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Leaving the probability tables behind in their appropriate historical dust while fully embracing the power of modern classroom technology to enhance my students’ statistical learning and understanding, I’m convinced I made the right decision to start this school year. They know more, understand the foundations of statistics better, and as a group feel much more confident and flexible. Whether their scores on next month’s AP exam will reflect their growth, I can’t say, but they’ve definitely learned more statistics this year than any previous statistics class I’ve ever taught.
COMPLETE FILES FROM MY 2015 T3 PRESENTATION
If you are interested, you can download here the PowerPoint file for my entire Nspired Statistics and CAS presentation from last week’s 2015 T3 International Conference in Dallas, TX. While not the point of this post, the presentation started with a non-calculus derivation/explanation of linear regressions. Using some great feedback from Jeff McCalla, here is an Nspire CAS document creating the linear regression computation updated from what I presented in Dallas. I hope you found this post and these files helpful, or at least thought-provoking.
## Calculus Humor
Completely frivolous post.
OK, my Halloween “costume” at school this year was pretty lame, but I actually did put a minute amount of thought into it.
In case you can’t read the sign, it says $\int 3(ice)^2d(ice)$. If you remember some calculus and treat ice as your variable, that works out to $ice^3$–An Ice Cube! Ha!
But it gets better. As there weren’t any bounds, adding the random constant of integration makes it $ice^3+C$ or “Ice Cube + C”, or maybe “Ice Cube + Sea”–I was really dressed up as an Iceberg. Ha! Ha! Having no idea how to dress like an iceberg, I wore a light blue shirt for the part of the iceberg above the water and dark blue pants for the part below the water. I tried to be clever even if the underlying joke was just “punny”. | {
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Then a colleague posted another integral joke I’d seen sometime before. It has some lovely extensions, so I’ll share that, too.
What is $\int \frac{d(cabin)}{cabin}$?
At first glance, it’s a “log cabin.” Funny.
But notationally, the result is actually $ln(cabin)$, so the environmentalists out there will appreciate that the answer is really a “natural log cabin.” Even funnier.
The most correct solution is $ln(cabin)+C$. If you call the end “+ Sea”, then the most clever answer is that $\int \frac{d(cabin)}{cabin}$ is a “Houseboat”. Ha!
Hope you all had some fun.
## Traveling Dots, Parabolas, and Elegant Math
Toward the end of last week, I read a description a variation on a paper-folding strategy to create parabolas. Paraphrased, it said:
1. On a piece of wax paper, use a pen to draw a line near one edge. (I used a Sharpie on regular copy paper and got enough ink bleed that I’m convinced any standard copy or notebook paper will do. I don’t think the expense of wax paper is required!)
2. All along the line, place additional dots 0.5 to 1 inch apart.
3. Finally, draw a point F between 0.5 and 2 inches from the line roughly along the midline of the paper toward the center of the paper.
4. Fold the paper over so one of the dots on line is on tope of point F. Crease the paper along the fold and open the paper back up.
5. Repeat step 4 for every dot you drew in step 2.
6. All of the creases from steps 4 & 5 outline a curve. Trace that curve to see a parabola. | {
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I’d seen and done this before, I had too passively trusted that the procedure must have been true just because the resulting curve “looked like a parabola.” I read the proof some time ago, but I consumed it too quickly and didn’t remember it when I was read the above procedure. I shamefully admitted to myself that I was doing exactly what we insist our students NEVER do–blindly accepting a “truth” based on its appearance. So I spent part of that afternoon thinking about how to understand completely what was going on here.
What follows is the chronological redevelopment of my chain of reasoning for this activity, hopefully showing others that the prettiest explanations rarely occur without effort, time, and refinement. At the end of this post, I offer what I think is an even smoother version of the activity, freed from some of what I consider overly structured instructions above.
CONIC DEFINITION AND WHAT WASN’T OBVIOUS TO ME
A parabola is the locus of points equidistant from a given point (focus) and line (directrix).
What makes the parabola interesting, in my opinion, is the interplay between the distance from a line (always perpendicular to some point C on the directrix) and the focus point (theoretically could point in any direction like a radius from a circle center).
What initially bothered me about the paper folding approach last week was that it focused entirely on perpendicular bisectors of the Focus-to-C segment (using the image above). It was not immediately obvious to me at all that perpendicular bisectors of the Focus-to-C segment were 100% logically equivalent to the parabola’s definition. | {
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1. I knew without a doubt that all parabolas are similar (there is a one-to-one mapping between every single point on any parabola and every single point on any other parabola), so I didn’t need to prove lots of cases. Instead, I focused on the simplest version of a parabola (from my perspective), knowing that whatever I proved from that example was true for all parabolas.
2. I am quite comfortable with my algebra, geometry, and technology skills. Being able to wield a wide range of powerful exploration tools means I’m rarely intimidated by problems–even those I don’t initially understand. I have the patience to persevere through lots of data and explorations until I find patterns and eventually solutions.
I love to understand ideas from multiple perspectives, so I rarely quit with my initial solution. Perseverance helps me re-phrase ideas and exploring them from alternative perspectives until I find prettier ways of understanding.
In my opinion, it is precisely this willingness to play, persevere, and explore that formalized education is broadly failing to instill in students and teachers. “What if?” is the most brilliant question, and the one we sadly forget to ask often enough.
ALGEBRAIC PROOF
While I’m comfortable handling math in almost any representation, my mind most often jumps to algebraic perspectives first. My first inclination was a coordinate proof.
PROOF 1: As all parabolas are similar, it was enough to use a single, upward facing parabola with its vertex at the origin. I placed the focus at $(0,f)$, making the directrix the line $y=-f$. If any point on the parabola was $(x_0,y_0)$, then a point C on the directrix was at $(x_0,-f)$.
From the parabola’s definition, the distance from the focus to P was identical to the length of CP:
$\sqrt{(x_0-0)^2-(y_0-f)^2}=y_0+f$
Squaring and combining common terms gives
$x_0 ^2+y_0 ^2-2y_0f+f^2=y_0 ^2+2y_0f+f^2$
$x_0 ^2=4fy$ | {
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Squaring and combining common terms gives
$x_0 ^2+y_0 ^2-2y_0f+f^2=y_0 ^2+2y_0f+f^2$
$x_0 ^2=4fy$
But the construction above made lines (creases) on the perpendicular bisector of the focus-to-C segment. This segment has midpoint $\displaystyle \left( \frac{x_0}{2},0 \right)$ and slope $\displaystyle -\frac{2f}{x_0}$, so an equation for its perpendicular bisector is $\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)$.
Finding the point of intersection of the perpendicular bisector with the parabola involves solving a system of equations.
$\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)=\frac{x^2}{4f}$
$\displaystyle \frac{1}{4f} \left( x^2-2x_0x+x_0 ^2 \right) =0$
$\displaystyle \frac{1}{4f} \left( x-x_0 \right) ^2 =0$
So the only point where the line and parabola meet is at $\displaystyle x=x_0$–the very same point named by the parabola’s definition. QED
Proof 2: All of this could have been brilliantly handled on a CAS to save time and avoid the manipulations.
Notice that the y-coordinate of the final solution line is the same $y_0$ from above.
MORE ELEGANT GEOMETRIC PROOFS
I had a proof, but the algebra seemed more than necessary. Surely there was a cleaner approach.
In the image above, F is the focus, and I is a point on the parabola. If D is the midpoint of $\overline{FC}$, can I conclude $\overline{ID} \perp \overline{FC}$, proving that the perpendicular bisector of $\overline{FC}$ always intersects the parabola?
PROOF 3: The definition of the parabola gives $\overline{FI} \cong \overline{IC}$, and the midpoint gives $\overline{FD} \cong \overline{DC}$. Because $\overline{ID}$ is self-congruent, $\Delta IDF \cong \Delta IDC$ by SSS, and corresponding parts make the supplementary $\angle IDF \cong \angle IDC$, so both must be right angles. QED | {
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PROOF 4: Nice enough, but it still felt a little complicated. I put the problem away to have dinner with my daughters and when I came back, I was able to see the construction not as two congruent triangles, but as the single isosceles $\Delta FIC$ with base $\overline{FC}$. In isosceles triangles, altitudes and medians coincide, automatically making $\overline{ID}$ the perpendicular bisector of $\overline{FC}$. QED
Admittedly, Proof 4 ultimately relies on the results of Proof 3, but the higher-level isosceles connection felt much more elegant. I was satisfied.
