $x=\frac{1}{5}$ and $y=-2$
(iv) $\frac{5}{(x-1)}+\frac{1}{(y-2)}=2$ ...(i)
\frac{6}{(x-1)}-\frac{3}{(y-2)}=1 ...(ii)
Take, $\frac{\mathbf{1}}{(\mathbf{x}-\mathbf{1})}=\mathrm{a}$ and $\frac{\mathbf{1}}{(\mathbf{y}-\boldsymbol{z})}=\mathrm{b}$
$5 a+b=2$ ..(iii)
$6 a-3 b=1$ ...(iv)
Multiplying (iii) by 3 , we get
$15 a+3 b=6$
$6 a-3 b=1$
Adding both we get
$21 a=7$
$a=1 / 3$
So, by solving, $b=1 / 3$
As, $a=\frac{1}{3}=\frac{1}{x-1} \Rightarrow x-1=3 \Rightarrow x=4$
and $\mathrm{b}=\frac{\mathbf{1}}{\mathbf{3}}=\frac{\mathbf{1}}{\mathbf{v}-\mathbf{2}} \Rightarrow \mathrm{y}-2=3$
$\Rightarrow y=5$
(v) $\frac{7 x-2 y}{x y}=5, \frac{8 x+7 y}{x y}=15$
By solving, we get
$\frac{7}{y}-\frac{2}{x}=5$ (i)
$\frac{8}{y}+\frac{7}{x}=15$ ...(ii)
Taking $\frac{\mathbf{1}}{\mathbf{y}}=\mathrm{u}, \frac{\mathbf{1}}{\mathbf{x}}=\mathrm{v}$
$7 u-2 v=5$ ...(iii)
$8 u+7 v=15$ ...(iv)
Multiplying (iii) by 7 and (iv) by 2 , we get
$49 u-14 v=35$
$16 u+14 v=30$
$65 u=65$
$u=1$
By solving, we get $\mathrm{v}=1$
As, $u=1=\frac{\mathbf{1}}{\mathbf{y}} \Rightarrow y=1$
and $\mathrm{v}=1=\frac{\mathbf{1}}{\mathbf{x}} \Rightarrow \mathrm{x}=1$
(vi) $6 x+3 y=6 x y$
$2 x+4 y=5 x y$
By solving, we get
$\frac{6}{v}+\frac{3}{z}=6$ ...(i)
$\frac{2}{y}+\frac{4}{x}=5$ ..(ii)
Take, $\frac{1}{\mathbf{y}}=\mathrm{u}, \frac{\mathbf{1}}{\mathbf{x}}=\mathrm{v}$, we get
$6 u+3 v=6$ ..(iii)
$2 u+4 v=5$ ..(iv)
Multiply (iv) by 3 , we get
$6 u+12 v=15$ ...(v)
Subtract (iii) from (v), we get
$9 v=9 v=1$
By solving we get $\mathrm{u}=1 / 2$
$\mathrm{As}, \frac{\mathbf{1}}{\mathbf{x}}=\mathrm{v}=1 \Rightarrow \mathrm{x}=1$
and $\frac{\mathbf{1}}{\mathbf{y}}=\mathrm{u}=\frac{\mathbf{1}}{\mathbf{2}} \Rightarrow \mathrm{y}=2$
(vii) $\frac{10}{(x+y)}+\frac{2}{(x-y)}=4, \frac{15}{(x+y)}-\frac{5}{(x-y)}=-2$
Take $\frac{\mathbf{1}}{\mathbf{x}+\mathbf{y}}=\mathrm{u}$ and $\frac{\mathbf{1}}{\mathbf{x}-\mathbf{y}}=\mathrm{v}$
$10 u+2 v=4$ ...(i)
$15 u-5 v=-2$ ...(ii)
Multiply (i) by 5 and (ii) by 2 , we get
$50 u+10 v=20$
$30 \mathrm{u}-10 \mathrm{v}=-4$
$80 \mathrm{u}=16$
$\mathrm{u}=1 / 5$
By solving, we get $\mathrm{v}=1$
$\mathrm{As}, \frac{1}{\mathrm{x}+\mathrm{y}}=\mathrm{u}=\frac{1}{5}$
$\Rightarrow x+y=5$ ...(iii)
and $\frac{\mathbf{1}}{\mathbf{x}-\mathbf{y}}=\mathrm{v}=1$
$\Rightarrow x-y=1$ ..(iv)
Adding (iii) and (iv), we get
$2 x=6$
$\Rightarrow x=3$ and $y=2$
Hence, $x=3, y=2$
(viii) $\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}$ ...(i)
$\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}$ ...(ii)
Let, $\frac{\mathbf{1}}{\mathbf{3 x}+\mathbf{y}}=\mathrm{u}, \quad \frac{\mathbf{1}}{\mathbf{3 x}-\mathbf{y}}=\mathrm{v}$
$u+v=\frac{3}{4}$ ..(iii)
$\frac{\mathbf{u}}{\mathbf{2}}-\frac{\mathbf{v}}{\mathbf{2}}=-\frac{\mathbf{1}}{\mathbf{8}}$ ...(iv)
From (iv), we get
$u-v=-\frac{1}{4}$ ...(v)
Solving (iii) and (v), we get
$2 \mathrm{u}=\frac{1}{2}$
$u=\frac{1}{4}, \quad v=\frac{1}{2}$
So, $\frac{1}{3 x+y}=\frac{1}{4} \Rightarrow 3 x+y=4$
$\frac{\mathbf{1}}{\mathbf{3 x}-\mathbf{y}}=\frac{\mathbf{1}}{\mathbf{2}} \Rightarrow 3 x-y=2$
Solving, we get
$x=1, y=1$
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New Zealand
Level 6 - NCEA Level 1
# Graph quadratics in general form
Lesson
We've looked at how to expand and factorise quadratics. Both these skills can be used to help us graph quadratics (parabolas).
## Key Features to Graph
There are some key features of a parabola that we need to know to graph it, namely, the $x$x and $y$y intercepts, the concavity, the axis of symmetry and the turning point.
##### the factorised form
The factorised form of a quadratic is: $y=\left(x+a\right)\left(x+b\right)$y=(x+a)(x+b). Use the factorised form to find:
• $x$x-intercept(s): the points where the line crosses the $x$x-axis
• $y$y-intercept: the point where the line crosses the $y$y-axis
##### the general form
The general (expanded) form of a quadratic is $ax^2+bx+c$ax2+bx+c. Use the general form to find:
• Concavity: concave up or concave down
• Equation of the axis of symmetry: used to find the turning point
• Turning point: a pair of coordinates that is also the maximum or minimum value
#### Worked Examples
##### Question 1
Consider the parabola $y=\left(x+1\right)\left(x-3\right)$y=(x+1)(x3).
1. Find the $y$y value of the $y$y-intercept.
2. Find the $x$x values of the $x$x-intercepts.
Write all solutions on the same line separated by a comma.
3. State the equation of the axis of symmetry.
4. Find the coordinates of the vertex.
Vertex $=$=$\left(\editable{},\editable{}\right)$(,)
5. Graph the parabola.
##### Question 2
Consider the equation $y=25-\left(x+2\right)^2$y=25(x+2)2.
1. Does $y$y have a minimum or maximum value?
Maximum
A
Minimum
B
Maximum
A
Minimum
B
2. What is the maximum value of $y$y?
##### Question 3
Determine the $y$y-intercept of $y=3x^2-6x-3$y=3x26x3.
##### Question 4
Determine the value of $c$c if the parabola $y=x^2+4x+c$y=x2+4x+c has exactly one $x$x-intercept.
A quadratic graph has the shape known as a parabola. Here is an example:
The graph shows how two quantities that can vary are related to one another. That is, it shows how the choice of a quantity on the horizontal axis leads to a particular quantity on the vertical axis.
For example, the number $2$2 on the horizontal axis is related to $0$0 on the vertical axis, and $3$3 on the horizontal axis is related to $2$2 on the vertical axis.
We see that $1.5$1.5 on the horizontal axis is related to $-0.25$0.25 on the vertical scale and this appears to be the least value possible for the quantity displayed on this axis.
##### Example 1
In the graph above, what is the value on the vertical axis when the horizontal value is $1$1? What vertical value corresponds to the horizontal value $4$4? What vertical value corresponds to the horizontal value $-1$1?
We can write the horizontal and vertical corresponding numbers as coordinate pairs $(x,y)$(x,y). So, we have $(1,0)$(1,0), $(4,6)$(4,6) and $(-1,6)$(1,6).
##### Example 2
Quadratic graphs arise when we have two quantities, we could call them $x$x and $y$y, that are related in a way that involves multiplying $x$x by itself to obtain the corresponding value of $y$y
The diagram represents five rectangles that have the same perimeter but different areas.
Suppose we have $100$100m of fencing material that we wish to use to make a rectangular enclosure. The sides of the enclosure could look like any one of rectangles in the diagram or like any rectangle in between. We note that although the total of the side lengths is fixed at $100$100m, the area enclosed by the fences varies.
We could have sides $49$49m, $49$49m, $1$1m, and $1$1m, adding up to $100$100m and the area would be $49\times1=49$49×1=49 m2. Or, we could have sides $30$30m, $30$30m, $20$20m, and $20$20m, again adding up to $100$100m but in this case enclosing an area of $20\times30=600$20×30=600 m2.
We could find many other possible measurements for the sides and then calculate the area in each case. The goal would be to plot a graph that shows how the area of a rectangle is related to its width. To find the measurements we would probably go through a thought process something like the following:
• The total $100$100m length of material is the same as $2$2 lots of the width of the rectangle and $2$2 lots of the length.
• So, half the total length is the same as the width plus the length.
• Therefore, the length must be $50$50m less the width.
• The area is the width multiplied by the length. So, this must be the width multiplied by $50$50m less the width.
At this point, it may be apparent that all this can be said more clearly and in less space with the use of some mathematical symbols. We could let $T=100$T=100 be the total length of fencing material, let $W$W be the width and $L$L be the length of the various rectangles, and let $A$A be the area enclosed. Then we can write:
$100$100 $=$= $2W+2L$2W+2L $50$50 $=$= $W+L$W+L $L$L $=$= $50-W$50−W
Now that we have an expression for the length $L$L in terms of the width $W$W, we can find an expression for the area $A$A in terms of $W$W:
$A$A $=$= $W\times L$W×L $=$= $W\left(50-W\right)$W(50−W) $=$= $-W^2+50W$−W2+50W
This formula allows us to calculate $A$A for each value of $W$W chosen from the range $0$0m to $50$50m. The graph looks like the following:
The graph has the parabola shape because, in the formula for area, two terms involving $W$W have been multiplied together making it a quadratic formula.
We can see from the graph that the maximum area is obtained when the width is $25$25m. For this measurement, the rectangle is actually a square, since $2\times25+2\times25=100$2×25+2×25=100.
Note that the area of a rectangle is zero when the width $W$W is zero and also when $w=50$w=50.
#### Worked Examples
##### Question 1
In a room of $n$n people, if everyone shakes hands with everyone else, the total number of handshakes is given by $H=\frac{n\left(n-1\right)}{2}$H=n(n1)2.
1. Here is the graph of $H$H:
What are the values of $n$n where the graph of $H$H intercepts the horizontal axis? Write your answers on the same line, separated by a comma.
2. The vertex of this parabola occurs when $n=0.5$n=0.5, as indicated below:
Select all true statements:
The value of $n=0.5$n=0.5 is not valid since you can't have half a person.
A
When there are $0.5$0.5 people there are a negative number of handshakes.
B
The minimum number of handshakes that can occur is $0.5$0.5.
C
The value of $n=0.5$n=0.5 is not valid since you can't have half a person.
A
When there are $0.5$0.5 people there are a negative number of handshakes.
B
The minimum number of handshakes that can occur is $0.5$0.5.
C
3. Here is the graph of $H$H with only the values corresponding to whole numbers $n\ge2$n2:
These points are the only ones with useful interpretations.
Why is the first useful point, closest to the origin, at $\left(2,1\right)$(2,1)?
It takes at least $2$2 people to make $1$1 handshake.
A
$1$1 person can only perform $2$2 handshakes.
B
It takes at least $2$2 people to make $1$1 handshake.
A
$1$1 person can only perform $2$2 handshakes.
B
##### Question 2
An object is released $700$700 metres above ground and falls freely. The distance the object is from the ground is modelled by the formula $d=700-16t^2$d=70016t2, where $d$d is the distance in metres that the object falls and $t$t is the time elapsed in seconds. This equation is graphed below.
1. When $t=0$t=0, we get $d=700$d=700.
What does this mean?
The object is initially at ground level.
A
Initially, the object has not fallen any distance.
B
$t=0$t=0 is the time when the object lands on the ground.
C
When the object hits the ground, time stops.
D
The object is initially at ground level.
A
Initially, the object has not fallen any distance.
B
$t=0$t=0 is the time when the object lands on the ground.
C
When the object hits the ground, time stops.
D
2. Which of the following statements are true? Select all that apply.
The graph is always decreasing.
A
The graph is increasing for some values of $t$t, and decreasing for others.
B
The graph is neither increasing nor decreasing.
C
The graph is always increasing.
D
The graph is always decreasing.
A
The graph is increasing for some values of $t$t, and decreasing for others.
B
The graph is neither increasing nor decreasing.
C
The graph is always increasing.
D
The object only loses height until it hits the ground.
A
The object falls faster and faster until it hits the ground.
B
The object gains height until it reaches $700$700 metres above ground.
C
The object only loses height until it hits the ground.
A
The object falls faster and faster until it hits the ground.
B
The object gains height until it reaches $700$700 metres above ground.
C
##### Question 3
A rectangle is to be constructed with $80$80 metres of wire. The rectangle will have an area of $A=40x-x^2$A=40xx2, where $x$x is the length of one side of the rectangle. The graph of this function is given below.
1. What are the $x$x-intercepts of this graph?
Enter each $x$x-value on the same line, separated by commas.
2. What is the $x$x-value of the vertex?
3. What does the height of the parabola at $x=20$x=20 represent?
The smallest rectangular area that can be fenced off with $80$80m of wire.
A
The largest side of a rectangle that can fence off $80$80m2 in area.
B
The largest rectangular area that can be fenced off with $80$80m of wire.
C
The smallest side of a rectangle that can fence off $80$80m2 in area.
D
The smallest rectangular area that can be fenced off with $80$80m of wire.
A
The largest side of a rectangle that can fence off $80$80m2 in area.
B
The largest rectangular area that can be fenced off with $80$80m of wire.
C
The smallest side of a rectangle that can fence off $80$80m2 in area.
D
### Outcomes
#### NA6-7
Relate graphs, tables, and equations to linear, quadratic, and simple exponential relationships found in number and spatial patterns
#### 91028
Investigate relationships between tables, equations and graphs
|
# What is the exact value of tan 30 degrees?
#### Understand the Problem
The question is asking for the exact value of the tangent of 30 degrees, which is a trigonometric function value that can be derived from known angles in trigonometry.
The exact value of $\tan(30^\circ)$ is $\frac{\sqrt{3}}{3}$.
The exact value of $\tan(30^\circ)$ is $\frac{\sqrt{3}}{3}$.
#### Steps to Solve
1. Identify the Relationship with Trigonometric Functions
To find the tangent of 30 degrees, we can use the relationship between tangent and sine/cosine. The tangent function is defined as:
$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$
1. Find Sine and Cosine Values
Using known values from the unit circle, we know:
$$\sin(30^\circ) = \frac{1}{2}$$ $$\cos(30^\circ) = \frac{\sqrt{3}}{2}$$
1. Calculate the Tangent Value
Now we can substitute these values into the formula:
$$\tan(30^\circ) = \frac{\sin(30^\circ)}{\cos(30^\circ)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$$
This simplifies to:
$$\tan(30^\circ) = \frac{1}{\sqrt{3}}$$
1. Rationalize the Denominator
To present the final answer in a more standard form, we can rationalize the denominator:
$$\tan(30^\circ) = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$
The exact value of $\tan(30^\circ)$ is $\frac{\sqrt{3}}{3}$.
The tangent function is often used in right triangle problems. The value of $\tan(30^\circ) = \frac{\sqrt{3}}{3}$ is an important reference that is derived from the properties of a 30-60-90 triangle, where the lengths of the sides have a specific ratio.
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Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 30
The correct option is (a)
Hint:
If a function f is continuous at x = a, then
$\lim _{x \rightarrow a+} f(x)=\lim _{x \rightarrow a-} f(x)=f(a)$
Given:
the function
$f(x)=\frac{x^{3}+x^{2}-16 x+20}{x-2}$
is not defined for x = 2. In order to make f(x) continuous at x = 2
Step 1: Take
$f(x)={x^{3}+x^{2}-16 x+20}$
and calculate further
$\therefore f(x)={x^{3}+x^{2}-16 x+20}$
\begin{aligned} &=x^{2}(x-2)+3 x(x-2)-10(x-2)\\ \end{aligned}
Simplify further,
\begin{aligned} &=(x-2)(x-2)(x+5)\\ &=(x-2)^{2}(x+5) \end{aligned}
Therefore, the function can be rewritten as
\begin{aligned} &\therefore f(x)=\frac{(x-2)^{2}(x+5)}{(x-2)} \\ &f(x)=(x-2)(x+5) \end{aligned}
Step 2: Understand that if f(x) is continuous at x = 2
Therefore,
\begin{aligned} &\therefore \lim _{h \rightarrow 0} f(x)=f(x) \\ &=\lim _{h \rightarrow 0}(x-2)(x+5)=f(x) \\ &f(2)=0 \end{aligned}
Hence, the correct answer is option (a)
|
# Graphing the Solution Region of System of Linear Inequalities
Example:
Graph the solution set of the system of linear inequalities.
Solution:
We have
For the corresponding equations of inequalities (A), (B) and (C), we get
For x–Intercepts:
Put ${\text{y}} = 0$in Eq (1), Eq (2) and Eq (3) and we get
$\Rightarrow 4{\text{x}} - 5\left( {\text{0}} \right) = 20$ $\Rightarrow 3{\text{x}} + 2\left( {\text{0}} \right) = 6$ $\Rightarrow 3{\text{x}} + {\text{0}} = 2$ $\Rightarrow 4{\text{x}} = 20$ $\Rightarrow 3{\text{x}} = 6$ $\Rightarrow 3{\text{x}} = 2$ $\Rightarrow {\text{x}} = 5$ $\Rightarrow {\text{x}} = 2$ $\Rightarrow {\text{x}} = \frac{2}{3}$ $\therefore \left( {5,0} \right)$ $\therefore \left( {2,0} \right)$ $\therefore \left( {\frac{2}{3},0} \right)$
For y–Intercepts:
Put ${\text{x}} = 0$in Eq (1), Eq (2) and Eq (3) and we get
$\Rightarrow 4\left( {\text{0}} \right) - 5{\text{y}} = 20$ $\Rightarrow 3\left( {\text{0}} \right) + 2{\text{y}} = 6$ $\Rightarrow 3\left( {\text{0}} \right) + {\text{y}} = 2$ $\Rightarrow - 5{\text{y}} = 20$ $\Rightarrow 2{\text{y}} = 6$ $\Rightarrow {\text{y}} = 2$ $\Rightarrow {\text{y}} = - 4$ $\Rightarrow {\text{y}} = 3$ $\Rightarrow {\text{y}} = 2$ $\therefore \left( {0, - 4} \right)$ $\therefore \left( {0,3} \right)$ $\therefore \left( {0,2} \right)$
Test:
Put the origin as test point in inequalities (A), (B) and (C).
$0 - 0 < 20$ $\Rightarrow 0 + 0 < 6$ $\Rightarrow 0 + {\text{0}} < 2$ $\Rightarrow 0 < 20$ (True) $\Rightarrow 0 < 6$ (True) $\Rightarrow 0 < 2$ (True) $\therefore$ Solution set towards origin side. $\therefore$ Solution set towards origin side. $\therefore$ Solution set towards origin side.
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# Lesson video
In progress...
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Hi, I'm Mrs Dennett.
In this lesson, we're going to be using trigonometry to find the perpendicular height of a triangle.
In order for us to start calculating the perpendicular height of a triangle, you need to be able to identify it on a diagram.
Perpendicular means that we have two lines meeting each other at a right angle.
The height will split a right angle to the base.
In this question, we're told that the base is length b.
So y must be the perpendicular height, as it is perpendicular to the base, that's at right angles to the base and touches the top, the apex of the triangle formed by the other two sides.
Now we can start to calculate the perpendicular height of a triangle.
This is the perpendicular height of this triangle.
We label it with a lowercase h.
You can see, we have formed two right angle triangles, either side of the line, h.
We are given a side length and an angle for the triangle on the left.
So I can redraw this triangle and use trigonometry to find the heights.
We now label the sides with capital H for the hypotenuse, capital O for the opposite and a capital A for the adjacent.
Take care when labelling to use capital letters for these sides.
So as not to confuse the hypotenuse, which is a capital H, with the heights, which is a little h and is the opposite side.
So in this triangle, we want to find the length of the opposite side, our height and we are given the length of the hypotenuse.
So we can use the sine ratio to work out the height.
We solve this equation using our calculators.
So I multiply 19.
5 by sine 30, and we get the height, which is equal to 9.
75 centimetres.
Here's a different triangle.
It is isosceles.
I know this because of the small lines or hatch marks on the sides, which aren't labelled with a length.
For isosceles triangles, the perpendicular height is right in the centre of the base.
Again, we've got two right angle triangles here, we formed two right angle triangles, either side of the line h.
So we can use either triangle here as both angles adjacent to the base at 28 degrees.
But we'll use the left triangle as it's labelled with the angle already.
Now, with this triangle, we're only using half the base.
Remember the perpendicular height line, split the triangle, the larger triangle in half.
So we need to half the base to get 21 millimetres.
We then label with H, O and A.
Hypotenuse, opposite and adjacent.
So we can use trigonometry to find the perpendicular height.
So we can see in this triangle, we're going to be using the opposite and the adjacent sides, O an A.
So this indicates we have to substitute into the tangent ratio.
So we get tan 28 equals the height divided by 21.
We solve this equation using our calculators to get the height, which is 11.
17 millimetres to two decimal places.
Remember the units, they are in millimetres.
So we've got 11.
17 millimetres for the perpendicular height.
Here are some questions for you to try.
Pause the video to complete the task and restart when you were finished.
Here are the answers.
These are all isosceles triangles.
So you need to use the hatch marks, which show the equivalent side lines to help you in part A.
For part B, the base length needs to be halfed to six centimetres when using trigonometry to find the height.
In the last question, part C it is instead the angle, which needs to be halved to 35 degrees before applying trigonometry to find the perpendicular height.
Here's some questions for you to try, pause the video to complete the tasks and restart when you are finished.
Here are the answers.
For part A, you need to use the sine ratio to find the height and for part B, you needed to use the tan ratio.
We now want to work out the area of the triangle.
Can you remember how to do this? Here's a reminder.
We need to find the base length and the perpendicular height.
We use trigonometry to work out the height.
In a previous example, it was 9.
75 centimetres.
The base length is 17 plus four, which is 21 centimetres in total.
We now substitute into the formula and work out the area.
This is 102.
4 to one decimal place.
Remember to include the units, the area which are centimetre squared for this triangle.
Let's apply what we have just learned about area.
So this isosceles triangle from earlier too.
We want the area.
So we first need the perpendicular height.
Let's remind ourselves how we found this.
We halved the base and labelled with O and A because we didn't need the hypotenuse.
So we use the tan ratio to get the height, which was 11.
1658, et cetera.
Notice I'm not rounding my answer just yet as I want the area calculation to be as accurate as possible.
So we use 42 millimetres for the base.
Make sure you use the full base length here and we get 234.
5 centimetres when we round the area to one decimal place using our calculator.
Here's a question for you to try.
Pause the video to complete the task and restart when you are finished.
Here are the answers.
Remember to include the units for area in each of your answers for these questions.
If you find that your answer seems very close to the one given, check that you didn't round your answer too soon after finding the perpendicular height.
Here's a question for you to try.
Pause the video to complete the task and restart when you are finished.
Here's the answer, in this question, the triangle is equilateral.
So all of the sides are six centimetres.
The perpendicular height is in the centre of the base of the triangle.
So when using trigonometry to find the height, half the base, three centimetres.
All the angles will be 60 degrees as they're all equal because the triangle is equilateral.
You can use the sine or the tangent ratio to find the height of the triangle here.
Or you may have even spotted an opportunity to use Pythagoras theorem too.
The height is 5.
196152423, et cetera.
Or root 27, if you use Pythagoras.
So it's a little bit easier and potentially more accurate and a lot more accurate to use Pythagoras here.
Now, remember not to round the decimal.
If you got 5.
19, et cetera, until you have multiplied that decimal by a half times the base, or you will lose the accuracy for the area.
Only round to one decimal place at the very end.
That's all for this lesson.
Thank you for watching.
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# Vedic Maths to Check Whether A Number is Prime
0
93
A Prime Number is a number which is divisible by only 1 and itself.
1 Digit Primes
• 1 is neither Prime nor composite.
• 2 is the only even Prime.
• 2 and 3 are known as fundamental prime numbers.
• The other two single digit prime numbers are 5 and 7.
Thus, there exist four single-digit prime numbers out of 9 single digits – 2, 3, 5 and 7.
2 Digit Primes
• There are 90 two-digit numbers from 10 to 99.
• Out of these 90, there are 45 odd numbers and 45 even numbers.
• All even numbers are Composite since they are multiples of 2. So, all the 45 even numbers of two-digits are Composite.
• It is evident that an odd number is a number which ends with 1 or 3 or 5 or 7 or 9. So, in the 45 odd number, each set has 9 odd numbers having two-digits.
• All numbers ending with 0 or 5 are always divisible by 5. Thus all the 9 two-digit numbers ending with 5 are Composite. Now the remaining 36 may be Prime numbers, i. e, the two-digit numbers ending with 1, 3, 7 or 9 may be Prime.
To check whether a given two-digit number is Prime or not.
1. Step 1: Check if the given two-digit number is ending with 1 or 3 or 7 or 9. If it is, then Go to Step 2. Otherwise it is Composite.
2. Step 2: Check if the number is either 49 or 77 or 91 (7×7 or 7×11 or 7×13, 7 multiples of consecutive primes 7, 11 and 13). If the answer is ‘ No’, then Go to Step 3. Else it is Composite.
3. Step 3(a): Check if the given number is ending in 1 or 7 – if it is, check the tenth digit of the given number. If that tenth digit is other than 2 or 5 or 8, then we conclude that the given number is definitely Prime. (21, 27, 51, 57, 81 and 87 are composite with 3 multiples)
4. Step 3(b): Check if the given number is ending with 3 or 9 – if it is, check the tenth digit of the given number. If that tenth digit is other than 3 or 6 or 9, then we conclude that the given number is Prime number. (33, 39, 63, 69, 93 and 99 are composite with 3 multiples )
Is 73 Prime?
1. Step 1: 73 ends with 3 which means it may be Prime. Move on to Step 2.
2. Step 2: 73 is not in the category of numbers 49 or 77 or 91. Move to Step 3.
3. Step 3: The tenth digit of 73 is other than 3 or 6 or 9 {Using Step 3(b)}
We conclude that 73 is definitely a prime number.
Digital Extract Concept
In the above case of determining the nature of the number 73, instead of Step 3, we can use the concept of Digital Extract. If the Digital Extract of any given number is other than 3 or 6 or 9, the number 73 is definitely a prime number. Here DE (73) = 7 + 3 -> 10 -> 1 + 0 = 1 which is other than the digits 3 or 6 or 9. Thus, 73 is a prime number.
Is 83 Prime?
• Step 1: 87 ends with 7 which means it may be Prime. Move to Step 2.
• Step 2: 87 is not in the category of numbers 49 or 77 or 91. Move to Step 3.
• Step 3: The tenth digit of 87 is 8. Using Step 3(b), we can conclude that 87 is Composite. Hence 87 is not Prime.
• Using Digital extract, DE(87) = 6 which helps us conclude that 87 is not Prime.
Thus, to check whether the given two-digit number is prime or not, we are not using any Multiplication Table or Divisibility Rules.
Arushi Sana is the Founder of Santerra Living and Co-Founder of NYK Daily. She was awarded the Times Power Women of the Year 2022 and Times Digital Entrepreneur of the Year 2023. Arushi is also a Sustainability Consultant for organisations looking to reduce their carbon footprint and also works with brands on social media to help them carve a presence in that niche. She holds a Degree in Computer Science Engineering from VIT University and a Diploma in Marketing Analytics from IIM Nagpur. Her interest in Sustainable Living and Interior Design led her to start a Sustainable e-Marketplace where customers can buy eco-furniture and eco-friendly products for everyday use. Arushi is a writer, political researcher, a social worker, a farmer and a singer with an interest in languages. Travel and nature are the biggest spiritual getaways for her, and she aims to develop a global community of knowledge and journalism par excellence through this News Platform.
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# What is the equation of the line that goes through (5, -6) and is perpendicular to y=9?
Feb 28, 2017
See the entire solution explanation below:
#### Explanation:
$y = 9$ is a vertical line because it has a value of $9$ for each and every value of $x$.
Therefore, a line perpendicular to the will be a horizontal line and $x$ will have the same value for each and every value of $y$.
The equation for a horizontal line is $x = a$.
In this case we are given the point $\left(5 , - 6\right)$ which has a value of $5$ for $x$.
Therefore, the equation for the line in this problem is:
$x = 5$
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# How to Conquer the Times Table, Part 3
Photo of Javier times 4, by Javier Ignacio Acuña Ditzel, via flickr.
If you remember, we are in the middle of an experiment in mental math. We are using the world’s oldest interactive game — conversation — to explore multiplication patterns while memorizing as little as possible. Talk through these patterns with your student. Work many, many, many oral math problems together. Discuss the different ways you can find each answer, and notice how the number patterns connect to each other.
So far, we have mastered the times-1 and times-10 families and the Commutative Property (that you can multiply numbers in any order).
## The Doubles
What else is relatively simple? Does your student know the doubles? Doubles are often considered easy, because children do so much counting and addition with numbers less than 20. Even if your child finds the doubles tricky, a little focused practice should fix these facts in mind.
Practice doubling big numbers, too. Use silly numbers to help:
38 $\times$ 2
= double 3 tens + double 8
= “sixty-sixteen”
= 76
And:
2 $\times$ 56
= double 5 tens + double 6
= “tenty-twelve”
= 112
Go back and forth, inventing double-puzzles for each other:
$\times$ 2
= double 3 + double ½
= 6 + 1
= 7
And:
2 $\times$ 47,000
= (double 4 tens + double 7) thousand
= “eighty-fourteen” thousand
= 94,000
## Review Game: Once Through the Deck
The best way to practice the math facts is through the give-and-take of conversation, orally quizzing each other and talking about how you might figure the answers out. But occasionally you may want a simple, solitaire method for review. Here’s how:
• Shuffle a deck of math cards and place it face down on the table in front of you.
• Flip the cards face up, one at a time.
• For each card, say (out loud) the product of that number times the number you want to practice.
• Don’t say the whole equation, just the answer.
• Go through the deck as fast as you can.
• But don’t try to go so fast that you have to guess! If you are not sure of the answer, stop and figure it out.
Brian at The Math Mojo Chronicles demonstrates the game in this video, which my daughter so thoroughly enjoyed that she immediately ran to find a deck of cards and practiced her times-4 facts. (It’s funny, sometimes, what will catch a child’s interest.)
You can use Once Through the Deck as a final check that your student knows the fact family well enough to mark it on your chart. Remember to mark both the row and the column!
## The Times-4 Family
Notice that the answers in the times-4 row are exactly double the answers in the times-2 row. Can your child see why that makes sense? If you have two of something, and you replicate two more of it, then you would have four of that thing, whether it is minions or cookies or numbers.
This means you do not need to memorize the times-4 facts. Just double the number to get the times-2 answer, and then double it again. For example, $7 \times 4$ would mean seven doubled, which is 14, and then that answer doubled again:
7 $\times$ 4
= 7 $\times$ 2 $\times$ 2
= 14 $\times$ 2
= 28
Practice double-doubling a bunch of numbers. Can you use the double-double trick to figure out something like $53 \times 4$ ?
53 $\times$ 4
= 53 $\times$ 2 $\times$ 2
= 106 $\times$ 2
= 212
This may take a little more time to practice — but that is okay! Quiz each other with unusual numbers, using double-doubling to get the answer. Test yourself with Once Through the Deck, and when you are ready, mark the chart.
## The Times-8 Family
In the same way that times-4 was the double of times-2, it makes sense that times-8 is the double of times-4. If you have four of something, and you replicate four more of it, then you will have eight of the thing in all. It doesn’t matter whether the thing is books or aliens or numbers. Even fractions: If you have four 1/16 size slices of pizza, and you take four more pieces, then you will have 8/16 of the pizza.
If you need to calculate something times eight, you can double the something, then double again — that makes four times — then double it once more for your final answer:
6 $\times$ 8
= 6 $\times$ 2 $\times$ 2 $\times$ 2
= 12 $\times$ 2 $\times$ 2
= 24 $\times$ 2
= 48
I find it helpful to count on three fingers, to make sure I don’t forget any of the doublings. Remember to experiment with big numbers, too. Can you double-double-double to figure out $132 \times 8$ ?
132 $\times$ 8
= 132 $\times$ 2 $\times$ 2 $\times$ 2
= 264 $\times$ 2 $\times$ 2
= 528 $\times$ 2
= 1056
It may take a few days or even weeks of practice before you and your student feel comfortable with these. Take all the time you need, and when you both are able to mentally double-double-double almost any number the other can pose, mark off the times-8 column and row on the chart.
## The Times-5 Family
Your daughter can probably count by fives, but many children get confused when trying to skip-count large multiplication problems. A more reliable number pattern for times-5 calculations uses the doubles in reverse. Two fives make ten, so any even number of fives will make exactly half that number of tens:
6 fives = 3 tens
18 fives = 9 tens
24 fives = 12 tens
450 fives = 225 tens
All the odd numbers times five will come out “somethingty-five,” and you can predict what the “somethingty” will be by looking at the next lower even number.
7 fives = (6 + 1) fives = 3 tens + 5
25 fives = (24 + 1) fives = 12 tens + 5
109 fives = (108 + 1) fives = 54 tens + 5
Practice intensively on these until you can both get the answer right every time. Test yourself with a round of Once Through the Deck, and then mark them off.
Wow! Look back and see how much you have learned. You started with 144 facts, and you have narrowed it down to only 21 — and all you had to memorize so far was the doubles! Your child could probably memorize the last 21 facts without too much trouble, but let’s see if we can find a few more patterns to make them easier.
This is the fourth post in my Times Table Series. To be continued…[Go to part 4.]
Claim your two free learning guide booklets, and be one of the first to hear about new books, revisions, and sales or other promotions.
## 4 thoughts on “How to Conquer the Times Table, Part 3”
1. I absolutely love the information in your blog. I’m a math tutor that developed my own method of teaching multiplication tables drawing from various sources. I’m constantly adding to it, and have added your tips on extending the doubling principal for times 4 and times 8 (when I think about it, this is the technique Greg Tang shows in his book “The best of times” but I never employed it before. Just like students, teachers need something new to spark their interest).
I’m working with a student right now that seems to have become stuck on times 8. We’ve been working on it for over a month now (off and on) and he just doesn’t seem to be able to recall that when he can’t remember a times 8 fact, that he can double 3 times. I’m just wondering if you (or anyone else with similar experience) have advice on whether to continue working on it until he gets it, or if it’s better to skip it, move on, then come back to try it again later.
In addition to “learning facts” I do practical applications, like multiplication war and DAMULT dice. He always gets stuck on times 8 no matter the application where it comes up. He resorts to guessing, or thinks that he has to double 2 times or 4 times when prompted “remember how we multiply by 8?”. Any further advice or suggestions would be appreciated!
2. I have had success with the “backing off” method when my kids hit a wall. Sometimes just forgetting about the topic for awhile lets it simmer in the student’s unconscious mind, and when we come back to it in a month or two, everything flows easily.
But I’m not sure that would work so well in this case. I’m assuming the student needs to use times-8 facts in his schoolwork and can’t ignore them. Is he mature enough to catch himself when he starts guessing, so that if you give him a fool-proof technique, he could slow down and force himself to use it?
If so, here’s the technique:
* Write down 8 copies of whatever number you are multiplying times 8, arranged like the dots on a domino. You are going to add these numbers up in pairs to find out how much all 8 of them are worth.
To calculate 13 times 8, I would write:
13 13
13 13
13 13
13 13
* Then circle the numbers in pairs, writing their sum right next to the circle. This is the first doubling. Now you only have 4 numbers:
26
26
26
26
* Now circle those numbers in pairs, and write their sum next to the circle. This is the second doubling, and it leaves you with only 2 numbers to add:
52
52
It takes some scratch paper and the discipline to resist wild guessing, but it will always work!
3. Russian Peasant Multiplication would also work. It’s quicker, but it’s more abstract, which means it may be harder for a child to remember. In each step, you double one number and cut the other in half:
13 times 8 =
26 times 4 =
52 times 2 =
104 times 1 = 104
4. Thanks for the tips! The student is in grade 6, but I’m not covering school work with him, I went back to just going over basic facts. Like most students I tutor, adding/subtracting isn’t a problem, but (basic) multiplication is, and doesn’t become noticeable until they start on long division (which is when the parent gives me a call to get help).
He’s a high energy little guy, and definitely a visual/auditory learner. Trying to use beads and cups to show grouping or having him write number bonds went disastrously. I like your first tip, and I’ll give it a try. If he rebels, I’m going to skip times 8 for now. We still have to work on times 3, 6 and 7. When an 8 times fact comes up that he’s stuck on, I’ll do what I always do, which is patiently explain it to him again. It’ll sink in eventually.
Thanks for all the resources on your blog, I can get lost in it reading everything and following the links. I look forward to your book coming out!
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# Question: What are the odds of rolling 3 pairs with 6 dice?
Contents
## What is the probability of rolling 3 pairs with 6 dice?
Calculation Details: Probability for Everyone!
Number of Dice Number of Possibilities
3 216
4 1296
5 7776
6 46656
## What are the odds of rolling all ones with 6 dice?
Assuming (1), you have a 1 in 6^6(46656) chance of getting six ones on one roll, so 1 in 15552 (. 006%) for 3 rolls.