TWO DYNAMIC GEOMETRY SOFTWARE VARIATIONS
Thinking how I could prompt students along this path, I first considered a trace on the perpendicular lines from the initial procedure above (actually tangent lines to the parabola) using to trace the parabolas. A video is below, and the Geogebra file is here.
http://vimeo.com/89759785
It is a lovely approach, and I particularly love the way the parabola appears as a digital form of “string art.” Still, I think it requires some additional thinking for users to believe the approach really does adhere to the parabola’s definition.
I created a second version allowing users to set the location of the focus on the positive y-axis and using a slider to determine the distances and constructs the parabola through the definition of the parabola. [In the GeoGebra worksheet (here), you can turn on the hidden circle and lines to see how I constructed it.] A video shows the symmetric points traced out as you drag the distance slider.
A SIMPLIFIED PAPER PROCEDURE | {
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A SIMPLIFIED PAPER PROCEDURE
Throughout this process, I realized that the location and spacing of the initial points on the directrix was irrelevant. Creating the software versions of the problem helped me realize that if I could fold a point on the directrix to the focus, why not reverse the process and fold F to the directrix? In fact, I could fold the paper so that F touched anywhere on the directrix and it would work. So, here is the simplest version I could develop for the paper version.
1. Use a straightedge and a Sharpie or thin marker to draw a line near the edge of a piece of paper.
2. Place a point F roughly above the middle of the line toward the center of the paper.
3. Fold the paper over so point F is on the line from step 1 and crease the paper along the fold.
4. Open the paper back up and repeat step 3 several more times with F touching other parts of the step 1 line.
5. All of the creases from steps 3 & 4 outline a curve. Trace that curve to see a parabola.
This procedure works because you can fold the focus onto the directrix anywhere you like and the resulting crease will be tangent to the parabola defined by the directrix and focus. By allowing the focus to “Travel along the Directrix”, you create the parabola’s locus. Quite elegant, I thought.
As I was playing with the different ways to create the parabola and thinking about the interplay between the two distances in the parabola’s definition, I wondered about the potential positions of the distance segments. | {
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1. What is the shortest length of segment CP and where could it be located at that length? What is the longest length of segment CP and where could it be located at that length?
2. Obviously, point C can be anywhere along the directrix. While the focus-to-P segment is theoretically free to rotate in any direction, the parabola definition makes that seem not practically possible. So, through what size angle is the focus-to-P segment practically able to rotate?
3. Assuming a horizontal directrix, what is the maximum slope the focus-to-P segment can achieve?
4. Can you develop a single solution to questions 2 and 3 that doesn’t require any computations or constructions?
CONCLUSIONS
I fully realize that none of this is new mathematics, but I enjoyed the walk through pure mathematics and the enjoyment of developing ever simpler and more elegant solutions to the problem. In the end, I now have a deeper and richer understanding of parabolas, and that was certainly worth the journey.
## The Value of Counter-Intuition
Numberphile caused quite a stir when it posted a video explaining why
$\displaystyle 1+2+3+4+...=- \frac{1}{12}$
Doug Kuhlman recently posted a great follow-up Numberphile video explaining a broader perspective behind this sum.
It’s a great reminder that there are often different ways of thinking about problems, and sometimes we have to abandon tradition to discover deeper, more elegant connections.
For those deeply bothered by this summation result, the second video contains a lovely analogy to the “reality” of $\sqrt{-1}$. From one perspective, it is absolutely not acceptable to do something like square roots of negative numbers. But by finding a way to conceptualize what such a thing would mean, we gain a far richer understanding of the very real numbers that forbade $\sqrt{-1}$ in the first place as well as opening the doors to stunning mathematics far beyond the limitations of real numbers. | {
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On the face of it, $\displaystyle 1+2+3+...=-\frac{1}{12}$ is obviously wrong within the context of real numbers only. But the strange thing in physics and the Zeta function and other places is that $\displaystyle -\frac{1}{12}$ just happens to work … every time. Let’s not dismiss this out of hand. It gives our students the wrong idea about mathematics, discovery, and learning.
There’s very clearly SOMETHING going on here. It’s time to explore and learn something deeper. And until then, we can revel in the awe of manipulations that logically shouldn’t work, but somehow they do.
May all of our students feel the awe of mathematical and scientific discovery. And until the connections and understanding are firmly established, I hope we all can embrace the spirit, boldness, and fearless of Euler.
## Base-x Numbers and Infinite Series
In my previous post, I explored what happened when you converted a polynomial from its variable form into a base-x numerical form. That is, what are the computational implications when polynomial $3x^3-11x^2+2$ is represented by the base-x number $3(-11)02_x$, where the parentheses are used to hold the base-x digit, -11, for the second power of x?
So far, I’ve explored only the Natural number equivalents of base-x numbers. In this post, I explore what happens when you allow division to extend base-x numbers into their Rational number counterparts.
Level 5–Infinite Series:
Numbers can have decimals, so what’s the equivalence for base-x numbers? For starters, I considered trying to get a “decimal” form of $\displaystyle \frac{1}{x+2}$. It was “obvious” to me that $12_x$ won’t divide into $1_x$. There are too few “places”, so some form of decimals are required. Employing division as described in my previous post somewhat like you would to determine the rational number decimals of $\frac{1}{12}$ gives | {
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Remember, the places are powers of x, so the decimal portion of $\displaystyle \frac{1}{x+2}$ is $0.1(-2)4(-8)..._x$, and it is equivalent to
$\displaystyle 1x^{-1}-2x^{-2}+4x^{-3}-8x^{-4}+...=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$.
This can be seen as a geometric series with first term $\displaystyle \frac{1}{x}$ and ratio $\displaystyle r=\frac{-2}{x}$. It’s infinite sum is therefore $\displaystyle \frac{\frac{1}{x}}{1-\frac{-2}{x}}$ which is equivalent to $\displaystyle \frac{1}{x+2}$, confirming the division computation. Of course, as a geometric series, this is true only so long as $\displaystyle |r|=\left | \frac{-2}{x} \right |<1$, or $2<|x|$.
I thought this was pretty cool, and it led to lots of other cool series. For example, if $x=8$,you get $\frac{1}{10}=\frac{1}{8}-\frac{2}{64}+\frac{4}{512}-...$.
Likewise, $x=3$ gives $\frac{1}{5}=\frac{1}{3}-\frac{2}{9}+\frac{4}{27}-\frac{8}{81}+...$.
I found it quite interesting to have a “polynomial” defined with a rational expression.
Boundary Convergence:
As shown above, $\displaystyle \frac{1}{x+2}=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ only for $|x|>2$.
At $x=2$, the series is obviously divergent, $\displaystyle \frac{1}{4} \ne \frac{1}{2}-\frac{2}{4}+\frac{4}{8}-\frac{8}{16}+...$.
For $x=-2$, I got $\displaystyle \frac{1}{0} = \frac{1}{-2}-\frac{2}{4}+\frac{4}{-8}-\frac{8}{16}+...=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-...$ which is properly equivalent to $-\infty$ as $x \rightarrow -2$ as defined by the convergence domain and the graphical behavior of $\displaystyle y=\frac{1}{x+2}$ just to the left of $x=-2$. Nice. | {
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I did find it curious, though, that $\displaystyle \frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ is a solid approximation for $\displaystyle \frac{1}{x+2}$ to the left of its vertical asymptote, but not for its rotationally symmetric right side. I also thought it philosophically strange (even though I understand mathematically why it must be) that this series could approximate function behavior near a vertical asymptote, but not near the graph’s stable and flat portion near $x=0$. What a curious, asymmetrical approximator.
Maclaurin Series:
Some quick calculus gives the Maclaurin series for $\displaystyle \frac{1}{x+2}$ : $\displaystyle \frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+...$, a geometric series with first term $\frac{1}{2}$ and ratio $\frac{-x}{2}$. Interestingly, the ratio emerging from the Maclaurin series is the reciprocal of the ratio from the “rational polynomial” resulting from the base-x division above.
As a geometric series, the interval of convergence is $\displaystyle |r|=\left | \frac{-x}{2} \right |<1$, or $|x|<2$. Excluding endpoint results, the Maclaurin interval is the complete Real number complement to the base-x series. For the endpoints, $x=-2$ produces the right-side vertical asymptote divergence to $+ \infty$ that $x=-2$ did for the left side of the vertical asymptote in the base-x series. Again, $x=2$ is divergent.
It’s lovely how these two series so completely complement each other to create clean approximations of $\displaystyle \frac{1}{x+2}$ for all $x \ne 2$.
Other base-x “rational numbers” | {
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Other base-x “rational numbers”
Because any polynomial divided by another is absolutely equivalent to a base-x rational number and thereby a base-x decimal number, it will always be possible to create a “rational polynomial” using powers of $\displaystyle \frac{1}{x}$ for non-zero denominators. But, the decimal patterns of rational base-x numbers don’t apply in the same way as for Natural number bases. Where $\displaystyle \frac{1}{12}$ is guaranteed to have a repeating decimal pattern, the decimal form of $\displaystyle \frac{1}{x+2}=\frac{1_x}{12_x}=0.1(-2)4(-8)..._x$ clearly will not repeat. I’ve not explored the full potential of this, but it seems like another interesting field.
CONCLUSIONS and QUESTIONS
Once number bases are understood, I’d argue that using base-x multiplication might be, and base-x division definitely is, a cleaner way to compute products and quotients, respectively, for polynomials.
The base-x division algorithm clearly is accessible to Algebra II students, and even opens the doors to studying series approximations to functions long before calculus.
Is there a convenient way to use base-x numbers to represent horizontal translations as cleanly as polynomials? How difficult would it be to work with a base-$(x-h)$ number for a polynomial translated h units horizontally?
As a calculus extension, what would happen if you tried employing division of non-polynomials by replacing them with their Taylor series equivalents? I’ve played a little with proving some trig identities using base-x polynomials from the Maclaurin series for sine and cosine.
What would happen if you tried to compute repeated fractions in base-x?
It’s an open question from my perspective when decimal patterns might terminate or repeat when evaluating base-x rational numbers.
I’d love to see someone out there give some of these questions a run!
## Number Bases and Polynomials | {
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## Number Bases and Polynomials
About a month ago, I was working with our 5th grade math teacher to develop some extension activities for some students in an unleveled class. The class was exploring place value, and I suggested that some might be ready to explore what happens when you allow the number base to be something other than 10. A few students had some fun learning to use their basic four algorithms in other number bases, but I made an even deeper connection.
When writing something like 512 in expanded form ($5\cdot 10^2+1\cdot 10^1+2\cdot 10^0$), I realized that if the 10 was an x, I’d have a polynomial. I’d recognized this before, but this time I wondered what would happen if I applied basic math algorithms to polynomials if I wrote them in a condensed numerical form, not their standard expanded form. That is, could I do basic algebra on $5x^2+x+2$ if I thought of it as $512_x$–a base-x “number”? (To avoid other confusion later, I read this as “five one two base-x“.)
Following are some examples I played with to convince myself how my new notation would work. I’m not convinced that this will ever lead to anything, but following my “what ifs” all the way to infinite series was a blast. Read on!
If I wanted to add $(3x+5)$$(2x^2+4x+1)$, I could think of it as $35_x+241_x$ and add the numbers “normally” to get $276_x$ or $2x^2+7x+6$. Notice that each power of x identifies a “place value” for its characteristic coefficient.
If I wanted to add $3x-7$ to itself, I had to adapt my notation a touch. The “units digit” is a negative number, but since the number base, x, is unknown (or variable), I ended up saying $3x-7=3(-7)_x$. The parentheses are used to contain multiple characters into a single place value. Then, $(3x-7)+(3x-7)$ becomes $3(-7)_x+3(-7)_x=6(-14)_x$ or $6x-14$. Notice the expanding parentheses containing the base-x units digit. | {
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The last example also showed me that simple multiplication would work. Adding $3x-7$ to itself is equivalent to multiplying $2\cdot (3x-7)$. In base-x, that is $2\cdot 3(-7)_x$. That’s easy! Arguably, this might be even easier that doubling a number when the number base is known. Without interactions between the coefficients of different place values, just double each digit to get $6(-14)_x=6x-14$, as before.
What about $(x^2+7)+(8x-9)$? That’s equivalent to $107_x+8(-9)_x$. While simple, I’ll solve this one by stacking.
and this is $x^2+8x-2$. As with base-10 numbers, the use of 0 is needed to hold place values exactly as I needed a 0 to hold the $x^1$ place for $x^2+7$. Again, this could easily be accomplished without the number base conversion, but how much more can we push these boundaries?
Level 3–Multiplication & Powers:
Compute $(8x-3)^2$. Stacking again and using a modification of the multiply-and-carry algorithm I learned in grade school, I got
and this is equivalent to $64x^2-48x+9$.
All other forms of polynomial multiplication work just fine, too.
From one perspective, all of this shifting to a variable number base could be seen as completely unnecessary. We already have acceptably working algorithms for addition, subtraction, and multiplication. But then, I really like how this approach completes the connection between numerical and polynomial arithmetic. The rules of math don’t change just because you introduce variables. For some, I’m convinced this might make a big difference in understanding.
I also like how easily this extends polynomial by polynomial multiplication far beyond the bland monomial and binomial products that proliferate in virtually all modern textbooks. Also banished here is any need at all for banal FOIL techniques.
Level 4–Division: | {
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Level 4–Division:
What about $x^2+x-6$ divided by $x+3$? In base-x, that’s $11(-6)_x \div 13_x$. Remembering that there is no place value carrying possible, I had to be a little careful when setting up my computation. Focusing only on the lead digits, 1 “goes into” 1 one time. Multiplying the partial quotient by the divisor, writing the result below and subtracting gives
Then, 1 “goes into” -2 negative two times. Multiplying and subtracting gives a remainder of 0.
thereby confirming that $x+3$ is a factor of $x^2+x-6$, and the other factor is the quotient, $x-2$.
Perhaps this could be used as an alternative to other polynomial division algorithms. It is somewhat similar to the synthetic division technique, without its significant limitations: It is not limited to linear divisors with lead coefficients of one.
For $(4x^3-5x^2+7) \div (2x^2-1)$, think $4(-5)07_x \div 20(-1)_x$. Stacking and dividing gives
So $\displaystyle \frac{4x^3-5x^2+7}{2x^2-1}=2x-2.5+\frac{2x+4.5}{2x^2-1}$.
CONCLUSION
From all I’ve been able to tell, converting polynomials to their base-x number equivalents enables you to perform all of the same arithmetic computations. For division in particular, it seems this method might even be a bit easier.