## How do you calculate chance?
Divide the number of events by the number of possible outcomes. This will give us the probability of a single event occurring.
## What is the probability of rolling 6?
So to get a 6 when rolling a six-sided die, probability = 1 ÷ 6 = 0.167, or 16.7 percent chance. So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0.0278, or 2.78 percent.
## What is the probability of rolling 6 sixes in a row?
Since there are six choices, then each time there is a 1/6 chance of rolling a six. The fact they are entirely independent of each other means we simply multiply each roll’s probability together: 1/6 × 1/6 × 1/6 × 1/6 = (1/6)4 = 1/1296 = 0.00077. Now let’s say you had already rolled double sixes.
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## What are the odds of throwing 5 sixes?
The probability of rolling five of a kind of any other number is also 1/7776. Since there are a total of six different numbers on a die, we multiply the above probability by 6. This means that the probability of a Yahtzee on the first roll is 6 x 1/7776 = 1/1296 = 0.08 percent.
## How do you find the probability of odds?
To convert odds to probability, take the player’s chance of winning, use it as the numerator and divide by the total number of chances, both winning and losing. For example, if the odds are 4 to 1, the probability equals 1 / (1 + 4) = 1/5 or 20%.
## How do you find the percentage of odds?
To write a percentage as an odds ratio, convert the percentage to a decimal x, then calculate as follows: (1/x) – 1 = first number in the odds ratio, while the second number in the odds ratio is 1. Substitute your result from Step 3 for X in the odds ratio X-to-1. In this example, the result from Step 3 is 1.5.
## How do you find the percentage of an odds?
To convert from a probability to odds, divide the probability by one minus that probability. So if the probability is 10% or 0.10 , then the odds are 0.1/0.9 or ‘1 to 9’ or 0.111. To convert from odds to a probability, divide the odds by one plus the odds.
## What are the odds of 1 in 100?
Number Converter
1 in __ Decimal Percent
1 in 20 0.05 5.0%
1 in 25 0.04 4.0%
1 in 50 0.02 2.0%
1 in 100 0.01 1.0%
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## How do you find the probability of A or B?
The probability of two disjoint events A or B happening is: p(A or B) = p(A) + p(B).
## How do you find the probability of an event?
The probability of an event is the number of favorable outcomes divided by the total number of outcomes.
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# What is difference between means in statistics?
## What is difference between means in statistics?
The mean difference, or difference in means, measures the absolute difference between the mean value in two different groups. In clinical trials, it gives you an idea of how much difference there is between the averages of the experimental group and control groups.
## How do you find the difference in means?
Given these assumptions, we know the following.
1. The expected value of the difference between all possible sample means is equal to the difference between population means. Thus,
2. The standard deviation of the difference between sample means (σd) is approximately equal to: σd = sqrt( σ12 / n1 + σ22 / n2 )
How do you know if two means are statistically different?
Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used. Test method. Use the two-sample t-test to determine whether the difference between means found in the sample is significantly different from the hypothesized difference between means.
How do you find a statistical difference?
Subtract the group two mean from the group one mean. Divide each variance by the number of observations minus 1. For example, if one group had a variance of 2186753 and 425 observations, you would divide 2186753 by 424. Take the square root of each result.
### How do you compare two means?
Comparison of Means Techniques
1. Independent Samples T-Test. Use the independent samples t-test when you want to compare means for two data sets that are independent from each other.
2. One sample T-Test.
3. Paired Samples T-Test.
4. One way Analysis of Variance (ANOVA).
### What is p-value in statistics?
In statistics, the p-value is the probability of obtaining results at least as extreme as the observed results of a statistical hypothesis test, assuming that the null hypothesis is correct. A smaller p-value means that there is stronger evidence in favor of the alternative hypothesis.
What does p-value tell you?
A p-value is a measure of the probability that an observed difference could have occurred just by random chance. The lower the p-value, the greater the statistical significance of the observed difference. P-value can be used as an alternative to or in addition to pre-selected confidence levels for hypothesis testing.
What is the z value that is used for a 95% confidence interval?
Z=1.96
The Z value for 95% confidence is Z=1.96.
#### What statistical test should I use to compare two groups?
The two most widely used statistical techniques for comparing two groups, where the measurements of the groups are normally distributed, are the Independent Group t-test and the Paired t-test. The Independent Group t-test is designed to compare means between two groups where there are different subjects in each group.
#### Which test is used to compare two means?
The compare means t-test is used to compare the mean of a variable in one group to the mean of the same variable in one, or more, other groups. The null hypothesis for the difference between the groups in the population is set to zero. We test this hypothesis using sample data.
What statistical analysis should I use to compare two groups?
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# Thread: Simple eigenvalue problem (x'=[2x2]x)
1. ## Simple eigenvalue problem (x'=[2x2]x)
Hi, I've got a homework problem that should be very simple, the previous problem was identical (different numbers) and was a breeze, but this one is giving me problems. Here's what we're asked for:
Apply the eigenvalue method to find the particular solution to the system of differential equations
$\displaystyle x'=\begin{bmatrix} 4 & 4\\ 3 & 5 \end{bmatrix}x$
which satifies the initial conditions
$\displaystyle x(0)=\begin{bmatrix} -4\\ 2 \end{bmatrix}$
So I begin by finding the eigenvalues from the characteristic equation:
$\displaystyle (4-\lambda)(5-\lambda)-12=(\lambda^2-9\lambda+8)=(\lambda-1)(\lambda-8)$
$\displaystyle \lambda=1,8$
To get the matrices:
$\displaystyle \begin{bmatrix} 3 & 4\\ 3 & 4 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=0$
and
$\displaystyle \begin{bmatrix} -4 & 4\\ 3 & -3 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=0$
To get
$\displaystyle \begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} -4\\ 3 \end{bmatrix}$ and $\displaystyle \begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} 1\\ 1 \end{bmatrix}$
Which gives the particular solution $\displaystyle x_p=a\begin{bmatrix} -4\\ 3 \end{bmatrix}e^t+b\begin{bmatrix} 1\\ 1 \end{bmatrix}e^{8t}$
But at this point, no values for a and/or b will satisfy $\displaystyle x(0)=\begin{bmatrix} -4\\ 2 \end{bmatrix}$
Can anyone help me out, please?
Thanks!
2. Don't you just need to solve the system
$\displaystyle -4a+b=-4,$
$\displaystyle 3a+b=2?$
That system has one unique solution, as far as I can tell.
3. Oh, yes, you're right. I got:
$\displaystyle (-4)6/7e^t-4/7e^{8t}$
and
$\displaystyle (3)6/7e^t-4/7e^{8t}$
Which are both correct.
Thank you!
4. You're very welcome. Have a good one!
5. there HAS to be a solution. at t = 0, the exponential functions are both 1. and since (-4,3) and (1,1) are linearly independent, they span a 2-dimensional subspace of R^2, which MUST be R^2 itself.
hint: a and b are not integers.
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# What`s the surface area formula for a rectangular pyramid?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
22
mason m Share
Jun 12, 2018
$\text{SA} = l w + l \sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}} + w \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$
#### Explanation:
The surface area will be the sum of the rectangular base and the $4$ triangles, in which there are $2$ pairs of congruent triangles.
Area of the Rectangular Base
The base simply has an area of $l w$, since it's a rectangle.
$\implies l w$
Area of Front and Back Triangles
The area of a triangle is found through the formula $A = \frac{1}{2} \left(\text{base")("height}\right)$.
Here, the base is $l$. To find the height of the triangle, we must find the slant height on that side of the triangle.
The slant height can be found through solving for the hypotenuse of a right triangle on the interior of the pyramid.
The two bases of the triangle will be the height of the pyramid, $h$, and one half the width, $\frac{w}{2}$. Through the Pythagorean theorem, we can see that the slant height is equal to $\sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}}$.
This is the height of the triangular face. Thus, the area of front triangle is $\frac{1}{2} l \sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}}$. Since the back triangle is congruent to the front, their combined area is twice the previous expression, or
$\implies l \sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}}$
Area of the Side Triangles
The side triangles' area can be found in a way very similar to that of the front and back triangles, except for that their slant height is $\sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$. Thus, the area of one of the triangles is $\frac{1}{2} w \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$ and both the triangles combined is
$\implies w \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$
Total Surface Area
Simply add all of the areas of the faces.
$\text{SA} = l w + l \sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}} + w \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$
This is not a formula you should ever attempt to memorize. Rather, this an exercise of truly understanding the geometry of the triangular prism (as well as a bit of algebra).
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## Estimating and Measuring Amounts: When Students Ask
• Why do I need to learn about both customary units and metric units of measurement?
Have students identify real-life situations in which they used customary units of measurement. Then have them identify real-life situations in which they used metric units of measurement. Although some students may only come up with situations for customary units, the experiences their classmates have had using metric units may help them better understand why they are learning both systems. Also, point out to students that in most countries metric units of measurement are used most often.
• What is capacity and how do I measure it?
Explain to students that capacity is the amount that a container can hold. Have students draw upon their own experiences with cooking and shopping to name some of the units of capacity they are familiar with. The customary units of capacity they will learn this year are pint (pt), cup (c), quart (qt), and gallon (gal). Give students the opportunity to investigate these units using pint, cup, quart, and gallon containers. For example, have them estimate and then measure the number of cups in one pint, pints in one quart, and quarts in one gallon.
2 cups (c) =1 pint (pt)
2 pints (pt) = 1 quart (qt)
4 quarts (qt) = 1 gallon (gal)
You can also have students use different-size containers such as a milk carton, water bottle, spoon, or eyedropper to estimate which container holds less than, more than, or about 1 customary unit. For example, a water bottle holds more than 1 pint. A spoon holds less than 1 cup. This activity can be adapted to metric units of capacity: milliliters (mL) and liters (L). A milk bottle holds about 1 liter. An eyedropper holds about 1 milliliter.
• What is mass?
Explain to students that mass is the amount of matter in an object. The metric units of mass are gram (g) and kilogram (kg). Have students explore the mass of different objects using a scale. Identify one object that has a mass of about 1 kilogram. Have students pick up other objects and estimate which objects have a mass that is less than, more than, or about 1 kilogram. Then have students use a scale to find the actual mass of each object. This activity is easily adaptable to customary units of weight: pound (lb) and ounces (oz). Lead students to discover that large objects may have a small mass or weight and small objects may have a great mass or weight.
• Why are there so many different units of measurement?
Students will discover the answer to this question when you respond with questions such as this: If feet and yards did not exist, how would we measure the height of our school? If the only unit of capacity was milliliters, what would it be like to measure the amount of water in a bathtub?
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# square and square roots
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• 1. a qu Ss reand Squar Roo e ts
2. We will: Understand the meaning of squaring a number and finding the square root of a number Find squares of numbers Raise a whole number to a whole number power Find square roots of perfect squares Find approximate square roots of nonperfect squares 3. Shade in graph paper to make each of the shapes below. Each shape is a square.Count and write the number of square tiles in each of the larger squares below. 1. 2. 3. 4. Continue to draw larger squares. Make one that is 7 tiles wide and 7 tiles high; then make one that is 8 wide and 8 high. Count the number of squares in each shape.4. 7 by 7 = ____ 5. 8 by 8 = ____ 5. Talk to the person next to you about the following questions. Be prepared to discuss them if I call on you.The numbers 1, 4, 9, 16, 25, etc. are known as perfect squares. Why do you think they are called perfect squares? How are the width and the height of the squares related? How are they related to the total number of tiles? How could you find the next numbers that are perfect squares without tiles? 6. If a square measures 4 inches on each side, how would you find its area? 4 inches4 inches4 inches4 inches 7. Remember how to calculate the area? 4 inches4 inchesA=lw A = 4 in. 4 in. A = 16 in.2 8. A square must have the same length and width. 4 inches4 inchesA=lw A = 4 in. 4 in. 9. Square Number Also called a perfect square A number that is the square of a whole number Can be represented by arranging objects in a square. 10. Square Numbers 11. To square a number means to multiply it by itself. 5 squared means 5 x 5 Take the number 5 And square it! 12. There is a shorter way to write 5 x 5. Say: 5 squared or 5 to the 2nd power. 13. These are the parts. This is the baseThis is the exponent 14. This is another formula for finding the area of a square.sA=s2s is the side length 15. Evaluate it!A. 14B. 49 16. Try these: Number Squared 2 911 2 32102Factors Standard Form9 9 11 113 3 10 1081 121 9100 17. These numbers are called perfect squares 18. If a number is a perfect square, then you can find its exact square root. A perfect square is simply a number that can be written as the square of another number. 19. What are the first 10 whole numbers that are perfect squares? 2222222221 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 101, 4, 9, 16, 25, 36, 49, 64, 81, 1002 20. -Whats the opposite operation to addition? -Whats the opposite operation to multiplication? -The opposite operation to squaring a number is taking the square root. 62 = 3636 = 6 21. The symbol used to indicate a root is the radical symbol - 22. Every radical expression has three parts Radical symbol Index Radicand 23. Every radical expression has Radical three parts Index249 Radicand 24. The index of a radical is a whole number greater than or equal to 2. 25. The index of a square root is always 2. 26. The square root of 49 could 2 be written as 49 but is normally written as 49 . 27. What does square root mean? 28. The square root of a number is another number which when multiplied by itself gives back the original number. 29. What is a square root??the measure of the side of the square16 = 4 30. What is the square root of 36?36 = 6 31. Example:49 = 7because7 7 = 7 = 49 2Also49 = 7 because ( 7 )( 7 ) = ( 7 ) = 49 2 32. Find the two square roots of each number.A. 49 49 = 7 49 = 7B. 100 100 = 10 100 = 107 is a square root, since 7 7 = 49. 7 is also a square root, since 7 7 = 49. 10 is a square root, since 10 10 = 100. 10 is also a square root, since 10 10 = 100. 33. Find the two square roots of each number.A. 25 25 = 5 25 = 5B. 144 144 = 125 is a square root, since 5 5 = 25. 5 is also a square root, since 5 5 = 25. 12 is a square root, since 12 12 = 144. 144 = 12 12 is also a square root, since 12 12 = 144. 34. Open your books to page 72 so we can try some problems together
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## Shortest Path
### Learning Outcomes
• Find the shortest path through a graph using Dijkstra’s Algorithm
When you visit a website like Google Maps or use your Smartphone to ask for directions from home to your Aunt’s house in Pasadena, you are usually looking for a shortest path between the two locations. These computer applications use representations of the street maps as graphs, with estimated driving times as edge weights.
While often it is possible to find a shortest path on a small graph by guess-and-check, our goal in this chapter is to develop methods to solve complex problems in a systematic way by following algorithms. An algorithm is a step-by-step procedure for solving a problem. Dijkstra’s (pronounced dike-stra) algorithm will find the shortest path between two vertices.
### Efficiency
Efficiency is a hallmark of mathematical practice. When mathematicians seek a proof for something they have conjectured to be true, the most elegant proof is often the most efficient one: an argument that packs the most information in the fewest words, so to speak.
The ideas explored in graph theory are frequently applied to computing algorithms: the language and instructions of software. Since resources are limited (time, computing power), mathematicians and computer scientists seek the most efficient ways to compute. Graph theory helps them find the shortest path from A to B.
### Dijkstra’s Algorithm
1. Mark the ending vertex with a distance of zero. Designate this vertex as current.
2. Find all vertices leading to the current vertex. Calculate their distances to the end. Since we already know the distance the current vertex is from the end, this will just require adding the most recent edge. Don’t record this distance if it is longer than a previously recorded distance.
3. Mark the current vertex as visited. We will never look at this vertex again.
4. Mark the vertex with the smallest distance as current, and repeat from step 2.
### EXAMPLE
Suppose you need to travel from Tacoma, WA (vertex T) to Yakima, WA (vertex Y). Looking at a map, it looks like driving through Auburn (A) then Mount Rainier (MR) might be shortest, but it’s not totally clear since that road is probably slower than taking the major highway through North Bend (NB). A graph with travel times in minutes is shown below. An alternate route through Eatonville (E) and Packwood (P) is also shown.
Step 1: Mark the ending vertex with a distance of zero. The distances will be recorded in [brackets] after the vertex name
Step 2: For each vertex leading to Y, we calculate the distance to the end. For example, NB is a distance of 104 from the end, and MR is 96 from the end. Remember that distances in this case refer to the travel time in minutes.
Step 3 & 4: We mark Y as visited, and mark the vertex with the smallest recorded distance as current. At this point, P will be designated current. Back to step 2.
Step 2 (#2): For each vertex leading to P (and not leading to a visited vertex) we find the distance from the end. Since E is 96 minutes from P, and we’ve already calculated P is 76 minutes from Y, we can compute that E is 96+76 = 172 minutes from Y.
Step 3 & 4 (#2): We mark P as visited, and designate the vertex with the smallest recorded distance as current: MR. Back to step 2.
Step 2 (#3): For each vertex leading to MR (and not leading to a visited vertex) we find the distance to the end. The only vertex to be considered is A, since we’ve already visited Y and P. Adding MR’s distance 96 to the length from A to MR gives the distance 96+79 = 175 minutes from A to Y.
Step 3 & 4 (#3): We mark MR as visited, and designate the vertex with smallest recorded distance as current: NB. Back to step 2.
Step 2 (#4): For each vertex leading to NB, we find the distance to the end. We know the shortest distance from NB to Y is 104 and the distance from A to NB is 36, so the distance from A to Y through NB is 104+36 = 140. Since this distance is shorter than the previously calculated distance from Y to A through MR, we replace it.
Step 3 & 4 (#4): We mark NB as visited, and designate A as current, since it now has the shortest distance.
Step 2 (#5): T is the only non-visited vertex leading to A, so we calculate the distance from T to Y through A: 20+140 = 160 minutes.
Step 3 & 4 (#5): We mark A as visited, and designate E as current.
Step 2 (#6): The only non-visited vertex leading to E is T. Calculating the distance from T to Y through E, we compute 172+57 = 229 minutes. Since this is longer than the existing marked time, we do not replace it.
Step 3 (#6): We mark E as visited. Since all vertices have been visited, we are done.
From this, we know that the shortest path from Tacoma to Yakima will take 160 minutes. Tracking which sequence of edges yielded 160 minutes, we see the shortest path is T-A-NB-Y.
Dijkstra’s algorithm is an optimal algorithm, meaning that it always produces the actual shortest path, not just a path that is pretty short, provided one exists. This algorithm is also efficient, meaning that it can be implemented in a reasonable amount of time. Dijkstra’s algorithm takes around V2 calculations, where V is the number of vertices in a graph[1]. A graph with 100 vertices would take around 10,000 calculations. While that would be a lot to do by hand, it is not a lot for computer to handle. It is because of this efficiency that your car’s GPS unit can compute driving directions in only a few seconds.
[1] It can be made to run faster through various optimizations to the implementation.
In contrast, an inefficient algorithm might try to list all possible paths then compute the length of each path. Trying to list all possible paths could easily take 1025 calculations to compute the shortest path with only 25 vertices; that’s a 1 with 25 zeros after it! To put that in perspective, the fastest computer in the world would still spend over 1000 years analyzing all those paths.
### EXAMPLE
A shipping company needs to route a package from Washington, D.C. to San Diego, CA. To minimize costs, the package will first be sent to their processing center in Baltimore, MD then sent as part of mass shipments between their various processing centers, ending up in their processing center in Bakersfield, CA. From there it will be delivered in a small truck to San Diego.
The travel times, in hours, between their processing centers are shown in the table below. Three hours has been added to each travel time for processing. Find the shortest path from Baltimore to Bakersfield.
Baltimore Denver Dallas Chicago Atlanta Bakersfield Baltimore * 15 14 Denver * 18 24 19 Dallas * 18 15 25 Chicago 15 18 18 * 14 Atlanta 14 24 15 14 * Bakersfield 19 25 *
While we could draw a graph, we can also work directly from the table.
Step 1: The ending vertex, Bakersfield, is marked as current.
Step 2: All cities connected to Bakersfield, in this case Denver and Dallas, have their distances calculated; we’ll mark those distances in the column headers.
Step 3 & 4: Mark Bakersfield as visited. Here, we are doing it by shading the corresponding row and column of the table. We mark Denver as current, shown in bold, since it is the vertex with the shortest distance.
Baltimore Denver[19] Dallas[25] Chicago Atlanta Bakersfield[0] Baltimore * 15 14 Denver * 18 24 19 Dallas * 18 15 25 Chicago 15 18 18 * 14 Atlanta 14 24 15 14 * Bakersfield 19 25 *
Step 2 (#2): For cities connected to Denver, calculate distance to the end. For example, Chicago is 18 hours from Denver, and Denver is 19 hours from the end, the distance for Chicago to the end is 18+19 = 37 (Chicago to Denver to Bakersfield). Atlanta is 24 hours from Denver, so the distance to the end is 24+19 = 43 (Atlanta to Denver to Bakersfield).
Step 3 & 4 (#2): We mark Denver as visited and mark Dallas as current.
Baltimore Denver[19] Dallas[25] Chicago[37] Atlanta[43] Bakersfield[0] Baltimore * 15 14 Denver * 18 24 19 Dallas * 18 15 25 Chicago 15 18 18 * 14 Atlanta 14 24 15 14 * Bakersfield 19 25 *
Step 2 (#3): For cities connected to Dallas, calculate the distance to the end. For Chicago, the distance from Chicago to Dallas is 18 and from Dallas to the end is 25, so the distance from Chicago to the end through Dallas would be 18+25 = 43. Since this is longer than the currently marked distance for Chicago, we do not replace it. For Atlanta, we calculate 15+25 = 40. Since this is shorter than the currently marked distance for Atlanta, we replace the existing distance.
Step 3 & 4 (#3): We mark Dallas as visited, and mark Chicago as current.
Baltimore Denver[19] Dallas[25] Chicago[37] Atlanta[40] Bakersfield[0] Baltimore * 15 14 Denver * 18 24 19 Dallas * 18 15 25 Chicago 15 18 18 * 14 Atlanta 14 24 15 14 * Bakersfield 19 25 *
Step 2 (#4): Baltimore and Atlanta are the only non-visited cities connected to Chicago. For Baltimore, we calculate 15+37 = 52 and mark that distance. For Atlanta, we calculate 14+37 = 51. Since this is longer than the existing distance of 40 for Atlanta, we do not replace that distance.
Step 3 & 4 (#4): Mark Chicago as visited and Atlanta as current.
Baltimore[52] Denver[19] Dallas[25] Chicago[37] Atlanta[40] Bakersfield[0] Baltimore * 15 14 Denver * 18 24 19 Dallas * 18 15 25 Chicago 15 18 18 * 14 Atlanta 14 24 15 14 * Bakersfield 19 25 *
Step 2 (#5): The distance from Atlanta to Baltimore is 14. Adding that to the distance already calculated for Atlanta gives a total distance of 14+40 = 54 hours from Baltimore to Bakersfield through Atlanta. Since this is larger than the currently calculated distance, we do not replace the distance for Baltimore.
Step 3 & 4 (#5): We mark Atlanta as visited. All cities have been visited and we are done.
The shortest route from Baltimore to Bakersfield will take 52 hours, and will route through Chicago and Denver.
### Try It
• Find the shortest path between vertices A and G in the graph below.
https://www.myopenmath.com/multiembedq.php?id=6888&theme=oea&iframe_resize_id=mom5
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# Can you express correctly 0.000 1465 in scientific notation as?
Scientific notation is a way of expressing very large or very small numbers in a concise and standardized format. In scientific notation, a number is written in the form of “a x 10^b,” where “a” is a number between 1 and 10, and “b” is an integer representing the power of 10.
In the case of the number 0.000 1465, it can indeed be expressed in scientific notation. To do so, we first move the decimal point to the right until there is only one nonzero digit to the left of the decimal. This results in 1.465 x 10^-4, where “a” is 1.465 and “b” is -4.
## Understanding Scientific Notation: A look at 0.000 1465
Let us delve deep into the world of scientific notation, a handy tool utilized by scientists, mathematicians, and students worldwide to express large or minute numbers with precision and simplicity. Our key focus will be to understand if the number 0.000 1465 can be accurately expressed in scientific notation.
### What is Scientific Notation?
To begin, understanding scientific notation is essential. It is a mathematical approach used to denote numbers that are either too large or too small to be written practically in standard decimal notation. Significant figures are a crucial element of this approach, which can make working with numbers, especially in scientific computations, considerably easier.
The format of a number in scientific notation is a product of a number between 1 and 10 (including 1 but excluding 10) and a power of 10. It carries the format a × 10^b, wherein ‘a’ is a number, or a significand, between 1 and 10, and ‘b’ is the exponent indicating the actual place of the decimal point.
### Conversion to Scientific Notation: Process
To convert a number into scientific notation involves moving the decimal point to a position where there’s only one non-zero digit left before the decimal. You then count the places moved, and this forms your exponent, ‘b’.
If you’re moving the decimal point to the right, as in small decimals, the exponent ‘b’ will be negative. Conversely, if you’re moving the decimal point to the left like in large numbers, the exponent ‘b’ will be positive.
### Expressing 0.000 1465 in Scientific Notation
Now, let’s take a closer examination of the number 0.000 1465 and whether it can be accurately expressed in scientific notation.
When we follow the conversion rules as discussed above, we move the decimal point four places to the right, which yields the number 1.465. This implies an exponent of -4 because the decimal has been relocated to the right. Therefore, the scientific notation for 0.000 1465 is 1.465 × 10^-4.
### Verifying Accuracy in the Conversion
The act of converting a number to scientific notation should not change its inherent value. Thus, to verify the accuracy of our conversion, if we perform the reverse process on our obtained scientific notation, we should regain the original number, 0.000 1465.
Executing this, 1.465 × 10^-4 essentially moves the decimal point of 1.465 four places to the left, as indicated by the negative exponent. And surely, this gives us the original value 0.000 1465.
Therefore, we can confidently proclaim that yes, the number 0.000 1465 can be correctly and accurately expressed in scientific notation as 1.465 × 10^-4.
Understanding methodologies such as these, aided by real examples, only serves to better comprehend and appreciate the scientific disciplines and mathematical procedures we employ, elevating our proficiency and fluency in these fields.
The number 0.0001465 can be correctly expressed in scientific notation as 1.465 x 10^-4.
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## How to Add Four-Digit Numbers
Let's review with an example.
2,341 + 4,537 = ?
1️⃣ First, write the numbers in column form.
2️⃣ Then, add the digits from right to left.
Let's try it.
First, we write the numbers in column form.
Let's start by adding the Ones column first.
Next, we add the Tens digits.
Next, we add the Hundreds digits.
Lastly, we add the Thousands digits.
The sum of 2,341 and 4,537 is 6,878.
Great work! 🎉
Let's try one more problem together.
1,234 + 2,766 = ?
First, we write the numbers in column form.
The sum in the Ones place is greater than 9!
This means we need to regroup.
To regroup means to turn 10 from one column into 1 in the next column.
We can regroup 10 Ones into 1 Ten.
We show regrouping by writing a small 1 at the top of the Tens place column, called a carry over.
Now, let's add the Tens, including the carry over.
The sum in this column, 10, is greater than 9 again. So we need to regroup 10 Tens into 1 Hundreds.
Awesome. Can you guess the next step?
Yes, we add the Hundreds column, including the new carry over.
1 + 2 + 7 = 10, a two-digit number.
Looks like we need to regroup again. No worries. 🏄♀️
Finally, we add the Thousands column.
You got the answer, 4,000. How?
1️⃣ You used column form.
2️⃣ You added column by column, starting with the Ones.
3️⃣ You regrouped when the sum in any column was 10 or more.
Great job!
Adding multi-digit numbers is no sweat if you follow the steps carefully.
Now, let's try the practice.
Complete the practice to earn 1 Create Credit
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# Proof: Why There Is No Rational Number Whose Square is 2?
This article is focused on discussing the proof that there is no rational number whose square is 2. Before starting the proof, let’s get familiar with the basic terms-
Rational Numbers
A number that can be expressed in the form of p/q, where p and q are integers and q ≠0, is known as a rational number. Examples are 0, 1, -1, 5/2, etc.
Problem Statement:
There is no rational number whose square is 2.
Solution:
Let’s get started with the proof of the above problem statement. Proof in this article will be done using a mathematical technique called Proof By Contradiction.
It is a mathematical technique, in which first, the assumption that the proposition that needs to be proved is false and then deduces a result using the false proposition, which comes out to be contradictory either to the assumption, or any other commonly known mathematical result. Thus, proving the validity of the proposition. That’s why this technique is known as Proof by Contradiction.
Proof-
In this proof, the Proof by Contradiction technique is being used, where first it is assumed that there exists a rational number, whose square equals 2, and then deduce a result using this assumption, which will come out to be contradictory with our assumption. So. Let’s get started with the proof-
1. Assume that there exists a rational number, X = p/q whose square is equal to 2 such that p and q are in their simplest form, i.e. they don’t have any common factor.
```X2 = 2 (Assumption)
(p/q)2 = 2 (Since X is a rational number)```
2. This implies,
```p2/q2 = 2
p2 = 2q2 ---(1)```
3. From the above equation, it can be said that p2 is an even integer as it can be expressed in the form of 2k, where k = q2. Now, it is known that the square of an odd integer is always odd, which means that p cannot be an odd integer, thus p is also an even integer. Hence, p can be expressed in the form 2k, where k is some integer i.e.
`p = 2k, for some integer k ---(2)`
3. Now, after substituting the value of p from equation (2) in equation (1), the following equation is obtained-
```(2k)2 = 2q2
4k2 = 2q2 ---(3)```
3. Dividing both sides by 2, in equation (3), the following equation is obtained-
```2k2 = q2
q2 = 2k2 ---(4)```
4. Again, it can be said that q2 is an even integer, and since it is known that the square of an odd integer is always odd, thus q cannot be an odd integer. This implies that q is also an even integer.
5. Now, from the above discussion, it can be concluded that p and q are both even integers i.e. they have a common factor of at least 2, but this statement contradicts the assumption at the beginning of this proof that p and q are in their simplest form i.e. they don’t have any common factor.
This contradiction means that the assumption that there exists a rational number, X = p/q whose square is equal to 2 such that p and q are in their simplest form, is false. Thus, it proves that there exists no rational number, whose square is equal to 2.
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# CBSE Class 7 Maths worksheet for chapter-Fraction and Decimals
Sep 29, 2022, 16:45 IST
## Introduction to Fraction
The word fraction is derived from the Latin word "Fractus" means broken. Represents the total component, which includes the number of equal parts out of a whole.
### Representation of Fractions
A fraction is represented by 2 more than one number, separated by a line. The number above is a numerator and the number below is a denominator. Example: 64 which means 6 out of 4 equal divisions.
#### Multiplication of Fractions
Multiplying a fraction by a whole number:
Example 1: 7 × (1/2) = 7/2
Example 2: 5 × (4/45) = 20/45, Dividing a number by a denominator by 5, we get 7/9
Fraction multiplication by fraction is actually a numerical product / product of denominators.
#### Solved Examples
Q1. If x= 9/11 and y= 4/15, then the value of x - y is
1. 5/165
2. 36/165
3. 91/165
4. 98/165
x-y = 9/11 - 4/15
= 135-44/165
= 91/165
Q2. Reciprocal of 7 is
1. -7
2. 1
3. 7
4. 1/7
We know that the reciprocal of x (or inverse of x)
= 1/x
Therefore, the reciprocal of 7 = 1/7
Q3. If the cost of 12 bananas is Rs 26.50, then the cost of 6 such bananas will be
1. Rs 10.50
2. Rs 12
3. Rs 13.25
4. Rs 20
Given, the cost of 12 bananas = Rs 26.50 The cost of 1 banana = Rs 26.50/12
Therefore, the cost of 6 bananas = Rs (26.50/12) x 6
= Rs 13.25.
Q4. If the side of a square is 12.5 cm, then the area of square will be
1. 156.25 cm2
2. 160 cm2
3. 173.5 cm2
4. 180 cm2
Given, side of a square = 12.5 cm
Area of square = side x side = 12.5 x 12.5 cm2
= 156.25 cm2.
Q5. Product of 0.04 x 1.1 is
1. 0.044
2. 0.44
3. 4.04
4. 4.4
0.04 x 1.1 = 0.044.
Q6. Decimal form of 15 rupees 5 paise is
1. Rs 15
2. Rs 15.5
3. Rs 155.0
4. Rs 15.05
Convert 5 paise into rupee
5 paise = Rs 5/100 = Rs 0.05
Thus, 15 rupees 5 pasie = Rs 15.05.
Q7. The value of 4kg 625g in gm will be
1. 4.620 g.
2. 4625 g
3. 40625 g
4. 400625 g
Ans. Correct option is 2
4kg 625g = 4 kg + 625 g
= (4 x 1000 + 625) g [1kg=1000g]
= (4000 + 625) g
= 4625 g
Q8. If the length of a rectangle is 7.1 cm and its breadth is 2.5 cm, then the area of rectangle will be
1. 4.6 cm2
2. 4.8 cm2
3. 8.875 cm2
4. 17.75 cm2
Area of rectangle = length x breadth
= 7.1 cm x 2.5 cm
= 17.75 cm2
Q9. If a car covers a distance of 95.5 km in 2.5 hours, then the average distance covered by it in 1 hour will be
1. 38 km
2. 38.2 km
3. 38.3 km
4. 39 km
Ans. Correct option is 1
Distance covered by the car = 95.5 km
And, the time required to cover this distance = 2.5 hours
Therefore, distance covered by it in 1 hour = 95.5/2.5 = 955/25
= 38.2 km
Q10. If we shift the decimal point two places towards right of a number 0.045, then the new number is
1. 0.0045
2. 4.5
3. 45
4. 450
Ans. Correct option is 2
Shifting the decimal points two places towards right of 0.045 means 0.045 x 100, which gives the number 4.5.
This page consists of CBSE Class 7 Maths Worksheet for chapter-Fraction and Decimals prepared by Physics Wallah. All questions of Fraction and Decimals are with detail solutions for reference. Find out all Worksheet for Class 7 Maths. Do follow NCERT Solutions for Class 7 Maths prepared by Physics Wallah.
For additional information related to the subject you can check the Maths Formula section.
### Also Check
Find Below PDF of CBSE Worksheet for Class 7 Maths Chapter 2 Fraction and Decimals
|
# Solving Word Problems with Plotted Data on a Coordinate Plane
Coming up next: How to Use Scatter Plots to Solve Word Problems
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#### Recommended Lessons and Courses for You
Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After watching this video lesson, you will be able to solve word problems that involve plotted points on a coordinate plane. Learn how to choose the right points that will answer your question.
## Plotted Data
In this lesson, you will learn about solving word problems that involve plotted data on a coordinate plane. Your word problems are math problems written with words, and your coordinate plane is the plane made up of the horizontal x-axis and the vertical y-axis. Points on this plane are written as (x, y), where x is the x-coordinate of the point and y is the y-coordinate of the point. Being able to solve these types of problems will help you on your math tests. Also, if you decide to become a scientist who does research and lab work, then you may also need to know how to solve these types of problems.
Your plotted data can be random, or it can form a pattern on your coordinate plane. Your plotted data will look like a collection of points on your coordinate plane. Each point represents one piece of data. For example, this coordinate plane includes plotted data that shows the number of cats and dogs that each surveyed household has. If you were doing research, you just might see something like this:
The x-axis represents the number of dogs and the y-axis represents the number of cats. So, the point (3, 0) represented by the green dot means that this particular household has 3 dogs and 0 cats.
## A Word Problem
When dealing with word problems, you might come across something like this on a test:
From the data shown on this coordinate plane, what is the highest number of dogs owned and the highest number of cats owned?
Whenever you come across a word problem that involves plotted data, you will most often also see the coordinate plane with the points on it. It's your job to make sense of the plotted data to find your answer.
To find your answer, you need to fully understand your problem and what it is asking of you. Reading over your problem again, you see that your problem is actually asking for two different things. The first is the highest number of dogs owned and the second is the highest number of cats owned. The answer could be from one point or from two separate points. You will find out by looking at the plotted data.
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# 7.4 The other trigonometric functions (Page 5/14)
Page 5 / 14
## Period of a function
The period $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ of a repeating function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is the number representing the interval such that $\text{\hspace{0.17em}}f\left(x+P\right)=f\left(x\right)\text{\hspace{0.17em}}$ for any value of $\text{\hspace{0.17em}}x.$
The period of the cosine, sine, secant, and cosecant functions is $\text{\hspace{0.17em}}2\pi .$
The period of the tangent and cotangent functions is $\text{\hspace{0.17em}}\pi .$
## Finding the values of trigonometric functions
Find the values of the six trigonometric functions of angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ based on [link] .
Find the values of the six trigonometric functions of angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ based on [link] .
$\begin{array}{l}\mathrm{sin}t=-1,\mathrm{cos}t=0,\mathrm{tan}t=\text{Undefined}\\ \mathrm{sec}t=\text{Undefined,}\mathrm{csc}t=-1,\mathrm{cot}t=0\end{array}$
## Finding the value of trigonometric functions
If $\text{\hspace{0.17em}}\mathrm{sin}\left(t\right)=-\frac{\sqrt{3}}{2}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{cos}\left(t\right)=\frac{1}{2},\text{find}\text{\hspace{0.17em}}\text{sec}\left(t\right),\text{csc}\left(t\right),\text{tan}\left(t\right),\text{cot}\left(t\right).$
$\text{\hspace{0.17em}}\mathrm{sin}\left(t\right)=\frac{\sqrt{2}}{2}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)=\frac{\sqrt{2}}{2},\text{find}\text{\hspace{0.17em}}\text{sec}\left(t\right),\text{csc}\left(t\right),\text{tan}\left(t\right),\text{and}\text{\hspace{0.17em}}\text{cot}\left(t\right)$
$\mathrm{sec}t=\sqrt{2},\mathrm{csc}t=\sqrt{2},\mathrm{tan}t=1,\mathrm{cot}t=1$
## Evaluating trigonometric functions with a calculator
We have learned how to evaluate the six trigonometric functions for the common first-quadrant angles and to use them as reference angles for angles in other quadrants. To evaluate trigonometric functions of other angles, we use a scientific or graphing calculator or computer software. If the calculator has a degree mode and a radian mode, confirm the correct mode is chosen before making a calculation.