In my next post, I push the exploration of these base-x numbers into infinite series. | {
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The hypotheses areH1 - the coin is two headed, and H2 the coin is fair. What is the probability mass function of X? What is the expected value of X?. probability of flipping nine heads on two headed coin = 1 probability of picking two headed coin and then flipping 9 straight heads =. a: a "biased" coin). You randomly take one of them out of your pocket without looking at it. 01, what is his posterior belief ? Note. Furthermore, if one wanted to determine whether the coin was fair or weighted, it would be difficult to do that without using inferential methods derived from measure theory. G is surprised to find that he loses the first ten times they play. (d) There are two coins in a box. What is the probability of getting two heads and a four? What is the probability of getting two heads and a four? Answer by stanbon(75874) ( Show Source ):. You reach one coin at random, toss it, and it lands up heads. One of these coins is selected at random and tossed three times. Find the probability that the coin is heads. So there is a probability of one that either of these will happen. Given that you see 10 heads, what is the probability that the next toss of that coin is also a head?. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. a coin is selected at random and tossed. There are three coins One is two headed coin, another is biased coin that comes up tails 25% of the times and the other is unbiased coin One of the three coins is chosen at random and tossed , it shows head What is the probability that - Math - Probability. All k times the coin landed up heads. There is a probability of 0. Problem 43: There are 3 coins in a box. In the case of the coins, we understand that there's a $$\frac{1}{3}$$ chance we have a normal coin, and a $$\frac{2}{3}$$ chance it's a two-headed coin. The probability that the two-headed coin is selected out of the box is $P(E_1)=1/3$. A probability of | {
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probability that the two-headed coin is selected out of the box is $P(E_1)=1/3$. A probability of zero means that an event is impossible. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. Consider an urn containing one fair coin and one two-headed coin. Let V denote the probability of heads of the selected coin, and Y the number of heads. This is, however, wrong, because given that heads came, it is more likely that the two-headed coin was chosen. You shouldn't believe or disbelieve anything. One is a two-headed coin (having a head on both faces), another is a biased coin that comes up heads 75% of the times, and third is also a biased coin that comes up tails 40% of the times. We use Wi to denote the event that the ith ball is white and Ri to denote the event that the ith ball is red. Furthermore, if one wanted to determine whether the coin was fair or weighted, it would be difficult to do that without using inferential methods derived from measure theory. What is the probability that the selected coin was the two-headed coin? Add to solve later. You flip it and it comes up "heads". M3070 - FALL 2003 - Quiz 2 NAME: Problem 1. A coin is selected at random and tossed twice. "Mathematical Expectation" is one of those few topics that is rarely discussed in details in any curriculum, but is nevertheless very important. Call the number of ways of flipping coins and not receiving any consecutive heads. enum Coin { Fair, DoubleHeaded } Now let's suppose we have a bag of coins. Online virtual coin toss simulation app. One of the most interesting aspects of blackjack is the probability math involved. b:if heads appear on the second toss,then it also appears on the first toss. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. 1 The while loop does Kerrich’s entire experiment in a split second! 2. The first coin is a fair coin, the second | {
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loop does Kerrich’s entire experiment in a split second! 2. The first coin is a fair coin, the second coin is a biased coin such that P(T) = 0:15, and the third coin is a two headed coin (a) What is the probability the coin lands on tails? (b) Given that the coin landed on heads, what is the probability it was the fair coin? 29. Instead of probability distributions, we use probability densities and integrate over ranges of possible. If the first 50 tosses of the coin are heads, what is the probability that it is the 2-headed coin. After all, real life is rarely fair. The correct reasoning is to calculate the conditional probability p= P(two-headed coin was chosen|heads came up) = P(two-headed coin was chosen and heads. There are three coins in a box. Given that you see 10 heads, what is the probability that the next toss of that coin is also a head?. What is the probability that it is the fair coin?. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. Case 2: One head. (a) What is the sample space of the experiment? (b) Given that both flips produce heads, what is the probability that Alice drew the two-headed coin from the urn?. What is the probability that it was the two-headed coin? 43. He is then either shouted at or not. If it is heads, he is willing. The first coin is two-headed. Problem 1 (20%) There are three coins in a box. A probability of one means that the event is certain. One is a two-headed coin; another is a fair coin; and the third is a biased coin that comes up heads 75% of the time. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and the third is also a biased coin that comes up tails 40% of the time. The probability of getting any number face on the die is no way influences the probability of getting a head or a tail on the coin. One coin is fair, the other has heads on both sides. There are three coins in a | {
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a tail on the coin. One coin is fair, the other has heads on both sides. There are three coins in a box. In contrast, a process in which the. Here we will learn how to find the probability of tossing two coins. Probability. The first coin is two-headed. " A coin is selected at random, flipped n times and in all flips it falls heads up. There is a probability of 0. Suppose a woman has two coins in her handbag. From the root, draw two branches showing the first ball drawn. Given that heads comes up, what is the probability that you flipped the two headed coin?. For the set of two fair coins the probability of getting a head is 1/2 in one toss. So the probability of choosing double head coin and get head is 1/3, while choosing single head coin and get head is 1/6. It is equally likely to be a fair coin, to be a two-headed coin, to be a two-tailed coin, or any mixture of alloy that has one side heavier than the other. Byju's Coin Toss Probability Calculator is a tool which makes calculations very simple and interesting. A coin is chosen at random from the bag and tossed 2 times. Let H 1 first coin flip is heads H 2 second coin flip is heads The likelihood of a coin flip coming up heads is 0. , flipping a two-headed coin. b:if heads appear on the second toss,then it also appears on the first toss. the opposite face is either heads or tails, the desired probability is 1/2. (a) what is the probability that the lower face of the coin is a head?. I'm not a mathematician so please bear with me. To find the probability of two independent events occuring, we simply multiply together the probabilities associated with two individual events. A box contains three coins. Odds & Probability in Blackjack. What is the probability that it was the two-headed coin?. There are three coins. what is the probability that this is the two headed coin?. 3 There are three coins in a bag: ordinary, two-headed, and two-tailed. At level 1 we toss it. 2- There are three coins in a box. All k times the | {
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and two-tailed. At level 1 we toss it. 2- There are three coins in a box. All k times the coin landed up heads. There are three coins in a box. The double-headed arrows (↔) in the statements above are read “if and only if” and show the equivalence of the statements on either side of the arrows. Pick one of the coins at random. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. He selects one of the coins at random; when he flips it, it shows heads. If G’s prior belief is that the chance of R having a two headed coin is 0. What should be your probability that it's the two-headed one--(a) 1/2, since it can only be two-headed or normal? (b) 2/3, because the other side could be the tail of the normal coin, or either side of the two-headed one?. Given that heads comes up, what is the probability that you flipped the two headed coin?. What is the probability that it is the fair coin?. My $\Pr(B)$ is the probability of flipping 10 heads, which is 1 in $2^{10}$. It comes up heads each time. You know that he had a two-headed coin and a regular coin in his pocket, and you believe that it is equally likely for him to have chosen either coin. I've read that if you flip a coin 10 times and it comes up heads every time then it's still a 50/50 chance to be a heads or tails on the 11th flip. Prisoner A asks the jailer to tell him. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two head. But the coin comes up heads 8 times in a row. Given that heads show both times, what is the probability that the coin is the two-headed one?. (a) A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random; when he flips it, it shows heads. Exercise 2: How many consecutive coin flips will it take for the subjective probability that it is a 2-headed coin to be creater than the subjective probability that it is a fair coin? Question: | {
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is a 2-headed coin to be creater than the subjective probability that it is a fair coin? Question: What happens to the posterior if you observe a single tail - even after a long string of heads?. EDIT #2: By "no retosses" I mean that your algorithm for obtaining the 1/3 probability can not have a "retoss until you get 1/3" rule which can theoretically cause you to toss infinitely many times. The probability of A and B is 1/100. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. Solution: a) A tree diagram of all possible outcomes. It is equally likely to be a fair coin, to be a two-headed coin, to be a two-tailed coin, or any mixture of alloy that has one side heavier than the other. Three of the outcomes show a head, of which two are from the double-headed coin. If X = 0, then the coin has to be the 1-headed one. b) Find the probability that of the first 4 marbles selected, exactly two are black. He selects one of the coins at random; when he flips it, it shows heads. two coins and one six sided number cube are tossed together. If heads appears both times, what is the probability that the coin is two-head. I think hugin is right: the probability of a 6th head is just the combination of the probability you have the double-headed coin (0. (a) What is the sample space of the experiment? (b) Given that both flips produce heads, what is the probability that Alice drew the two-headed coin from the urn?. The second one is a fair coin. You know that the coins you are given to test are either unfair coins with heads probability 1 4; unfair coins with heads probability 3 4; and fair coins. for the two-headed coin, the probability of getting two heads in two flips is 1. coin is chosen at random and flipped, and comes up heads. 1 Directed graphical models. A coin is chosen at random from the bag and tossed 2 times. If the experiment can be repeated potentially infinitely many times, | {
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from the bag and tossed 2 times. If the experiment can be repeated potentially infinitely many times, then the probability of an event can be defined through relative frequencies. Published on June 14, 2016. We do not return it to the bin. You flip it and it comes up "heads". (d) There are two coins in a box. The first coin is two-headed. A fair coin has a 50-50 probability of coming up heads or tails; a double-headed coin always comes up heads. Given that heads show both times, what is the probability that the coin is the two-headed one?. There are three coins in a box. A two headed quarter is not something that was done at the mint, it is a novelty item, generally with high enough magnification you can see the seam that the two coins were joined together. Math Two-Headed Coin and Bayesian Probability Date: 04/21/2003 at 17:12:44 From: Maggie Subject: Probability In a box there are nine fair coins and one two-headed coin. What is the probability that there is a head on the OTHER side of this coin? Yes, it could be the fair coin or the two-headed coin, but they're not equally likely: because the fair coin COULD have come up tails, the two-headed coin is now twice as likely. Given that a tail appears on the third toss, then the probability that it is the two-headed coin is 0, so the probability that it is the fair coin is 1 in this case. I'm not a mathematician so please bear with me. I think hugin is right: the probability of a 6th head is just the combination of the probability you have the double-headed coin (0. Search Items > CARINA Grey Sleeveless Lace Dress & Cropped Jacket Formal/Wedding Outfit Size 12. There is a probability of 0. A gambler has in his pocket a fair coin and a two headed coin. It shows heads. It comes up heads every time. What is the probability that it is the fair coin?(b) Suppose that he flips the same coin a second time and, again, it shows heads. This already is a pretty good estimate of the real bias! But you might want an even better | {
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heads. This already is a pretty good estimate of the real bias! But you might want an even better estimate. Randomness as a tool: graph theory; scheduling; internet routing. Answer on Question #26655 - Math - Statistics and Probability A bag contains three coins, one of which is coined with two heads, while the other two coins are normal and not biased. So the probability of choosing double head coin and get head is 1/3, while choosing single head coin and get head is 1/6. b:if heads appear on the second toss,then it also appears on the first toss. 2 Conditioning Independence, Conditional Probability Hypotheses, Total Probability Examples: Queen of Spades, Manufacturing Bayes, and Bridged Circuit 3. The outcome of the tosses comes up heads, heads, and heads. Answer: $$\frac{41}{72}$$. There are 3 coins in a box. The 2-headed coin is H, H, and the normal coin is h,t. R tosses a coin, and wins $1 if it lands on H or loses$1 on T. The same coin (from part (b)) is ipped again and it shows heads. In the very first Two-Face story, after Two-Face captured Batman and Robin and released them unharmed because the coin said so, he captured Batman again. One has two heads, one has two tails, and the other is a fair coin with one head and one tail. If you get it wrong, only one more coin toss is required to choose between the remaining two doors. Suppose your materials science roommate managed to make a two-headed coin. You have two coins, one of which is fair and comes up heads with a probability 1/2, and the other which is biased and comes up heads with prob You have two coins in a bag: one fair coin and one trick coin that has heads on both sides. Step 2 of 3: (b) It is given that the gambler flips the coin for the second time and again he gets a head. G is surprised to find that he loses the first ten times they play. What is the probability that two headed coin was selected? Denote with Ak the event that randomly selected coin lands heads up k times. 5 is the probability that | {
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with Ak the event that randomly selected coin lands heads up k times. 5 is the probability that the selected coin is a fair coin. a) What is the probability that it is the fair coin?. 6 LAB 1: some elementary (but creative) extensions. What is now the probability that it is the fair coin?. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. So the probability that you get three heads with a randomly chosen coin is:. You give the probability of a two-headed coin a 5% chance and probability of a two-tailed coin also a 5% chance. what is the probability that this is the two headed coin?. Compute the proportion of. What is the probability that it shows heads? b. If heads appears both times, what is the probability that the coin is two-head. Flipping a coin many times and getting heads each time, however, seems to go against the 50% chance of a normal coin landing on heads, that's when you might wonder if the coin is two-headed, but I still seem to reject that it actually affects the probability. What is the probability of throwing a head on 1 toss? Throwing a tail on one toss? Answer: B. Exercise 2: How many consecutive coin flips will it take for the subjective probability that it is a 2-headed coin to be creater than the subjective probability that it is a fair coin? Question: What happens to the posterior if you observe a single tail - even after a long string of heads?. (a) A gambler has in his pocket a fair coin and a two-headed coin. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two heade. Answer on Question #26655 - Math - Statistics and Probability A bag contains three coins, one of which is coined with two heads, while the other two coins are normal and not biased. Given that you see 10 heads, what | {
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two heads, while the other two coins are normal and not biased. Given that you see 10 heads, what is the probability that the next toss of that coin is also a head?. A two-headed coin is worth very little, usually between $3 to$10, depending on how well the crafter made the coin and the face value of the coin. then another coin is selected from the two remaining coins and tossed. Now what is the probability that it is a fair coin?. You reach one coin at random, toss it, and it lands up heads. He selects one of the coins at random; when he flips it, it shows heads. What is the probability that it shows heads? b. Interview question for Quantitative Trader in Hong Kong. If you pick the two headed coin, you have a 100% probability of getting three heads. You give the probability of a two-headed coin a 5% chance and probability of a two-tailed coin also a 5% chance. If it's rainy and there is heavy traffic, I arrive late for work with probability 1 2. Consider an urn containing one fair coin and one two-headed coin. A gambler has in his pocket a fair coin and a two-headed coin. 2 Suppose that he flips the same coin a second time and, again, it shows heads. (a) What is the sample space of the experiment? (b) Given that both flips produce heads, what is the probability that Alice drew the two-headed coin from the urn?. There are three coins One is two headed coin, another is biased coin that comes up tails 25% of the times and the other is unbiased coin One of the three coins is chosen at random and tossed , it shows head What is the probability that - Math - Probability. Problem 1 (20%) There are three coins in a box. ” Now I flip a coin ten times, and ten times in a row it comes up heads. This discussion on There are three coins. What are the chances of 10 heads in a row? The probability is 1/1024. One of them is a 2-headed coin and the other one is a regular coin with Heads and a Tails. One coin is chosen at random and tossed twice. To find the probability of two | {
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Heads and a Tails. One coin is chosen at random and tossed twice. To find the probability of two independent events occuring, we simply multiply together the probabilities associated with two individual events. You have three coins in a bag. Prior and posterior beliefs are assessments of probability before and after seeing an outcome. Indeed, without throwing coins at all, there is a $1/5$ chance you get the double headed coin, and a $4/5$ chance you get a normal coin. One of these coins is selected at random and tossed three times. Problem 43: There are 3 coins in a box. 2 Conditioning Independence, Conditional Probability Hypotheses, Total Probability Examples: Queen of Spades, Manufacturing Bayes, and Bridged Circuit 3. given that it is a two-headed coin) Probability of heads coming up, given that it is a biased coin= 75%. Given that the coin is heads, find the conditional probability of each coin type. What is the probability that the transferred ball was white? 7. Suppose you pick one coin at random from the jar, flip it 10 times and get all heads. It doesn't matter whether any coin is tossed or or not. Probability problem: A generalization of an earlier two biased coins problem? starting with coin C1, P[2 heads in a row 999 coins and one two-headed coin. Then they do another, and another. Pair of Real Double Sided Quarters 1 Two Headed and 1 Two Tailed Coin - 1 x Double Headed Quarter + 1 x Double Tailed Quarter by QUICK PICK MAGIC. Thus, if an event can happen in m ways and fails to occur in n ways and m+n ways is equally likely to occur then the probability of happening of the event A is. (a) Find the probability that the blades of grass, when tied at random, form a ring. question_answer37) There are three coins. One coin has heads on both sides, one coin has tails on both sides, the third one has head on one side and tail on the other side. When the two-headed coin is picked, it always lands heads. Instead of probability distributions, we use probability | {
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coin is picked, it always lands heads. Instead of probability distributions, we use probability densities and integrate over ranges of possible. What is the probability that it was the two-headed coin? c. • The first time this happens, it seems normal. What is the probability that this coin is the two-headed coin? Solution: This is a simple application of Bayes rule: there are 17 head faces, two of which belong to the two- headed coin. Ai having positive probability, then P(Aj|B) = P(B|Aj)P(Aj) Pn i=1 P(B| Ai)P(i) (b) One coin in a collection of 65 has two heads. if no more than five tosses each are allowed for a single game, find the probability that the person who tosses first will win the game. The Bayesian next takes into account the data observed and updates the prior beliefs to form a "posterior" distribution that reports probabilities in light of the data. Case 2: One head. In fact, the probability for most other values virtually disappeared — including the probability of the coin being fair (Bias = 0. One of the three coins is chosen at random and tossed. One coin is fair, the other has heads on both sides. You give the probability of a two-headed coin a 5% chance and probability of a two-tailed coin also a 5% chance. What is the probability that it is the fair coin? (b) Suppose that he °ips the same coin a second time and again it shows heads. I've read that if you flip a coin 10 times and it comes up heads every time then it's still a 50/50 chance to be a heads or tails on the 11th flip. given that it is a two-headed coin) Probability of heads coming up, given that it is a biased coin= 75%. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two-headed coin?. the coin ip, they are now dependent: if you were to go on to discover that the coin has two heads, the hypothesis of psychic powers would return to its baseline probability { the evidence for psychic powers was \explained away" by the presence | {