Evaluating a tangent function with a scientific calculator as opposed to a graphing calculator or computer algebra system is like evaluating a sine or cosine: Enter the value and press the TAN key. For the reciprocal functions, there may not be any dedicated keys that say CSC, SEC, or COT. In that case, the function must be evaluated as the reciprocal of a sine, cosine, or tangent.
If we need to work with degrees and our calculator or software does not have a degree mode, we can enter the degrees multiplied by the conversion factor $\text{\hspace{0.17em}}\frac{\pi }{180}\text{\hspace{0.17em}}$ to convert the degrees to radians. To find the secant of $\text{\hspace{0.17em}}30°,$ we could press
Given an angle measure in radians, use a scientific calculator to find the cosecant.
1. If the calculator has degree mode and radian mode, set it to radian mode.
2. Enter: $\text{\hspace{0.17em}}1\text{/}$
3. Enter the value of the angle inside parentheses.
4. Press the SIN key.
5. Press the = key.
Given an angle measure in radians, use a graphing utility/calculator to find the cosecant.
• If the graphing utility has degree mode and radian mode, set it to radian mode.
• Enter: $\text{\hspace{0.17em}}1\text{/}$
• Press the SIN key.
• Enter the value of the angle inside parentheses.
• Press the ENTER key.
## Evaluating the cosecant using technology
Evaluate the cosecant of $\text{\hspace{0.17em}}\frac{5\pi }{7}.$
For a scientific calculator, enter information as follows:
Evaluate the cotangent of $\text{\hspace{0.17em}}-\frac{\pi }{8}.$
$\approx -2.414$
Access these online resources for additional instruction and practice with other trigonometric functions.
Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1
|
# GSEB Solutions Class 7 Maths Chapter 1 Integers Ex 1.4
Ā Ā Ā
Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 1 Integers Ex 1.4 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 7 Maths Chapter 1 Integers Ex 1.4
Question 1.
Evaluate each of the following:
(a) (- 30) Ć· 10
(b) 50 Ć· (- 5)
(c) (- 36) Ć· (- 9)
(d) (- 49) Ć· (49)
(e) 13 Ć· [(- 2) +1]
(f) 0 Ć· (- 12)
(g) (- 31) Ć· [(- 30) + (- 1)]
(h) [(- 36) + 12] + 3
(i) [(- 6) + 5] Ć· [(- 2) + 1]
Solution:
(a) (- 30) Ć· 10 = $$\frac { – 30 }{ 10 }$$
(b) 50 Ć· (- 5) = $$\frac { 50 }{ (- 5) }$$
(c) (- 36) Ć· (- 9) = $$\frac { – 36 }{ (- 9) }$$
= 4
(d) (- 49) Ć· 49 = $$\frac { ( – 49) }{ 49 }$$
= 1
(e) 13 Ć· [(- 2) + 1] = 13 Ć· [- 1]
= $$\frac { 13 }{ (- 1) }$$ = – 13
(f) 0 Ć· (- 12) = 0
(g) (- 31) Ć· [(- 30) + (- 1)] = (- 31) Ć· (- 31)
= $$\frac { – 31 }{ (- 31) }$$ = 1
(h) [(- 36) Ć· 12] Ć· 3 = [ $$\frac { (- 36) }{ 12 }$$ ] + 3
= [- 3] Ć· 3 = – 1
(i) [ (- 6) + 5] Ć· [(- 2) + 1] = [- 1] + [- 1]
= $$\frac { (- 1) }{ (- 1) }$$ = 1
Question 2.
Verify that a Ć· (6 + c) ā (a Ć· b) + (a Ć· c) for each of the following values of a, b and c.
(a) a = 12, b = – 4, c = 2
(b) a = – 10, b = 1, c = 1
Solution:
(a) Given that a = 12, b = – 4 and c = 2
L.H.S. = a Ć· (b + c) = 12 + (- 4 + 2)
= 12 Ć· (- 2) = – 6
R.H.S. = (a Ć· b) + (a Ć· c)
= [12 Ć· (- 4)] + (12 Ć· 2)
= [- 3] + 6 = 3
āµ – 6 ā 3
ā“ L.H.S. ā R.H.S.
Hence, a Ć· (b + c) ā (a Ć· b) + (a Ć· c)
(b) Given that a = – 10, b = 1, c = 1
L.H.S. = a + (b + c)
= (- 10) + (1 + 1) = (- 10) Ć· 2 = – 5
R.H.S. = (a Ć· b) + (a Ć· c)
= [(- 10) Ć· 1] + [(- 10) Ć· 1]
= (- 10) + (- 10) = – 20
āµ – 5 ā 20
ā“ L.H.S. ā R.H.S.
Hence, a Ć· (b + c) ā (a Ć· b) + (a Ć· c)
Question 3.
Fill in the blanks:
(a) 369 Ć· ____= 369
(b) (- 75)Ć· ____ = – 1
(c) (- 206) Ć· ____ = 1
(d) – 87 Ć· ____ = 87
(e) ____Ć· 1 = – 87
(f) ____ Ć· 48 = – 1
(g) 20 Ć· ____ = – 2
(h) ____ Ć· 4 = – 3
Solution:
(a) āµ a + 1 Ć· aĀ Ā Ā Ā Ā Ā Ā Ā ā“ 369 + 1 = 369
(b) āµ (- a) Ć· a = – 1Ā Ā Ā Ā Ā ā“ (- 75) Ć· 75 = – 1
(c) āµ (- a) Ć· (- a) = 1Ā Ā Ā Ā ā“ (- 206) Ć· (- 206)= 1
(d) āµ (- a) Ć· (- 1) = aĀ Ā Ā Ā ā“ (- 87) Ć· (- 1) = 87
(e) āµ (- a) Ć· 1 = – aĀ Ā Ā Ā Ā ā“ (- 87) Ć· 1 = – 87
(f) āµ (- a) Ć· a = – 1Ā Ā Ā Ā Ā ā“ (- 48) Ć· 48 = – 1
(g) āµ 20 Ć· 10 = 2Ā Ā Ā Ā Ā Ā Ā ā“ 20 Ć· (-10) = -2
(h) āµ 12 Ć· 4 = 3Ā Ā Ā Ā Ā Ā Ā Ā ā“ (- 12) Ć· 4 = – 3
Question 4.
Write five pairs of integers (a, b) such that a Ć· b = – 3. One such pair is (6, – 2) because 6 Ć· (- 2) = (- 3).
Solution:
I. Since, 3 Ć· 1 = 3
3 + (- 1) = – 3
Compairing it with a Ć· b = – 3 we have a = 3 and b = (- 1)
ā“ The required pair of integers = (3, – 1)
II. Since, 3 Ć· 1 = 3
ā“ (- 3) – 1 = – 3
Compairing with a + b = – 3 we have a = (- 3) and b = 1
ā“ The required pair of integers = (- 3, 1)
III. Since, 9 Ć· 3 = 3
ā“ 9 Ć· (- 3) = – 3
Compairing it with a + b = – 3 we have a = 9 and b = – 3
ā“ The required pair of integers = (9, – 3)
IV. Since, 12 Ć· 4 = 3
ā“ (- 12) Ć· 4 = – 3
Comparing it with a Ć· b = – 3 we have a = – 12 and b = 4
ā“ The required pair of integers = [- 12, 4].
V. Since, 12 Ć· 4 = 3
ā“ 12 + (- 4) = – 3
Comparing it with a + b = – 3 we have
a = – 12 and b = – 4
ā“ The required pair of integers = [12, (- 4)]
Question 5.
The temperature at 12 noon was 10Ā°C above zero. If it decreases at the rate of 2Ā°C per hour , until mid-night, at what time would the temperature be 8Ā°C below zero? What would be the temperature at mid-night?
Solution:
Temperature at 12 noon = +10Ā°C Rate of change in temperature = – 2Ā°C per hour
Number of hours from 12 noon to mid- night =12
ā“ Change in temperature in 12 hours = 12Ā° x (- 2) = – 24Ā°C
ā“ Temperature at mid- night (i.e. 12 hours after 12 noon) = + 10Ā°C + (- 24Ā°C) = – 14Ā°C
Thus, temprature at mid- night = – 14Ā°C
Now, temperature difference between + 10Ā°C and – 8Ā°C = +10Ā°C – (- 8Ā°C) = 18Ā°C
āµ $$\frac { 18 }{ 2 }$$ = 9
ā“ Temperature change of 18Ā°C will take place in 9 hours from 12 noon. Time after 9 hours from 12 noon = 9 p.m. Thus, the temperature 8Ā°C below 0Ā° (- 8Ā°C) would be at 9 p.m.
Question 6.
In a class test (+3) marks are given for every correct answer and (- 2) marks are given for every incorrect answer and no marks for not attempting any Question uestion.
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
(ii) Mohini scores – 5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution:
Marks for every correct answer = +3
Marks for every incorrect answer = – 2
(i) Total marks scored by Radhika = 20
Marks scored for correct answers = 12 x 3 = 36
ā“ Marks given for incorrect answers = 20 – 36 = – 16
Number of incorrect answers = (- 16) Ć· (- 2) = 8
Thus, Radhika attempted 8 questions incorrectly.
(ii) Marks obtained by Mohini = – 5
Marks obtained for 7 correct answers = 7 x 3 = 21
ā“ Marks obtained for incorrect answers = – 5 – 21 = – 26
ā“ Number of incorrect answers = – 26 Ć· (- 2) = 13
Question 7.
An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 ml
Solution:
Present position of the elevator is at 10 m above the ground level.
Distance to be moved by the elevator below the ground level = 350 m
ā“ Total distance to be moved by the elevator = 350 m + 10 m = 360 m
The rate of descent = 6 m/min
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# How To Divide Multiply Add And Subtract Fraction
## What is a fraction?
### 1- Definition
A fraction is a number, which is obtained by dividing an integer into equal parts. For example, when we say a quarter of the cake, we are dividing the cake into four parts and consider one of them.
A fraction is described in a scientific discipline by numbers that are written one on another and that are separated by a horizontal straight line denominated fractional line.
The fraction consists of two terms: the numerator and the denominator. The numerator is the number that is on the fractional line and the denominator is the number that is under the fractional line.
The numerator is the number of parts that is considered the unit or total.
The denominator is the number of equal parts in which the unit or total has been divided.
All fractions are given a specific name, can be read as such, according to the numerator and denominator they have.
The number in the numerator reads the same, but not the denominator. When the denominator goes from 2 to 10, it has a specific name (if it is 2 is “means”, if 3 is “thirds”, if 4 is “quarter”, if 5 is “fifth”, if 6 is “sixth,” if 7 is “seventh,” if 8 is “eighth,” if 9 is “ninth”, if 10 is “tenth”), however, when it is greater than 10.
### 3- The meanings of the fractions in the different contexts of use
3.1 The fraction as an expression that links the part with the whole
In this case, it is used to indicate “the fracture” or “division into parts”, answering the question what part is it? of the integer in question or as considered parts of a collection of equal objects. It is agreed that the denominator of the fraction indicates the number of parts in which said integer is divided and the numerator the parts considered.
Issue:
From a basket of 36 flowers, 1/3 are pink roses; 1/4 are margaritas and rest are chocolates. How many flowers of each class are there?
To calculate the fraction of a number n, in this case, flowers, you can divide the number n by the denominator of the fraction and then multiply it by the numerator, or multiply the numerator of the fraction by n and divide the result by the denominator.
So in our problem:
– 1/3 of 36 are pink = 36: 3 = 12 x 1 = 12
Therefore of the 36 flowers that are in the basket: 12 are roses
Therefore of the 36 flowers that are in the basket: 9 are margaritas.
– If the rest of the flowers of the basket are chocolates, we must subtract the total of flowers, the sum of the other two.
roses + daisies = 12 + 9 = 21
36 – 21 = 15
Then we have that there are 15 chocolates.
Answer: Of the 36 flowers in the basket, 12 are roses, 9 are daisies and 15 are Chocolates.
3.2- The fraction as an equitable distribution
Answering the question, how much does each one have?
Similarly, if I am to distribute 3 chocolate bars between 4 children, each will receive 3/4 bar.
To make it clear, we will see another example:
– A group of 4 friends gets together to eat. They have 3 pizzas, which they will distribute in equal parts. What fraction of pizza corresponds to each?
Since the 3: 4 division is not accurate, we must do the following:
We will divide each pizza into 4 equal parts that is in quarters.
Then split the 12 pieces between the 4 friends
3.3- The fraction as reason including
– Two different sets, for example, the ratio or relation between a number of books in the class and the
number of students. Thus, 13 books for 26 students can be expressed as 13/26
reading “13 to 26” or what is the same, “1 for every 2”.
– A set and a subset of it, for example, the ratio of the 21 students in total and the male students (11) of a class can be expressed as 11/21 or “11 to 21”. A special case is a probability defined as the number of favorable cases on the number of possible cases of a given event. For example, in the roll of a die, the probability or odds ratio of a 2 “is one to 6” which is indicated as 1/6.
To divide two or more fractions, multiply “in the cross”.
This is the numerator (top number) of the first fraction by the denominator (number down) of the second fraction, so we get the numerator. To obtain the denominator, we have to multiply the denominator (number of down) of the first fraction by the numerator (number of above) of the second fraction.
In this post, we will learn how to do a division of fractions.
### Fraction division method 1: Multiply in cross
This method consists of multiplying the numerator of the first fraction by the denominator of the second fraction and writing the result into the numerator of the resulting fraction.
On the other hand, we multiply the denominator of the first fraction by the numerator of the second and the result we write it in the denominator of the resulting fraction.
Finally, the fraction is simplified.
For example, to divide fractions 3/4 between 6/10.
We calculate the numerator by the denominator of the second. Now, we have the numerator of the final fraction 3 × 10 = 30
On the other hand, we multiply the denominator of the first (4) by the numerator of the second (6). In this way, we have the denominator of the final fraction 4 × 6 = 24
The last step is to simplify the fraction.
30: 6 = 5
24: 6 = 4
Hence, the outcome of the division is 5/4
### Fraction division method 2: Invert and multiply
Step 1: Change the second fraction, ie alter the numerator by the denominator and contrariwise.
Step 2. Simplify any numerator with any denominator.
For example, let’s split 12/5 between 6/4.
Step 1: Change the second fraction 6/4. This becomes 4/6
Step 2: We simplify numerators with denominators.
The numerators are:
12 = 2x2x3
4 = 2 × 2
The denominators are:
5 = 5
6 = 2 × 3
We can simplify both the numerator and the denominator by a 2 and a 3. Thus we have, 2x2x2 / 5
Step 3: Multiply online: 2x2x2 / 5 = 8/5
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There are two cases:
While of making the fractions we can find two cases: Fractions with same denominator and Fractions with different denominator
Fractions with the same denominator:
The sum of the fractions with the same denominator is very simple, just add all the numerators and divide by common denominator. However, if, the fractions have a different denominator. Then follow the below steps-
1. Multiply in a cross. The numerator of the first fraction is multiplied by the denominator of the second, and the denominator of the first by the numerator of the second. Both multiplications add up.
2. Multiply the denominators of the two fractions. The denominators of the two fractions are multiplied.
3. We resolve all operations.
We divide them by that number.
In this case, it is an improper fraction because the denominator (4) is smaller than the numerator (5).
Fraction with different denominator can also be divided through alternative ways, such as
1. between the two denominators, there is always least common multiple.
2. Calculate the numerator with the formula: old numerator x common denominator (the drawn with the common minimum multiple) and divided by the old denominator.
3. Once the denominator is equal, the fractions are added as in the first case (since the fractions have the same denominator).
1. We calculate the least common multiple (l.c. m.). The minimum common divisor (m.c.m) of 4 and 2 is 4.
2. Calculate the numerators. We work out the numerators with the formula. old numerator x common denominator (the drawn with the common minimum multiple) and divided by the old denominator.
Numerator of the first fraction: 3 x 4: 4 = 3
Numerator of the second fraction: 4 x 4: 2 = 8
3. Once the denominators are the same the operations are performed. The result of these operations is:
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## Subtraction of Fractions
There are two cases:
In the subtraction of fractions we can find two different cases:
Fractions with the same denominator
with the different denominator
Fractions with the same denominator.
The subtraction fractions with the same denominator are very plain, just work out the numerators and leave the denominator.
2. Multiply the denominators of the two fractions. The denominators of the two fractions are multiplied.
We divide the denominator and the numerator by this number.
Another way to do it:
The subtraction of two or more fractions with different denominator is a bit less simple. Let’s go step by step:
1 The minimum common multiple of the two denominators
2 Calculate the numerator with the formula: old numerator x common denominator and divided by old denominator
3 Proceed as in the first case (since the fractions have the same denominator)
Example:
6/4 – 1/2
1 We calculate the least common multiple (L.c. m.). (4, 2) = 4.
2 We calculate the numerators.
Numerator of the first fraction: 6 x 4: 4 = 6
Numerator of the second fraction: 1 x 4: 2 = 2
3 Thus we have a fraction that is:
6/4 – 2/4
As the denominators are identical, we can subtract it as in case 1
## Multiplication of fractions
The multiplication of fractions is very simple.
The multiplication of two or more fractions is done “in line”. That is, the numerator of the first fraction by the numerator of the second and the denominator of the first fraction by the denominator of the second.
## Divide fractions
When dividing an integer, multiply by the reciprocal of its divisor. In the example of painting where you need 3 gallons of paint to apply a layer and you have 6 gallons of paint, you can find the total number of layers you can paint by dividing 6 between 3, 6 ÷ 3 = 2. You can also multiply 6 by the reciprocal of 3, which is, then the multiplication problem becomes.
If you have a sweet that needs to split in half, you can divide by 2, or you can multiply it to find the amount you want.
Similarly, with a mixed number, you can either divide between the whole number or you can multiply by the reciprocal. Suppose you have pizzas that you want to divide equally among 6 people.
Dividing a fraction of a whole number is the same as multiplying by its reciprocal, so you can always use the multiplication of fractions to solve problems of divisions.
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# Video: CBSE Class X • Pack 3 • 2016 • Question 2
CBSE Class X • Pack 3 • 2016 • Question 2
02:22
### Video Transcript
What value of 𝑘 will make 𝑘 plus nine, two 𝑘 minus one, and two 𝑘 plus seven consecutive terms of an arithmetic progression?
An arithmetic progression is a sequence of numbers such that the difference between any two consecutive terms is constant. Remember, consecutive simply means one after the other. For example, one, three, five, seven is an example of an arithmetic progression. It has a constant common difference of two. And the first two consecutive terms are one and three. This means that the difference between the first two terms, that’s 𝑘 plus nine and two 𝑘 minus one, must be equal to the difference between the second and the third term. That’s two 𝑘 minus one and two 𝑘 plus seven.
Remember, when we find the difference, we subtract one from the other. So the difference between the first two terms is two 𝑘 minus one minus 𝑘 plus nine. And the difference between the second and third term is two 𝑘 plus seven minus two 𝑘 minus one. Let’s simplify each of these expressions.
On the left-hand side, we get two 𝑘 minus one minus 𝑘 minus nine. It’s really important to be careful when multiplying out the second bracket. Technically, what we’re doing is multiplying everything inside this bracket by negative one, which is why we end up with negative nine. A common mistake is to forget this fact and simply write plus nine.
Similarly, on the right-hand side, we’re multiplying the second bracket by negative one. That gives us negative two 𝑘 plus one. When we simplify the left-hand side of this expression, we get 𝑘 minus 10. Since two 𝑘 minus 𝑘 is 𝑘 and negative one minus nine is negative 10. Two 𝑘 minus two 𝑘 is zero. So we end up with eight on the right-hand side.
And we can solve this expression for 𝑘 by adding 10 to both sides. Eight plus 10 is 18. And we can see that the value of 𝑘 that will make the terms consecutive terms of an arithmetic progression is 18.
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# 1950 AHSME Problems/Problem 31
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
John ordered $4$ pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by $50\%$. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:
$\textbf{(A)}\ 4:1 \qquad \textbf{(B)}\ 2:1 \qquad \textbf{(C)}\ 1:4 \qquad \textbf{(D)}\ 1:2 \qquad \textbf{(E)}\ 1:8$
## Solution
Let the number of blue socks be represented as $b$. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of blue socks is $1$. That means the price of one pair of black socks is $2$.
Now from the third and fourth sentence, we see that $1.5(2(4)+1(b))=1(4)+2(b)$. Simplifying gives $b=16$. This means the ratio of the number of pairs of black socks and the number of pairs of blue socks is $\boxed{\textbf{(C)}\ 1:4}$.
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## Calculus 10th Edition
Published by Brooks Cole
# Chapter 1 - Limits and Their Properties - 1.2 Exercises - Page 56: 33
#### Answer
$$L=8$$ and $$\delta=\frac{1}{300}$$
#### Work Step by Step
1) From the definition of limit, the statement $$\lim_{x\to c}f(x)=L$$ means for each $\varepsilon\gt0$ there exists a $\delta\gt0$ such that if $$0\lt|x-c|\lt\delta$$ then $$|f(x)-L|\lt\varepsilon$$ 2) $$\lim_{x\to2}(3x+2)$$ To find $L$, we plug $x=2$ into $f(x)$ $$\lim_{x\to2}(3x+2)=3\times2+2=8$$ So $L=8$. Next, the question asks to find $\delta\gt0$ such that if $|f(x)-L|\lt0.01$ whenever $0\lt|x-c|\lt\delta$. Here, $c=2$, $L=8$ and $f(x)=3x+2$. Therefore, $$|x-c|=|x-2|$$ and $$|f(x)-L|=|3x+2-8|=|3x-6|=3|x-2|$$ Since $|f(x)-L|\lt0.01$, that means $3|x-2|\lt0.01$ Therefore as we need to find $\delta$ for $0\lt|x-2|\lt\delta$, we can pick $\delta=\frac{0.01}{3}=\frac{1}{300}$. 3) Try $\delta=\frac{1}{300}$ back again For $0\lt|x-2|\lt\frac{1}{300}$, we have $$|f(x)-L|=3|x-2|\lt\Big(3\times\frac{1}{300}\Big)=\frac{1}{100}=0.01$$ So $\delta=\frac{1}{300}$ is a right choice.
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# Question #91306
Feb 6, 2017
$= 12 \sqrt{2} \times {m}^{3} {k}^{3}$
#### Explanation:
Note one of the laws of indices:
$\sqrt[q]{{x}^{p}} = {x}^{\frac{p}{q}}$
In words this will say " to find a root, divide the index by the root "
So $\sqrt{{x}^{10}} = {x}^{5} \text{ } \sqrt[3]{{x}^{9}} = {x}^{3}$
It is useful to write the numbers as the product of the prime factors.
$324 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 = {2}^{2} \times {3}^{4}$
$64 = {2}^{6}$
Use these factors instead of $324 \mathmr{and} 64$
$\sqrt[4]{324 {m}^{12}} \times \sqrt[3]{64 {k}^{9}}$
$= \sqrt[4]{{2}^{2} \times {3}^{4} \times {m}^{12}} \times \sqrt[3]{{2}^{6} \times {k}^{9}}$
Now divide each index by the root you want to find.
$= {2}^{\frac{1}{2}} \times \textcolor{b l u e}{3} \times {m}^{3} \times \textcolor{b l u e}{{2}^{2}} \times {k}^{3} \text{ } \leftarrow$ multiply the numbers
$= \textcolor{b l u e}{12} \sqrt{2} \times {m}^{3} {k}^{3} \text{ } \left({2}^{\frac{1}{2}} = \sqrt{2}\right)$
Hope this helps?
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# Solve the following :$\frac{-2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}$
Given :
The given expression is $\frac{-2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}$
To do :
We have to solve the expression
Solution :
Distributive Property:
The distributive property of multiplication states that when a factor is multiplied by the sum or difference of two terms, it is essential to multiply each of the two numbers by the factor, and finally perform the addition or subtraction operation.
This property is symbolically stated as:
$a (b+c) = a\times b + a \times c$
$a (b-c) = a \times b - a \times c$
Therefore,
$\frac{-2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}=\frac{-2}{3} \times \frac{3}{5} - \frac{3}{5} \times \frac{1}{6}+ \frac{5}{2}$
$= \frac{-3}{5} (\frac{2}{3} + \frac{1}{6}) + \frac{5}{2}$
$= \frac{-3}{5} (2\times2+\frac{1}{6}) + \frac{5}{2}$
[LCM of 3 and 6 is 6]
$= \frac{-3}{5} (\frac{5}{6}) + \frac{5}{2}$
$= \frac{-1}{1}(\frac{1}{2}) + \frac{5}{2}$
$= \frac{-1}{2} + \frac{5}{2}$
$= \frac{(5-1)}{2}$
$= \frac{4}{2}$
$= 2$.
The value of $\frac{-2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}$ is 2.
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## Math Trick 3: The Answer is Always 1089
This is the third part of the Math and Multimedia Math Trick Series. The first two tricks are multiplying by 11 and squaring numbers ending in 5.
As I have promised, I will teach you more math tricks that will impress your friends. The most exciting part, however, is not actually the trick but why the trick works. In this post, we are going to learn math trick which we will call magic 1089, a trick I learned at BasicMathematics.com.
#### The 1089 Math Trick
Step 1: Think of a 3-digit number where its digits are decreasing.
Step 2: Reverse the order of the digits.
Step 3: Subtract the number in step 2 from the number in step 1.
Step 4: Reverse the order of the difference in step 3.
Step 5: Add the numbers in step 3 and step 4.
The result is 1089. » Read more
## Math Puzzle: Where is The Missing Square?
If you have already found answer to the missing area puzzle, you might want to try the mysterious missing square below. The largest triangle in the first figure is made up of two smaller triangles and two L-shaped polygons.
The largest triangle below (well, technically it’s not really a triangle) is also made up of the same polygons mentioned above. » Read more
## Math Puzzle: Where is the Missing Area?
The square below is made up of two triangles and two trapezoids. The dimensions of the square is $13$ by $13$, which means that the area is $169$ square units.
The same polygons above are used to create the rectangle below. The dimensions of the rectangle is $8$ by $21$, which means that the area is $168$. » Read more
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## Why do we perceive mathematics as a difficult subject?
Math seems difficult because it takes time and energy. Many people don’t experience sufficient time to “get” math lessons, and they fall behind as the teacher moves on. Many move on to study more complex concepts with a shaky foundation. We often end up with a weak structure that is doomed to collapse at some point.
## What is the use of mathematics in daily life?
People use math knowledge when cooking. For example, it is very common to use a half or double of a recipe. In this case, people use proportions and ratios to make correct calculations for each ingredient. If a recipe calls for 2/3 of a cup of flour, the cook has to calculate how much is half or double of 2/3 of a cup.
## How is mathematics used in nature?
A few examples include the number of spirals in a pine cone, pineapple or seeds in a sunflower, or the number of petals on a flower. The numbers in this sequence also form a a unique shape known as a Fibonacci spiral, which again, we see in nature in the form of shells and the shape of hurricanes.
## What are the two types of pattern in nature in mathematics?
Symmetry – includes two types of patterns: radial and bilateral. Radial symmetry references the numerical symmetry referred to as the Fibonacci sequence (1, 2, 3, 5, 8, 13, 21, 34, 55, 89 . . .) If you counted the seeds within a sunflower, you would find the number of seeds is equal to a Fibonacci number.
## What is the importance of patterns?
Pattern is fundamental to our understanding of the world; it is an important element in every mathematics curriculum. The importance of patterns usually gets lost in a repeating pattern of two dimensional shapes. Patterns in mathematics are much more than a repeating pattern of shapes.
## What are patterns in maths?
In Mathematics, a pattern is a repeated arrangement of numbers, shapes, colours and so on. The Pattern can be related to any type of event or object. If the set of numbers are related to each other in a specific rule, then the rule or manner is called a pattern. Sometimes, patterns are also known as a sequence.
## How do you introduce a pattern?
There are many different levels to teaching and learning pattern skills, here’s the developmental sequence for teaching patterning skills to your Preschool or Pre-K students.
1. Stage 1: Recognize a pattern.
2. Stage 2: Describe a pattern.
3. Stage 3: Copy a pattern.
4. Stage 4: Extend a pattern.
5. Stage 5: Create a pattern.
## What is ABAB pattern?
The patterns are encoded by letters of the alphabet. Lines designated with the same letter rhyme with each other. For example, the rhyme scheme ABAB means the first and third lines of a stanza, or the “A”s, rhyme with each other, and the second line rhymes with the fourth line, or the “B”s rhyme together.
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Many household projects require that you be able to calculate area or volume. Whether you're painting your kid's room, mulching your flowerbed for the winter, planting grass seed in the spring, or tackling any other project for which you need to find out how much of something you need to cover, knowing how to figure area and volume is a time- and money-saver.
## Calculating area
To calculate area, you simply use this formula:
area = length × width
a = lw
From there, you can get the other information you need:
• Figuring how much seed you need: Use the area formula to calculate the area you want to cover and then calculate the number of pounds of seed you need. Seed companies indicate how much coverage a pound of seed gives you. Simply divide the area you need to cover by the amount of area that 1 pound of seed covers. For example, if you want to cover an area of 1,625 square feet, and 1 pound of seed covers 600 square feet, you need about 3 pounds of seed (1,625 divided by 600 is 2.7)
• Calculating area to be painted: You can calculate the area of each wall and then add the values together, or you can determine the perimeter of the room (add the length of each wall) and multiply that value by the room's height. A room that has a 42-foot perimeter and 8-foot ceilings has an area of 336 feet.
• Calculating how many square yards of carpeting you need: First calculate the area of the room and then convert that number from square feet to square yards by dividing the area by 9 (a square yard is 3 feet wide and 3 feet long).
## Calculating volume
When the area you want to cover has depth as well as height and width, you need to determine volume rather than area. Mulching a flowerbed is one such project: To protect your flowers, you need to spread a few inches of mulch over the whole bed.
To calculate volume, you use this formula:
volume = length × width × height
v = lwh
With that info, you can then do additional calculations to figure out how many bags of mulch to buy.
Say, for example, that you want to cover 6.68 cubic feet with mulch (that's the volume you need for a small bed, mulching to a depth of 3 inches). Simply divide the volume you need by the number of cubic feet in a bag of mulch. If each bag holds 3 cubic feet, you need 2.23 bags. So buy 3 to be safe.
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# Systems Using Substitution
## Solve for one variable, substitute the value in the other equation
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Solving Linear Systems with Substitution
Rex and Carl are making a mixture in science class. They need to have 12 ounces of a 60% saline solution. To make this solution they have a 20% saline solution and an 80% saline solution. How many ounces of each do they need to make the correct mixture?
### Guidance
In the substitution method we will be looking at the two equations and deciding which variable is easiest to solve for so that we can write one of the equations as x=\begin{align*}x=\end{align*} or y=\begin{align*}y=\end{align*}. Next we will replace either the x\begin{align*}x\end{align*} or the y\begin{align*}y\end{align*} accordingly in the other equation. The result will be an equation with only one variable that we can solve
#### Example A
Solve the system using substitution:
2x+y3x5y=12=11\begin{align*}2x+y &= 12\\ -3x-5y &= -11\end{align*}
Solution: The first step is to look for a variable that is easy to isolate. In other words, does one of the variables have a coefficient of 1? Yes, that variable is the y\begin{align*}y\end{align*} in the first equation. So, start by isolating or solving for y\begin{align*}y\end{align*}: y=2x+12\begin{align*}y=-2x+12\end{align*}
This expression can be used to replace the y\begin{align*}y\end{align*} in the other equation and solve for x\begin{align*}x\end{align*}:
3x5(2x+12)3x+10x607x607xx=11=11=11=49=7\begin{align*}-3x-5(-2x+12) &= -11\\ -3x+10x-60 &= -11\\ 7x-60 &= -11\\ 7x &= 49\\ x &= 7\end{align*}
Now that we have found x\begin{align*}x\end{align*}, we can use this value in our expression to find y\begin{align*}y\end{align*}:
yyy=2(7)+12=14+12=2\begin{align*} y &= -2(7)+12\\ y &= -14+12\\ y &= -2\end{align*}
Recall that the solution to a linear system is a point in the coordinate plane where the two lines intersect. Therefore, our answer should be written as a point: (7, -2). You can check your answer by substituting this point into both equations to make sure that it satisfies them:
2(7)+23(7)5(2)=142=12=21+10=11\begin{align*}2(7)+ -2 &= 14-2=12\\ -3(7)-5(-2) &= -21+10 = -11 \end{align*}
#### Example B
Solve the system using substitution:
2x+3yx+5y=13=4\begin{align*}2x+3y &= 13\\ x+5y &= -4\end{align*}
Solution: In the last example, y\begin{align*}y\end{align*} was the easiest variable to isolate. Is that the case here? No, this time, x\begin{align*}x\end{align*} is the variable with a coefficient of 1. It is easy to fall into the habit of always isolating y\begin{align*}y\end{align*} since you have done it so much to write equation in slope-intercept form. Try to avoid this and look at each system to see which variable is easiest to isolate. Doing so will help reduce your work.
Solving the second equation for x\begin{align*}x\end{align*} gives: x=5y4\begin{align*}x = -5y-4\end{align*}.
This expression can be used to replace the x\begin{align*}x\end{align*} in the other equation and solve for y\begin{align*}y\end{align*}:
2(5y4)+3y10y8+3y7y87yy=13=13=13=21=3\begin{align*}2(-5y-4)+3y &= 13\\ -10y-8+3y &= 13\\ -7y-8 &= 13\\ -7y &= 21\\ y &= -3\end{align*}
Now that we have found y\begin{align*}y\end{align*}, we can use this value in our expression to find x\begin{align*}x\end{align*}:
xxx=5(3)4=154=11\begin{align*}x &= -5(-3)-4\\ x &= 15-4\\ x &= 11\end{align*}
So, the solution to this system is (11, -3). Don’t forget to check your answer:
2(11)+3(3)11+5(3)=229=13=1115=4\begin{align*}2(11)+3(-3) &= 22-9=13\\ 11+5(-3) &= 11-15= -4 \end{align*}
#### Example C
Solve the system using substitution:
4x+3y6x2y=4=19\begin{align*}4x+3y &=4\\ 6x-2y &= 19\end{align*}
Solution: In this case, none of the variables have a coefficient of 1. So, we can just pick on to solve for. Let’s solve for the x\begin{align*}x\end{align*} in equation 1:
4xx=3y+4=34y+1\begin{align*}4x &= -3y +4\\ x &= - \frac{3}{4}y+1\end{align*}
Now, this expression can be used to replace the x\begin{align*}x\end{align*} in the other equation and solve for y\begin{align*}y\end{align*}:
6(34y+1)2y184y+62y92y42y132y(213)(132)yy=19=19=13=13=13(213)=2\begin{align*}6 \left(- \frac{3}{4}y+1 \right)-2y &= 19\\ - \frac{18}{4}y+6-2y &= 19\\ - \frac{9}{2}y- \frac{4}{2}y &= 13\\ - \frac{13}{2}y &= 13\\ \left(- \frac{2}{13} \right)\left(- \frac{13}{2} \right)y &= 13\left(- \frac{2}{13} \right)\\ y &= -2\end{align*}
Now that we have found y\begin{align*}y\end{align*}, we can use this value in our expression find x\begin{align*}x\end{align*}:
xxxx=(34)(2)+1=64+1=32+22=52\begin{align*}x &= \left(- \frac{3}{4} \right)(-2)+1\\ x &= \frac{6}{4}+1 \\ x &= \frac{3}{2}+\frac{2}{2}\\ x &= \frac{5}{2}\end{align*}
So, the solution is (52,2)\begin{align*}\left( \frac{5}{2}, -2 \right)\end{align*}. Check your answer:
4(52)+3(2)6(52)2(2)=106=4=15+4=19\begin{align*}4\left(\frac{5}{2}\right)+3(-2) &= 10-6=4\\ 6\left(\frac{5}{2}\right)-2(-2) &= 15+4 =19 \end{align*}
Intro Problem Revisit Let’s try to make the word problem easier by organizing our information into a “picture” equation as shown below:
In this picture, we can see that we will be mixing x\begin{align*}x\end{align*} ounces of the 20% solution with y\begin{align*}y\end{align*} ounces of the 80% solution to get 12 ounces of the 60% solution. The two equations are thus:
0.2x+0.8yx+y=0.6(12)=12\begin{align*}0.2x+0.8y &= 0.6(12)\\ x+y &= 12\end{align*}
Now we can solve the system using substitution. Solve for y\begin{align*}y\end{align*} in the second equation to get:y=12x\begin{align*}y = 12-x\end{align*}.
Now, substitute and solve in the first equation:
0.2x+0.8(12x)0.2x+9.60.8x0.6xx=0.6(12)=7.2=2.4=4\begin{align*}0.2x+0.8(12-x) &= 0.6(12)\\ 0.2x+9.6-0.8x &= 7.2\\ -0.6x &= -2.4\\ x &= 4\end{align*}
Now we can find y\begin{align*}y\end{align*}:
yyy=12x=124=8\begin{align*}y &= 12-x\\ y &= 12-4\\ y &= 8\end{align*}
Therefore, Rex and Carl need 4 ounces of the 20% saline solution and 8 ounces of the 80% saline solution to make the correct mixture.
### Guided Practice
Solve the following systems using the substitution method.
1. 3x+4yx=13=2y9\begin{align*}3x+4y &= -13\\ x &=-2y-9\end{align*}
2. 2x5yx+3y=39=24\begin{align*}-2x-5y &= -39\\ x+3y &=24\end{align*}
3. yy=12x21=2x+9\begin{align*}y &= \frac{1}{2}x-21\\ y &= -2x+9\end{align*}
1. In this problem, the second equation is already solved for \begin{align*}x\end{align*} so we can use that in the first equation to find \begin{align*}y\end{align*}:
\begin{align*}3(-2y-9)+4y &= -13\\ -6y-27+4y &= -13\\ -2y-27 &= -13\\ -2y &= 14\\ y &= -7\end{align*}
Now we can find \begin{align*}x\end{align*}:
\begin{align*}x &= -2(-7)-9\\ x &= 14-9\\ x &= 5\end{align*}
Therefore the solution is (5, -7).
2. This time the \begin{align*}x\end{align*} in the second equation is the easiest variable to isolate: \begin{align*}x=-3y+24\end{align*}. Let’s use this in the first expression to find \begin{align*}y\end{align*}:
\begin{align*}-2(-3y+24)-5y &= -39\\ 6y-48-5y &= -39\\ y-48 &= -39\\ y &=9\end{align*}
Now we can find \begin{align*}x\end{align*}:
\begin{align*}x &= -3(9)+24\\ x &= -27+24\\ x &= -3\end{align*}
Therefore the solution is (-3, 9).
3. In this case, both equations are equal to \begin{align*}y\end{align*}. Since \begin{align*}y=y\end{align*}, by the Reflexive Property of Equality, we can let the right hand sides of the equations be equal too. This is still a substitution problem; it just looks a little different.
\begin{align*}\frac{1}{2}x-21 &= -2x+9\\ 2 \left( \frac{1}{2}x-21 \right . &= \left . -2x+9 \right )\\ x-42 &= -4x+18\\ 5x &= 60\\ x &= 12\end{align*}
Now we can find \begin{align*}y\end{align*}:
\begin{align*}y &= \frac{1}{2}(12)-21 \qquad \quad \quad y = -2(12)+9\\ y &= 6-21 \qquad \quad or \qquad y = -24+9\\ y &= -15 \qquad \qquad \qquad \quad y = -15\end{align*}
Therefore our solution is (12, -15).
### Practice
1. .
\begin{align*}x+3y &= -1\\ 2x+9y &= 7\end{align*}
1. .
\begin{align*}7x+y &= 6\\ x-2y &= -12\end{align*}
1. .
\begin{align*}5x+2y &= 0\\ y &= x-7\end{align*}
1. .
\begin{align*}2x-5y &= 21\\ x &= -6y +2\end{align*}
1. .
\begin{align*}y &= x+3\\ y &= 2x-1\end{align*}
1. .
\begin{align*}x+6y &= 1\\ -2x-11y &= -4\end{align*}
1. .
\begin{align*}2x+y &= 18\\ -3x+11y &= -27\end{align*}
1. .
\begin{align*}2x+3y &= 5\\ 5x+7y &= 8\end{align*}
1. .
\begin{align*}-7x+2y &= 9\\ 5x-3y &= 3\end{align*}
1. .
\begin{align*}2x-6y &= -16\\ -6x+10y &= 8\end{align*}
1. .
\begin{align*}2x-3y &= -3\\ 8x+6y &= 12\end{align*}
1. .
\begin{align*}5x+y &= -3\\ y &= 15x+9\end{align*}
Set up and solve a system of linear equations to answer each of the following word problems.
1. Alicia and Sarah are at the supermarket. Alicia wants to get peanuts from the bulk food bins and Sarah wants to get almonds. The almonds cost $6.50 per pound and the peanuts cost$3.50 per pound. Together they buy 1.5 pounds of nuts. If the total cost is $6.75, how much did each girl get? Set up a system to solve using substitution. 2. Marcus goes to the department store to buy some new clothes. He sees a sale on t-shirts ($5.25) and shorts ($7.50). Marcus buys seven items and his total, before sales tax, is$43.50. How many of each item did he buy?
3. Jillian is selling tickets for the school play. Student tickets are $3 and adult tickets are$5. If 830 people buy tickets and the total revenue is \$3104, how many students attended the play?
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Unformatted text preview: oldhomewk 07 – YOO, HEE – Due: Jan 31 2008, 4:00 am 1 Question 1, chap 3, sect 3. part 1 of 1 10 points Four vectors, each of magnitude 89 m, lie along the sides of a parallelogram as shown in the figure. The angle between vector A and B is 59 ◦ . Hint: Remember you are adding four vec- tors. A 8 9 m C 89 m B 89 m D 8 9 m 59 ◦ What is the magnitude of the vector sum of the four vectors? Correct answer: 309 . 847 m (tolerance ± 1 %). Explanation: α = 59 ◦ and ℓ = 89 m . Basic Concepts: The components of a vector vector R = vector A + vector B + vector C + vector D are obtained by adding up the respective com- ponents of each vector in the sum. Solution: vector B = vector C = (89 m) ˆ ı vector A = vector D = (89 m) (cos A ) ˆ ı + (89 m)(sin α ) ˆ Thus, vector A + vector B + vector C + vector D = 2 [ ℓ ı + ℓ cos α ˆ ı ] + 2 [ ℓ sin α ˆ ] = 2 [(89 m) ı + (89 m) cos(59 ◦ ) ˆ ı ] + 2 [(89 m) sin(59 ◦ ) ˆ ] = (269 . 677 m) ı + (152 . 576 m) ˆ , where R x = 269 . 677 m and R y = 152 . 576 m . The magnitude of the vector sum is R = radicalBig R 2 x + R 2 y = radicalBig (269 . 677 m) 2 + (152 . 576 m) 2 = 309 . 847 m . Alternative Solution: Rotate your figue clockwise by one-half the angle between vec- tors A and B . The resulant vector must lie along the horizontal line connecting the tails of vectors A and B and the heads of vectors C and D . The four components along this line are ℓ times the cosine of the half angle. R = 4 ℓ cos parenleftBig α 2 parenrightBig = 4 (89 m) cos (29 . 5 ◦ ) = 309 . 847 m . Question 2, chap 3, sect 3. part 1 of 1 10 points A novice golfer on the green takes three strokes to sink the ball. The successive dis- placements are 5 . 2 m to the north, 8 . 4 m northeast, and 3 . 6 m 71 ◦ west of south. Start- ing at the same initial point, an expert (lucky) golfer could make the hole in a single displace- ment. What is the magnitude of this single dis- placement? Correct answer: 10 . 2852 m (tolerance ± 1 %). Explanation: Take the sum of these three displacements: d = d 1 + d 2 + d 3 = d x ˆ ı + d y ˆ where d x = d 2 cos 45 ◦ − d 3 sin30 ◦ d y = d 1 + d 2 sin45 ◦ − d 3 cos 30 ◦ oldhomewk 07 – YOO, HEE – Due: Jan 31 2008, 4:00 am 2 so, the magnitude of d is: | d | = radicalBig d 2 x + d 2 y Question 3, chap 3, sect 3....
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# Warm-Up: December 15, 2011 Divide and express the result in standard form.
## Presentation on theme: "Warm-Up: December 15, 2011 Divide and express the result in standard form."— Presentation transcript:
Warm-Up: December 15, 2011 Divide and express the result in standard form
Homework Questions?
Quadratic Functions Section 2.2
Quadratic Functions A quadratic function is any function that can be written in the form The graph of a quadratic function is a parabola. Every parabola has a vertex at either its minimum or its maximum. Every parabola has a vertical axis of symmetry that intersects the vertex.
Example Graphs Vertex Axis of Symmetry
Standard Form of a Quadratic Function Vertex is at (h, k) Axis of symmetry is the line x=h If a>0, the parabola opens upward, U If a<0, the parabola opens downward,
Graphing Quadratics in Standard Form 1. Determine the vertex, (h, k) 2. Find any x-intercepts by replacing f(x) with 0 and solving for x 3. Find the y-intercept by replacing x with 0 4. Plot the vertex, axis of symmetry, and y-intercepts and connect the points. Draw a dashed vertical line for the axis of symmetry. 5. Check the sign of “a” to make sure your graph opens in the right direction.
Example 1 Graph the quadratic function. Give the equation of the parabola’s axis of symmetry. Determine the graph’s domain and range.
You-Try #1 Graph the quadratic function. Give the equation of the parabola’s axis of symmetry. Determine the graph’s domain and range
Graphing Quadratics in General Form General form is The vertex is at x-intercepts can be found by quadratic formula (or sometimes by factoring and zero product property) y-intercept is at (0, c) Graph the parabola using these points just as we did before.
Example 3 Graph the quadratic function. Give the equation of the parabola’s axis of symmetry. Determine the graph’s domain and range
You-Try #3 Graph the quadratic function. Give the equation of the parabola’s axis of symmetry. Determine the graph’s domain and range
Minimum and Maximum Consider If a>0, then f has a minimum If a<0, then f has a maximum The maximum or minimum occurs at The maximum or minimum value is
Example 4 (Page 266 #44) A football is thrown by a quarterback to a receiver 40 yards away. The quadratic function models the football’s height above the ground, s(t), in feet, when it is t yards from the quarterback. How many yards from the quarterback does the football reach its greatest height? What is that height?
You-Try #4 (Page 266 #43) Fireworks are launched into the air. The quadratic function models the fireworks’ height, s(t), in feet, t seconds after they are launched. When should the fireworks explode so that they go off at the greatest height? What is that height?
Assignment Page 264 #1-8 ALL (use your graphing calculator for 5-8), #9-41 Every Other Odd
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# Emma's Dilemma
Extracts from this document...
Introduction
Emma’s Dilemma
The Problem
Emma is looking at the different arrangements of her name.
1) Find the different arrangements of Emma’s name.
2) Emma has a friend named Lucy; find the different arrangements of her name.
3) Investigate the number of different arrangements of the letters of names you have chosen.
Investigating the Problem
To investigate this problem I am going to look at the different arrangements of a four letter name with two letters the same and a four letter name with all different letters. I am then going to try three and five letter names, with two letters the same and all different. I am then going to look at the results and predict the next set. I will then move on to three and four letters the same.
Four Letter Words
One/Two Letter Words
We know that a one letter word will only have 1 combination. For two letters we know that it only has 2 different combinations and 1 for both letters repeating:
1. A
1. AB
2. BA
1. AA
Three Letter Words
Five Letter Words
James is a five letter word with different letters.
...read more.
Middle
5
120
60
From the results we can already see a pattern. The combination for all different letters divided by two equals that of two letters the same in a word.
Prediction
By looking at the results I have found a pattern. The table below highlights the pattern.
1 1 - 2 2 1 3 6 3 4 24 12 5 120 60
To predict the result for 6 letters:
120 X 6 = 720
To check this:
Let’s take ABCDEF. We know that by putting the A at the front of the word, we are only changing a five letter word. The results for a five letter word with different letters in 120.
120 X 6 = 720
This is the same sum as the pattern that I found. So, we can predict the results for the rest of the numbers by using this rule.
Factorial
The factorial button on a calculator [ ! ] proved very helpful in finding out the rule to Emma’s Dilemma.
Factorial means multiplying itself by all the numbers lower than itself until reaching 1.
e.g. 3! = 3 X 2 X 1 = 6
4! = 4 X 3 X 2 X 1 = 24
By using the factorial rule, the number of combinations is easier to work out.
...read more.
Conclusion
( n ! / 2 ) / 3 simplified n!/6 because 2 X 3 = 6
I predict that for four letters the same the formula would be:
( ( n ! / 2) / 3 ) / 4 simplified n!/24 because 2 X 3 X 4 = 24
The Formula
There is in fact a link between the formulas shown above where you divide n! by 2, 6, 24 e.t.c. and the factorial numbers itself. By looking at the numbers we can see that 2, 6, 24 is in fact 2!, 3!, 4! and so on.
( n ! / 2 ) / 3 = n!/6 = n!/2!
( ( n ! / 2) / 3 ) / 4 = n!/24 = n!/3!
Let’s put this in a table:
n n!/1! n!/2! n!/3! Letters All different 2 the same 3 the same
Notice that the 2! or 3! that n! is divided by equates to the number of letters the same in each word. By using the information on this page we can find a general formula for the ‘dilemma’.
N!
L!
Where…
N is the number of letters in the word;
L is the number of letters the same.
This formula is a general rule for all combinations containing any number of letters with a number of letters the same (excluding combination of letters that have more than one letter repeated, i.e. AABB cannot use this formula).
...read more.
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# Related GCSE Emma's Dilemma essays
1. ## Emma's Dilemma
I created a table but instead replaced the starting letter i.e. I replaced the L for E, U for M, C for M and Y for A. the diagram below displays what I mean: L U C Y E M M A letter 1 letter 2 letter 3 letter 4
2. ## Emma's Dilemma
= 6 x 5 x 4 x 3 x 2 x 1 = 720 Working out the different arrangements for words with repeated letters is different to words with no repeated letters. I predict that words with 1 letter repeated twice will have fewer arrangements than words with no repeated letters.
1. ## Emma's Dilemma
A 1 POSSIBILITY There was only 1 possibility and hence it was very easy to arrange the word. I did not need to split it up either as it was very straight forward. I will now be investigating how many possible outcomes there are for a 2 lettered word, with no repeats.
2. ## Emma's Dilemma
If I make the double- letter different then there will be the same amount of arrangements as Ben. I will now investigate two-letter names. The first name I will investigate is TJ. TJ JT I have found 2 arrangements for the name TJ.
1. ## Emma's Dilemma
There were 6 different arrangements of the letters in that too so the results above are credible: DOM ODM MDO DMO OMD MOD There seemed to be a pattern developing but there was not yet enough evidence to make it conclusive.
2. ## Emma's Dilemma
Now I am going to investigate the number of arrangements in Emma's name: 1. EAMM 2. EMAM 3. EMMA 4. MEAM 5. MMEA 6. MMAE 7. MAEM 8. MAME 9. AMME 10. AEMM 11. AMEM 12. MEMA To work this out without writing all the combinations down I will have
1. ## Emma's Dilemma Question One: Investigate the number of different arrangements of the letters
This means that the total number of arrangements possible is reduced by different amounts, according to the total number of letters, the number of repeated letters, and the number of times each one is repeated. For example For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
2. ## I have been given a problem entitled 'Emma's Dilemma' and I was given the ...
/ 3! = The Number of permutations and 5! / 3! = 150 / 6 = 25. This is the wrong answer as there were only 10 arrangements. So again my previous equation is proven wrong. AAABBB: This word is made up of six letters, including two sets of three repeated letters.
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MCQs Questions for Class 9 Maths Chapter 13 Surface Areas and Volumes
# MCQs Questions for Class 9 Maths Chapter 13 Surface Areas and Volumes
## MCQs Questions for Class 9 Maths Chapter 13 Surface Areas and Volumes
In this 21st century, Multiple Choice Questions (MCQs) play an important role to prepare for a competitive examination. Now-a-days, MCQs questions are asked in CBSE board examinations as well. In most of the competitive examinations, only MCQ Questions are asked.
In future, if you want to prepare for competitive examination, then you should focus on the MCQs questions. Thus, let’s solve these MCQs questions to make our foundation strong.
In this post, you will find 20 MCQs questions for class 9 maths chapter 13 surface area and volume.
## MCQs Questions for Class 9 Maths Chapter 13 Surface Areas and Volumes
1. The formula to find surface area of a cuboid of length (l), breadth (b) and height (h) is:
(a) lb + bh + hl
(b) 2(lb + bh + hl)
(c) 2(lbh)
(d) lbh/2
2. If the perimeter of one of the faces of a cube is 40 cm, then its volume is:
(a) 6000 cm³
(b) 1600 cm³
(c) 1000 cm³
(d) 600 cm³
## 3. If slant height of the cone is 21 cm and diameter of base is 24 cm. The total surface area of cone is:
(a) 1200.77 sq. cm
(b) 1177 sq. cm
(c) 1222.77 sq. cm
(d) 1244.57 sq. cm
4. The radius of a cylinder is doubled and the height remains the same. The ratio between the volumes of the new cylinder and the original cylinder is
(a) 1 : 2
(b) 3 : 1
(c) 4 : 1
(d) 1 : 8
5. The surface area of a cube whose edge equals to 3 cm is:
(a) 62 sq. cm
(b) 30 sq. cm
(c) 54 sq. cm
(d) 90 sq. cm
## 6. A rectangular sand box is 5 m wide and 2 m long. How many cubic metres of sand are needed to fill the box up to a depth of 10 cm?
(a) 1
(b) 10
(c) 100
(d) 1000
7. The length of diagonals of a cube of side a cm is
(a) √2a cm
(b) √3a cm
(c) 3a cm
(d) 1 cm
8. The surface area of cuboid-shaped box having length = 80 cm, breadth = 40 cm and height = 20 cm is:
(a) 11200 sq. cm
(b) 13000 sq. cm
(c) 13400 sq. cm
(d) 12000 sq. cm
## 9. In a cylinder, radius is doubled and height is halved, curved surface area will be
(a) halved
(b) doubled
(c) same
(d) four time
10. Volume of hollow cylinder
(a) Ï€(R² – r²)h
(b) Ï€R²h
(c) Ï€r²h
(d) Ï€r²(h1 – h1)
11. The volume of hemisphere whose radius is r, is:
(a) 4/3 πr3
(b) 4Ï€r3
(c) 2Ï€r3
(d) ⅔ Ï€r3
## 12. The length of the longest pole that can be put in a room of dimension (10 m × 10 m × 5 m) is
(a) 15 m
(b) 16 m
(c) 10 m
(d) 12 m
13. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 2 : 1
14. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. The diameter of the base is:
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 6 cm
15. The diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. The curved surface area is:
(a) 150 sq. cm
(b) 165 sq. cm
(c) 177 sq. cm
(d) 180 sq. cm
16. The lateral surface area of a cube is 256 m³. The volume of the cube is
(a) 512 m³
(b) 64 m³
(c) 216 m³
(d) 256 m³
17. If slant height of the cone is 21 cm and diameter of base is 24 cm, the the total surface area of cone is:
(a) 1200.77 sq. cm
(b) 1177 sq. cm
(c) 1222.77 sq. cm
(d) 1244.57 sq. cm
18. A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a shape. The radius of the sphere is
(a) 4.2 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 1.6 cm
19. The surface area of a sphere of radius 14 cm is:
(a) 1386 sq. cm
(b) 1400 sq. cm
(c) 2464 sq. cm
(d) 2000 sq. cm
|
## HR. CardColm
Colm Mulcahy, of Spelman College in Atlanta, joins us to share his ice cream trick from his CardColm mathematical card trick column on the MAA website! You’re invited to explain how this works in the comments below.
Colm also shares a quick puzzle, tweeted on his What Would Martin Gardner Tweet feed @WWMGT. And finally we touch on the Gathering For Gardner and the Celebration of Mind, held all over the world around the time of Martin Gardner’s birthday, October 21.
And at last we get around to answering our quiz from a few weeks ago. There are indeed two solutions for correctly filling in the blanks in:
The number of 1’s in this paragraph is ___; the number of 2’s is ___; the number of 3’s is ____; and the number of 4’s is ___.
[spoiler] namely (3,1,3,1) and (2,3,2,1) [/spoiler]
We can vary this puzzle at will, asking
The number of 1’s in this paragraph is ___; the number of 2’s is ___; ….. and the number of N’s is ___.
For N=2 or 3, there are no solutions (Asking that all the numbers we fill in are between 1 and N); for N=4 there are two. For N=5 there is just one, for N=6 there are none and beyond that there is just one. I think we’ll let the commenters explain that.
But here’s the cool thing.
One way to approach the problem is to try filling in any answer at all, and then counting up what we have, filling that in, and repeating. Let’s illustrate, but first stipulate that we’ll stick with answers that are at least plausible– you can see that the total of all the numbers we fill in the blanks has to be 2N (since there are 2N total numbers in the paragraph).
So here’s how this works. Suppose our puzzle is:
There are ___ 1’s;___ 2’s; ___ 3’s; ___ 4’s; ___ 5’s
Let’s pick a (bad) solution that totals 10, say, (2,4,1,2,1). So we fill in:
There are __2_ 1’s; __4_ 2’s; _1__ 3’s; __2_ 4’s; _1__ 5’s
That’s pretty wrong! There are actually three 1’s in that paragraph, three 2’s; at least there is just one 3, and two 4’s and one 5. In any case this gives us another purported solution to try: (3,3,1,2,1). Let’s fill that in:
There are __3_ 1’s; __3_ 2’s; _1__ 3’s; __2_ 4’s; _1__ 5’s
That attempt actually does have three 1’s; but has only two 2’s; it does have three 3’s but only one 4 and one 5. So let’s try (3,2,3,1,1):
There are __3_ 1’s; __2_ 2’s; _3__ 3’s; __1_ 4’s; _1__ 5’s
Lo and behold that works! We do in fact have three 1’s; two 2’s; three 3’s and yes, one 4 and one 5.
So we can think of it this way: filling in a purported solution and reading off what we actually have gives another purported solution.
In this case (2,4,1,2,1) -> (3,3,1,2,1) -> (3,2,3,1,1) -> (3,2,3,1,1) etc,
We can keep following this process around, and if we ever reach a solution that gives back itself, we have a genuine answer, as we did here.
So here’s an interesting thing to think about.
First, find, for N>=7, a correct solution; and a pair of purported solutions A,B that cycle back and forth A->B->A->B etc.
Second, find a proof that this is all that can happen (unless I’m mistaken)– any other purported solution eventually leads into the correct one or that cycle.
## HM. Five Cards
Let’s see: First, the “Big News“, a discussion of Carlos May, and another puzzle (a pretty easy one)
And still more 2012 facts! From Primepuzzles.net, we learn that
2012 = (1+2-3+4)*(5-6+7*8*9)
and there’s still more amazing stuff there that we didn’t try to read on the air.
## Yoak: Pirate Treasure Map
Our band of intrepid pirates, having resolved previous squabbles over distributing booty amongst themselves and other issues have come across a treasure map fragment. The picture has been destroyed, but the following text can be read:
Stand upon the gravesite and you’ll see two great palms towering above all others on the island. Count paces to the tallest of them and turn 90 degrees clockwise and count the same number of paces and mark the spot with a flag. Return to the gravesite and count paces to the second-tallest of the trees, turn 90 degrees counter-clockwise and count off that number of paces, marking the spot with a second flag. You’ll find the treasure at the mid-point between the two flags.
Fortunately, our pirates knew which island the map referred to. Sadly, upon arriving at the island, the pirates discovered that all evidence of a gravesite had faded. The captain was preparing to order his men to dig up the entire island to find the fabled treasure when one of the more geometrically inclined pirates walked over to a particular spot and began to dig. The treasure was quickly unearthed on that very spot.
How did the pirate know where to dig?
## GL. Math 2033
### Standard Podcast Play Now | Play in Popup | DownloadpodPressShowHidePlayer('3', 'http://mathfactor.uark.edu/podpress_trac/play/893/0/167%20Math%202033%20_Math_Factor_2009_11_19.mp3', 290, 24, 'false', 'http://mathfactor.uark.edu/wp-content/plugins/podpress//images/vpreview_center.png', 'Standard Podcast', 'The Math Factor Podcast');
So, I’m teaching a new course, Math 2033, Mathematical Thought, and it’s going great! I’d like to take a moment to write about it!
(This is one reason the MF has been kinda slow lately; another is that I’m chair) When it’s fully up and running, we’ll have about 150 students in one large section each semester (we’re starting with about 100). In a nutshell, it’s the Math Factor, as a course.
## Yoak: Lewis Carroll – Some Chance I’m Being Obtuse
This will be the last of my Lewis Carroll posts. In Pillow Problems, Carroll writes:
Three Points are taken at random on an infinite Plane. Find the chance of their being the vertices of an obtuse-angled Triangle.
Note: An obtuse-angled triangle is one that has an angle measuring more than 90 degrees.
## Harriss: Mathematical Sculpture
Strange appearance in the North Atlantic:
## Yoak: A Rather Odd Car Trip
Here’s a puzzle that sounds a little like those, “A train leaves…” questions we were all prepared for but rarely saw on the SAT, but with a twist.
You are going to take a drive from City A to City B and back, but in a rather unusual car. When travelling uphill, the car always moves at exactly 56 miles per hour. On level ground, it travels at 63 miles per hour and finally when travelling downhill it travels at 72 miles per hour. Assume that it transitions from one speed to another instantaneously and all of those other “mathematically perfect” qualities that make questions like this answerable.
You find that travelling from City A to City B takes exactly 4 hours of travel time. On the return trip, driving time sums to 4 hours and 40 minutes.
How far apart are Cities A and B?
## Harriss: Rabbit Sequence
There has been a theme in some of the recent posts and problems. It’s a little buried but almost enough to say its another of those Mathfactor agendas when we try to sneak some knowledge to you buried in the fun. Never one to miss such an opportunity I will jump in with a post, and a problem. This is a slight change to a classic problem that comes out of the work of one of my mathematical heroes: Leonardo of Pisa, also known as Fibonacci. He is responsible for changing how we count! Not many people can claim that. He introduced the system base value, also known as Arabic numerals that we still use today into Europe. He is more famous however for talking about rabbits:
Imagine that you have immortal rabbits, Bugs Bunny’s version of Olympus perhaps. Even if they are immortal however rabbits are famous for one thing. They breed like, well rabbits. Some of the rabbits are children and some adults and are divided into pairs. Each month any child pairs become adults and any adult pairs breed to produce a new child pair. They are immortal so no pair ever dies. These rabbits are also a little odd. They live on a line (don’t complain, this is no more ludicrous than that they are immortal!), but can shuffle along. Also if you are worried about inbreeding, the male rabbit leave the family hutch and shuffle along the line past others until they find a suitably unrelated mate. Why we would be worried about inbreeding in immortal rabbits living on a line escapes me!
Anyway we start with one pair of children. Lets put a c. After a month they become adults, a. Another month passes and they now have a pair of children, but are still there themselves. We therfore have the original pair and a pair of children: ac. Next month the adults have another pair of children and the children become adults: aca. Can you see how this will work? Each month the children become adults so we replace every c with and a, each pair of adults has a new pair of children but stays as adults, so we replace every a with ac. We can continue to get longer and longer sequences of rabbits on this line:
aca to acaac to acaacaca to acaacacaacaac….
Now some puzzles. Given a line with 21 adult pairs and 13 child pairs, how many pairs of adults and children would there be after one month?
Given p adults and q children how many adults and children will there be after one month?
Finally a more difficult one. How will the ratio of adults to children behave month on month? Will it
a) Get closer and closer to a particular number?
b) Keep on changing without pattern?
In either case can you say more?
## Yoak: Mountain Climbing
Here’s a quick puzzle that I don’t think has appeared here previously.
A man leaves at exactly 6:00 AM to climb a mountain. He may not climb at a constant rate. In fact, he may stop to rest, or even backtrack a bit to see something interesting. He arrives at the summit at exactly 6:00 PM and camps for the night.
The next day, he starts down at exactly 6:00 AM, again meandering unevenly, but following exactly the same path and reaching the bottom at exactly 6:00 PM.
The question is, was there some point on the path he was sure to have visited at exactly the same time on the two trips?
## Harriss: Algebraic Surfaces
I have just published a (rather long) article on mathematical surfaces, their models and links to art over at Maxwell’s Demon. Here is a sneak preview.
Minimal Möbius, Benjamin Storch
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# Ambiguous case of the law of sines
The ambiguous case of the law of sines happens when two sides and an angle opposite one of the two sides are given. We can shorten this situation with SSA.
Since the length of the third side is not known, we don't know if a triangle will be formed or not. That is the reason we call this case ambiguous.
In fact, this kind of situation or SSA can give the following 4 scenarios.
• No triangle
• 1 triangle
• 1 right triangle
• 2 triangles
## The first scenario of the ambiguous case of the law of sines occurs when a < h.
For example, take a look at the triangle below where only two sides are given. These two given sides are a and b. An angle opposite to one side is also given. Angle A is the angle that is opposite to side a or one of the two sides.
Note that SSA in this case means side a side b angle A in that order.
Because a is shorter than h, a is not long enough to form a triangle. In fact, the number of possible triangles that can be formed in the SSA case depends on the length of the altitude or h.
Notice that sin A =
h / b
Once you multiply both sides of the equation above by b, we get h = b sin A.
An example showing that no triangle can be formed
Method #1
Suppose A = 74°, a = 51, and b = 72.
h = 72 × sin (74°) = 72 × 0.9612 = 68.20
Since 51 or a is less than h or 69.20, no triangle will be formed.
Method #2
We can also show that no triangle exists by using the law of sines.
a / sin A = b / sin B
The ratio a / sin A is known since a / sin A = 51 / sin 74°
Since we also know the length of b, the missing quantity in the law of sines is sin B. It is logical then to look for sin B and see what we end up with.
51 / sin 74° = 72 / sin B
51 sin B = 72 sin 74°
sin B = (72 sin 74°) / 51
sin B = (72 × 0.9612) / 51
sin B = (69.2064) / 51
sin B = 1.3569
Since the sine of an angle cannot be bigger than 1, angle B does not exist. Therefore, no triangle can be formed with the given measurements.
## The second scenario of the ambiguous case of the law of sines occurs when a = h.
When a = h, the resulting triangle will always be a right triangle.
An example showing that a right triangle can be formed
Method #1
Suppose A = 30°, a = 25, and b = 50.
h = 50 × sin (30°) = 50 × 0.5 = 25
Since 25 or a is equal to h or 25, 1 right triangle will be formed.
Method #2
Again, we can use the law of sines to show that this time sin B exists and it is equal to 90 degrees.
a / sin A = b / sin B
25 / sin 30° = 50 / sin B
25 sin B = 50 sin 30°
sin B = (50 sin 30°) / 25
sin B = (50 × 0.5) / 25
sin B = (25) / 25
sin B = 1
B = sin-1(1) = 90 degrees.
## The third scenario of the ambiguous case of the law of sines occurs when a > h and a > b.
When a is bigger than h, again a triangle can be formed. However, since a is bigger than b, we can only have one triangle. Try to make a triangle where a is bigger than b, you will notice that there can only be 1 such triangle.
An example showing that exactly 1 triangle can be formed
Suppose A = 30°, a = 50, and b = 40.
h = 40 × sin (30°) = 40 × 0.5 = 20
Since 50 or a is bigger than both h (or 20) and b (or 40), 1 triangle will be formed.
## Last situation: a > h and a < b
When a is less than b, 2 triangles can be formed as clearly illustrated below. The two triangles are triangle ACD and triangle AED.
An example showing that exactly 2 triangles can be formed
Suppose A = 30°, a = 40, and b = 60
h = 60 × sin (30°) = 60 × 0.5 = 30
Since 40 or a is bigger than h and a is smaller than b or 60, 2 triangles will be formed.
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# 0.11 Independent and dependent events (Page 2/2)
Page 2 / 2
## Use of a venn diagram
We can also use Venn diagrams to check whether events are dependent or independent.
Independent events
Events are said to be independent if the result or outcome of one event does not affect the result or outcome of the other event. So P(A/C)=P(A), where P(A/C) represents the probability of event A after event C has occured.
Dependent events
Two events are dependent if the outcome of one event is affected by the outcome of the other event i.e. $P\left(A/C\right)\ne P\left(A\right)$
. Also note that $P\left(A/C\right)=\frac{P\left(A\cap C\right)}{P\left(C\right)}$ . For example, we can draw a Venn diagram and a contingency table to illustrate and analyse the following example.
A school decided that its uniform needed upgrading. The colours on offer were beige or blue or beige and blue. 40% of the school wanted beige, 55% wanted blue and 15% said a combination would be fine. Are the two events independent?
1. Beige Not Beige Totals Blue 0,15 0,4 0,55 Not Blue 0,25 0,2 0,35 Totals 0,40 0,6 1
2. P(Blue)=0,4, P(Beige)=0,55, P(Both)=0,15, P(Neither)=0,20
Probability of choosing beige after blue is:
$\begin{array}{ccc}\hfill P\left(Beige/Blue\right)& =& \frac{P\left(Beige\cap Blue\right)}{P\left(Blue\right)}\hfill \\ & =& \frac{0,15}{0,55}\hfill \\ & =& 0,27\hfill \end{array}$
3. Since $P\left(Beige/Blue\right)\ne P\left(Beige\right)$ the events are statistically dependent.
## Applications of probability theory
Two major applications of probability theory in everyday life are in risk assessment and in trade on commodity markets. Governments typically apply probability methods in environmental regulation where it is called “pathway analysis”, and are often measuring well-being using methods that are stochastic in nature, and choosing projects to undertake based on statistical analyses of their probable effect on the population as a whole. It is not correct to say that statistics are involved in the modelling itself, as typically the assessments of risk are one-time and thus require more fundamental probability models, e.g. “the probability of another 9/11”. A law of small numbers tends to apply to all such choices and perception of the effect of such choices, which makes probability measures a political matter.
A good example is the effect of the perceived probability of any widespread Middle East conflict on oil prices - which have ripple effects in the economy as a whole. An assessment by a commodity trade that a war is more likely vs. less likely sends prices up or down, and signals other traders of that opinion. Accordingly, the probabilities are not assessed independently nor necessarily very rationally. The theory of behavioral finance emerged to describe the effect of such groupthink on pricing, on policy, and on peace and conflict.
It can reasonably be said that the discovery of rigorous methods to assess and combine probability assessments has had a profound effect on modern society. A good example is the application of game theory, itself based strictly on probability, to the Cold War and the mutual assured destruction doctrine. Accordingly, it may be of some importance to most citizens to understand how odds and probability assessments are made, and how they contribute to reputations and to decisions, especially in a democracy.
Another significant application of probability theory in everyday life is reliability. Many consumer products, such as automobiles and consumer electronics, utilize reliability theory in the design of the product in order to reduce the probability of failure. The probability of failure is also closely associated with the product's warranty.
## End of chapter exercises
1. In each of the following contingency tables give the expected numbers for the events to be perfectly independent and decide if the events are independent or dependent.
1. Brown eyes Not Brown eyes Totals Black hair 50 30 80 Red hair 70 80 150 Totals 120 110 230
2. Point A Point B Totals Busses left late 15 40 55 Busses left on time 25 20 45 Totals 40 60 100
3. Durban Bloemfontein Totals Liked living there 130 30 160 Did not like living there 140 200 340 Totals 270 230 500
4. Multivitamin A Multivitamin B Totals Improvement in health 400 300 700 No improvement in health 140 120 260 Totals 540 420 960
2. A study was undertaken to see how many people in Port Elizabeth owned either a Volkswagen or a Toyota. 3% owned both, 25% owned a Toyota and 60% owned a Volkswagen. Draw a contingency table to show all events and decide if car ownership is independent.
3. Jane invested in the stock market. The probability that she will not lose all her money is 0,32. What is the probability that she will lose all her money? Explain.
4. If D and F are mutually exclusive events, with P(D ${}^{\text{'}}$ )=0,3 and P(D or F)=0,94, find P(F).
5. A car sales person has pink, lime-green and purple models of car A and purple, orange and multicolour models of car B. One dark night a thief steals a car.
1. What is the experiment and sample space?
2. Draw a Venn diagram to show this.
3. What is the probability of stealing either a model of A or a model of B?
4. What is the probability of stealing both a model of A and a model of B?
6. The probability of Event X is 0,43 and the probability of Event Y is 0,24. The probability of both occuring together is 0,10. What is the probability that X or Y will occur (this inculdes X and Y occuring simultaneously)?
7. P(H)=0,62, P(J)=0,39 and P(H and J)=0,31. Calculate:
1. P(H ${}^{\text{'}}$ )
2. P(H or J)
3. P(H ${}^{\text{'}}$ or J ${}^{\text{'}}$ )
4. P(H ${}^{\text{'}}$ or J)
5. P(H ${}^{\text{'}}$ and J ${}^{\text{'}}$ )
8. The last ten letters of the alphabet were placed in a hat and people were asked to pick one of them. Event D is picking a vowel, Event E is picking a consonant and Event F is picking the last four letters. Calculate the following probabilities:
1. P(F ${}^{\text{'}}$ )
2. P(F or D)
3. P(neither E nor F)
4. P(D and E)
5. P(E and F)
6. P(E and D ${}^{\text{'}}$ )
9. At Dawnview High there are 400 Grade 12's. 270 do Computer Science, 300 do English and 50 do Typing. All those doing Computer Science do English, 20 take Computer Science and Typing and 35 take English and Typing. Using a Venn diagram calculate the probability that a pupil drawn at random will take:
1. English, but not Typing or Computer Science
2. English but not Typing
3. English and Typing but not Computer Science
4. English or Typing
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
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Other chapter Q/A we can ask
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# Write the Following Relations as Sets of Ordered Pairs and Find Which of Them Are Functions:(C) {(X, Y) : X + Y = 3, X, Y, ∈ [0, 1, 2, 3]} - CBSE (Science) Class 11 - Mathematics
#### Question
Write the following relations as sets of ordered pairs and find which of them are functions:
{(xy) : x + y = 3, xy, ∈ [0, 1, 2, 3]}
#### Solution
Given:
{(xy) : x + y = 3, xy, ∈ [0, 1, 2, 3]}
x + y = 3
∴ y = 3 – x
On substituting x = 0,1, 2, 3 in y, we get:
y = 3, 2, 1, 0, respectively.
∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)}
Hence, we observe that each element of the given set has appeared as the first component in one and only one ordered pair in R . So, R is a function in the given set.
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Solution for Mathematics Class 11 (2019 to Current)
Chapter 3: Functions
Ex.3.10 | Q: 8.3 | Page no. 8
Solution Write the Following Relations as Sets of Ordered Pairs and Find Which of Them Are Functions:(C) {(X, Y) : X + Y = 3, X, Y, ∈ [0, 1, 2, 3]} Concept: Sets - Ordered Pairs.
S
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# Page 1 of 7
1.4
What you should learn
GOAL 1 Use angle postulates.
Angles and Their Measures
GOAL 1 USING ANGLE POSTULATES
C vertex sides A
Classify angles as acute, right, obtuse, or straight.
GOAL 2
An angle consists of two different rays that have the same initial point. The rays are the sides of the angle. The initial point is the vertex of the angle. The angle that has sides AB and AC is denoted by ™BAC, ™CAB, or ™A. The point A is the vertex of the angle.
Æ ˘ Æ ˘
Why you should learn it
To solve real-life problems about angles, such as the field of vision of a horse wearing blinkers in Example 2. AL LI
F FE
RE
B
EXAMPLE 1
Naming Angles
Name the angles in the figure.
SOLUTION
P q S R
There are three different angles.
• • •
™PQS or ™SQP ™SQR or ™RQS ™PQR or ™RQP
You should not name any of these angles as ™Q because all three angles have Q as their vertex. The name ™Q would not distinguish one angle from the others. .......... The measure of ™A is denoted by m™A. The measure of an angle can be approximated with a protractor, using units called degrees (°). For instance, ™BAC has a measure of 50°, which can be written as m™BAC = 50°.
80 90 100 11 01 70 80 7 60 110 100 0 6 20 1 30 0 0 120 5 0 13
70 180 60 1 0 1 0 10 0 15 2 0 0 14 0 3 4
0 10 20 180 170 1 3 60 1 0 50 40 14 0
B
1
2
3
4
5
6
A
C
Angles that have the same measure are called congruent angles. For instance, ™BAC and ™DEF each have a measure of 50°, so they are congruent.
50
MEASURES ARE EQUAL. ANGLES ARE CONGRUENT.