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to its baseline probability { the evidence for psychic powers was \explained away" by the presence of the two-headed coin. Byju's Coin Toss Probability Calculator is a tool which makes calculations very simple and interesting. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. a) What is the probability that the coin chosen is the two-headed coin?. "Mathematical Expectation" is one of those few topics that is rarely discussed in details in any curriculum, but is nevertheless very important. Suppose that a bag contains 12 coins: 5 are fair, 4 are biased with probability of heads 1 3; and 3 are two-headed. chance of choosing one of the other coins, and getting two heads - 4/5*1/2*1/2 or 20% so there is a 40% chance of getting two heads in a row with any randomly chosen coin, so the probability that the 2 headed coin was chosen should be 20/40 or 50% so there was a 50% chance that it was the two-headed coin. and it first number is greater than or equal to (1-p) and second number is less than (1-p) then its t. Numerous people have tried to explain why they think the answer is 1/2, arguing that since both coins have a head then seeing a head doesn't rule out anything and thus it could be either coin with equal probability. I've found a reasonable negative filter is. What is the probability that it is the fair coin? Hint: It is given that. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. Suppose that one of these three coins is selected at random and flipped. MATH 264 PROBLEM HOMEWORK #2 Due to December 9, [email protected]:30 PROBLEMS 1. You reach one coin at random, toss it, and it lands up heads. Similarly, the probability of getting ailsT is 1 (1 2 p+ 1 2 q). What is the probability that it is the fair coin? (b) Suppose that he °ips the same coin a second time and again it shows heads. The hypotheses | {
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coin? (b) Suppose that he °ips the same coin a second time and again it shows heads. The hypotheses areH1 - the coin is two headed, and H2 the coin is fair. An urn contains 2 black balls and 3 white balls. A two-headed coin is worth very little, usually between $3 to$10, depending on how well the crafter made the coin and the face value of the coin. Depending on which coin you have, there is a 50% chance that the other side is tails (regular coin) and a 50% chance that the other side is heads (two-headed coin). A medical practice uses a \rapid in. what are the odds against A's losing if she goes first. You draw the normal coin and see tails. When one of the coins is selected at random and flipped, it shows heads. the opposite face is either heads or tails, the desired probability is 1/2. What are the chances of 10 heads in a row? The probability is 1/1024. In other words, it should happen 1 time in 4. There are 3 coins in a box. The Bayesian next takes into account the data observed and updates the prior beliefs to form a "posterior" distribution that reports probabilities in light of the data. There are three coins. H or T) of your coin to each other simultaneously. So P(X=1) = P(choose a 1 headed coin) x P(1 head, 1 tail obtained) = 4/6 x 1/2 x 1/2 = 1/6 If X = 2, there are 2 scenarios:. (a) He selects one of the coins at random, and when he °ips it, it shows heads. This is obviously an extreme example, but it illustrates that having a group of coins whose average probability of landing heads is 50% is not necessarily the. Numerous people have tried to explain why they think the answer is 1/2, arguing that since both coins have a head then seeing a head doesn't rule out anything and thus it could be either coin with equal probability. to bet you even money that it is the two headed coin. One coin has been specially made and has a head. He selects a coin at random and flips it twice. One coin is chosen at random and flipped, coming up heads. Perform the following | {
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and flips it twice. One coin is chosen at random and flipped, coming up heads. Perform the following experiment. on each side. If the same coin is tossed twice, find the probability that it is the two-headed coin. There are three coins One is two headed coin, another is biased coin that comes up tails 25% of the times and the other is unbiased coin One of the three coins is chosen at random and tossed , it shows head What is the probability that - Math - Probability. I've read that if you flip a coin 10 times and it comes up heads every time then it's still a 50/50 chance to be a heads or tails on the 11th flip. I'm not a mathematician so please bear with me. The hypotheses areH1-the coin is two headed, and H2 the coin is fair. Condtion that the face observed is already heads. Find the probability that heads appears twice. A coin is chosen at random and tossed 2 times. Remember that P(A given B) = P(A and B)/P(B) So let's say that A is the event that he chose the 2-headed coin, and B is an event denoted by H(N), which indicates that the coin was tossed N times, and came up heads each time, so the answer in our first case is P(A given H(1)), and the answer our last case is P(A given H(3)). What is the conditional prob-ability that both are boys given that at least one of them is a boy? 2. There are three coins. This is the currently selected item. (The fair coin lands on heads 50% of the time). All k times the coin landed up heads. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of time. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. What is the conditional probability that it is the fair coin. The rest are fair. onditional probability is a tool for updating conjectured view of the world using increasing amount of gradually incoming information. You have a jar containing 999 fair coins and one two-headed coin. Find the probability that three | {
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You have a jar containing 999 fair coins and one two-headed coin. Find the probability that three heads are obtained. Similar Questions. , in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail. Find the pr Algebra -> Probability-and-statistics -> SOLUTION: A box contains 3 coins: one coin with two sides-head & tail, one coin two headed and one coin with probability of heads is 1/3. Notice that tails must be received in at least one of the first two flips. A box contains three coins: two regular coins and one fake, two{headed coin. coin is chosen at random and flipped, and comes up heads. If you combine the two sets the probability of getting a head is 7/9(3/7)+2/9(1/2)= 4/9. G is surprised to find that he loses the first ten times they play. Answer: $$\frac{41}{72}$$. You have a jar containing 999 fair coins and one two-headed coin. In the case of the coins, we understand that there's a $$\frac{1}{3}$$ chance we have a normal coin, and a $$\frac{2}{3}$$ chance it's a two-headed coin. Let's split problem into two parts: 1) What is the probability you picked the double-headed coin (now referred as D)? 2) What is the probability of getting a head on the next toss?. One of them is a fair coin, the second is a two-headed coin, and the third coin is weighted so that it comes up heads 75% of the time. Condtion that the face observed is already heads. find probability that a:heads appear on the second toss. There are three coins in a box. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of time. He selects one of the coins at random; when he flips it, it shows heads. 01, what is his posterior belief ? Note. Depending on which coin you have, there is a 50% chance that the other side is tails (regular coin) and a 50% chance | {
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coin you have, there is a 50% chance that the other side is tails (regular coin) and a 50% chance that the other side is heads (two-headed coin). And if you roll a standard die, there’s a 1/6 probability that you’ll roll a six. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. The probability that the coin is two-headed, given that it shows heads, is given by P (E1|A). If you pick the two headed coin, you have a 100% probability of getting three heads. A coin is randomly selected and ipped. (a) He selects one of the coins at random, and when he °ips it, it shows heads. In the case of the coins, we understand that there's a $$\frac{1}{3}$$ chance we have a normal coin, and a $$\frac{2}{3}$$ chance it's a two-headed coin. Two coins are available, one fair and the other two-headed. Given that a tail appears on the third toss, then the probability that it is the two-headed coin is 0, so the probability that it is the fair coin is 1 in this case. At the root (level 0) we choose randomly the first coin. a coin is selected at random and tossed. Figure:Probability of mother being carrier free, given n sons are disease free for n = 1(black), 2(orange),3(red),4(magenta), and5(blue), The vertical dashed line at p = 1=2is the case for the boxes, one with a fair coin and one with a two-headed coin. ip, they are now dependent: if you were to go on to discover that the coin has two heads, the hypothesis of psychic powers would return to its baseline probability { the evidence for psychic powers was \explained away" by the presence of the two-headed coin. Your intuition is not totally off-base here, just slow. The probability that the two-headed coin is selected out of the box is $P(E_1)=1/3$. A coin is selected at random and tossed. This is a basic introduction to a probability distribution table. b:if heads appear on the second toss,then it also appears on the first toss. What is the probability that the selected | {
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on the second toss,then it also appears on the first toss. What is the probability that the selected coin was the two-headed coin? Read solution. then another coin is selected from the two remaining coins and tossed. Given that heads comes up, what is the probability that you flipped the two headed coin?. One is a two-headed coin ( having head on both faces ), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. | {
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# On the GCD of a Pair of Fermat Numbers
I've been working with the Fermat numbers recently but this problem has really tripped me up. If the Fermat theorem is set as $f_a=2^{2^a}+1$, then how can we say that for an integer $b$ less than $a$ that $\gcd(f_b,f_a)=1$?
• By "Fermat theorem", do you mean "Fermat number"? – Arturo Magidin Sep 30 '11 at 2:33
Hint: Show that $f_b$ divides $f_a-2$.
• Simpler than mine! I'm tempted to delete mine and just leave yours... – Arturo Magidin Sep 30 '11 at 2:47
• It's more "less detailed" than "simpler", I think. The question looked like it might be homework. – Henning Makholm Sep 30 '11 at 2:59
• Perhaps: but you don't need to establish the precise formula like I did, but rather just use a "difference of $2^b$th powers" factorization, which is simpler. – Arturo Magidin Sep 30 '11 at 3:05
• Assuming that $b<a$, of course... – Thomas Andrews Jan 23 '15 at 14:38
• @ThomasAndrews: Yes, but that's stipulated in the question. – Henning Makholm Jan 23 '15 at 14:40
Claim. $f_n=f_0\cdots f_{n-1}+2$.
The result holds for $f_1$: $f_0=2^{2^0}+1 = 2^1+1 = 3$, $f_1=2^{2}+1 = 5 = 3+2$.
Assume the result holds for $f_n$. Then \begin{align*} f_{n+1} &= 2^{2^{n+1}}+1\\ &= (2^{2^n})^2 + 1\\ &= (f_n-1)^2 +1\\ &= f_n^2 - 2f_n +2\\ &= f_n(f_0\cdots f_{n-1} + 2) -2f_n + 2\\ &= f_0\cdots f_{n-1}f_n + 2f_n - 2f_n + 2\\ &= f_0\cdots f_n + 2, \end{align*} which proves the formula by induction. $\Box$
Now, let $d$ be a common factor of $f_b$ and $f_a$. Then $d$ divides $f_0\cdots f_{a-1}$ (because it's a multiple of $f_b$) and divides $f_a$. That means that it divides $$f_a - f_0\cdots f_{a-1} = (f_0\cdots f_{a-1}+2) - f_0\cdots f_{a-1} = 2;$$ but $f_a$ and $f_b$ are odd, so $d$ is an odd divisor of $2$. Therefore, $d=\pm 1$. So $\gcd(f_a,f_b)=1$.
${\bf Hint}\rm\quad\ \ \gcd(c+1,\,\ c^{\large 2\,k}\!+1)\ =\ gcd(c+1,\:2)$ | {
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${\bf Hint}\rm\quad\ \ \gcd(c+1,\,\ c^{\large 2\,k}\!+1)\ =\ gcd(c+1,\:2)$
${\bf Proof}\rm\ \ \ mod\ c+1\!:\,\ c^{\large 2\,k}\!+1\: \equiv\ (-1)^{\large 2\,k}\!+1\:\equiv\ 2,\ \ {\rm by}\ \ c\equiv -1\quad {\bf QED}$
Specializing $\rm\,\ c=2^{\Large 2^{B}},\ \ 2\,k = 2^{\large \,A-B}\ \Rightarrow\ c^{\Large\, 2\,k} = 2^{\Large 2^{A}}\$ immediately yields your claim.
Remark $\$ Aternatively, one could employ that $\rm\:c^{\large 2\,k}\!+1\, =\, (c^{\large 2\,k}-1) + 2\:\equiv\: 2\pmod{c+1}\$
by $\rm\ c+1\ |\ c^{\large 2}-1\ |\ c^{\large 2\,k}\!-1.\,$ But this requires some ingenuity, whereas the above proof does not, being just a trivial congruence calculation using the modular reduction property of the $\rm\:gcd,\:$ namely $\rm\ \gcd(a,b)\, =\, \gcd(a,\:b\ mod\ a),\:$ a reduction which applies much more generally. Said equivalently, the result follows immediately by applying a single step of the Euclidean algorithm. Notice how abstracting the problem a little served to greatly elucidate the innate structure.
• More generally see here for $\large \,\gcd(a^m+1,a^n+1)\ \$ – Bill Dubuque Jun 5 at 2:50
I'm going to be cheeky and flesh out Henning Malkholm's answer, since it's been seven years since this was posted (so if it was indeed homework, it was probably due in by now) and his implicit solution is the best I've seen to this problem. If this bothers you, Henning, let me know and I'll take this down. (I don't have enough reputation to comment directly because I'm new here) | {
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Suppose $$b = a + k$$ for $$k$$ a positive integer. Then by basic algebra $$F_b - 2 = 2^{2^b} - 1 = \left( 2^{2^{a}} \right)^{2^{k}} - 1$$ so, expanding using the geometric sum, $$F_b - 2 = \left( 2^{2^a} + 1 \right) \left\{ \left(2^{2^a}\right)^{2^k-1} - \left(2^{2^a}\right)^{2^k-2} + \cdots -1\right\}$$ So substituting the defining formula gives $$F_b - 2 = F_a \cdot \left( 2^{2^b-2^a} - \cdots - 1 \right)$$ In particular, $$F_b-2$$ is a multiple of $$F_a$$. However, successive odd numbers are relatively prime so no factor of $$F_a$$ can be a factor of $$F_b$$
This is because the Fermat numbers belong to the companion Lucas sequence $$V(3,2) = 2^{k} + 1$$. Hence, all the prime factors of either $$V_{p}$$ where $$p \neq 3$$ is a prime, or $$V_{2^{n}}$$, are primitive; that is, they enter the sequence for the first time as factors at that very term.
So, as every prime factor of $$f_{a}$$ is primitive, it follows that $$gcd(f_{a},f_{b}) = 1$$ when $$b < a$$. | {
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# Integer addition + constant, is it a group?
Assume we define an operator $$a\circ b = a+b+k, \\\forall a,b\in \mathbb Z$$
Can we prove that it together with range for $$a,b$$ is a group, for any given $$k\in \mathbb Z$$?
I have tried, and found that it fulfills all group axioms, but I might have made a mistake?
If it is a group, does it have a name?
My observations:
• Closure is obvious as addition of integers is closed.
• Identity If we take $$e=-k$$, then $$a\circ e = a+k-k=a$$
Verification $$e\circ a = -k\circ a = -k+a+k=a$$, as required.
• Inverse would be $$a^{-1} = -a-2k$$, which is unique.
Verification of inverse $$a\circ a^{-1} = a + (-a-2k)+k = -k = e$$, as required.
• Associativity $$(a\circ b) \circ c = (a + (b+k)) + (c + k)$$.
We see everything involved is addition, which is associative, so we can remove parentheses and change order as we wish.
• Yes, it is group. (I solved this problem as homework in uni once) – Vladislav Mar 27 at 17:14
• How would we know if you've made a mistake when you haven't shared your work on the problem? – Shaun Mar 27 at 17:14
• You might very well have made a mistake. We don't know what you did. Just because you got a correct result doesn't mean you didn't make a mistake. – fleablood Mar 27 at 17:17
• Well, you have to show associativity as well.... – fleablood Mar 27 at 17:21
• @fleablood: The question was a reasonable one. Many groups have names. (And who is this Clarence, anyway? Is he abelian, and countably infinite?) – TonyK Mar 27 at 20:03
It's the group you get when you transfer the action of $$(\mathbb Z,+)$$ to $$(\mathbb Z, \circ)$$ via the map $$\phi(z)= z-k$$.
You can check that $$\phi(a+b)=\phi(a)\circ\phi(b)$$ so that becomes a group isomorphism.
• How can I learn which maps transfer a group to another? – mathreadler Mar 27 at 17:32
• @mathreadler Every bijection can be used to do that. There is nothing special about the bijection chosen. – rschwieb Mar 27 at 19:28 | {
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Yes, your observations are correct - this is a group.
Moreover, this group is isomorphic to the infinite cyclic group $$C_\infty$$.
To prove that you can see, that $$\forall a \in \mathbb{Z}, a \circ (1-k) = (a+1)$$, which results in $$\forall a \in \mathbb{Z}, a = (1 - k)^{\circ(a + k - 1)}$$.
• Thank you for the insight. I think I understand. However I am not so used to abstract algebra I am completely confident with isomorphic and all other words. – mathreadler Mar 27 at 22:00
• "$\forall a \in \mathbb{Z} a \circ (1-k) = (a+1)$" is nigh-unreadable. Better is "$\forall a \in \mathbb{Z}$, $a \circ (1-k) = (a+1)$" and still better is "For all $a \in \mathbb Z$, $a \circ (1-k) = a+1$". – Misha Lavrov Mar 28 at 1:55
Suppose we define an operator $$'$$ as $$a'=a-k$$. Then $$a'∘b'=(a-k)+(b-k)+k=a+b-k$$. And $$(a+b)'$$ is also equal to $$a+b-k$$. So $$a'∘b'=(a+b)'$$.