D
E
F
m™BAC = m™DEF “is equal to”
26 Chapter 1 Basics of Geometry
™BAC £ ™DEF “is congruent to”
the vision for the left eye alone measures 40°. The rays of the form OA can be matched one to one with the real numbers from 0 to 180. VISION Each eye of a horse wearing blinkers has an angle of vision that measures 100°. The measure of ™AOB is equal to the absolute value of the difference between the real Æ ˘ Æ ˘ numbers for OA and OB . 70 180 60 1 0 1 0 10 0 15 2 0 0 14 0 3 4 A 1 2 3 4 5 6 O B A point is in the interior of an angle if it is between points that lie on each side of the angle. The angle of vision that is seen by both eyes measures 60°. then m™RSP + m™PST = m™RST. A point is in the exterior of an angle if it is not on the angle or in its interior.Page 2 of 7 P O S T U L AT E POSTULATE 3 Protractor Postulate 80 90 100 11 01 70 80 7 60 110 100 0 6 20 1 3 0 0 120 5 0 50 0 13 Logical Reasoning 0 10 20 180 170 1 3 60 1 0 50 40 14 0 Consider a point A on one side of ¯ ˘ Æ ˘ OB . exterior D interior A E P O S T U L AT E POSTULATE 4 Angle Addition Postulate S måRST R P If P is in the interior of ™RST. Subtract m™2 from each side.4 Angles and Their Measures 27 FE 1 2 3 . Substitute 60° for m™2. So. T P O S T U L AT E måRSP måPST EXAMPLE 2 STUDENT HELP RE L AL I Calculating Angle Measures Study Tip As shown in Example 2. 1. m™2 + m™3 = 100° m™3 = 100° º m™2 m™3 = 100° º 60° m™3 = 40° Total vision for left eye is 100°. SOLUTION region seen by both eyes You can use the Angle Addition Postulate. Subtract. Find the angle of vision seen by the left eye alone. it is sometimes easier to label angles with numbers instead of letters.
a. right. and P(2. Angles have measures greater than 0° and less than or equal to 180°. m™LMP = 180° c. 28 25 0 180 0 180 6 1 2 3 4 5 6 R S R S Chapter 1 Basics of Geometry . obtuse. M(º1. 35 STUDENT HELP INT NE ER T b. Angles are classified as acute. ™LMP c. ™NMQ d. which has four right angles.. P (2. obtuse.. m™LMQ = 135° right angle straight angle acute angle obtuse angle L ( 4. Then use a protractor to measure each angle. Q(4. 4) 1) .com for extra examples. 80 90 100 110 70 12 01 60 30 0 5 65 80 90 100 110 70 12 01 60 30 0 5 T 65 0 10 2 0 40 P 14 40 0 10 2 0 3 0 30 14 1 50 01 1 50 01 HOMEWORK HELP P 1 2 3 4 5 T 60 1 7 60 1 7 Visit our Web site www. º4). according to their measures. right. ™LMQ Begin by plotting the points... Two angles are adjacent angles if they share a common vertex and side. º1). or straight. 2) N (2. N(2. Then measure and classify the following angles as acute. and straight.. A Acute angle 0° < m™A < 90° A Right angle m™A = 90° A Obtuse angle 90° < m™A < 180° A Straight angle m™A = 180° EXAMPLE 3 Classifying Angles in a Coordinate Plane Plot the points L(º4. 2). 2). 2) x M ( 1..Page 3 of 7 GOAL 2 STUDENT HELP CLASSIFYING ANGLES Study Tip The mark used to indicate a right angle resembles the corner of a square. 1) œ (4.mcdougallittell.. EXAMPLE 4 Drawing Adjacent Angles Use a protractor to draw two adjacent acute angles ™RSP and ™PST so that ™RST is (a) acute and (b) obtuse. MEASURE CLASSIFICATION y a. ™LMN SOLUTION b. m™NMQ = 45° d.. SOLUTION a. but have no common interior points. º1). m™LMN = 90° b.
4 Angles and Their Measures 29 . L M N 11. A B. 17–22 Exs.Page 4 of 7 GUIDED PRACTICE Vocabulary Check Match the angle with its classification. 9. E U D T A HOMEWORK HELP 21. 23–34 Exs. C B A C. 39 B C 22. Example 1: Example 2: Example 3: Example 4: Exs. A D. H 12. K N E 19. 35–43 Exs. m™C = 100° 14. 18. Is ™DEG £ ™HEG? 7. Is ™DEF £ ™FEG? 6. or straight. right. R S q STUDENT HELP NAMING ANGLES Write two names for the angle. acute 1. obtuse 2. Then estimate its measure. 13. P S 1. 38. m™D = 45° PRACTICE AND APPLICATIONS STUDENT HELP NAMING PARTS Name the vertex and sides of the angle. Are ™DEF and ™FEH adjacent? 8. 20. Are ™GED and ™DEF adjacent? D F 45 E H 45 G Skill Check Name the vertex and sides of the angle. straight 4. 17. obtuse. Explain your answers. m™B = 90° 16. D F E 10. m™A = 180° 15. 803 and 804. A. X T F Extra Practice to help you master skills is on pp. J K R S T Classify the angle as acute. 5. right 3. C B A B C B C Concept Check Use the diagram at the right to answer the questions.
1) B(5. ™DBA is a right angle. plot the points and sketch ™ABC. 39. ™DBE is a straight angle. º4) Chapter 1 Basics of Geometry B(3. E is in the interior of ™DAF. º1) B(º2. A. m™ADC + m™CDE = 120°. H K 37. N D F G M L LOGICAL REASONING Draw five points. Find m™FAB. Find m™BAE. and use a protractor to measure it to the nearest degree. m™ABC = ? RE L AL I 27. º4) B(3. Find m™FAD. D is in the interior of ™BAE. or straight.mcdougallittell. INT NE ER T www. obtuse. Then estimate its measure. º1) 43. º2) C(4. E 36. A(3. º2) C(0. m™BAC = 130° m™EAC = 100° m™BAD = m™EAF = m™FAC 31. 23. º1) 42. 26. D. A 24. extend its sides. 30. which can measure angles to the nearest 1/3600 of a degree.com 30 FE A 45 60 B D 60 C D 120 E F R CAREER LINK LOGICAL REASONING Draw a sketch that uses all of the following information. right. A(5. F Z B C X D E ANGLE ADDITION Use the Angle Addition Postulate to find the measure of the unknown angle. and E so that all three statements are true. A(5. CLASSIFYING ANGLES State whether the angle appears to be acute. ™ABC is a straight angle. 40. 35. m™DEF = ? 28. m™PQR = ? q S 20 160 P SURVEYOR Surveyors use a tool called a theodolite. A(º3.Page 5 of 7 FOCUS ON CAREERS MEASURING ANGLES Copy the angle. xy USING ALGEBRA In a coordinate plane. 32. Find m™DAE. º1) C(4. Y 25. Classify the angle. º2) 41. C. 4) . F is in the interior of ™EAC. 29. 33. ™CDB is a straight angle. C is in the interior of ™ADE. 34. 2) C(º1. Find m™BAD. Find m™FAC. B. Write the coordinates of a point that lies in the interior of the angle and the coordinates of a point that lies in the exterior of the angle. 38.
AO = 4 in. NWT 120 90 60 Clyde River. A H G F O E D B C a. Name eight right angles. The score of a toss is indicated by numbers around 135 the board. Yukon Territory Tuktoyaktuk. Angmagssalik. m™BOA = 90°. Canada 0 B 48. AO = 3 in. Greenland 49. Name eight congruent acute angles. and A is the city. Greenland Old Crow. 51. c. Clyde River. Old Crow. 171 63 5 20 1 1 8 12 A 117 99 81 45 9 4 27 9 B 13 6 50. Name two adjacent angles that combine to form a straight angle.Page 6 of 7 GEOGRAPHY For each city on the polar map. 1. AO = 6. m™BOA = 35°. m™BOA = 160°. Tuktoyaktuk. NWT PLAYING DARTS In Exercises 50–53. Canada 180 46. AO = 5 in. Fairbanks.4 Angles and Their Measures 31 . Reykjavik. Only the top half of the dart board is shown. use the following information to find the score for the indicated dart toss landing at point A. Iceland 150 Fairbanks. 52. Name eight congruent obtuse angles. The score is doubled if a dart lands in the double ring and tripled if 153 it lands in the triple ring. where B is on the Prime Meridian (0° longitude). Canada 45. Iceland North Pole O Reykjavik. MULTI-STEP PROBLEM Use a piece of paper folded in half three times and labeled as shown. Alaska 30 Angmagssalik. m™BOA = 60°. 9 no score double ring triple ring O 5 8 31 8 3 8 21 8 3 8 23 8 Test Preparation 54. 1 14 A dartboard is 18 inches across. 53. It is divided into twenty wedges of equal size. Alaska 47.5 in. 44. d. estimate the measure of ™BOA. b. O is the North Pole.
º2). º3).5) 74. Find the measure of ™2. 67. 9) 79. 3 36 www. H(0.com MIXED REVIEW STUDENT HELP xy USING ALGEBRA Solve for x. 69. x+4 = º4 2 º9 + x = º7 2 x + (º3) = º4 2 EVALUATING STATEMENTS Decide whether the statement is true or false.mcdougallittell.2) 70. F(4. You can think of bearings between 180° and 360° as angles that are “bigger” than a straight angle. 72. T. 61. Because a full circle contains 360°. 64. p. N W S E 15 18 HOMEWORK HELP Bearings are measured around a circle. 66. B(º2. for 1.3 for 1. L(0. 5+x =5 2 x+7 = º10 2 8+x = º1 2 63. C(0. 0) 32 Chapter 1 Basics of Geometry . use the diagram of Ronald Reagan Washington National Airport and the information about runway numbering on page 1. see p. An airport runway is named by dividing its bearing (the angle measured clockwise from due north) by 10. Find the measure of ™1. 59. E(º3. 10). 71. EXTRA CHALLENGE Writing Explain why the difference between the numbers at the opposite ends of a runway is always 18. (Skills Review.790. Q. and P are coplanar. S. (Review 1.5) Skills Review For help solving equations. K(7. D(º8. S. runway numbers range from 1 to 36. 55. 56. A(3. so they can have values larger than 180°. º2) 76. M(º3. SR and TS are opposite rays. 57. 8). What is the number of the unlabeled 1 ? 4 3 33 2 runway in the diagram? 60. 65. 58. 68. UQ and PT intersect. 3) 77.Page 7 of 7 5 Challenge STUDENT HELP AIRPORT RUNWAYS In Exercises 55–60. 790. 7). x+3 =3 2 º8 + x = 12 2 x + (º1) =7 2 62. Æ ˘ Æ ˘ ¯ ˘ ¯ ˘ U R q S T P DISTANCE FORMULA Find the distance between the two points. and Q are collinear. 4) 78. Find the measure of ™4. U. 5) 75. Find the measure of ™3. J(5. 11). 73. (Review 1. G(10.
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# Transitive Relation on a Set
A relation is a subset of the cartesian product of a set with another set. A relation contains ordered pairs of elements of the set it is defined on. To learn more about relations refer to the article on “Relation and their types“.
## What is a Transitive Relation?
A relation R on a set A is called transitive relation if and only if
∀ a, b, c ∈ A, if (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R,
where R is a subset of (A x A), i.e. the cartesian product of set A with itself.
This means if an ordered pair of elements “a” to “b” (aRb) and “b” to “c” (bRc) is present in relation R, then an ordered pair of elements “a” to “c” (aRC) should also be present in the relation R. If any such aRc is not present for any aRb & bRc in R then R is not a transitive relation.
Example:
Consider set A = {a, b, c}
R = {(a, b), (b, c), (a, c)} is transitive relation but
R = {(a, b), (b, c)} is not transitive relation
## Properties of Transitive Relation
1. Empty relation on any set is always transitive.
2. Universal relation on any set is always transitive.
## How to verify a Transitive Relation?
To verify transitive relation:
• Firstly, find the tuples of form aRb & bRc in the relation.
• For every such pair check if aRc is also present in R.
• If any of the tuples does not exist then the relation is not transitive else it is transitive.
Follow the below illustration for a better understanding
Illustration:
Consider set R = {(1, 2), (1, 3), (2, 3), (3, 4)}
For the pairs (1, 2) and (2, 3):
=> The relation (1, 3) exists
=> This satisfies the condition.
For the pairs (1, 3) and (3, 4):
=> The relation (1, 4) does not exist
=> This does not satisfy the condition.
So the relation is not transitive.
Below is the code implementation of the idea:
## C++
`#include ` `using` `namespace` `std;` `class` `Relation {` `public``:` ` ``bool` `checkTransitive(set > R)` ` ``{` ` ``// Property 1` ` ``if` `(R.size() == 0) {` ` ``return` `true``;` ` ``}` ` ``// Create a dictionary to store tuple as key value` ` ``// pair` ` ``map<``int``, set<``int``> > tup;` ` ``// Creating dictionary of relation where (a) is key` ` ``// and (b) is value` ` ``for` `(``auto` `i = R.begin(); i != R.end(); i++) {` ` ``if` `(tup.find(i->first) == tup.end()) {` ` ``set<``int``> temp;` ` ``temp.insert(i->second);` ` ``tup.insert(` ` ``pair<``int``, set<``int``> >(i->first, temp));` ` ``}` ` ``else` `{` ` ``tup.at(i->first).insert(i->second);` ` ``}` ` ``}` ` ``for` `(``auto` `a = tup.begin(); a != tup.end(); a++) {` ` ``// Set of all b's related with a` ` ``set<``int``> all_b_in_aRb = tup.at(a->first);` ` ``// Taking all c's from each b one by one` ` ``for` `(``int` `b : all_b_in_aRb) {` ` ``if` `(tup.find(b) != tup.end()` ` ``&& a->first != b) {` ` ``// Set of all c's related with b` ` ``set<``int``> all_c_in_bRc = tup.at(b);` ` ``// All c's related with each b must be` ` ``// subset of all b's related with a` ` ``for` `(``int` `c : all_c_in_bRc) {` ` ``if` `(all_b_in_aRb.find(c)` ` ``== all_b_in_aRb.end()) {` ` ``return` `false``;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// For all aRb and bRc there exist aRc in relation R` ` ``return` `true``;` ` ``}` `};` `int` `main()` `{` ` ``set > R;` ` ``// Inserting tuples in relation R` ` ``R.insert(make_pair(1, 1));` ` ``R.insert(make_pair(1, 2));` ` ``R.insert(make_pair(2, 1));` ` ``R.insert(make_pair(2, 2));` ` ``R.insert(make_pair(1, 3));` ` ``R.insert(make_pair(2, 3));` ` ``R.insert(make_pair(3, 4));` ` ``R.insert(make_pair(1, 4));` ` ``Relation obj;` ` ``// R is not transitive as (2, 4) tuple is not present` ` ``if` `(obj.checkTransitive(R)) {` ` ``cout << ``"Transitive Relation"` `<< endl;` ` ``}` ` ``else` `{` ` ``cout << ``"Not a Transitive Relation"` `<< endl;` ` ``}` ` ``return` `0;` `}`
## Java
`// Java code to implement the approach` `import` `java.io.*;` `import` `java.util.*;` `class` `pair {` ` ``int` `first, second;` ` ``pair(``int` `first, ``int` `second)` ` ``{` ` ``this``.first = first;` ` ``this``.second = second;` ` ``}` `}` `class` `GFG {` ` ``static` `class` `Relation {` ` ``boolean` `checkTransitive(Set R)` ` ``{` ` ``// Property 1` ` ``if` `(R.size() == ``0``) {` ` ``return` `true``;` ` ``}` ` ``// Create a hashmap to store tuple as key value` ` ``// pair` ` ``HashMap > tup` ` ``= ``new` `HashMap<>();` ` ``// Creating hashmap of relation where (a) is key` ` ``// and (b) is value` ` ``for` `(pair i : R) {` ` ``if` `(!tup.containsKey(i.first)) {` ` ``Set temp = ``new` `HashSet<>();` ` ``temp.add(i.second);` ` ``tup.put(i.first, temp);` ` ``}` ` ``else` `{` ` ``Set temp = ``new` `HashSet<>();` ` ``temp = tup.get(i.first);` ` ``temp.add(i.second);` ` ``tup.put(i.first, temp);` ` ``}` ` ``}` ` ``for` `(Integer a : tup.keySet()) {` ` ``// Set of all b's related with a` ` ``Set all_b_in_aRb = tup.get(a);` ` ``// Taking all c's from each b one by one` ` ``for` `(``int` `b : all_b_in_aRb) {` ` ``if` `(tup.containsKey(b) && a != b) {` ` ``// Set of all c's related with b` ` ``Set all_c_in_bRc` ` ``= tup.get(b);` ` ``// All c's related with each b must` ` ``// be subset of all b's related with` ` ``// a` ` ``for` `(Integer c : all_c_in_bRc) {` ` ``if` `(all_b_in_aRb.contains(c)) {` ` ``return` `false``;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// For all aRb and bRc there exist aRc in` ` ``// relation R` ` ``return` `true``;` ` ``}` ` ``}` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``// Creating relation R` ` ``Set R = ``new` `HashSet<>();` ` ``// Inserting tuples in relation R` ` ``R.add(``new` `pair(``1``, ``1``));` ` ``R.add(``new` `pair(``1``, ``2``));` ` ``R.add(``new` `pair(``2``, ``1``));` ` ``R.add(``new` `pair(``2``, ``2``));` ` ``R.add(``new` `pair(``1``, ``3``));` ` ``R.add(``new` `pair(``2``, ``3``));` ` ``R.add(``new` `pair(``3``, ``4``));` ` ``R.add(``new` `pair(``1``, ``4``));` ` ``Relation obj = ``new` `Relation();` ` ``// R is not transitive as (2, 4) tuple is not` ` ``// present` ` ``if` `(obj.checkTransitive(R)) {` ` ``System.out.println(``"Transitive Relation"``);` ` ``}` ` ``else` `{` ` ``System.out.println(``"Not a Transitive Relation"``);` ` ``}` ` ``}` `}` `// This code is contributed by lokeshmvs21.`
## Python3
`class` `Relation:` ` ``def` `checkTransitive(``self``, R):` ` ``# Property 1` ` ``if` `len``(R) ``=``=` `0``:` ` ``return` `True` ` ``# Create a dictionary to store tuple as key value pair` ` ``tup ``=` `dict``()` ` ``# Creating dictionary of relation where (a) is key and (b) is value` ` ``for` `i ``in` `R:` ` ``if` `tup.get(i[``0``]) ``is` `None``:` ` ``tup[i[``0``]] ``=` `{i[``1``]}` ` ``else``:` ` ``tup[i[``0``]].add(i[``1``])` ` ``for` `a ``in` `tup.keys():` ` ``# Set of all b's related with a` ` ``all_b_in_aRb ``=` `tup.get(a)` ` ``if` `all_b_in_aRb ``is` `not` `None``:` ` ``# Taking all c's from each b one by one` ` ``for` `b ``in` `all_b_in_aRb:` ` ``# Set of all c's related with b` ` ``all_c_in_bRc ``=` `tup.get(b)` ` ``if` `a !``=` `b ``and` `all_c_in_bRc ``is` `not` `None``:` ` ``if` `not` `all_c_in_bRc.issubset(all_b_in_aRb):` ` ``# All c's related with each b must be` ` ``# subset of all b's related with a` ` ``return` `False` ` ``# For all aRb and bRc there exist aRc in relation R` ` ``return` `True` `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` ` ``# Creating relation R` ` ``R ``=` `{(``1``, ``1``), (``1``, ``2``), (``2``, ``1``), (``2``, ``2``), (``1``, ``3``), (``2``, ``3``), (``3``, ``4``), (``1``, ``4``)}` ` ``obj ``=` `Relation()` ` ``# R is not transitive as (2, 4) tuple is not present` ` ``if` `obj.checkTransitive(R):` ` ``print``(``"Transitive Relation"``)` ` ``else``:` ` ``print``(``"Not a Transitive Relation"``)`
## C#
`// C# code to implement the approach` `using` `System;` `using` `System.Collections.Generic;` `class` `pair {` ` ``public` `int` `first, second;` ` ``public` `pair(``int` `first, ``int` `second)` ` ``{` ` ``this``.first = first;` ` ``this``.second = second;` ` ``}` `}` `public` `class` `GFG {` ` ``class` `Relation {` ` ``public` `bool` `checkTransitive(HashSet R)` ` ``{` ` ``// Property 1` ` ``if` `(R.Count == 0) {` ` ``return` `true``;` ` ``}` ` ``// Create a hashmap to store tuple as key value` ` ``// pair` ` ``Dictionary<``int``, HashSet<``int``> > tup` ` ``= ``new` `Dictionary<``int``, HashSet<``int``> >();` ` ``// Creating hashmap of relation where (a) is key` ` ``// and (b) is value` ` ``foreach``(pair i ``in` `R)` ` ``{` ` ``if` `(!tup.ContainsKey(i.first)) {` ` ``HashSet<``int``> temp = ``new` `HashSet<``int``>();` ` ``temp.Add(i.second);` ` ``tup[i.first] = temp;` ` ``}` ` ``else` `{` ` ``HashSet<``int``> temp = ``new` `HashSet<``int``>();` ` ``temp = tup[i.first];` ` ``temp.Add(i.second);` ` ``tup[i.first] = temp;` ` ``}` ` ``}` ` ``foreach``(``var` `a ``in` `tup)` ` ``{` ` ``// Set of all b's related with a` ` ``HashSet<``int``> all_b_in_aRb = tup[a.Key];` ` ``// Taking all c's from each b one by one` ` ``foreach``(``int` `b ``in` `all_b_in_aRb)` ` ``{` ` ``if` `(tup.ContainsKey(b) && a.Key != b) {` ` ``// Set of all c's related with b` ` ``HashSet<``int``> all_c_in_bRc = tup[b];` ` ``// All c's related with each b must` ` ``// be subset of all b's related with` ` ``// a` ` ``foreach``(``int` `c ``in` `all_c_in_bRc)` ` ``{` ` ``if` `(all_b_in_aRb.Contains(c)) {` ` ``return` `false``;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// For all aRb and bRc there exist aRc in` ` ``// relation R` ` ``return` `true``;` ` ``}` ` ``}` ` ``static` `public` `void` `Main()` ` ``{` ` ``// Code` ` ``// Creating relation R` ` ``HashSet R = ``new` `HashSet();` ` ``// Inserting tuples in relation R` ` ``R.Add(``new` `pair(1, 1));` ` ``R.Add(``new` `pair(1, 2));` ` ``R.Add(``new` `pair(2, 1));` ` ``R.Add(``new` `pair(2, 2));` ` ``R.Add(``new` `pair(1, 3));` ` ``R.Add(``new` `pair(2, 3));` ` ``R.Add(``new` `pair(3, 4));` ` ``R.Add(``new` `pair(1, 4));` ` ``Relation obj = ``new` `Relation();` ` ``// R is not transitive as (2, 4) tuple is not` ` ``// present` ` ``if` `(obj.checkTransitive(R)) {` ` ``Console.WriteLine(``"Transitive Relation"``);` ` ``}` ` ``else` `{` ` ``Console.WriteLine(``"Not a Transitive Relation"``);` ` ``}` ` ``}` `}` `// This code is contributed by lokesh`
## Javascript
`class Relation {` ` ``constructor() {}` ` ``checkTransitive(R) {` ` ``// Property 1` ` ``if` `(R.size === 0) {` ` ``return` `true``;` ` ``}` ` ``// Create a dictionary to store tuple as key value` ` ``// pair` ` ``const tup = ``new` `Map();` ` ``// Creating dictionary of relation where (a) is key` ` ``// and (b) is value` ` ``for` `(const i of R) {` ` ``if` `(!tup.has(i[0])) {` ` ``const temp = ``new` `Set();` ` ``temp.add(i[1]);` ` ``tup.set(i[0], temp);` ` ``} ``else` `{` ` ``tup.get(i[0]).add(i[1]);` ` ``}` ` ``}` ` ``for` `(const a of tup) {` ` ``// Set of all b's related with a` ` ``const all_b_in_aRb = tup.get(a[0]);` ` ``// Taking all c's from each b one by one` ` ``for` `(const b of all_b_in_aRb) {` ` ``if` `(tup.has(b) && a[0] !== b) {` ` ``// Set of all c's related with b` ` ``const all_c_in_bRc = tup.get(b);` ` ``// All c's related with each b must be` ` ``// subset of all b's related with a` ` ``for` `(const c of all_c_in_bRc) {` ` ``if` `(!all_b_in_aRb.has(c)) {` ` ``return` `false``;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// For all aRb and bRc there exist aRc in relation R` ` ``return` `true``;` ` ``}` `}` `function` `main() {` ` ``const R = ``new` `Set();` ` ``// Inserting tuples in relation R` ` ``R.add([1, 1]);` ` ``R.add([1, 2]);` ` ``R.add([2, 1]);` ` ``R.add([2, 2]);` ` ``R.add([1, 3]);` ` ``R.add([2, 3]);` ` ``R.add([3, 4]);` ` ``R.add([1, 4]);` ` ``const obj = ``new` `Relation();` ` ``// R is not transitive as (2, 4) tuple is not present` ` ``if` `(obj.checkTransitive(R)) {` ` ``console.log(``"Transitive Relation"``);` ` ``} ``else` `{` ` ``console.log(``"Not a Transitive Relation"``);` ` ``}` `}` `main();` `// This code is contributed by akashish__.`
Output
`Not a Transitive Relation`
Time Complexity: O(N * K * log N) where N is the number of tuples in relation and K is the maximum number of tuples (a, b) for which a are same
Auxiliary Space: O(N)
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# Nth root calculator
This nth root calculator will compute the nth root of any number with just the click of a button.
Enter the number you want to take the root of:
What root do you want to take?:
## Guidelines to follow when using the nth root calculator to find the nth root of a number
1. Calculate the cube root of 27 (∛27):
Enter 27 in the box that says,"Enter the number you want to take the root of"
Enter 3 in the box that says, "What root do you want to take?"
The big box will show 3 as an answer.
2. Calculate the fourth root of 625 (∜625):
Enter 625 in the box that says,"Enter the number you want to take the root of"
Enter 4 in the box that says, "What root do you want to take?"
The big box will show 5 as an answer.
3. Calculate 6th root of 4096:
Enter 4096 in the box that says, "Enter the number you want to take the root of"
Enter 6 in the box that says, "What root do you want to take?"
The big box will show 3.99999996 as an answer.
Important note:
If the answer for a root looks like 3.99999996 or 4.99999999 or 67.99999996, then you know the answer is 4, 5, or 68 respectively.
## Key concept
### Square root of a number
Square root of 4 is 2 because 2 times 2 is 4
Square root of 64 is 8 because 8 times 8 is 64
### Cube root of a number
Cube root of 27 is 3 because 3 times 3 times 3 = 27
Cube root of 125 is 5 because 5 times 5 times 5 = 125
### Fourth root of a number
The fourth root of 16 is 2 because 2 times 2 times 2 times 2 = 16
The fourth root of 625 is 5 because 5 times 5 times 5 times 5 = 625
### Fifth root of a number
The fifth root of 243 is 3 because 3 times 3 times 3 times 3 times 3 = 243
The fifth root of 1024 is 4 because 4 times 4 times 4 times 4 times 4 = 1024
In many cases, you may get a real number when looking for the square root.
For example, use the square root calculator below to find the square root of 7
The result includes 2.64 with lots of other numbers after the decimal point.
Ready for big time challenge? Just like long division, learn how to compute the square root without a calculator for any number that is not a perfect square. I promise you will not sweat too much!
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
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# How do you find the inverse of A=((1, 2, 1), (2, 5, 4), (1, 4, 9)) ?
Mar 2, 2017
The inverse is $= \left(\begin{matrix}\frac{29}{4} & - \frac{7}{2} & \frac{3}{4} \\ - \frac{7}{2} & 2 & - \frac{1}{2} \\ \frac{3}{4} & - \frac{1}{2} & \frac{1}{4}\end{matrix}\right)$
#### Explanation:
For a matrix $A$ to be invertible, the determinant $\det A \ne 0$
Let's calculate
$\det A = | \left(1 , 2 , 1\right) , \left(2 , 5 , 4\right) , \left(1 , 4 , 9\right) |$
$= 1 \cdot | \left(5 , 4\right) , \left(4 , 9\right) | - 2 \cdot | \left(2 , 4\right) , \left(1 , 9\right) | + 1 \cdot | \left(2 , 5\right) , \left(1 , 4\right) |$
$= 1 \cdot \left(45 - 16\right) - 2 \cdot \left(18 - 4\right) + 1 \cdot \left(8 - 5\right)$
$= 29 - 28 + 3$
$= 4$
As $\det A \ne 0$, matrix $A$ is invertible
We calculate the matrix of cofactors
$C = \left(\begin{matrix}| \left(5 4\right) & \left(4 9\right) | & - | \left(2 4\right) & \left(1 9\right) | & | \left(2 5\right) & \left(1 4\right) | \\ - | \left(2 1\right) & \left(4 9\right) | & | \left(1 1\right) & \left(1 9\right) | & - | \left(1 2\right) & \left(1 4\right) | \\ | \left(2 1\right) & \left(5 4\right) | & - | \left(1 1\right) & \left(2 4\right) | & | \left(1 2\right) & \left(2 5\right) |\end{matrix}\right)$
$= \left(\begin{matrix}29 & - 14 & 3 \\ - 14 & 8 & - 2 \\ 3 & - 2 & 1\end{matrix}\right)$
The transpose of $C$ is
${C}^{T} = \left(\begin{matrix}29 & - 14 & 3 \\ - 14 & 8 & - 2 \\ 3 & - 7 & 1\end{matrix}\right)$
The inverse is
${A}^{-} 1 = \frac{1}{\det} A \cdot {C}^{T}$
$= \frac{1}{4} \left(\begin{matrix}29 & - 14 & 3 \\ - 14 & 8 & - 2 \\ 3 & - 2 & 1\end{matrix}\right)$
$= \left(\begin{matrix}\frac{29}{4} & - \frac{14}{4} & \frac{3}{4} \\ - \frac{14}{4} & \frac{8}{4} & - \frac{2}{4} \\ \frac{3}{4} & - \frac{2}{4} & \frac{1}{4}\end{matrix}\right)$
Verification
$A \cdot {A}^{-} 1 = \left(\begin{matrix}1 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 4 & 9\end{matrix}\right) \cdot \left(\begin{matrix}\frac{29}{4} & - \frac{14}{4} & \frac{3}{4} \\ - \frac{14}{4} & \frac{8}{4} & - \frac{2}{4} \\ \frac{3}{4} & - \frac{2}{4} & \frac{1}{4}\end{matrix}\right)$
$= \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$
$= I$
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# How to Work out the Square Root and Cube Root of a Number?
Updated on February 25, 2019
First of all let’s take a look at square roots.
If you want to find a square root of a number then you need to look for a number when multiplied by itself to give the number that you are square rooting.
So if you have √a then ? × ? = a
Let’s cover a few questions:
Question 1
Work out √16 (this means the square root of 16)
Since 4 × 4 = 16 then the square root of 16 is 4.
Question 2
Work out √49
Since 7 × 7 = 49 then the square root of 49 is 7.
Question 3
Work out √225
Since 15 × 15 = 225 then the square root of 225 is 15.
Question 4
Work out √1
Since 1 × 1 = 1 then the square root of 1 is 1.
Question 5
Work out √144
Since 12 × 12 = 144 then the square root of 144 is 12.
Let’s now look at cube roots. This time you are looking for a number such that ? × ? × ? gives the number that you are cube rooting. Let’s go over a few questions of cube roots:
Question 6
Work out ³√27 (that is the cube root of 27)
So you are looking for a number such that ? × ? × ? = 27. The number that you looking for is 3 because 3 × 3 × 3 = 27 (3 × 3 = 9 so 9 × 3 = 27). Therefore the cube root of 27 is 3.
Question 7
Work out ³√8 (that is the cube root of 8)
So you are looking for a number such that ? × ? × ? = 8. The number that you looking for is 8 because 2 × 2 × 2 = 8 (2 × 2 = 4 so 4 × 2 = 8). Therefore the cube root of 8 is 2.
Question 8
Work out ³√64 (that is the cube root of 64)
So you are looking for a number such that ? × ? × ? = 64. The number that you looking for is 4 because 4 × 4 × 4 = 64 (4 × 4 = 16 so 16 × 4 = 64). Therefore the cube root of 64 is 4.
Question 9
Work out ³√125 (that is the cube root of 125)
So you are looking for a number such that ? × ? × ? = 125. The number that you looking for is 5 because 5 × 5 × 5 = 125 (5 × 5 = 25 so 25 × 5 = 125). Therefore the cube root of 125 is 5.
Question 10
Work out ³√-8 (that is the cube root of negative 8)
So you are looking for a number such that ? × ? × ? = -8. The number that you looking for is -2 because -2 × -2 × -2 = -8 (-2 × -2 = 4 so 4 × -2 = -8). Therefore the cube root of -8 is -2.
So to summarise if you are taking the square root of a number then you are looking for a number such that ? × ? = the number you are rooting. And if you are taking the cube root of a number then you are looking for a number such that ? × ? × ? = the number that you are cube rooting.
## What happens if there is no obvious answer?
Sometimes you will find that you will not be able to square root a number in your head because the answer will be a decimal answer. In this case you will need to use a scientific calculator. On a Casio calculator the button is usually located near the top of the calculated and has a tick with a box underneath. So for example, to work out the square root of 21, then you will need to use the root key on your calculator to give 4.582575695, which you can round off to 4.58 to 2 decimal places. Again, you can check the answer by multiplying 4.58 by 4.58 to give 20.9764. Although there is a slight inaccuracy due to rounding off the numbers to two decimal places.
Similarly, with cube roots, there might not be an obvious answer so you will need a scientific calculator for tricky one. The cube root key on a Casio calculator is usually the second function key on the square root key. So for example, the cube root of 85 would not have an integer answer so you will need to use your calculator. The answer will be 4.396829672 or 4.40 rounded to 2 decimal places.
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# How do you implicitly differentiate e^(3x)=sin(x+2y) ?
$y ' = \frac{3 {e}^{3 x}}{2 \cdot \cos \left(x + 2 y\right)} - \frac{1}{2}$
#### Explanation:
From the given equation, just differentiate both sides of the equation with respect to $x$
${e}^{3 x} = \sin \left(x + 2 y\right)$
$\frac{d}{\mathrm{dx}} \left({e}^{3 x}\right) = \frac{d}{\mathrm{dx}} \left(\sin \left(x + 2 y\right)\right)$
${e}^{3 x} \frac{d}{\mathrm{dx}} \left(3 x\right) = \cos \left(x + 2 y\right) \cdot \frac{d}{\mathrm{dx}} \left(x + 2 y\right)$
$\left({e}^{3 x}\right) \left(3\right) = \cos \left(x + 2 y\right) \left(1 + 2 y '\right)$
$\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} = \frac{\cancel{\cos} \left(x + 2 y\right) \left(1 + 2 y '\right)}{\cancel{\cos} \left(x + 2 y\right)}$
$\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} = 1 + 2 y '$
$\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} - 1 = 2 y '$
$\frac{\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} - 1}{2} = \frac{2 y '}{2}$
$\frac{\frac{3 {e}^{3 x}}{\cos \left(x + 2 y\right)} - 1}{2} = \frac{\cancel{2} y '}{\cancel{2}}$
$y ' = \frac{3 {e}^{3 x}}{2 \cdot \cos \left(x + 2 y\right)} - \frac{1}{2}$
God bless....I hope the explanation is useful.
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Question Video: Arranging Given Numbers | Nagwa Question Video: Arranging Given Numbers | Nagwa
# Question Video: Arranging Given Numbers Mathematics • Second Year of Primary School
Arrange the numbers 197, 971, 791, 179, 917 to complete the following: ＿< ＿< ＿< ＿< ＿.
03:50
### Video Transcript
Arrange the numbers 197, 971, 791, 179, 917 to complete the following.
The first thing we notice about the order they are asking us to put these numbers in is that we’re using all “less than” symbols, which means the digit on the far left will be the smallest. And from left to right, we’ll move to the largest value. This question is asking us to arrange five digits from least to greatest. To do this I’m gonna stack numbers on top of each other in a line. Starting with 197, next up 971, then we have 791, 179, and finally 917.
To compare these values, we’ll start by comparing the largest place values, the digits in the largest place values. In our case, those digits fall in the hundreds place. And remember, while we’re comparing, we’re looking for the smallest. So let’s look for the smallest digit in the hundreds column.
In the hundreds column, we have two numbers that have one in the hundreds place. Since those values are the same — they’re the smallest, but they’re the same — we’ll need to check the tens place on these values. Again remember that we’re looking for the smallest value, so our first row has one and then a nine. Our second number we’re looking at has a one and then a seven.
Seven is smaller than nine, so 179 would be smaller than 197, which makes 179 the smallest value and the first value we would write down. After that, we can go ahead and write 197 because it’s the second smallest. Now that we’ve used 179 and 197, we’ll cross these two values out and we’ll look for the next smallest hundreds value. The next smallest hundreds value is a seven, the number 791 would be the next highest value.
As we look back at our hundreds again, we see that we have two nines. Our hundreds place is not going to help us compare. We’ll need to move to the tens place. Comparing the tens place for each of these values, we see we have a seven and a one. The one is smaller than the seven, which means 917 is less than 971. 917 comes next in our ordering; we cross that out, leaving us with 971 as our largest value.
From smallest to largest, our values are 179, 197, 791, 917, and 971.
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# Copyright © 2007 Pearson Education, Inc. Slide 5-2 Chapter 5: Exponential and Logarithmic Functions 5.1 Inverse Functions 5.2 Exponential Functions 5.3.
## Presentation on theme: "Copyright © 2007 Pearson Education, Inc. Slide 5-2 Chapter 5: Exponential and Logarithmic Functions 5.1 Inverse Functions 5.2 Exponential Functions 5.3."— Presentation transcript:
Copyright © 2007 Pearson Education, Inc. Slide 5-2 Chapter 5: Exponential and Logarithmic Functions 5.1 Inverse Functions 5.2 Exponential Functions 5.3 Logarithms and Their Properties 5.4 Logarithmic Functions 5.5 Exponential and Logarithmic Equations and Inequalities 5.6 Further Applications and Modeling with Exponential and Logarithmic Functions
Copyright © 2007 Pearson Education, Inc. Slide 5-3 5.1 Inverse Functions Example Also, f [g(12)] = 12. For these functions, it can be shown that for any value of x. These functions are inverse functions of each other.
Copyright © 2007 Pearson Education, Inc. Slide 5-4 Only functions that are one-to-one have inverses. 5.1 One-to-One Functions A function f is a one-to-one function if, for elements a and b from the domain of f, a b implies f (a) f (b).
Copyright © 2007 Pearson Education, Inc. Slide 5-5 5.1 One-to-One Functions Example Decide whether the function is one-to-one. (a) (b) Solution (a) For this function, two different x-values produce two different y-values. (b)If we choose a = 3 and b = –3, then 3 –3, but
Copyright © 2007 Pearson Education, Inc. Slide 5-6 5.1 The Horizontal Line Test Example Use the horizontal line test to determine whether the graphs are graphs of one-to-one functions. (a) (b) If every horizontal line intersects the graph of a function at no more than one point, then the function is one-to-one. Not one-to-one One-to-one
Copyright © 2007 Pearson Education, Inc. Slide 5-7 5.1 Inverse Functions Example are inverse functions of each other. Let f be a one-to-one function. Then, g is the inverse function of f and f is the inverse of g if
Copyright © 2007 Pearson Education, Inc. Slide 5-8 5.1 Finding an Equation for the Inverse Function Notation for the inverse function f -1 is read “f-inverse” Finding the Equation of the Inverse of y = f(x) 1. Interchange x and y. 2. Solve for y. 3. Replace y with f -1 (x). Any restrictions on x and y should be considered.
Copyright © 2007 Pearson Education, Inc. Slide 5-9 5.1 Example of Finding f -1 (x) ExampleFind the inverse, if it exists, of Solution Write f (x) = y. Interchange x and y. Solve for y. Replace y with f -1 (x).
Copyright © 2007 Pearson Education, Inc. Slide 5-10 5.1 The Graph of f -1 (x) f and f -1 (x) are inverse functions, and f (a) = b for real numbers a and b. Then f -1 (b) = a. If the point (a,b) is on the graph of f, then the point (b,a) is on the graph of f -1. If a function is one-to-one, the graph of its inverse f -1 (x) is a reflection of the graph of f across the line y = x.
Copyright © 2007 Pearson Education, Inc. Slide 5-11 5.1 Finding the Inverse of a Function with a Restricted Domain ExampleLet SolutionNotice that the domain of f is restricted to [–5, ), and its range is [0, ). It is one-to-one and thus has an inverse. The range of f is the domain of f -1, so its inverse is
Copyright © 2007 Pearson Education, Inc. Slide 5-12 5.1 Important Facts About Inverses 1.If f is one-to-one, then f -1 exists. 2.The domain of f is the range of f -1, and the range of f is the domain of f -1. 3.If the point (a,b) is on the graph of f, then the point (b,a) is on the graph of f -1, so the graphs of f and f -1 are reflections of each other across the line y = x.
Copyright © 2007 Pearson Education, Inc. Slide 5-13 5.1 Application of Inverse Functions Example Use the one-to-one function f (x) = 3x + 1 and the numerical values in the table to code the message BE VERY CAREFUL. A1F6K 11P 16U21 B 2G 7L 12Q 17V22 C 3H 8M 13R 18W23 D4I 9N 14S 19X24 E 5J 10O 15 T 20Y 25 Z 26 SolutionBE VERY CAREFUL would be encoded as 7 16 67 16 55 76 10 4 55 16 19 64 37 because B corresponds to 2, and f (2) = 3(2) + 1 = 7, and so on.
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# What is 51/64 as a decimal?
## Solution and how to convert 51 / 64 into a decimal
51 / 64 = 0.797
To convert 51/64 into 0.797, a student must understand why and how. Both are used to handle numbers less than one or between whole numbers, known as integers. The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. If we need to convert a fraction quickly, let's find out how and when we should.
## 51/64 is 51 divided by 64
Converting fractions to decimals is as simple as long division. 51 is being divided by 64. For some, this could be mental math. For others, we should set the equation. Fractions have two parts: Numerators and Denominators. This creates an equation. We must divide 51 into 64 to find out how many whole parts it will have plus representing the remainder in decimal form. This is our equation:
### Numerator: 51
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. 51 is one of the largest two-digit numbers you'll have to convert. The bad news is that it's an odd number which makes it harder to covert in your head. Values closer to one-hundred make converting to fractions more complex. Time to evaluate 64 at the bottom of our fraction.
### Denominator: 64
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. 64 is one of the largest two-digit numbers to deal with. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Overall, two-digit denominators are no problem with long division. Next, let's go over how to convert a 51/64 to 0.797.
## Converting 51/64 to 0.797
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 64 \enclose{longdiv}{ 51 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 64 \enclose{longdiv}{ 51.0 }$$
Uh oh. 64 cannot be divided into 51. So that means we must add a decimal point and extend our equation with a zero. This doesn't add any issues to our denominator but now we can divide 64 into 510.
### Step 3: Solve for how many whole groups you can divide 64 into 510
$$\require{enclose} 00.7 \\ 64 \enclose{longdiv}{ 51.0 }$$
Since we've extended our equation we can now divide our numbers, 64 into 510 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply by the left of our equation (64) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.7 \\ 64 \enclose{longdiv}{ 51.0 } \\ \underline{ 448 \phantom{00} } \\ 62 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 51/64 into 0.797 If there is a remainder, extend 64 again and pull down the zero
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. And the same is true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But they all represent how numbers show us value in the real world. Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 51/64 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.797 to 51/64 as a fraction
Meal Prep - Body builders need to count macro calories. One of the ways of doing this is measuring every piece of food consumed. This is through halves and quarters in order to keep it consistent.
### Practice Decimal Conversion with your Classroom
• If 51/64 = 0.797 what would it be as a percentage?
• What is 1 + 51/64 in decimal form?
• What is 1 - 51/64 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.797 + 1/2?
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# Application Of Multivariable Calculus
Application Of Multivariable Calculus I’m using Calculus Over Logic to illustrate my problem with multivariable calculus. To start with, I have defined multiplication as the name of a function from 1 to n, where n is a positive integer. The problem with this definition is that it requires that the series of terms in the expansion be considered as sums of factor functions. As a result, it turns out that the multiplication of n is not the same as the sum of factors in the expansion of a series. So, to finish the problem, I have to find a way to do this. First, we need to define the term x in terms of the term why not try this out in terms of z. The two terms are equal in terms of both z and x. The first term is the sum of the terms of the expansion of x. This term is the product of the terms in the series of z and x, while the second term is the difference between the series of x and z. These terms are given by A series of factors is a sum of terms in a series. The terms of which you are interested are the terms of which the series is taken. For example, if we have a series of factors, we want to find the terms of terms where the series is the sum or difference of the terms. Here is my attempt to solve this problem: In the following, I am using the notation that the expression x1 represents the product of a series of terms. That is, x1 represents x1, x2 represents x2, and x3 represents the series of the terms x3,x4,x5. So, for example, if I were to take x1 and x2 as the sum and x3 as the difference, then I would get a series of the form a1 = a1 – a2 b1 = b1 – b2 c1 = c1 – c2 d1 = d1 – d2 x1 = x1 + 1 x2 = x2 + 1 (x3) = (x3 – x1) + 1 (x4) = ( x3 – x2) + 1 (x5) = ( (x4 – x3) + 1) + 1 In this example, I am trying to find the following series of terms, where the terms of x1 and y1 represent the terms of n. The terms are given as follows: (n) = x10 = 10 + 5 x20 = 20 + 10 x30 = 30 + 10 x1 = x5 + 1 x2 = x15 + 1 (x15) = ( 10 + 5 ) + 5 x3 = x20 + 1 p1 = p20 + x5 (p20) = p20 – 2 (p20) (x30) = ( 30 + 10) + 5 I have tried to find the solutions but I am not sure how to do this, and I am not getting the result I want. And this is where I am stuck! A: If you want to find all of the terms for $n$, you might want to use $x\wedge y$ to represent the sum of terms. The term x is the sum over all possible terms. For example: $$x = \sum_{n=0}^{n-1} (n+1)(n+2)(n+3) \ldots$$ A for example: $x = \frac{1}{2}(x+1)$. A is the sum.
## Pay To Take Online Class
Ferentzi. Oxford and London: Clarendions, 1980. 4. Introduction to Multivariable calculus. By David E. Ferentzer. London and New York; Oxford: Clarerdons, 1982. 5. Introduction to Cauchy-Riemann Geometry. By David H. Koeppe. Cambridge: Cambridge University Press, 1995. 6. Introduction to Discrete Geometry and Its Applications. By Peter D. McNeill. Cambridge: Harvard University Press, 1994. 7. Introduction to Quantum Mechanics. By Peter H.
## Take My Test Online For Me
Simon. Cambridge: MIT Press, 1997. 8. Introduction to Geometry.by Peter H.Simon. Cambridge: Chapman and Hall, 1991. 9. Introduction to Combinatorics and Its Applications by Peter H.Muller. Cambridge: Wiley, 1991. An Introduction to Calculative Mathematics. By David J. Ferentze. London and Oxford: Wiley, 1980. Introduction to Calculus and its Applications Part II: Calculus and Its Applications Chapter 1: The Basics Chapter 2: A General Theory of Evolutionary Calculus. Chapter 3: Applications If you can try these out are starting with a Calculus of Differentiation, then we need a differentiable function for the application. The Calculus of Evolutionary Differential Equations (also known as Newton-Penrose Calculus) is the natural generalization of Newton-Pen rose calculus to the calculus of differential equations, and will be discussed in section 4. A Calculus of differential equations is a mathematical system of differential equations on a set of variables, and may be viewed as a generalization of differential calculus. It is a set of equations on a space of visit here which may be represented by a differential operator.
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Integration by Parts
If we have studied the integral substitution method consisting of two versions, it is incomplete if we do not study the partial integral method. Because both of these methods are often used in solving integral problems both integral and uncertain integrals.
If integration uses substitution fails, we might be able to use double substitution, which is better known as partial integral. This method is based on the integration of the formula for derivatives of the product of two functions.
Suppose that $u = u (x)$ and $v = v (x)$, then:
$D_x [uv] = uv '+ vu'$
or
$uv '= D_x [uv] -vu'$
By integrating the two segments of the equation, we obtain
$\int uv ' \ dx = uv - \int vu' \ dx$
Because $dv = v ' \ dx$ and $du = u' \ dx$, the above equation is usually written with the symbol as follows.
$$\int u \ dv = uv – \int v \ du$$
The formulas that correspond to the integral of course are:
$$\int_{a}^{b} u \ dv = [uv]_{a}^{b} – \int_{a}^{b} v \ du$$
Question example:
Look for $\int x \cos x \ dx$
Settlement:
The question cannot indeed be done by using a substitution method. We want to write $x \cos x \ dx$ as $u \ dv$. The possibility is to suppose that $u = x$ and $dv = \cos x \ dx$ or $u = \cos x$ and $dv = x \ dx$. We must be smart to choose between the two which are more appropriate. To determine which one has to be $u$ and $dv$, you can read the way in another article titled Select u and dv in Integral Partial, I can.
For example, $u = x$ and $dv = \cos x \ dx$
Then,
$du = dx$ and
\begin{align} v & = \int \cos x \ dx \ & = \sin x \end{align}
So that,
\begin{align} & \int \underbrace{x}_u \underbrace{cos x dx}_{dv} \\ & = \underbrace{x}_{u} \underbrace{sin x}_{v} – \int \underbrace{sin x}_{v} \underbrace{dx}_{du} \\ & = x \sin x – (- \cos x) + C \\ & = x \sin x + \cos x + C \end{align}
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# FRACTIONS MODULE Part I
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1 FRACTIONS MODULE Part I I. Basics of Fractions II. Rewriting Fractions in the Lowest Terms III. Change an Improper Fraction into a Mixed Number IV. Change a Mixed Number into an Improper Fraction BMR.Fractions Part I Page 1
2 I. Fraction: Basics Introduction: This is the first of four parts on working with fractions. You re going to review a few definitions regarding fractions. Afterwards, you re going to try to do some problems on your own. There will be 20 problems for you to practice. After you re successful in doing the practice problems, try the short quiz. The answers can be found at the end of each section. Definitions: The numerator is written above the fraction bar and the denominator is written under the fraction bar. Another way to look at defining a fraction is written as a part of the whole. The part is always written in the numerator. The whole is always written in the denominator. A) Example: What part of the bar is shaded? Step 1: Count how many pieces the bar contains. The whole bar is broken up into 6 pieces. Therefore the whole = 6 pieces. Step 2: Count how many pieces of the bar is shaded. The part of the bar that is shaded is 5 pieces. Therefore, the part = 5 pieces. Step 3: Put all the information from Step 1 and Step 2 together to get a fraction. The numerator is 5 and the denominator is 6. Answer: of the bar is shaded BMR.Fractions Part I Page 2
4 D) You Try: 1. In the figure: a. What part of the figure is shaded? b. What part of the figure is not shaded? 2. In the figure: a. What part of the figure is shaded? b. What part of the figure is not shaded? 3. BONUS. In the figure: a. What part of the figure is shaded? b. What part of the figure is not shaded? Answers to D You Try : 1a. 3/6 or 1/2, 1b. 3/6 or 1/2, 2a. 2/5, 2b. 3/5, 3a. 1 2/5, 3b. 3/5 BMR.Fractions Part I Page 4
5 E) Identifying the Types of Fractions. Three Types of Fractions 1. Proper Fraction 2. Improper Fraction 3. Mixed Number 1) Proper Fraction Definition: A proper fraction is a fraction with a numerator that is smaller than the denominator. Examples:,,,, 2) Improper Fraction Definition: An improper fraction is a fraction with a numerator that is larger than the denominator. Examples:,,, 3) Mixed Number Definition: A number that is made up of the sum of a whole number and a proper fraction. Examples:,,, BMR.Fractions Part I Page 5
6 F) PRACTICE PROBLEMS. (20) Write a fraction that represents the shaded part of the figure. 1) 2) 3) 4) 5) Identify what type of fraction for the following: 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) BMR.Fractions Part I Page 6
7 Answers to F Practice Problems : 1) ; 2) ; 3) ; 4) ; 5) ; 6) proper fraction; 7) mixed number; 8)proper fraction; 9) improper fraction; 10) improper fraction; 11) mixed number; 12) proper fraction; 13) improper fraction ; 14) mixed number; 15) mixed number; 16) proper fraction; 17) proper fraction; 18) mixed number; 19) mixed number; 20) improper fractio G) QUIZ. (5 questions) Write a fraction that represents the shaded part of the figure. 1) 2) Identify the type of fraction for the following: 3) 4) 5) Answers to G Quiz : 1) ; 2) ; 3) improper fraction; 4) proper fraction; 5) mixed number BMR.Fractions Part I Page 7
8 II. Rewriting Fractions in Lowest Terms. This is the second of four parts on understanding fractions. You re going to look at examples on how to rewrite fractions in lowest terms. Afterwards, you re going to try to do some problems on your own. There will be 20 problems for you to practice. After you re successful in doing the practice problems, try the short quiz. The answers can be found at the end of each section. A) Method 1: Prime Factorization Example: Rewrite in lowest terms. STEP 1: Find the prime factorization for 63. Recall that prime numbers are numbers that can only be divided by the number and one. The number 1 is not a prime number. Here is a list of a first ten prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Use a prime factorization tree to break down 63 into its prime numbers. 63 The prime factorization of 63 = 3 x 3 x (3 x 21 = 63; 3 is a prime number) 3 7 (3 x 7 = 21; 3 & 7 are prime numbers) STEP 2: Find the prime factorization for 81 Use a prime factorization tree to break down 81 into its prime numbers (3 x 27 = 81; 3 is a prime number) 3 9 (3 x 9 = 27; 3 is a prime number) 3 3 (3 x 3 = 9; 3 & 3 are a prime numbers) The prime factorization for 81 = 3 x 3 x 3 x 3 STEP 3: Rewrite the fraction using the prime factorizations of both 63 and 81. BMR.Fractions Part I Page 8
9 STEP 4: Cancel like factors. STEP 5: Multiply the left over factors to get your new fraction reduced to lowest terms. B) Method 2: Guess and Reduce Example: Rewrite in lowest terms. STEP 1: Guess what number can be divided into the numerator and then the denominator without a remainder. Since both numbers are even, guess 2. STEP 2: Divide numerator and denominator by 2. STEP 3: Guess what number can be divided into the numerator and then the denominator of the fraction in STEP 2 without a remainder. Try dividing each number by 3. BMR.Fractions Part I Page 9
10 C) You Try: a) b) c) Answers to C You Try : a) b) c) BMR.Fractions Part I Page 10
11 D) PRACTICE PROBLEMS. Rewrite the following fractions in lowest terms. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) Answers to D Practice Problems : 1) ; 2) ; 3) ; 4) ; 5) ; 6) ; 7) ; 8) ; 9) ; 10) ; 11) ; 12) ; 13) ; 14) ; 15) ; 16) ; 17) ; 18) ; 19) ; 20) BMR.Fractions Part I Page 11
12 E) QUIZ. Rewrite the following fractions in lowest terms. 1) 2) 3) 4) 5) Answers to E Quiz : 1) ; 2) ; 3) ; 4) ; 5) BMR.Fractions Part I Page 12
13 III. Changing an Improper Fraction into a Mixed Number. Introduction: This is the third of four parts on understanding fractions. You re going to look at some examples on how to change an improper fraction into a mixed number. Afterwards, you re going to try to do some problems on your own. There will be 20 problems for you to practice. After you re successful in doing the practice problems, try the short quiz. The answers can be found at the end of each section. A) Example 1: Change into a mixed number. Step 1: Rewrite the fraction as a division problem. The denominator becomes the divisor and the numerator becomes the dividend. Divisor Quotient Dividend 7 15 Step 2: Divide 7 into (2 x 7 = 14) 1 Remainder Step 3: Use the different parts of the division problem to construct the mixed number. The parts of a mixed number: The parts of the mixed number replaced with the parts of a division problem: The quotient is 2; the remainder is 1; and the divisor is 7. Substituting these values in the proper place, the end result is a mixed number. BMR.Fractions Part I Page 13
14 B) Example 2: Change into a mixed number. Step 1: Rewrite the fraction as a division problem. The denominator becomes the divisor and the numerator becomes the dividend. Divisor Quotient Dividend 6 27 Step 2: Divide 6 into (4 x 6 = 24) 3 Remainder Step 3: Use the different parts of the division problem to construct the mixed number. The parts of a mixed number: The parts of the mixed number replaced with the parts of a division problem: The quotient is 4; the remainder is 3; and the divisor is 6. Substituting these values in the proper place, the end result is a mixed number. Step 4: Make sure the fraction part is in lowest terms. The fraction part is BMR.Fractions Part I Page 14
15 Need to reduce the numerator and denominator by the same value. The numerator and denominator can be divided by 3. Step 5: Rewrite the reduced mixed number. C) Example 3: Change into a mixed number. Step 1: Rewrite the fraction as a division problem. The denominator becomes the divisor and the numerator becomes the dividend. Divisor Quotient Dividend 8 56 Step 2: Divide 8 into (7 x 8 = 56) 0 Remainder Step 3: Use the different parts of the division problem to construct the mixed number. The parts of a mixed number: BMR.Fractions Part I Page 15
16 The parts of the mixed number replaced with the parts of a division problem: The quotient is 7; the remainder is 0; and the divisor is 8. Substituting these values in the proper place, the end result is a mixed number. Rewrite the mixed fraction as an addition problem. Simplify the fraction part: = 0 Do the addition. Answer: 7 D) You try: a) b) c) Answers to D You Try : a) b) c) 6 BMR.Fractions Part I Page 16
17 E) PRACTICE PROBLEMS. Change the following improper fractions into mixed numbers: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) Answers to E Practice Problems : 1) ; 2) ; 3) ; 4) ; 5) ; 6) ; 7) ; 8) ; 9) ; 10) ; 11) ; 12) ; 13) ; 14) ; 15) ; 16) ; 17) ; 18) ; 19) ; 20) BMR.Fractions Part I Page 17
18 F) QUIZ. Change the following improper fractions into mixed numbers: 1) 2) 3) 4) 5) Answers to F Quiz : 1) ; 2) ; 3) ; 4) ; 5) BMR.Fractions Part I Page 18
19 IV. Changing a Mixed Number Into an Improper Fraction Taken from Treff, A. & Jacobs, D.,Life Skills Mathematics,Media Materials, Inc. Baltimore, Maryland 1983, p 275. Introduction: This is the fourth of four parts on understanding fractions. You re going to look at an example on how to change a mixed number into an improper fraction. Afterwards, you re going to try to do some problems on your own. There will be 20 problems for you to practice. After you re successful in doing the practice problems, try the short quiz. The answers can be found at the end of each section. A) Example: Write the following fraction as an improper fraction: REVIEW: STEP 1: Multiply the whole number by the denominator. STEP 2: Add the numerator to the product from STEP 1. STEP 3: Write the sum over the original denominator. FIRST RULE OF MULTIPLYING, DIVIDING, ADDING & SUBTRACTING FRACTIONS: CHANGE ALL MIXED NUMBERS INTO IMPROPER FRACTIONS. B) You Try: a) b) c) Answers to B You Try : a) b) c) BMR.Fractions Part I Page 19
20 C) PRACTICE PROBLEMS. Change the following mixed numbers into improper fractions: 1) 2) 1 3) 2 4) 1 5) 3 6) 6 7) 4 8) 5 9) 1 10) 1 11) 12) 4 13) 14) 15) 2 16) 17) 2 18) 2 19) 3 20) 4 Answers to Practice Problems : 1) ; 2) ; 3) ; 4) ; 5) ; 6) ; 7) ; 8) ; 9) ; 10) ; 11) ; 12) ; 13) ; 14) ; 15) ; 16) ; 17) ; 18) ; 19) ; 20) BMR.Fractions Part I Page 20
21 D) QUIZ. Change the following mixed numbers into improper fractions: 1) 1 2) 2 3) 3 4) 5) Answers to D Quiz : 1) ; 2) ; 3) ; 4) ; 5) BMR.Fractions Part I Page 21
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### Multiplication and Division with Rational Numbers
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### 1004.6 one thousand, four AND six tenths 3.042 three AND forty-two thousandths 0.0063 sixty-three ten-thousands Two hundred AND two hundreds 200.
Section 4 Decimal Notation Place Value Chart 00 0 0 00 000 0000 00000 0. 0.0 0.00 0.000 0.0000 hundred ten one tenth hundredth thousandth Ten thousandth Hundred thousandth Identify the place value for
### is identically equal to x 2 +3x +2
Partial fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as 1 + 3
### Solution Guide Chapter 14 Mixing Fractions, Decimals, and Percents Together
Solution Guide Chapter 4 Mixing Fractions, Decimals, and Percents Together Doing the Math from p. 80 2. 0.72 9 =? 0.08 To change it to decimal, we can tip it over and divide: 9 0.72 To make 0.72 into a
### Chapter 5. Rational Expressions
5.. Simplify Rational Expressions KYOTE Standards: CR ; CA 7 Chapter 5. Rational Expressions Definition. A rational expression is the quotient P Q of two polynomials P and Q in one or more variables, where
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Math 1710
Topics for second exam
Chapter 2: Derivatives
§ 7:
Related Rates
Idea: If two (or more) quantities are related (a change in one value means a change in others), then their rates of change are related, too.
xyz = 3 ; pretend each is a function of t, and differentiate (implicitly).
General procedure:
Draw a picture, describing the situation; label things with variables.
Which variables, rates of change do you know, or want to know?
Find an equation relating the variables whose rates of change you know or want to know.
Differentiate!
Plug in the values that you know.
Chapter 3: Applications of Derivatives
§ 1:
Extreme Values
c is an (absolute) maximum for a function f(x) if f(c) ³ f(x) for every other x
d is an (absolute) minimum for a function f(x) if f(d) £ f(x) for every other x
max or min = extremum
Extreme Value Theorem: If f is a continuous function defined on a closed interval
[a,b], then f actually has a max and a min.
Goal: figure out where they are!
c is a relative max (or min) if f(c) is ³ f(x) (or £ f(x)) for every x near c. Rel max or min = rel extremum.
An absolute extremum is either a rel extremum or an endpoint of the interval.
c is a critical point if f¢(c) = 0 or does not exist.
A rel extremum is a critical point.
So absolute extrema occur either at critical points or at the endpoints.
So to find the abs max or min of a function f on an interval [a,b] :
(1) Take derivative, find the critical points.
(2) Evaluate f at each critical point and endpoint.
(3) Biggest value is maximum value, smallest is minimum value.
§ 2:
The Mean Value Theorem
You can (almost) recreate a function by knowing its derivative
Mean Value Theorem: if f is containous on [a,b] and differentiable on (a,b), then there is at least one c in (a,b) so that
f¢(c) = [(f(b)-f(a))/(b-a)]
Consequences:
Rolle's Theorem: f(a) = f(b) = 0; between two roots there is a critical point.
So: If a function has no critical points, it has at most one root!
A function with f¢(x)=0 is constant.
Functions with the same derivative (on an interval) differ by a constant.
f is increasing on an interval if x > y implies f(x) > f(y)
f is decreasing on an interval if x > y implies f(x) < f(y)
If f¢(x) > 0 on an interval, then f is increasing
If f¢(x) < 0 on an interval, then f is decreasing
§ 3:
The First Derivative Test
Local max's / min's occur at critical points; how do you tell them apart?
Near a local max, f is increasing, then decreasing; f¢(x) > 0 to the left of the critical point, and f¢(x) < 0 to the right.
Near a local min, the opposite is true; f¢(x) < 0 to the left of the critical point, and f¢(x) > 0 to the right.
If the derivative does not change sign as you cross a critical point, then the critical point is not a rel extremum.
Basic use: plot where a function is increasing/decreasing: plot critical points; in between them, sign of derivative does not change.
§ 4:
Graphing
when we look at a graph, we see where function is increasing/decreasing. We also see:
f is concave up on an interval if f¢¢(x) > 0 on the interval
Means: f¢ is increasing; f is bending up.
f is concave down on an interval if f¢¢(x) < 0 on the interval
Means: f¢ is decreasing; f is bending down.
A point where the concavity changes is called a point of inflection
Graphing:
Find where f¢(x) and f¢¢(x) are 0 or DNE
Plot on the same line.
In between points, derivative and second derivative don't change sign, so graph looks like one of:
Then string together the pieces!
Second derivative test: If c is a critical point and
f¢¢(c) > 0, then c is a rel min (smiling!)
f¢¢(c) < 0, then c is a rel max (frowning!)
§ 5:
Limits at infinity, asymptotes
Last bit of information for a graph: what happens at the ends?
limx®¥f(x) = L means f(x) is close of L when x is really large.
limx® -¥f(x) = M means f(x) is close of M when x is really large and negative.
Basic fact: limx®¥ [1/x] = limx® -¥ [1/x] = 0
More complicated functions: divide by the highest power of x in the denomenator.
f(x),g(x) polynomials, degree of f = n, degree of g = m
limx®±¥ [f(x)/g(x)] = 0 if n < m
limx®±¥ [f(x)/g(x)] =
(coeff of highest power in f)/(coeff of highest power in g) if n = m
limx®±¥ [f(x)/g(x)] = ±¥ if n > m
The line y = a is a horizontal asymptote for a function f if limx®¥f(x) or limx® -¥f(x) is equal to a.
I.e., the graph of f gets really close to y = a as x®¥ or a®-¥
The line x = b is a vertical asymptote for f if f®±¥ as x®b from the right or left.
If numerator and denomenator of a rational function have no common roots, then vertical asymptotes = roots of denom.
f®¥ or f®-¥ : can use f incr or decr on either side of b to decide (so long as you already know it is blowing up!)
§ 6:
Optimization
This is really just finding the max or min of a function on an interval, with the added complication that you need to figure out which function, and which interval! Solution strategy is similar to related rates:
Draw a picture; label things.
What do you need to maximize/minimize? Write down a formula for the quantity.
Use other information to eliminate variables, so your quantity depends on only one variable.
Determine the largest/smallest that the variable can reasonably be (i.e., find your interval)
Turn on the max/min machine!
File translated from TEX by TTH, version 0.9.
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# GRE Math : How to find the equation of a line
## Example Questions
← Previous 1
### Example Question #1 : How To Find The Equation Of A Line
What is the equation of the straight line passing through (–2, 5) with an x-intercept of 3?
y = –x – 3
y = –5x – 3
y = –5x + 3
y = –x + 3
y = –x + 3
Explanation:
First you must figure out what point has an x-intercept of 3. This means the line crosses the x-axis at 3 and has no rise or fall on the y-axis which is equivalent to (3, 0). Now you use the formula (y y1)/(x– x1) to determine the slope of the line which is (5 – 0)/(–2 – 3) or –1. Now substitute a point known on the line (such as (–2, 5) or (3, 0)) to determine the y-intercept of the equation y = –x + b. b = 3 so the entire equation is y = –x + 3.
### Example Question #2 : How To Find The Equation Of A Line
What is the equation of the line passing through (–1,5) and the upper-right corner of a square with a center at the origin and a perimeter of 22?
y = (–3/5)x + 22/5
y = –x + 5
y = (–1/5)x + 2.75
y = (–3/5)x + 28/5
y = (3/5)x + 22/5
y = (–3/5)x + 22/5
Explanation:
If the square has a perimeter of 22, each side is 22/4 or 5.5. This means that the upper-right corner is (2.75, 2.75)—remember that each side will be "split in half" by the x and y axes.
Using the two points we have, we can ascertain our line's equation by using the point-slope formula. Let us first get our slope:
m = rise/run = (2.75 – 5)/(2.75 + 1) = –2.25/3.75 = –(9/4)/(15/4) = –9/15 = –3/5.
The point-slope form is: y – y0 = m(x – x0). Based on our data this is: y – 5 = (–3/5)(x + 1); Simplifying, we get: y = (–3/5)x – (3/5) + 5; y = (–3/5)x + 22/5
### Example Question #3 : How To Find The Equation Of A Line
What is the equation of a line passing through the points and ?
Explanation:
Based on the information provided, you can find the slope of this line easily. From that, you can use the point-slope form of the equation of a line to compute the line's full equation. The slope is merely:
Now, for a point and a slope , the point-slope form of a line is:
Let's use for our point
This gives us:
Now, distribute and solve for :
### Example Question #4 : How To Find The Equation Of A Line
What is the equation of a line passing through the two points and ?
Explanation:
Based on the information provided, you can find the slope of this line easily. From that, you can use the point-slope form of the equation of a line to compute the line's full equation. The slope is merely:
Now, for a point and a slope , the point-slope form of a line is:
Let's use for our point
This gives us:
Now, distribute and solve for :
### Example Question #1 : How To Find The Equation Of A Line
What is the equation of a line passing through with a -intercept of ?
Explanation:
Based on the information that you have been provided, you can quickly find the slope of your line. Since the y-intercept is , you know that the line contains the point . Therefore, the slope of the line is found:
Based on this information, you can use the standard slope-intercept form to find your equation:
, where and
### Example Question #11 : How To Find The Equation Of A Line
Given the graph of the function below, find the equation of the line.
Explanation:
To solve this question, you could use two points such as (1.2,0) and (0,-4) to calculate the slope which is 10/3 and then read the y-intercept off the graph, which is -4.
### Example Question #61 : Lines
What line goes through the points (0, 6) and (4, 0)?
y = –3/2x + 6
y = –3/2 – 3
y = 1/5x + 3
y = 2/3 + 5
y = 2/3x –6
y = –3/2x + 6
Explanation:
P1 (0, 6) and P2 (4, 0)
First, calculate the slope: m = rise ÷ run = (y2 – y1)/(x– x1), so m = –3/2
Second, plug the slope and one point into the slope-intercept formula:
y = mx + b, so 0 = –3/2(4) + b and b = 6
Thus, y = –3/2x + 6
### Example Question #1 : How To Find The Equation Of A Line
What line goes through the points (1, 3) and (3, 6)?
–2x + 2y = 3
3x + 5y = 2
2x – 3y = 5
4x – 5y = 4
–3x + 2y = 3
–3x + 2y = 3
Explanation:
If P1(1, 3) and P2(3, 6), then calculate the slope by m = rise/run = (y2 – y1)/(x2 – x1) = 3/2
Use the slope and one point to calculate the intercept using y = mx + b
Then convert the slope-intercept form into standard form.
### Example Question #141 : Coordinate Geometry
What is the slope-intercept form of ?
Explanation:
The slope intercept form states that . In order to convert the equation to the slope intercept form, isolate on the left side:
### Example Question #3 : How To Find The Equation Of A Line
A line is defined by the following equation:
What is the slope of that line?
Explanation:
The equation of a line is
y=mx + b where m is the slope
Rearrange the equation to match this:
7x + 28y = 84
28y = -7x + 84
y = -(7/28)x + 84/28
y = -(1/4)x + 3
m = -1/4
← Previous 1
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# Video: AQA GCSE Mathematics Foundation Tier Pack 2 • Paper 1 • Question 9
Calculate 43 × 29.
03:37
### Video Transcript
Calculate 43 multiplied by 29.
There are lots of ways of working out this calculation. We will look at the grid method and long multiplication. For the grid method, we split 43 into 40 and three. We split 29 into 20 and nine. We then need to work out four different multiplication calculations: 40 multiplied by 20, 40 multiplied by nine, three multiplied by 20, and three multiplied by nine.
Four multiplied by two is equal to eight. This means that 40 multiplied by 20 is equal to 800. We add the two zeros. Four multiplied by nine is equal to 36. Therefore, 40 multiplied by nine is equal to 360. We add the zero. Three multiplied by two is equal to six. Adding the zero gives us 60. Therefore, three multiplied by 20 is 60. Finally, three multiplied by nine is equal to 27.
We now need to add our four answers: 800, 360, 60, and 27. Adding the numbers in the units column gives us seven. Adding the numbers in the tens column, six, six, and two, gives us 14. We put a four in the answer and carry the one. Eight plus three is equal to 11. Adding the one that we carried gives us 12. Therefore, our answer is 1247. 43 multiplied by 29 is equal to 1247.
We can check this answer using long multiplication. Our first step here is multiplying nine by three. This gives us 27. We then multiply nine by four. Nine multiplied by four is equal to 36. Adding the two that we carried gives us 38. Our next step is to multiply the three and the four by two. But the two is in the tens column. So in effect, we’re multiplying by 20. In order to do this, we put a zero in the units column first. Two multiplied by three is equal to six. And finally, two multiplied by four is equal to eight. This tells us that 43 multiplied by nine is 387 and 43 multiplied by 20 is 860.
We now need to add these two rows. Seven plus zero is equal to seven. Eight plus six is equal to 14. So we need to carry the one. Three plus eight is equal to 11. Adding the one that we carried gives us 12. Once again, we have proved that the answer to 43 multiplied by 29 is 1247. There are lots of other ways you could have worked out this calculation. And as long as you show your method, you will get the full marks.
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CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard
Class 10
Solutions of Sample Papers for Class 10 Boards
## A train covers a distance of 360 km at a uniform speed. Had the speed been 5 km/hour more, it would have taken 48 minutes less for the journey. Find the original speed of the train.
Note : This is similar to Ex 4.3, 8 of NCERT – Chapter 4 Class 10
### Transcript
Question 37 (OR 1st question) A train covers a distance of 360 km at a uniform speed. Had the speed been 5 km/hour more, it would have taken 48 minutes less for the journey. Find the original speed of the train. Let the speed of train be x km/hr Normal speed Distance = 360 km Speed = x km/hr Speed = π·ππ π‘ππππ/ππππ x = 360/ππππ Time = 360/π₯ Speed 5 km/h more Distance = 360 km Speed = (x + 5) km/hr Time = (360/π₯ " β " 48/60) hours Speed = π·ππ π‘ππππ/ππππ x + 5 = 360/((360/π₯ " β " 48/60) ) (x + 5) (360/π₯ " β " 48/60) = 360 From (1) (x + 5) (360/π₯ " β " 48/60) = 360 (x + 5) (360/π₯ " β " 4/5) = 360 (x + 5) ((5 Γ 360 β 4π₯)/5π₯) = 360 (x + 5) ((1800 β 4π₯)/5π₯) = 360 (x + 5) (1800 β 4x) = 360 Γ 5x x(1800 β 4x) + 5(1800 β 4x) = 1800x 1800x β 4x2 + 5(1800) β 20x = 1800x 1800x β 4x2 + 9000 β 20x = 1800x 1800x β 4x2 + 9000 β 20x β 1800x = 0 β 4x2 β 20x + 9000 = 0 4x2 + 20x β 9000 = 0 4(x2 + 5x β 2250) = 0 x2 + 5x β 2250 = 0 Comparing with ax2 + bx + c = 0 a = 1, b = 5, c = β2250 Roots of the equation are given by x = (β π Β± β(π^2 β 4ππ))/2π Putting values x = (β5 Β± β(5^2 β 4 Γ 1 Γ (β2250) ))/(2 Γ 1) x = (β5 Β± β(25 + 4 Γ 2250 ))/2 x = (β5 Β± β(25 + 9000))/2 x = (β5 Β± β9025)/2 x = (β5 Β± β(5^2Γγ19γ^2 ))/2 x = (β5 Β± 5Γ19)/2 x = (β5 Β± 95)/2 x = (β5 + 95)/2 x = 90/2 x = 45 x = (β5 β 95)/2 x = (β100)/2 x = β50 Hence x = 45, x = β50 are the roots of the equation We know that Speed of train = x So, x cannot be negative β΄ x = 45 is the solution So, Speed of train = x = 45 km/hr
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# Class 8 – Set Theory
Take practice tests in Set Theory
## Online Tests
Topic Sub Topic Online Practice Test
Set Theory
• Introduction to sets and subsets
• Operations on sets
Take Test See More Questions
Set Theory
• Cardinal property of sets
• Venn diagrams
Take Test See More Questions
## Study Material
Introduction: The history of set theory is rather different from the history of most other areas of mathematics. It is the creation of one person, George Cantor. Set: A well-defined collection of objects is called a set. The objects in a set are called its members or elements. By the term ‘ well – defined ’ we mean that it is defined in such a way that we are able to decide as to which object of the universe is there in our collection and which object is not there is our collection. Notation of a Set: We usually denote sets by capital letters and their elements by small letters. If x is an element of a set A, we write, , which means ‘ x belongs to A’ or that ‘ x is an element of A’. If x is not an element of set A, we say that ‘x does not belong to A’, and we write . It is customary to put the elements of a set within braces { }. Representation of a Set: There are two methods of representing a set. 1. Roster Method (or Tabulation Method): Under this method, we make a list of the objects in our collection and put them within braces { }. Example: i) Let A be the set of vowels in English alphabet. Then A = { a, e, i, o, u } ii) Let B be the set of natural numbers between 3 and 10. Then, B = {4, 5, 6, 7, 8, 9} 2. Rule Method (or Set-builder form): Under this method, we list the property or properties satisfied by the elements of a set. We write, {x/x satisfies the properties P} or {x : x satisfies the properties of P} which means ‘The set of all those elements x such that each x satisfies the properties of P. Each of the symbol ‘/’ and ‘:’ stands for ‘such that’ Example: i) Let A= {41, 43, 47, 51, 53, 59} Clearly A is the set of all prime numbers between 40 and 60. Thus in the set-builder form, we write: A= {x/x is a prime number, 40 < x < 60}
i) Let B= {8, 16, 24, 32, 40}. Then, in set-builder form, we write
C = {x/is a multiple of 8, x < 40} or
C = {x:x= 8n, and n < 6}
Types of Sets:
i) Finite set: A set in which the process of counting of elements surely comes to an end is called a finite set.
Example: The set of all vowels in the English alphabet is a finite set having 5 elements namely a, e, i, o, u.
ii) Infinite Set: A set which is not finite is called an infinite set. In other words, a set is called an infinite set if the process of counting of its elements does not come to an end.
Example: The set N = {1, 2, 3, 4, 5,6…….} of all natural numbers.
iii) Empty set or Null set: A set having no elements at all is called an empty set or null set or void set and we denote it by .
In Roster method, we denote it by { }
Example:
The empty set is a finite set, since the number of elements contained in an empty set is 0.
Singleton Set: A set containing only one element is called a singleton set.
Example: {x/is an even prime number} is a singleton set containing only one element namely 2.
Equal sets: Two sets A and B are said to be equal, if every element of A is in B and every element of B is in A and we write A = B.
Example: Let A = { x is an even between 13 and 19 number, x < 9} and B = { x is even,13 < x < 19}
Then A = {14, 16, 18} and B ={14, 16, 18}
Clearly every element of A is in B and every element of B is in A.
Cardinal number of a set: The number of distinct elements contained in a finite set A is called its cardinal number and is denoted by n(A)
Example: Let A = {21, 22, 23, 24, 25}. Then n(A)= 5
Cardinal number of empty set is zero. i.e., n(Ø) = 0
Equivalent sets: Two finite sets A and B are said to be equivalent, if they have the same number of elements and we write, A ⇔ B
Thus, whenever n(A) =n(B), then we have A ⇔ B
Example: Let A be the set of vowels in the English alphabet and B be the set of first five odd natural numbers. Then, A = {a, e,i, o, u} and B = {1, 3, 5, 7, 9}
Clearly n(A) = n(B)= 5 therefore A ⇔ B
Two equal sets are always equivalent but equivalent sets need not to be equal.
Sub sets: The collections are generally linked in a given context. If we think of ourselves, then we belong to a certain society, which in turn belongs to a state, which in turn belongs to a country and so on. In the context of a school, all students of a school belong to a school. Some of them belong to a certain class. If there are sections with in a class, then some of these belongs to a section.
The need to have a mathematical relationship between different collections of similar types lead to the evolution of “subset”.
Subset: A set “A” is a subset of set “ B ”, if each member of “ A ” is also a member of set “ B ”. We use symbol “ ⊂ ” to denote this relationship between a “subset” and a “set”. Hence,.
We read this symbolic representation as : set “A” is a subset of set “B”. We express the intent of relationship as:
It is evident that set “B” is larger of the two sets. This is sometimes emphasized by calling set “B” as the “superset” of “A”. We use the symbol “ ⊃ ” to denote this relation:
If set “A” is not a subset of “B”, then we write this symbolically as:
Important results/deductions: Some of the important characteristic and related deductions are presented here:
Equality of two sets: In case, all elements of “B” are also in “A”, then two sets are equal. We express this symbolically as:
If and , then A = B. This is true in other direction as well: If A = B, then and .
We can write two instances in a single representation as: and
The symbol “ ⇔ ”means that relation holds in either direction.
Relation with itself: Every set is a subset of itself. This is so because every element is present in itself.
Relation with Empty set: Empty set is a subset of every set. This deduction is a direct consequence of the fact that empty set has no elements. As such, this set is a subset of all sets.
Proper subset: We have seen from the deductions above that special circumstances of “equality” can blur the distinction between “set” and “subset”. In order to emphasize, mother-child relation between sets, we coin the term “proper subset”. If every element of set “ B ” is not present in set “ A” , then “A” is a “proper” subset of set“B”; otherwise not. This means that set “B” is a larger set, which besides other elements, also includes all elements of set “A”.
Set of vowels in English alphabet, “V”, is a “proper” subset of the set of English alphabet,
“E”. All elements of “V” are present in “E”, but not all elements of “E” are not present in “V”. Therefore, V⊂ E
Conventional differences. Some write a “proper” subset relation using symbol “ ⊂ ” and write symbol “ ⊆ ” to mean possibility of equality as well.
Number system: The number system is one such system, in which different number groups are related. Natural numbers are a subset of integers. Integers are a subset of rational numbers and rational numbers are a subset of real numbers. None of these sets are equal. Hence, relations are described by proper subsets.
The chain of relation among number sets is as follows: (where C is the set of complex numbers).However, irrational numbers are also subsets of real numbers, but irrational number is not the same as rational numbers. We represent this relation by emphasizing that rational numbers is not a subset of irrational
Q (rational number) ⊄ Q1 (Irrational numbers). But irrational number is subset of real numbers. The real numbers comprises of only two subsets at the highest – rational and irrational. Therefore, irrational numbers is the remaining collection after deducting rational numbers from real numbers.
Following the logic, we define set of irrational numbers as:
Q1 (irrational numbers) =
Power set: Power set is formed from of all possible subsets of a given set. It is denoted asP(A) (or), the collection of all subsets of a set “ A ” is called power set, P(A).
For example, consider a set given by:
A = {1, 3, 4}
What are the possible subsets? There are three subsets consisting of individual elements. { 1 }, { 3 } and { 4 }. Then, elements taken two at a time the subsets are: {1, 3}, {1, 4} and {3, 4}. Since order or sequence does not matter in set representation, there are only three subsets of two elements taken together. Now, the elements taken three at a time form only one subset: {1, 3,4}. Remember, a set is a subset of itself. Further, empty set (f) is a subset of any set. Hence, is also a subset of the given set “A”.
The set comprising of all possible subsets of a given set “A” is:
We note two important points from this representation of power set:
i) The elements of a power set are themselves a set. In other words, every element of a power set is a set.
ii) If the number of elements (cardinality) excludes empty set. It is, however, counted as members of power set. For a set having three elements, the total numbers of elements in the power set is:
iii) A set containing n elements has 2nsubsets but proper subsets.
Example: The finite set “A” and “B” have “m” and “n” number of elements respectively. The total numbers of subsets of “A” is 56 more than the total number of subsets of“B”. Find “m” and “n”.
Solution: According to the relation obtained for power set, the total number of subsets of “A” and“B” are”
We need to find two equations to find “m” and “n”. For this we seek expansion of“56” in terms of powers of “2”.
56 = 8 × 7 = 8(8 – 1) = 23(23-1)
In order to get this form, we rearrange the expression on the LHS of the earlier equation as:
Equating powers of similar base,
n = 3and m = 6
Verification
1. Types of subsets:
i) Proper subsets. ii) Empty sets.
2. Emptyset is a subset of every set.
3. The set itself is a subset of given set.
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A+ » VCE Blog » VCE Maths Methods Blog » Sampling and Estimation [Video Tutorial]
# Sampling and Estimation [Video Tutorial]
This tutorial covers material encountered in chapter 17 of the VCE Mathematical Methods Textbook, namely:
• Sampling and estimation methods
• The population, sample and their proportions
• Sampling distributions
• Point estimates
• The 95% confidence interval and the margin of error
## Worksheet
Q1. Triassic Park has 200 dinosaurs, 80 of them carnivores. A random sample of 30 dinosaurs was selected, and 14 of them were carnivores. In this example:
(a) What’s the population?
(b) What’s the population proportion?
(c) What’s the sample proportion? Do you think the random sample is representative of the entire population?
Q2. A standard six-sided die is cast 50 times, and n even numbers were observed.
(a) Give a point estimate for p, the probability of observing an even number when the die is cast.
(b) Give an expression for a 95% confidence interval for p.
Q3. To study the effectiveness of video games for reducing stress levels, a researcher measured the stress levels of 100 people who regularly play League of Legends and measured their stress levels at the end of a 4 hour gaming session.
(a) Do you think this sample is representative of the general population? Give a general explanation.
(b) How would you suggest the sample be chosen?
(c) Bonus: Do you think the subjects’ stress levels would generally decrease or increase?
Q4. A sample of k people were asked if they liked mangos, with 85% of people saying yes.
(a) Find the value of the sample proportion \hat{p}.
(b) Give an expression for the margin of error, M, for this estimate at the 95% confidence level in terms of k.
(c) If the number of people in the sample were tripled, what would be the effect on the margin of error M?
Q5. Suppose that 30 independent random samples were taken from a large population, and that a 95% confidence interval for the population proportion p was computed from each of these samples.
(a) How many of the 95% confidence intervals would you expect to contain the population proportion p?
(b) Give an expression for the probability that all 30 confidence intervals contain p.
Q6. Tom has determined that an approximate 95% confidence interval for the proportion of people subscribed to his blog who regularly read new entries is (0.60, 0.80).
(a) What \hat{p} value was used to determine this confidence interval?
(b) Write down an expression for the margin of error M.
(c) Comment on how Tom could increase the precision of his findings.
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# 2.10 Cubic Spline Interpolation
The method of least squares provides, among other things, an alternative to ordinary interpolation that avoids the problem of overfitting. Another alternative is spline interpolation, which encompasses a range of interpolation techniques that reduce the effects of overfitting. The method of cubic spline interpolation presented here is widely used in finance. It applies only in one dimension, but is useful for modeling yield curves, forward curves, and other term structures.
A cubic spline is a function f : → constructed by piecing together cubic polynomials pk(x) on different intervals [x[k], x[k+1]]. It has the form
[2.112]
Consider points (x[1], y[1]), (x[2], y[2]), … , (x[m], y[m]), with x[1] < x[2] < … < x[m]. We construct a cubic spline by interpolating a cubic polynomial pk between each pair of consecutive points (x[k], y[k]) and (x[k+1], y[k+1]) according to the following constraints:
1. Each polynomial passes through its respective end points:
[2.113]
2. First derivatives match at interior points:
[2.114]
3. Second derivatives match at interior points:
[2.115]
4. Second derivatives vanish at the end points:
[2.116]
The above conditions specify a system of linear equations that can be solved for the cubic spline. In practice, it makes little sense to fit a cubic spline to fewer than five points. However, for the purpose of illustration, let’s interpolate a cubic spline between just three points.
Consider the points (x[k], y[k]) = (1, 1), (2, 5), (3, 4). We seek to fit a cubic polynomial on the interval [1, 2] and another cubic polynomial on the interval [2, 3]. These take the forms
[2.117]
[2.118]
Our first condition requires
[2.119]
[2.120]
[2.121]
[2.122]
The second condition requires
[2.123]
The third condition requires
[2.124]
Finally, the last condition requires
[2.125]
[2.126]
We have eight equations in eight unknowns. These can be expressed as
[2.127]
which we solve to obtain
[2.128]
Our two polynomials are
[2.129]
[2.130]
The cubic spline, along with the three points upon which it is based, is shown in Exhibit 2.15.
Exhibit 2.15: A cubic spline interpolated between the points (1, 1), (2, 5) and (3, 4) is constructed from two cubic polynomials.
###### Exercises
2.17
Interpolate a cubic spline between the three points (0, 1), (2, 2) and (4, 0).
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## Everyday Mathematics 4th Grade Answer Key Unit 2 Multiplication and Geometry
Exploring Square Numbers
A square number is a number that can be written as the product of a number multiplied by itself. For example, the square number 9 can be written as 3 ∗ 3.
Question 1.
Fill in the missing factors and square numbers.
We will fill the blanks factors and square numbers.
Explanation:
As the square number is a number that can be written as the product of a number multiplied by itself. So, here we will fill the blanks factors and square numbers.
Question 2.
What pattern(s) do you see in the factors? In the products?
The pattern do we see in the factors is Prime Factorization.
Explanation:
Here, in the above table, we have used the Prime Factorization pattern. The Prime Factorization method is defined as a way of finding prime numbers, which are multipled together to get the original number.
Question 3.
What other pattern(s) do you see in the table?
The other pattern that we have seen in the table is the squares of the numbers.
Question 4.
Write an equation to describe each array.
a.
Equation: ___
The equation will be 4 × 9 = 36.
Explanation:
As we have 36 dots in all and there are 4 rows with 9 dots in each row. So the equation will be 4 × 9 which is 36 dots.
b.
Equation: ___
The equation will be 5 × 5 = 25.
Explanation:
As we have 25 dots in all and there are 5 rows with 5 dots in each row. So the equation will be 5 × 5 which is 25 dots.
Question 5.
a. Which of the arrays above shows a square number? ___________
b. Explain. ________
a. The above arrays that show a square number is array b.
b. As there are 5 rows of 5 dots which is 25 dots.
Explanation:
In the above arrays that show a square number is array b as there are 5 rows of 5 dots which is 25 dots.
Practice
Question 6.
32, 45, 58, ____, ____, ____ Rule: _____
32, 45, 58, 71, 84, 97.
Explanation:
Here, we need to add 13. If we check the difference between 45 and 32 we will get the difference as 13. So, we can see that in the given series, so 13 is added to the given numbers. So the rule is adding 13.
Question 7.
___, ___, ___, 89, 115, 141 Rule: ____
11,37,63,89,115,141
Rule: subtract 26.
Explanation:
Here, we need to subtract 26. If we check the difference between 141 and 115 we will get the difference as 26. So, we can see that in the given series, so 26 is subtracted to the given numbers. So the rule is subtracting 26.
Area of a Rectangle
Question 1.
Draw a rectangle that has length of 9 units and width of 4 units.
Equation: ___
Area = ___ square units
The area of the rectangle is 36 square units.
Explanation:
Here, we have constructed a rectangle that has a length of 9 units and a width of 4 units. And here we need to find the area of the rectangle, so the area of the rectangle is
= 9 × 4
= 36 square units.
So the area of the rectangle is 36 square units.
Question 2.
Draw a rectangle that has a length of 7 units and a width of 8 units.
Equation: ___
Area = ___ square units
The area of the rectangle is 56 square units.
Explanation:
Here, we have constructed a rectangle that has a length of 7 units and a width of 8 units. And here we need to find the area of the rectangle, so the area of the rectangle is
= 7 × 8
= 56 square units.
So the area of the rectangle is 56 square units.
Use the formula A = I ∗ w to find the area of each rectangle.
Question 3.
Equation: ___
Area = ___ square units
The area of the rectangle is 48 square units.
Explanation:
Here, we need to find the area of the rectangle, so the area of the rectangle is
= 8 × 6
= 48 square units.
So the area of the rectangle is 48 square units.
Question 4.
Equation: ___
Area = ___ square units
The area of the rectangle is 48 square units.
Explanation:
Here, we need to find the area of the rectangle, so the area of the rectangle is
= 8 × 6
= 48 square units.
So the area of the rectangle is 48 square units.
Question 5.
Riley’s dining room tabletop is 9 feet long and 6 feet wide. What is the area of the tabletop?
Equation: _____
Area = ___ square feet
The area of the tabletop is 54 square feet.
Explanation:
Riley’s dining table top is 9 feet long and width is 6 feet, so the area of the tabletop is
Area = length × width
= 9 × 6
= 54 square feet.
So the area of the tabletop is 54 square feet.
Practice
Question 6.
368 – 59 = ____
The subtraction of two numbers is 309.
Explanation:
By subtracting the given numbers 368 – 59 we will get the result as 309.
Question 7.
194 – 147 = ___
The subtraction of two numbers is 47.
Explanation:
By subtracting the given numbers 194 – 147 we will get the result as 47.
Question 8.
____ = 1,729 – 623
The subtraction of two numbers is 1,106.
Explanation:
By subtracting the given numbers 1,729 – 623 we will get the result as 1,106.
Working with Factor Pairs
Write equations to help you find all the factor pairs of each number below. Use dot arrays, if needed.
Practice
Question 2.
356 + 433 = ___
The addition of two numbers is 789.
Explanation:
The addition of the two given numbers 356 + 433 is 789.
Question 3.
___ = 2,167 + 696
The addition of two numbers is 2,863.
Explanation:
The addition of the two given numbers 2,167 + 696 is 2,863.
Question 4.
___ = 4,578 – 2,232
The subtraction of two numbers is 2,346
Explanation:
The subtraction of the two given numbers 4,578 – 2,232 is 2,346.
Question 5.
3,271 – 1,089 = ___
The subtraction of two numbers is 2,182.
Explanation:
The subtraction of the two given numbers 4,578 – 2,232 is 2,182.
Finding Multiples
Question 1.
List the first 5 multiples of 4. ____________
4 × 1 = 4,
4 × 2 = 8,
4 × 3 = 12,
4 × 4 = 16,
4 × 5 = 20.
Explanation:
The first 5 multiples of 4 are:
4 × 1 = 4,
4 × 2 = 8,
4 × 3 = 12,
4 × 4 = 16,
4 × 5 = 20.
Question 2.
List the first 10 multiples of 2. _______________
2 × 1 = 2,
2 × 2 = 4,
2 × 3 = 6,
2 × 4 = 8,
2 × 5 = 10,
2 × 6 = 12,
2 × 7 = 14,
2 × 8 = 16,
2 × 9 = 18,
2 × 10 = 20.
Explanation:
The first 10 multiples of 2 is
2 × 1 = 2,
2 × 2 = 4,
2 × 3 = 6,
2 × 4 = 8,
2 × 5 = 10,
2 × 6 = 12,
2 × 7 = 14,
2 × 8 = 16,
2 × 9 = 18,
2 × 10 = 20.
Question 3.
a. List the first 10 multiples of 3. _________
3 × 1 = 3,
3 × 2 = 6,
3 × 3 = 9,
3 × 4 = 12,
3 × 5 = 15,
3 × 6 = 18,
3 × 7 = 21,
3 × 8 = 24,
3 × 9 = 27,
3 × 10 = 30.
Explanation:
The first 10 multiples of 3 is
3 × 1 = 3,
3 × 2 = 6,
3 × 3 = 9,
3 × 4 = 12,
3 × 5 = 15,
3 × 6 = 18,
3 × 7 = 21,
3 × 8 = 24,
3 × 9 = 27,
3 × 10 = 30.
b. List the first 10 multiples of 5. _________
5 × 1 = 5,
5 × 2 = 10,
5 × 3 = 15,
5 × 4 = 20,
5 × 5 = 25,
5 × 6 = 30,
5 × 7 = 35,
5 × 8 = 40,
5 × 9 = 45,
5 × 10 = 50.
Explanation:
The first 10 multiples of 5
5 × 1 = 5,
5 × 2 = 10,
5 × 3 = 15,
5 × 4 = 20,
5 × 5 = 25,
5 × 6 = 30,
5 × 7 = 35,
5 × 8 = 40,
5 × 9 = 45,
5 × 10 = 50.
c. List the multiples of 3 that are also multiples of 5. ________
Multiples of 3 that are also multiples of 5 are 15, 30, 45, 60, 75, 90.
Explanation:
The multiples of 3 that are also multiples of 5 are 15, 30, 45, 60, 75, 90.
Question 4.
Is 28 a multiple of 7? ___ Explain. ______
Yes, 28 is a multiple of 7.
Explanation:
Yes, 28 is a multiple of 7. As 28 is divisible by 7 and when we divide 28 by 7 we will get the reminder as zero. So 28 is a multiple of 7.
Question 5.
Is 35 a multiple of 6? __ Explain. _____
No, 35 is not a multiple of 6.
Explanation:
No, 35 is not a multiple of 6. As 35 is not divisible by 6 and when we divide 35 by 6 we will not get the reminder as zero. So 35 is not divisible by 6.
Question 6.
a. List the factors of 15. List the multiples through 15 of each factor.
b. Is 15 a multiple of each of its factors? Explain.
Yes, 15 is a multiple of each of its factors.
Explanation:
Yes, 15 is a multiple of each of its factors. As 15 is divisible by 1, 3, 5, 15, and by dividing we will get the reminder as zero. So 15 is a multiple of each of its factors.
Practice
Question 7.
24, ____, 48, ___, 72, ___ Rule: ____
24, 36, 48, 72, 84.
Explanation:
Here, we will add 12 to the given number 24 then we will get 36 and then we will add 12 to the number 36 then we will get 48 and we will continue until we will get the last number. So here the rule is adding 12.
Question 8.
___, 108, 162, ___, 270, ___ Rule: ____
54, 108, 162, 216, 270, 324.
Explanation:
Here, we need to add 54. If we check the difference between 162 and 108 we will get the difference as 54. So, we can see that in the given series, so 54 is added to the given numbers. So the rule is adding 54.
Question 9.
86, ___, 52, ___, 18, Rule: ___
86, 69, 52, 35, 18, 1.
Rule: The difference between consecutive numbers is 17. 86, 69, 52, 35, 18, 1.
Explanation:
Here, we will subtraction 17 between the consecutive numbers.
Question 10.
425, ___, 339, ____, 253, ____ Rule: ___
425, 382, 339, 296, 253, 210.
Rule: The difference between each pair of numbers is 43.
Explanation:
Here, the difference between each pair of the numbers is 43. So 425, 382, 339, 296, 253, 210.
Prime and Composite Numbers
A prime number is a whole number that has exactly two different factors—1 and the number itself. A composite number is a whole number that has more than two different factors. For each number:
• List all of its factors.
• Write whether the number is prime or composite.
• Circle all of the factors that are prime numbers.
Practice
Solve.
Question 10.
841 + 527 = _____
841 + 527 = 1,368.
Explanation:
The addition of the given two numbers 841 + 527 is 1,368.
Question 11.
____ = 3,263 + 5,059
8,322 = 3,263 + 5,059
Explanation:
The addition of the given two numbers 3,263 + 5,059 is 8,322.
Question 12.
7,461 + 2,398 = _____
7,461 + 2,398 = 9,859
Explanation:
The addition of the given two numbers 7,461 + 2,398 is 9,859.
Question 13.
___ = 4,172 – 3,236
936 = 4,172 – 3,236
Explanation:
The difference between the given numbers 4,172 – 3,236 is 936.
Question 14.
8,158 = 5,071 + __
The missing number is 3,087.
Explanation:
Let the missing number be x and we will replace the missing number with x, then 8,158 = 5,071 + x. So x is
x = 8,158 – 5,071
= 3,087.
Question 15.
3,742 – 3,349 = ___
3,742 – 3,349 = 393.
Explanation:
The difference between the given numbers 3,742 – 3,349 is 393.
Using Multiplication
Home Market sells 3 grapefruits for $2. Question 1. Darius spent$6 on grapefruits. How many did he buy? Use words, numbers, or diagrams to show your reasoning.
Darius bought 9 grapefruits.
Explanation:
As Darius spent $6 on grapefruits, here home market sells 3 grapefruits for$2. So the number of grapefruits did Darius bought for $6 is 3 + 3 + 3 which is 9. So Darius bought 9 grapefruits. Question 2. Jana bought 15 grapefruits. How much did she spend? Use words, numbers, or diagrams to show your reasoning. Answer: Jana spends$10.
Explanation:
As Jana bought 15 grapefruits. here home market sells 3 grapefruits for $2. So the number of grapefruits did Darius bought for$6 is 2 + 2 + 2 + 2 + 2 which is 10. So Jana spends $10. Question 3. On the back of this page, write a multiplication number story about buying grapefruits at Home Market. Show how to solve your number story. Answer: Explanation: Practice Write these numbers using words. Question 4. 12,309 _________________ Answer: Twelve thousand three hundred and nine. Explanation: I have written the given numbers using words which is Twelve thousand three hundred and nine. Question 5. 30,041 _________ Answer: Thirty Thousand and forty-one. Explanation: I have written the given numbers using words which is Thirty Thousand and forty-one. Question 6. 600,780 _________ Answer: Six hundred thousand seven hundred eighty. Explanation: I have written the given numbers using words which is Six hundred thousand seven hundred eighty. Question 7. 9,090,506 ________ Answer: Ninety Lakh Ninety Thousand Five hundred Six. Explanation: I have written the given numbers using words which is Ninety Lakh Ninety Thousand Five hundred Six. ### Everyday Math Grade 4 Home Link 2.6 Answer Key Converting Units of Time Use the measurement scales to fill in the tables and answer the questions. Question 1. Answer: Filled the table with minutes for the given hours using the measurement scales. Explanation: Here, we have filled the table with minutes for the given hours using the measurement scales. Question 2. Answer: Filled the table with minutes for the given hours using the measurement scales. Explanation: Here, we have filled the table with seconds for the given minutes using the measurement scales. Question 3. Zac worked on his spelling for 9 minutes last night and 8 minutes this afternoon. How many seconds did he work? Answer: seconds Answer: Zac worked 1,020 seconds. Explanation: Here, Zac worked on his spelling for 9 minutes last night and 8 minutes this afternoon, so the total number of minutes did Zac worked is 9 + 8 which is 17 minutes. Now we will convert minutes to seconds 17 × 60 which is 1,020 seconds. So Zac worked 1,020 seconds. questions 4. Eton’s baby sister took a nap for 2 hours and 22 minutes yesterday and 1 hour and 35 minutes today. How many more minutes did she sleep yesterday than today? Answer: ___ minutes Answer: Eton’s baby sister took a nap yesterday then today is 47 minutes. Explanation: Here, Eton’s baby sister took a nap for 2 hours and 22 minutes yesterday, so the total number of minutes 120 + 22 which is 142 minutes. And 1 hour and 35 minutes today, so the total number of minutes 60 + 35 is 95 minutes. And the total number of minutes did Eton’s baby sister took a nap yesterday then today is 142 – 95 = 47 minutes. Try This Question 5. How many seconds did Eton’s baby sister sleep all together? Answer: ___ seconds Answer: The total number of seconds did Eton’s sister sleeps is 14,220 seconds. Explanation: Eton’s baby sister sleeps all together 142 + 95 which is 237 minutes and we need to convert minutes to seconds 237 × 60 = 14,220 seconds. So the total number of seconds did Eton’s sister sleeps is 14,220 seconds. Practice Question 6. 945 + 1,055 = ___ Answer: 945 + 1,055 = 2000 Explanation: Add the two given numbers to get the resultant sum. Hence the answer is 2000. Question 7. 2,953 + 4,471 = ___ Answer: 2,953 + 4,471 = 7424 Explanation: Add the two given numbers to get the resultant sum. Hence the answer is 7424. Question 8. 4,552 + 4,548 = ___ Answer: 4,552 + 4,548 = 9100 Explanation: Add the two given numbers to get the resultant sum. Hence the answer is 9100. Question 9. 3,649 + 3,649 = ___ Answer: 3,649 + 3,649 = 7298 Explanation: Add the two given numbers to get the resultant sum. Hence the answer is 7298. ### Everyday Math Grade 4 Home Link 2.8 Answer Key Multiplicative Comparisons Family Note In this lesson students used comparison statements and equations to represent situations in which one quantity is a number of times as much as another quantity. For example: José saved$5 over the summer. His sister saved 3 times as much. How much money did José’s sister save? In this number story students compare the amount of money José saved to the amount his sister saved. Students write the equation 3 ∗ 5 = 15 to represent this comparison and solve the problem:
José’s sister saved \$15. Because these comparison statements and equations involve multiplication, they are called multiplicative comparisons.
Complete the problems below. Write an equation with a letter for the unknown and solve.
Question 1.
What number is 7 times as much as 9?
Equation with unknown:
n = 7 × 9
63
Explanation:
The number which is 7 times as much as 9 is 63 and the equation is 7 × 9 which is 63.
Question 2.
What number is 5 times as much as 6?
Equation with unknown:
n = 5 * 6
30
Explanation:
The number which is 5 times as much as 6 is 30 and the equation is 5 × 6 which is 30.
Question 3.
32 is 4 times as much as what number?
a. Equation with unknown: ___
n = 32 * 4
128
Explanation:
The number which is 32 times as much as 4 is 128 and the equation is 32 × 4 which is 128.
Question 4.
Write an equation to represent this situation and solve.
Ameer worked 3 times as many hours as Simi each week during the summer.
If Simi worked 10 hours each week, how many hours did Ameer work each week?
a. Equation with unknown: ____
The number of hours did Ameer work each week is 30 hours.
Explanation:
Here, Ameer worked 3 times as many hours as Simi each week during the summer and Simi worked 10 hours each week. So the number of hours did Ameer work each week 3 × 10 which is 30 hours. So the number of hours did Ameer work each week is 30 hours.
Practice
Question 5.
7,482 – 7,083 = ___
7,482 – 7,083 = 399
Explanation:
By subtracting the given numbers 7,482 – 7,083 the answer we get is 399.
Question 6.
7,702 – 3,581 = ___
7,702 – 3,581 = 4121
Explanation:
By subtracting the given numbers 7,702 – 3,581 the answer we get is 4121.
Question 7.
5,201 – 3,052 = ___
5,201 – 3,052 = 2149
Explanation:
By subtracting the given numbers 5,201 – 3,052 the answer we get is 2149.
Question .8
8,002 – 5,403 = ____
8,002 – 5,403 = 2599
Explanation:
By subtracting the given numbers 8,002 – 5,403 the answer we get is 2599.
Solving Multiplicative Comparison Number Stories
Make a diagram or drawing and write an equation to represent the situation. Then find the answer.
Question 1.
Judith collected 9 marbles. Swen has 6 times as many. How many marbles does Swen have?
Diagram or drawing:
Equation with unknown: ____
54 marbles
Explanation:
Here the Swen has 54 marbles. As Judith collected 9 marbles and Swen has 6 times as many as Judith it means the Swen has 9 * 6 = 54 marbles.
Question 2.
Sol ran 4 times as many minutes as Jerry. Jerry ran 12 minutes. How many minutes did Sol run?
Diagram or drawing:
Equation with unknown: ____
48 minutes
Explanation:
Here, Sol ran 4 times as many minutes as Jerry and Jerry ran 12 minutes. So the number of minutes did Sol ran 12 × 4 which 48 minutes.
Insert quantities into the number story. Make a diagram and write an equation to represent the story.
Question 3.
Lola picked apples. Eilene picked apples. Eilene picked times as many apples as Lola.
Diagram or drawing:
Equation with unknown: _____
Practice
Write these numbers in expanded form.
Question 4.
3,830 _________
3000 + 800 + 30
Question 5.
56,037 ________
50,000 + 6000 + 30 + 7
Question 6.
800,700 __
800000 + 700
Question 7.
716,305 _____
700,000 + 10,000 + 6000 + 300 + 5
Identifying Triangles
Write the letter or letters that match each statement.
Question 1.
Has perpendicular line segments
C, D
Explanation:
The perpendicular line segments in the given question are C, D
Question 2.
Has an obtuse angle
E, F
Explanation:
E and F are the triangles which are obtuse angle.
Question 3.
Has right angles
C, D
Explanation:
C and D are the right angles.
Question 4.
Has acute angles
A, B
Explanation:
A and B are the acute angles given in the question.
Question 5.
Has more than one kind of angle
C, D, E, F
Explanation:
C, D, E, and F are the one kind of angle that is given in the question.
Question 6.
Has only one kind of angle
A, B
Explanation:
A and B are the only kind of angle.
Question 7.
Does NOT have any right angles
A, B, E, F
Explanation:
A, B, E, and F do not have any right angles in the given question.
Question 8.
Is a right triangle
C, D
Explanation:
C and D is the right triangle in the given question.
Practice
Question 9.
List all the factors of 12.
1, 2, 3, 4, 6, 12
Explanation:
The factors of 12 are 1, 2, 3, 4, 6, 12 because each of those divides 12 without leaving a remainder.
Question 10.
Name the next 4 multiples of 7. 35, _____, _____, ___, ____
42, 49, 56, 63.
Explanation:
The next 4 multiples of 7 are 42, 49, 56, 63.
Question 1.
A parallelogram is a quadrilateral that has 2 pairs of parallel sides. Draw a parallelogram.
I have drawn a parallelogram that has 2 pairs of parallel sides.
Explanation:
As asked in the question I have drawn a parallelogram below.
Question 2.
a. Can a parallelogram have right angles? Explain.
b. Could a quadrilateral have 4 obtuse angles? Explain.
c. Name a quadrilateral that has at least 1 pair of parallel sides.
a.
b. No, all the four angles of a quadrilateral cannot be obtuse. As the sum of the angles of a quadrilateral is 360∘, they may have a maximum of three obtuse angles.
c. Trapezoids
Trapezoids have only one pair of parallel sides.
Question 3.
Draw a quadrilateral that has at least 1 right angle.
Question 4.
Draw a quadrilateral that has 2 separate pairs of equal length sides but is NOT a parallelogram.
This is called a ____.
Practice
Question 5.
5 ∗ 30 = ___
150
Explanation:
By multiplying the given numbers the answer we get is 150.
Question 6.
__ = 40 ∗ 3
120
Explanation:
By multiplying the given numbers the answer we get is 120.
Question 7.
___ = 80 ∗ 6
480
Explanation:
By multiplying the given numbers the answer we get is 480.
Question 8.
6 ∗ 70 = ____
420
Explanation:
By multiplying the given numbers the answer we get is 420.
Drawing Lines of Symmetry
Question 1.
Draw the other half of each picture to make it symmetrical. Use a straightedge to form the line of symmetry.
Question 2.
Draw a line of symmetry for each figure.
Question 3.
List four items in your home that are symmetric. Pick one item and draw it below, including at least one line of symmetry.
Item: __ Item: ___ Drawing:
Item: __ Item: ___
Practice
Question 4.
___ = 2,767 + 3,254
6021
Explanation:
Add the two given numbers to get the resultant sum. Hence the answer is 6021.
Question 5.
193 + 6,978 = ___
7171
Explanation:
Add the two given numbers to get the resultant sum. Hence the answer is 7171.
Question 6.
7,652 − 5,388 = ___
5674
Explanation:
By subtracting the given numbers the answer we get is 5674.
Question 7.
___ = 4,273 − 1,678
2595
Explanation:
By subtracting the given numbers the answer we get is 2595.
Identifying Patterns
Question 1.
Complete.
What patterns do you see?
The pattern is multiples of 9.
Explanation:
The pattern is multiples of 9 in which 2 is in and 18 is out and that series continuous till the number 6.
Question 2.
Complete.
What patterns do you see?
Explanation:
The pattern is adding 11 in which 11 is in and 22 is out and that series continuous till the number 55.
Question 3.
Study the pattern
a. Draw the next step in the pattern. What patterns do you notice?
b. How many circles will be in the 6th step? __ In the 10thstep? ___
c. How did you figure out how many circles will be in the 10th step?
a. In this question the number of circles is odd and increases by 2 every time.
Explanation:
I have drawn the next step in the pattern below.
b. 11; 19
Explanation:
There will be 11 circles in the 6th step. And 19 circles in the 10th step.
c. In this, each step is the next odd number so I skip counted from 1 by 2 until I got to the 10th step.
Explanation:
We can figure it out by observing each step given in the pattern. As in each step, the next one is an odd number so we can skip counted from 1 by 2 until we got to the 10th step.
Practice
Question 4.
800,000 + 90 = ___
800,090
Explanation:
Add the two given numbers to get the resultant sum. Hence the answer is 800,090.
Question 5.
200,000 + 50,000 + 4 = ___
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# Bar Graphs
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Progress
Practice Bar Graphs
Progress
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Bar Graphs
Have you ever looked at a bar graph? Bar graphs are used all the time.
It is the first week of September and while there are still vegetables growing in Alex and Tania’s garden, there has been a lot of harvesting during the months of July and August. Tania and Alex have kept track of how many vegetables were harvested each month. Here is their data:
July August
30 carrots 60 carrots
10 tomatoes 20 tomatoes
25 zucchini 30 zucchini
15 squash 25 squash
10 potatoes 20 potatoes
Tania and Alex want to display their data. Tania wants to make a bar graph that shows the data for July. Alex is going to create a display for August.
### Guidance
We make bar graphs from a set of data. It is called a bar graph because it is a visual display of data using bars. The number of items tells us how many bars the graph will have. The amount of each item tells us how tall each bar will be. Let’s make a graph of the following data. It tells how many hours students in the fifth, sixth, seventh, and eighth grade classes volunteered in a month.
Class Number of Hours
$5^{th}$ 51
$6^{th}$ 88
$7^{th}$ 75
$8^{th}$ 39
You can see that this information has been written in the form of a frequency table. It shows us how many hours each class has worked.
Now we can take this and draw a bar graph to show us the information.
To make a bar graph, we draw two axes. One axis represents the items, and the other represents the amounts. The “items” in this case are each class. The amounts are the number of hours the classes worked. For this example, our axes might look like the graph below. Remember to label each axis!
Next, we need to choose scale for the amounts on the left side of the bar graph. We can use scales of 1, 2, 5, 10, 20, 50, 100, 1,000, or more. To choose the scale, look at the amounts you’ll be graphing, especially the largest amount. In our example, the greatest value is 88 . If we used a scale of 100, the scale marks on the left side of the graph would be 0, 100, 200, and so on. It would be very difficult to read most of our amounts on this scale because it is too big. Every amount would fall between 0 and 100, and we would have to guess to be more specific! On the other hand, if we used a small scale, such as 5, the graph would have to be very large to get all the way up to 90 (since our greatest value is 88).
It makes the most sense to use a scale that goes from 0 to 90 counting by 10’s. That way each value can easily represent the hours that each class worked.
Here is what the graph looks like with the scale filled in.
Now we can draw in the bars to represent each number of hours that the students worked.
Look at how easy it is to get a visual idea of which class worked the most hours and which class worked the least number of hours. We can use bar graphs to give us a visual sense of the data.
Now let's practice by using a bar graph to analyzing data.
#### Example A
Which state has the highest average price for gasoline?
Solution: Hawaii
#### Example B
Which state has the lowest average price?
Solution: Missouri
#### Example C
Which state has the second highest average price?
Solution: California
Tania and Alex want to display their data. They have decided that bar graphs are the best way to do that. Tania is going to make a bar graph that shows the vegetable counts for July.
Let’s start by helping Tania to make a bar graph to represent July’s harvest. Here are her counts.
July
30 carrots
10 tomatoes
25 zucchini
15 squash
10 potatoes
Now we can make the bar graph. We know that the amounts range from 10 to 30, so we can start our graph at 0 and use a scale that has increments of five. Here is the bar graph.
Next, Alex can create his bar graph for August. Here is his data.
August
60 carrots
20 tomatoes
30 zucchini
25 squash
20 potatoes
Notice that these numbers are different than the ones Tania had. Here our range is from 20 to 60. Because of this, we can use a scale of 0 to 60 in increments of five. Here is Alex’s bar graph.
### Guided Practice
Here is one for you to try on your own.
Based on this graph, how many seventh graders have a favorite activity of watching tv?
First, you can look at the column that refers to television. Then look at the vertical axis.
9 seventh graders have "watching tv" as their favorite activity.
### Explore More
Directions: Use the bar graph to answer the following questions.
1. How many students were asked if they have summer jobs?
2. What is the range of the data?
3. What are the three jobs that students have?
4. How many students do not have a summer job?
5. How many students babysit?
6. How many students do yard work in the summer?
7. How many students work at an ice cream stand in the summer?
8. If ten more students got a job this summer, how many students would have summer jobs?
9. If each category had double the number of students in it, how many students would have summer jobs?
10. How many students would babysit?
11. How many students would work at an ice cream stand?
12. How many students wouldn’t have a summer job?
13. What scale was used for this graph?
14. What interval was used in the scale?
15. What is the difference between working at an ice cream stand and doing yard work?
### Vocabulary Language: English
Analyze
Analyze
To analyze is to look at data and draw conclusions based on patterns or numbers.
bar graph
bar graph
A bar graph is a plot made of bars whose heights (vertical bars) or lengths (horizontal bars) represent the frequencies of each category, with space between each bar.
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Numbers
# Numbers - Notes | Study Mathematics Olympiad for Class 1 - Class 1
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Introduction
Zero
• Zero is the smallest counting number. It is represented as 0.
• The image below shows a jar with two balls – black and white
• Now if you remove the two balls, the jar contains 0 balls—
One-digit numbers
Numbers from 0 to 9 are one-digit numbers.
Number line
A number line can be defined as a straight line with numbers placed at equal distance along its length.
Place value
• The value of each digit in a number is known as place value.
• Starting from right, the first place is units or ones (O) and the second place is tens (T).
Example: The place value of the two-digit number 13 will be as follows:
Two-digit numbers
• Numbers from 10 to 99 are two-digit numbers.
• Here are some examples for two-digit numbers.
• How to get the number 18?
• What number do you get when 3 tens is added to 4 ones?
• What number do you get when 5 tens is added to 1 one?
• What number do you get when 7 tens is added to 2 ones?
• What number do you get when 8 tens is added to 6 ones?
• What number do you get when 9 tens is added to 8 ones?
• See the following examples.
Given that
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Greater than
• Greater than Which number is greater: 7 or 10?
• We can identify the greater number using a number line.
• 10 lies on the right side of 7, so 10 is greater than 7 and we write it as 10 > 7.
Less than
• Which number is smaller: 34 or 53?
• We compare the numbers using a number line.
• 34 lies on the left side of 53, so 34 is smaller than 53 and we write it as 34 < 53.
Equal to
• If the two numbers given are same, we say that the numbers are equal.
• We can say that 7 is equal to 7 and write it as 7 = 7
Example: Are the numbers 19 and 91 equal?
The given numbers are 19 - Nineteen and 91 - Ninety-one. They are not same. So, 19 and 91 are not equal.
Writing numbers in reverse order.
9, 8, 7, 6, 5, 4, 3, 2, 1, 0 is the backward order of one-digit numbers.
99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89, _______, 10 is the reverse order of two-digit numbers.
Ascending and descending orders.
• Ascending order means arrangement of numbers from small to big.
• Descending order means arrangement of numbers from big to small.
Examples:
• Arrange the given numbers in ascending order: 54, 28, 47, 87, 18, 64
• Arrange the numbers from the smallest to the biggest or largest.
So, the order is 87, 64, 54, 47, 28, and 18.
• Arrange the numbers 45, 12, 56, 84 in descending order.
• Arrange the numbers from the biggest or largest to the smallest.
So, the order is 84, 56, 45, 12.
Just before, Just after, and In-between
• If the numbers are 13, 14, and 15,
• 13 comes just before 14.
• 15 comes just after 14.
• 14 lies in between 13 and 15.
Skip counting
Numbers obtained by adding the same number.
Example:
• 2, 4, 6, 8, 10 are the numbers we get by skip count of 2.
• 3,6,9,12,15 are the numbers we get by skip count of 3.
• 4, 8, 12, 16, 20 are the numbers we get by skip count of 4.
• 5, 10, 15, 20, 25 are the numbers we get by skip count of 5.
• 10, 20, 30, 40, 50 are the numbers we get by skip count of 10.
The document Numbers - Notes | Study Mathematics Olympiad for Class 1 - Class 1 is a part of the Class 1 Course Mathematics Olympiad for Class 1.
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## Mathematics Olympiad for Class 1
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Home | | Maths 8th Std | Basic Arithmetic Operations on Rational Numbers
# Basic Arithmetic Operations on Rational Numbers
All the rules and principles that govern fractions in the basic operations apply to rational numbers also.
Basic Arithmetic Operations on Rational Numbers
All the rules and principles that govern fractions in the basic operations apply to rational numbers also.
There can be four different situations while doing addition.
Type 1 : Adding numbers that have same denominators
This is simply like adding like fractions and the result is the sum of the numerators divided by their common denominator.
Example 1.10
Solution:
Write the given rational numbers in the standard form and then add them.
Type 2 : Adding numbers that have different denominators
After writing the given rational numbers in the standard form, use the LCM of their denominators to convert the numbers into equivalent rational numbers with a common denominator so that this reduces to Type1.
Example 1.11
The additive inverse of a rational number is another rational number which when added to the given number, gives zero.
For example, 4/3 and -4/3 are additive inverses of each other, since their sum is zero.
Think
Is zero a rational number? If so, what is its additive inverse?
Solution: Yes zero is a national number. Additive inverse of zero is zero.
3. Subtraction
4. Multiplication
Product of two or more rational numbers is found by multiplying the corresponding numerators and denominators of the numbers and then writing them in the standard form.
5. Multiplicative Inverse
If the product of two rational numbers is 1, then each of them is said to be the reciprocal or the multiplicative inverse of the other.
Think
What is the multiplicative inverse of 1 and –1?
Solution: Multiplicative inverse of 1 is 1 and −1 is −1.
For the rational number a, its reciprocal is 1/a and vice versa since a × 1/a= 1/a × a = 1.
For the rational number a/b , its multiplicative inverse is b/a and vice versa since a/b × b/a = b/a × a/b = 1.
6. Division
The idea of reciprocals of fractions is extended to the division of rational numbers also. To divide a given rational number by another rational number, we have to multiply the given rational number by the reciprocal of the second rational number. That is, division is simply multiplying by the multiplicative inverse of the divisor.
Solution:
(i) −7/3 ÷ 5 = [−7/3] ÷ [5/1] = [−7/3] × [1/5] = −7/15
(ii) 5 ÷ (−7/3) = [5/1] × [3/−7] = 15/−7 = −2 (1/7)
(iii) −7/3 ÷ 35/62 = [−7/3] × [6/35] = −2/5
Tags : Numbers | Chapter 1 | 8th Maths , 8th Maths : Chapter 1 : Numbers
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
8th Maths : Chapter 1 : Numbers : Basic Arithmetic Operations on Rational Numbers | Numbers | Chapter 1 | 8th Maths
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# Find a recurrence relation satisfied by this sequence. a_{n}=n^{2}+n
Find a recurrence relation satisfied by this sequence.
$$\displaystyle{a}_{{{n}}}={n}^{{{2}}}+{n}$$
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Given:
$$\displaystyle{a}_{{{n}}}={n}^{{{2}}}+{n}$$
Let us first determine the first term by replacing n in the given expression for $$\displaystyle{a}_{{{n}}}$$ by 0:
$$\displaystyle{a}_{{{0}}}={0}^{{{2}}}+{0}={0}$$
Let us similarly determine the next few terms as well:
$$\displaystyle{a}_{{{1}}}={1}^{{{2}}}+{1}={2}={a}_{{{0}}}+{2}{\left({1}\right)}$$
$$\displaystyle{a}_{{{2}}}={2}^{{{2}}}+{2}={6}={a}_{{{1}}}+{2}{\left({2}\right)}$$
$$\displaystyle{a}_{{{3}}}={3}^{{{2}}}+{3}={12}={a}_{{{2}}}+{2}{\left({3}\right)}$$
$$\displaystyle{a}_{{{4}}}={4}^{{{2}}}+{4}={20}={a}_{{{3}}}+{2}{\left({4}\right)}$$
$$\displaystyle{a}_{{{5}}}={5}^{{{2}}}+{5}={30}={a}_{{{4}}}+{2}{\left({5}\right)}$$
$$\displaystyle{a}_{{{6}}}={6}^{{{2}}}+{6}={42}={a}_{{{5}}}+{2}{\left({6}\right)}$$
We note that each term is the previous term increased by 2n-1:
$$\displaystyle{a}_{{{n}}}={a}_{{{n}-{1}}}+{2}{n}-{1}$$
Thus a recurrence relation for $$\displaystyle{a}_{{{n}}}$$ is then:
$$\displaystyle{a}_{{{0}}}={0}$$
$$\displaystyle{a}_{{{n}}}={a}_{{{n}-{1}}}+{2}{n}-{1}$$
Note: There are infinitely many different recurence relations that satisfy any sequence.
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AMM Problem
## AMM Problem 11544
Max Alekseyev, Department of Computer Science and Engineering, University of South Carolina.
Frank Ruskey, Department of Computer Science, University of Victoria.
$\def\I#1{{[\![#1]\!]}} % Iversonian$
### Our solution to the problem:
We make two observations initially. First, for any function $A(m)$, if $A(0) = 0$ then $A(m) = \sum_{1 \le n \le m} \left( A(n) - A(n-1) \right)$ because all terms cancel each other except $A(m)$. Secondly, if $t (2k+1) = n+k$, then multiplying by 2 and subtracting $2k+1$ we obtain the middle equation below: $(2t-1)(2k+1) = 2t(2k+1) - (2k+1) = 2(n+k) -(2k+1) = 2n-1.$ Thus, since any divisor of an odd number must be odd, $\I{ 2k+1 \ \backslash \ n+k } = \I{ 2k+1 \ \backslash \ 2n-1 },$ where $\I{P}$ is $0$ or $1$ depending on whether $P$ is true or not and $\backslash$ is the "divides" symbol popularized in the book Concrete Mathematics as an alternative to $|$. We can now proceed as follows, where $A(m)$ is the sum that we are trying to evaluate: \begin{align*} A(m) & = \sum_{k=0}^{m-1} \phi(2k+1) \left\lfloor \frac{m+k}{2k+1} \right\rfloor \\ & = \sum_{n=1}^m \left( A(n) - A(n-1) \right) \ \ \ \ \ \text{Note that $A(0)=0$.} \\ & = \sum_{n=1}^m \left( \sum_{k=0}^{n-1} \phi(2k+1) \left\lfloor \frac{n+k}{2k+1} \right\rfloor - \sum_{k=0}^{n-2} \phi(2k+1) \left\lfloor \frac{n-1+k}{2k+1} \right\rfloor \right) \\ & = \sum_{n=1}^m \sum_{k=0}^{n-1} \phi(2k+1) \left( \left\lfloor \frac{n+k}{2k+1} \right\rfloor - \left\lfloor \frac{n+k-1}{2k+1} \right\rfloor \right) \\ & = \sum_{n=1}^m \sum_{k=0}^{n-1} \phi(2k+1) \I{ 2k+1 \ \backslash \ n+k } \\ & = \sum_{n=1}^m \sum_{k=0}^{n-1} \phi(2k+1) \I{ 2k+1 \ \backslash \ 2n-1 } \\ & = \sum_{n=1}^m \sum_{2k+1 \backslash 2n-1} \phi(2k+1) \\ & = \sum_{n=1}^m (2n-1) \\ & = m^2. \end{align*}
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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 6 Statistics Ex 6.3 Text Book Back Questions and Answers, Notes.
## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 6 Statistics Ex 6.3
Miscellaneous Practice Problems
Question 1.
Draw a pie chart for the given table.
Converting the area in percentage into components parts of 360°. we have.
Continental Area.
Question 2.
The data on modes of transport used by the students to come to school are given below. Draw a pie chart for the data.
Converting the percentage into components parts of 360°. we have
Mode of Transport by students.
Question 3.
Draw a histogram for the given frequency distribution.
The given distribution is discontinuous.
Lower boundary = lower limit – $$\frac { 1 }{ 2 }$$ (gap between the adjacent class interval)
= 41 – $$\frac { 1 }{ 2 }$$(1) = 40.5
Upper boundary = Upper limit + $$\frac { 1 }{ 2 }$$ (gap between the adjacent class interval)
= 45 + $$\frac { 1 }{ 2 }$$(1) = 45.5
Now continuous frequency table is as below
Question 4.
Draw a histogram and the frequency polygon in the same diagram to represent the following data.
The given distribution is discontinuous.
Lower boundary = lower limit – $$\frac { 1 }{ 2 }$$ (gap between the adjacent class interval)
= 50 – $$\frac { 1 }{ 2 }$$(1) = 49.5
Upper boundary = Upper limit + $$\frac { 1 }{ 2 }$$(gap between the adjacent class interval)
= 55 + $$\frac { 1 }{ 2 }$$(1) = 55.5
∴ The continuous frequency table is as below.
Challenging problems
Question 5.
Form a continuous frequency distribution table and draw histogram from the following data.
Age (in years) No of persons Under 5 1 Under 10 12 Under 15 19 Under 20 26 Under 25 27 Under 30 35 Under 35 38 Under 40 45 Under 45 48 Under 50 53
Converting into continuous distribution we have
Class interval No. of persons 0 – 5 1 5 – 10 11 10 – 15 7 15 – 20 7 20 – 25 1 25 – 30 8 30 – 35 3 35 – 40 7 40 – 45 3 45 – 50 5 Total 53
Question 6.
A rupee spent in a cloth manufacturing company is distributed as follows. Represent this in a pie chart.
Particulars Paise Farmer 20 Spinner 35 Dyer 15 Weaver 15 Printer 05 Salary 10
1 Rupee = 100 paise.
Expenditure of a cloth manufacturing company.
Question 7.
Draw a histogram for the following data.
Lower boundary lower limit – $$\frac { 1 }{ 2 }$$(gap between the adjacent class interval)
= 15 – $$\frac { 1 }{ 2 }$$(10) = 10
= 15 + $$\frac { 1 }{ 2 }$$(10) = 20
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# The different kinds of shapes in math
Written by catherine copeland
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Geometry classes cover the study of shapes. Students study different types of shapes in the different grade levels, eventually calculating their areas and perimeters. As students are promoted higher through the grade levels, they will study increasingly difficult and more complex shapes, often using a coordinate plane. The study of shapes is an ongoing process in higher math classes. The properties of triangles, for instance, are the basis of trigonometry.
A quadrilateral has four sides. The most common and basic quadrilateral is the square -- a four-sided figure with equal sides whose angles add up to 360 degrees. Other examples of quadrilaterals include rectangles, parallelograms, rhombuses, trapezoids and kites. A rectangle has four right angles. A rhombus is a parallelogram with all sides having equal length. Two of the trapezoid's four sides are parallel. Students often begin their studies of geometry with quadrilaterals, but these shapes will continue to make appearances during the coming school years.
## Triangles
Students will learn about triangles in beginning geometry, and will study them thoroughly in trigonometry. A triangle is a three-sided polygon whose interior angles add up to 180 degrees. The types of triangles include equilateral, scalene, obtuse, right and isosceles triangles. An equilateral triangle has all sides of the same length. In a scalene triangle, all sides are different lengths. An obtuse triangle has an angle greater than 90 degrees, while an acute triangle's angles are all less than 90 degrees.
## Other Polygons
A polygon is a figure closed in an all sides by connecting line segments. These include quadrilaterals and triangles. Other polygons, however, have more than four sides. Common polygons include the pentagon, with five sides; the hexagon, six sides; the heptagon, seven sides; and the octagon, with eight sides. In a regular polygon, all sides are of the same length, while the sides of an irregular polygon are not the same length.
## Circles
While it is a simple shape for beginning math students, it is a complex shape to study in later years. A circle is defined as a set of points on a plane that are all the same distance from a fixed point. A circle has no line segments or sides. Circles are often studied by a diameter and a radius. The properties of circles include the number pi. Pi, approximated as 3.14, represents the relationship between a circle's diameter and its circumference. The diameter is the width of a circle, and circumference is the distance around a circle.
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# How To Do Substitution??In math class we are doing these substitution questions and I don't understand how to do them at all. They are all like this: The sum of two numbers is 8. Three times the...
How To Do Substitution??
In math class we are doing these substitution questions and I don't understand how to do them at all. They are all like this:
The sum of two numbers is 8. Three times the first plus four times the second is 29. Find the numbers.
If you could explain how to do it in steps that would really help! Thanks!! :)
Asked on by wwjd
### 5 Answers |Add Yours
job518 | College Teacher | (Level 2) Adjunct Educator
Posted on
Since these are two pretty simple equations, you could also do the following:
For (1) a + b = 8 and (2) 3a + 4b = 29
solve (1) for either variable (in this case they both have a coefficient of one so both are just as easy). So subtract b from each side and get a = 8 - b.
Now plug in 8 - b for a in (2). You will get 3 ( 8 - b ) + 4b = 29. Now simplify and solve for b.
3 ( 8 - b ) + 4b = 24 - 3b + 4b = 24 + b = 29 => subtract 24 from both sides => b = 5
Now you can substitue b=5 in (1) and get a + 5 = 8 => a =3.
This approach is really best when you are given two simple equations without exponents and a lot of dividing. There are so many ways to do substitution that as long as you are doing legitimate math "moves" and being very careful in simplifying, you will get the same answers. Practice really is the best way to learn it. The link below should help a good deal.
skyler12 | eNotes Newbie
Posted on
well you use the key words that the expression says to help you for example
the word sum means addition or the word of means to multiply like that. I put some other words down below to help you a little more
Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian
Posted on
The sum of two numbers is 8. Three times the first plus four times the second is 29. Find the numbers.
First we must parse the sentence and write it algebraically.
The sum of two numbers is 8: a + b = 8
Three times the first (3a) plus four times the second (+ 4b) is 29 (=29) so:
3a + 4b = 29.
a + b = 8
Given our two equations, we can solve for a and b. In general, you need as many unique equations as you have variables.
We know a in terms of b: a = 8 - b
Plug this into the second equation:
3(8-b) + 4b = 29 Solve for b.
We know b in terms of a: b = 8 - a
Plug into second equation:
3a + 4(8-a) = 29 Solve for a.
neela | High School Teacher | (Level 3) Valedictorian
Posted on
Substitution and elimation are the techniques employed to used while solving two (or more) simultaneous equations with two (or more) unknwons.Once we detrmined the value of a variable, we substitute the value in any one of the equations to detrmine the other unknown.
We assume two unknown numbers x and y.There two conditions given which is obeyed by these numbers:
By first condition the sum of the numbers, x+y =8 .......... (1).
By 2nd condition, 3times first or 3x and 4times the 2nd or 4y add up 3x+4y and this is 29. So the the second codition is:
3x+4y = 29...............(2).
The two given conditone are the reation between two numbers x and y in the form of above 2equations in 2 variables.
Substitution is a technique, by which we reduce the two equations in two variables to one equation in one variable.
We can use from the first equation x+y = 8, x = 8-y in the 2nd equation and replace x in 2nd equation by 8-y, as 3(8-y)+4y = 29 and solve for y. You get now 24-3y+4y =29. Or 24+y = 29. Or y =5.
Using this already detrmined value of y=5 in any of the two equations, x+y=8 or 3x+4y = 29, we get x+5= 8 . Or x = 8-5 =3. Here also, when we determined one of the value of the two variables, we use the technique of substitution of that velue of the varible to find the value of the other variable.
krishna-agrawala | College Teacher | (Level 3) Valedictorian
Posted on
In the type of question there are two variables to be found out - the two numbers. To find value of two numbers using algebraic equation we need a pair of equation. These two equation can be formed by the two conditions given in the example. That is:
1. The sum of two numbers is 8.
2. Three times the first plus four times the second is 29.
Let us assume that the two numbers are x and y.
Then, using the first given condition we form the equation:
x + y = 8 .... (1)
And using the second given condition we form the equation:
3x + 4y = 29 .... (2)
We can sole these equation for x an y as follows.
Multiplying equation (1) by 3 we get:
3x + 3 y = 24 ... (3)
subtracting equation 3 from equation 2 we get:
3x - 3x + 4y - 3y = 29 -24
Simplifying the above equation we get:
y = 5.
Substituting this value of y in equation (1), that is using the number 5 instead of y, we get:
x + 5 = 8
Therefore x = 8 - 5 = 3
Therefore the two numbers are 3 and 5.
We’ve answered 319,379 questions. We can answer yours, too.
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# 51000 in words
51000 in words is written as Fifty One Thousand. In 51000, 5 has a place value of ten thousand and 1 has the place value of thousand. The article on Place Value gives more information. The number 51000 is used in expressions that relate to money, distance, length, population and many more. For example, “A new township has Fifty One Thousand people.” We can also use 51000 as “A Mango farm area was 5.1 hectares. It means the area was 51000 sq meters.”
51000 in words Fifty One Thousand Fifty One Thousand in Numbers 51000
## How to Write 51000 in Words?
We can convert 51000 to words using a place value chart. The number 51000 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.
Ten thousand Thousands Hundreds Tens Ones 5 1 0 0 0
Thus, we can write the expanded form as:
5 × Ten thousand + 1 × Thousand + 0 × Hundred + 0 × Ten + 0 × One
= 5 × 10000 + 1 × 1000 + 0 × 100 + 0 × 10 + 0 × 1
= 51000
= Fifty One Thousand
51000 is the natural number that is succeeded by 50999 and preceded by 51001.
51000 in words – Fifty One Thousand.
Is 51000 an odd number? – No.
Is 51000 an even number? – Yes.
Is 51000 a perfect square number? – No.
Is 51000 a perfect cube number? – No.
Is 51000 a prime number? – No.
Is 51000 a composite number? – Yes.
## Solved Example
1. Write the number 51000 in expanded form
Solution: 5 x 10000 + 1 x 1000 + 0 x 100 + 0 x 10 + 0 x 1
Or Just 5 x 10000 + 1 x 1000
We can write 51000 = 50000 + 1000 + 0 + 0 + 0
= 5 x 10000 + 1 x 1000 + 0 x 100 + 0 x 10 + 0 x 1
## Frequently Asked Questions on 51000 in words
### How to write 51000 in words?
51000 in words is written as Fifty One Thousand.
### State True or False. 51000 is divisible by 4?
True. 51000 is divisible by 4.
### Is 51000 divisible by 10?
Yes. 51000 is divisible by 10. It is also divisible by 2, 3, 4, 5, 6, 10, 100 and 1000.
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How to play hyper sudoku
# Logic for developing strategies in hypersudoku
### The rules for playing hypersudoku are very simple.
1. All rows must have all the numbers from 1 - 9 in them. (none can be repeated) There are 9 rows in the game.
2. All colums must have all the numbers from 1 - 9 in them. (none can be repeated) There are 9 colums in the game.
3. All squares must have all the numbers from 1 - 9 in them. (none can be repeated) There are 13 squares in the game.
### New Game
This is the first of a series of pages on Hypersudoku game logic for new comers and complete beginners. This logic is necessary to develop your strategies for the game of hypersudoku.
As you will see in the following diagrams there are many starting points for the game in figure 1. This is a fairly typical layout for an easy level hypersudoku game. These are only a few of the many starting points in this particular game.
Figure 1
### Example 1
Looking at the center square we can see that colum 4 has a 9 at the bottom(circled in red). This means that a nine can not appear again in this colum. Colum 5 also has a 9 in it. This time up in the top middle square.
This leaves colum 6 as the only option for a 9 to appear in the center square. As 6 and 4 take up 2 of the 3 squares available, there is only one square left. This is the one with the blue tick in it. This square will contain the 9.
Figure 2
### Example 2
In figure 3 if we look at the top right square, we can see that rows 2 and 3 already have a 5 (circled in red) in them.
This leaves the square with the blue tick as the only possibility for a 5 in this square.
Figure 3
### Example 3
Again the same starting game, but this time looking at the middle right square. Rows 4 and 6 contain a 4 already, leaving only row 3.
As the other 2 squares already contain numbers, the square with the blue tick is the only square that can contain a 4.
Figure 4
### Example 4
In figure 5 we can see that the top row and third row already have a 6 in them. The sixth colum also has a 6 in it.
This leaves the square with the blue tick as the only square that could contain a 6 in this block.
Figure 5
### Example 5
If we look at the bottom center square we can see that colums 5 and 6 both contain a 5 (circled in red).
This leaves only one square in this block that can contain a 5.(square with blue tick)
Figure 6
### Example 6
In figure 7 we see that colum 1 and 2 have a 2 in them already. Also note that row 5 has a 2 in it. (circled in red)
Again this only leaves the square with the blue tick in it as the only possible square in this block that could contain a 2.
Figure 7
As you can see there are many possible first moves to start the game of hypersudoku. These are certainly not the only starting moves for this particular game.
If you would like to try playing a game just click on this link.
# Play Now
The images used in this article are public domain and may be redistributed.
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# Gr 5_NF_FractionsMultiplication_Problem_Construct_AngelasIdea
I used this task to continue our 5th grade discussion about multiplying fractions. Prior to this talk frame, our class discussed the difference between multiplying whole numbers and multiplying fractions, and explored why fractions multiplied together result in a fraction that contains smaller pieces. Students struggled with the deeper conceptual understanding of why, and this talk frame supported that discussion.
Microsoft Word version: 5_NF_FractionsMultiplication_Problem_Construct_AngelasIdea
# Gr 5_OA_AlgebraPatterns_Problem_Construct_DotPattern
Dot Pattern is a scaffolded task in which students must create an expression to represent the number of dots in a figure at any given point in a pattern. Students are given a visual representation of the first three figures and asked to think about how the figures change each time, and how the next figures would look. Ultimately, students will generate an algebraic expression to represent any figure. Students must construct an argument in order to explain why an expression works and where it comes from.
Microsoft Word version: 5_OA_AlgebraPatterns_Problem_Construct_DotPattern
Getting There is a problem created for fifth graders on adding numbers in different combinations. Students are given a distance a frog hops in three hops and a set of questions asking them about three numbers that will add up to the distance. The task provides a variation to include more argumentation language, specifically to explain their thinking, as they construct a response.
# Gr 5_NF_FractionsMultiplicationDivision_Problem_Construct_MakingHotCocoa
Making Hot Cocoa is a task on fraction multiplication and division for a fifth-grade classroom. Students are given a certain amount of cocoa powder and the fraction amount needed for each cup. Through a series of questions containing argumentative language, students are asked to construct a response to how many cups of cocoa they can make. Students are to construct a solution through a variety of methods, including pictures, mental thought, multiplication, and division. Sample solutions and commentary are provided.
Microsoft Word version: 5_NF_FractionsMultiplicationDivision_Problem_Construct_MakingHotCocoa
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# Gr 5_NF_FractionsMultiplication_Problem_Construct
This task was used with 4th and 5th graders because it focused on multiplying fractions by whole numbers and converting common measurements. The main objective was to use a real world situation so that students would develop a deeper understanding of multiplying fractions by whole numbers as well as converting ounces and cups.
Microsoft Word version: 5_NF_FractionsMultiplication_Problem_Construct
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This task is designed for fifth graders working on fraction fluency. Students must know the vocabulary term “sum” and be able to add fractions, as they are asked to find the sum of two fractions with different denominators. Students must be able to determine if the fraction is closer to a half or whole by rounding, as well. Finding common denominators, creating equivalent fractions, and adding fractions are all skills necessary to complete this task. The problem allows space to solve as well as lines to justify a critique of student work.
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# Gr 5_NF_Fractions_Problem_Construct_IllustrativeMathematics
This is a task from Illustrative Mathematics. The students constructed an argument that described how they found equivalent fractions. The directions for this task were slightly modified for my fifth grade students. My fifth grade students we asked to come up with no more than six equivalent fractions for each of the diagrams on part b. Beyond what is already stated this task also met CSSSM:CCSS.MATH.CONTENT.4.NF.A.1:
Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b)
by using visual fraction models, with attention to how the number and
size of the parts differ even though the two fractions themselves are
the same size. Use this principle to recognize and generate equivalent
fractions.
Microsoft Word version: 5_NF_Fractions_Problem_Construct_IllustrativeMathematics
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This handy Math in Focus Grade 7 Workbook Answer Key Chapter 9 Statistics detailed solutions for the textbook questions.
## Math in Focus Grade 7 Course 2 B Chapter 9 Answer Key Statistics
### Math in Focus Grade 7 Chapter 9 Quick Check Answer Key
Find the mean of each set of data. Round your answer to 2 decimal places if it is not exact.
Question 1.
9, 11, 6, 29, 5
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data. It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x1, x2, x3, … xn, the mean of data is given as:
It can also be denoted as:
The above-given data: 9, 11, 6, 29, 5
Mean=9+11+6+29+5/5
Mean=60/5
Mean=12.
Question 2.
43, 88, 39, 10, 26, 17, 35
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data. It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x1, x2, x3, … xn, the mean of data is given as:
It can also be denoted as:
The above-given data: 43, 88, 39, 10, 26, 17, 35
Mean=43+88+39+10+26+17+35/7
Mean=258/7
Mean=36.85 (round to nearest number)
Mean=37.
Question 3.
53.6, 36.7, 88.5, 90.6
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data. It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x1, x2, x3, … xn, the mean of data is given as:
It can also be denoted as:
The above-given data: 53.6, 36.7, 88.5, 90.6
Mean=53.6+36.7+88.5+90.6/4
Mean=269.4/4
Mean=67.35
Mean=67 (rounded to nearest number)
Question 4.
0.14, 1.05, 3.1, 7.18, 4.3, 8
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data. It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x1, x2, x3, … xn, the mean of data is given as:
It can also be denoted as:
The above-given data: 0.14, 1.05, 3.1, 7.18, 4.3, 8
Mean=0.14+1.05+3.1+7.18+4.3+8/6
Mean=23.77/6
Mean=3.96
Mean=4
Question 5.
The heights, in inches, of 10 children are
54, 66, 52, 60.5, 61.25, 55, 58.75, 51.5, 53, 50.
Find the mean height of the children.
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data. It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x1, x2, x3, … xn, the mean of data is given as:
It can also be denoted as:
the above-given data: 54, 66, 52, 60.5, 61.25, 55, 58.75, 51.5, 53, 50.
Mean=54+66+52+60.5+61.25+55+58.75+51.5+53+50/10
Mean=562/10
Mean height of children=56.2
Find the median of each set of numbers.
Question 6.
41,29, 78, 12, 56, 30, 22
Explanation:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
The formula to calculate the median of the data set is given as follow.
An odd number of observations:
If the total number of observations given is odd, then the formula to calculate the median is:
Median = {(n+1)/2}thterm
where n is the number of observations
Even number of observations:
If the total number of observations is even, then the median formula is:
Median = [(n/2)th term + {(n/2)+1}th]/2
where n is the number of observations
How to calculate Median:
To find the median, place all the numbers in the ascending order and find the middle.
Put them in ascending order: 12, 22, 29, 30, 41, 56, 78
The middle number is 30, so the median is 30.
Question 7.
193, 121.5, 162.3, 125, 103.8, 149.6
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 103.8, 121.5, 125, 149.6, 162.3, 193
Median=274.6/2
Median=137.3
Thus, the median is 137.3
Question 8.
9, 2, 2, 4, 4, 4, 1, 3, 6, 5, 3, 6
Explanation:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 1, 2, 2, 3, 3, 4, 4, 4, 5, 6, 6, 9
Median=4+4/2
Median=8/2
Median=4
Thus, the median is 4.
Question 9.
1,011, 1,100, 1,001, 1,010, 1,110, 1,000, 1,011, 100
Explanation:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 100, 1000, 1001, 1010, 1011, 1011, 1100, 1110
Median=2021/2
Median=1010.5
Thus, the median is 1010.5
Question 10.
The daily low temperatures for the past 10 days were
30.6°F, 32.1 °F, 29.5°F, 30.2°F, 26.4°F, 34.3°F, 31.6°F, 32°F, 25.9°F, and 26.4°F.
What was the median daily low temperature?
The statistical concept of the median is a value that divides a data sample, population, or probability distribution into two halves. Finding the median essentially involves finding the value in a data sample that has a physical location between the rest of the numbers. Note that when calculating the median of a finite list of numbers, the order of the data samples is important. Conventionally, the values are listed in ascending order, but there is no real reason that listing the values in descending order would provide different results. In the case where the total number of values in a data sample is odd, the median is simply the number in the middle of the list of all values. When the data sample contains an even number of values, the median is the mean of the two middle values. While this can be confusing, simply remember that even though the median sometimes involves the computation of a mean, when this case arises, it will involve only the two middle values, while a mean involves all the values in the data sample. In the odd cases where there are only two data samples or there is an even number of samples where all the values are the same, the mean and median will be the same. Given the same data set as before, the median would be acquired in the following manner:
30.6°F, 32.1 °F, 29.5°F, 30.2°F, 26.4°F, 34.3°F, 31.6°F, 32°F, 25.9°F, and 26.4°F.
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 25.9, 26.4, 26.4, 29.5, 30.2, 30.6, 31.6, 32, 32.1, 34.3
Median=30.2+30.6/2
Median=60.8/2
Median=30.4
Thus, the median is 30.4
Summarize each of the following data sets in a frequency table and draw a dot plot.
Question 11.
The data show the number of misspelled words found in 15 essays.
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# More probability
We learned about a number of probability topics in AS Maths including Venn Diagrams, Tree Diagrams and the definitions of Mutual Exclusivity and Independence. In A2 Maths, we take a closer look at conditional probabilities and mutual exclusivity & independence with a particular focus on set notation and introduce some useful formulae.
## Conditional Probability
Conditional probabilities are probabilities that change given the occurrence of a previous event or given additional information about an outcome. We have seen conditional probabilities in tree diagrams:
The tree diagram above shows the different outcomes when drawing two counters without replacement from a bag containing 4 red and 6 blue counters – see more Tree Diagram examples. The probabilities change depending on the outcome of the first counter drawn – this is an example of conditional probability. Conditional probabilities may also be presented in a two-way table (or contingency table) – see Example 1. We represent a conditional probability as and we read it as ‘the probability that A occurs given that B has occurred’ or ‘the probability of A given information of B’. A conditional probability question for a tree diagram can also be seen in Example 1.
## Set Notation
We can also calculate conditional probabilities using Venn diagrams by using prior information of an outcome to restrict the sample space. Before we do so, we take a brief look at set notation. We saw the following definitions when learning about Venn diagrams in AS Maths. We now introduce the corresponding set notation:
or Â
Â
## Probability Formulae
If you colour in event A in one colour, then event B in another, you will notice that the intersection was coloured in twice. Hence, in order to find the probability of the union, one must add the respective probabilities of events A and B whilst removing a single probability corresponding to one of the intersections.
2. Multiplication Formula:
This can be seen in a Venn diagram – given the information that B has occurred, the new full sample space becomes the set B. The probability that A has occurred is then the part of A that lies in the intersection. It can be thought of as the chances of being in A given that you are in B is the probability of the intersection out of the probability of B. It follows from the formula that . This can be seen in a tree diagram – given that B has occurred follow event B along the first branch with probability . The probability that A then occurs is following event A along the second branch. Alternatively, , which can be seen from the original formula by swapping A and B around.
## Mutually Exclusive and Independent Events
We saw the definitions of mutual exclusivity and independence in AS Maths. If events are mutually exclusive, they do not intersect on a Venn diagram. It follows that…
for mutually exclusive events.
For independent events, – the fact that event B occurred does not affect the probability of A occurring. From the multiplication formation it follows that and so…
for independent events.
See Example 2 for a demonstration of how to use the formulae and how to determine if events are mutually exclusive or independent.
## Examples
The following two-way table shows the eye and hair colour of 100 university maths students:
Let be the event that a student has blue eyes, be the event that a student has brown eyes, if they have dark hair and if they have light hair.
a) For a student drawn at random, find the following probabilities:
i)
ii)
iii)
iv)
b) Show this information in a tree diagram.
c) Use the tree diagram to find and .
a)
i) According to the table, 24 out of 100 students have blue eyes and so .
ii) Similarly,
iii) Of the 24 people that have blue eyes, 10 have dark hair and so the probability of dark hair given blue eyes is , that is, .
iii) Of the 76 people that have brown eyes, 22 have light hair and so the probability of light hair given brown eyes is , that is, .
b)
c) By following the branches through the tree diagram, blue eyes then light hair, we see that . Similarly, not blue eyes (or brown eyes) then light hair, gives .
Given that , and ,
a) Find .
b) Determine whether the events and are mutually exclusive and/or independent or not.
c) Find .
d) Show the information in a Venn diagram and determine
i) and
ii)
a) We use the addition formula: or and rearrange to get .
b) Since , events and are not mutually exclusive. Events are independent if – in this case, whereas and so the events aren’t independent either.
c) Using the multiplication formula: , it follows that .
d)
To find we restrict the sample space to not B and look at the part of A that is not in B. This is 0.4 and since it is out of 0.75, we have .
To find we restrict the sample space to and look at the part of that lies in it. All of is in and so .
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