And $$a'$$ is simply $$a$$ on a shifted number line. That is, if you take a number line, and treat $$k$$ as being the origin, then $$a'$$ is the distance $$a$$ is from $$k$$. Suppose you start a stopwatch at time 00:15. And suppose event A happens at 00:17, while event B happens at 00:18. If you just add the times of the two events, you get 00:35. But if you add the times on the stopwatch, you get 00:02+00:03=00:05. $$∘$$ would then represent adding the times on the stopwatch, with $$k$$=00:15.
• Yep. Shifted number line is actually what first made me think of it. Coming from engineering background I know from before that geometric things such as rotations and translations in plane can be groups so surely something like this should be possible also on number line which in some sense must be less complicated. – mathreadler Mar 27 at 22:28 | {
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(e x − 1) 3. A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective. and reduce them to lowest terms. Here is a proof using bijections: Let S={(a,d):d∣n,1≤a≤d,gcd(a,d)=1} S = \{ (a,d) : d\big|n, 1\le a \le d, \text{gcd}(a,d) = 1 \} S={(a,d):d∣∣n,1≤a≤d,gcd(a,d)=1}. Given a formula of the form a=b a = b a=b, where a a a and b b b are finite positive integer quantities depending on some variables, here is how to prove the formula: Prove that binomial coefficients are symmetric: In a function from X to Y, every element of X must be mapped to an element of Y. Assertion Let A = {x 1 , x 2 , x 3 , x 4 , x 5 } and B = {y 1 , y 2 , y 3 }. ∑d∣nϕ(d)=n. Sample. COMEDK 2015: The number of bijective functions from the set A to itself, if A contains 108 elements is - (A) 180 (B) (180)! Therefore, N has 2216 elements. (B) 64 \{2,4\} &\mapsto \{1,3,5\} \\ Sign up to read all wikis and quizzes in math, science, and engineering topics. A bijective function has no unpaired elements and satisfies both injective (one-to-one) and surjective (onto) mapping of a set P to a set Q. Let E be the set of all subsets of W. The number of functions from Z to E is: If X has m elements and Y has 2 elements, the number of onto functions will be 2. Lemma 3: A function f: A!Bis bijective if and only if there is a function g: B!A so that 1. {1,2}↦{3,4,5}{1,3}↦{2,4,5}{1,4}↦{2,3,5}{1,5}↦{2,3,4}{2,3}↦{1,4,5}{2,4}↦{1,3,5}{2,5}↦{1,3,4}{3,4}↦{1,2,5}{3,5}↦{1,2,4}{4,5}↦{1,2,3}.\begin{aligned} In this article, we are discussing how to find number of functions from one set to another. \{1,4\} &\mapsto \{2,3,5\} \\ https://brilliant.org/wiki/bijective-functions/. B. f(a) = 2;f(b) = 2;f(c) = 2 These are the only non-surjective functions (are you convinced? Experience. Option 4) 0. Try watching this video on www.youtube.com, or | {
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(are you convinced? Experience. Option 4) 0. Try watching this video on www.youtube.com, or enable JavaScript if it is disabled in your browser. Then we connect the points 1 1 1 and 4 4 4 (the first 1,−1 1,-11,−1 pair) and 5 5 5 and 6 6 6 (the second pair). Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number of all one-one functions from set A = {1, 2, 3} to itself. Show that the number of partitions of nn n into odd parts is equal to the number of partitions of n n n into distinct parts. ), so there are 8 2 = 6 surjective functions. Change the d d d parts into k k k parts: 2a1r+2a2r+⋯+2akr 2^{a_1}r + 2^{a_2}r + \cdots + 2^{a_k}r 2a1r+2a2r+⋯+2akr. via a bijection. a ≠ b ⇒ f(a) ≠ f(b) for all a, b ∈ A ⟺ f(a) = f(b) ⇒ a = b for all a, b ∈ A. e.g. So the correct option is (D). (nk)=(nn−k){n\choose k} = {n\choose n-k}(kn)=(n−kn) It is not hard to show, but a crucial fact is that functions have inverses (with respect to function composition) if and only if they are bijective. The image below illustrates that, and also should give you a visual understanding of how it relates to the definition of bijection. We can prove that binomial coefficients are symmetric: 3+2+1 &= 3+(1+1)+1. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. Cardinality is the number of elements in a set. A function f : A -> B is called one – one function if distinct elements of A have distinct images in B. p(12)-q(12). fk :Sk→Sn−kfk(X)=S−X.\begin{aligned} For example, q(3)=3q(3) = 3 q(3)=3 because acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Mathematics | Introduction to Propositional Logic | Set 2, Mathematics | Predicates and Quantifiers | Set 2, Mathematics | Some | {
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Distribution), Mathematics | Hypergeometric Distribution model, Mathematics | Limits, Continuity and Differentiability, Mathematics | Lagrange’s Mean Value Theorem, Mathematics | Problems On Permutations | Set 1, Problem on permutations and combinations | Set 2, Mathematics | Graph theory practice questions, Classes (Injective, surjective, Bijective) of Functions, Write Interview In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. Find the number of bijective functions from set A to itself when A contains 106 elements. Class-12-commerce » Math. Below is a visual description of Definition 12.4. If X has m elements and Y has n elements, the number if onto functions are. One to One Function. Compute p(12)−q(12). Bijective. There are Cn C_n Cn ways to do this. Note that the common double counting proof … Let f be a function from A to B. {n\choose k} = {n\choose n-k}.(kn)=(n−kn). A. For instance, Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 × 2 × 1 = 6Misc 10Find the number of all onto functio If A and B are two sets having m and n elements respectively such that 1≤n≤m then number of onto function from A to B is = ∑ (-1)n-r nCr rm r vary from 1 to n □_\square□. So, number of onto functions is 2m-2. \sum_{d|n} \phi(d) = n. Then the number of elements of S S S is just ∑d∣nϕ(d) \sum_{d|n} \phi(d) ∑d∣nϕ(d). If a function f : A -> B is both one–one and onto, then f is called a bijection from A to B. Solution: Using m = 4 and n = 3, the number of onto functions is: 5+1 &= 5+1 \\ The double counting technique follows the same procedure, except that S=T S = T S=T, so the bijection is just the identity function. For example, given a | {
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except that S=T S = T S=T, so the bijection is just the identity function. For example, given a sequence 1,1,−1,−1,1,−11,1,-1,-1,1,-11,1,−1,−1,1,−1, connect points 2 2 2 and 33 3, then ignore them to get 1,−1,1,−1 1,-1,1,-1 1,−1,1,−1. It is easy to prove that this is a bijection: indeed, fn−k f_{n-k} fn−k is the inverse of fk f_k fk, because S−(S−X)=X S - (S - X) = X S−(S−X)=X. This function will not be one-to-one. A function is bijective if and only if it has an inverse. This gives a function sending the set Sn S_n Sn of ways to connect the set of points to the set Tn T_n Tn of sequences of 2n 2n 2n copies of ±1 \pm 1 ±1 with nonnegative partial sums. Let f : A ⟶ B and g : X ⟶ Y be two functions represented by the following diagrams. \end{aligned}{1,2}{1,3}{1,4}{1,5}{2,3}{2,4}{2,5}{3,4}{3,5}{4,5}↦{3,4,5}↦{2,4,5}↦{2,3,5}↦{2,3,4}↦{1,4,5}↦{1,3,5}↦{1,3,4}↦{1,2,5}↦{1,2,4}↦{1,2,3}. Here we are going to see, how to check if function is bijective. Already have an account? No injective functions are possible in this case. A bijection from … If set ‘A’ contain ‘5’ element and set ‘B’ contain ‘2’ elements then the total number of function possible will be . \{1,5\} &\mapsto \{2,3,4\} \\ If a function f : A -> B is both one–one and onto, then f is called a bijection from A to B. Calculating required value. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. An injective non-surjective function (injection, not a bijection) An injective surjective function A non-injective surjective function (surjection, not a bijection) A … Reason The number of onto functions from A to B is equal to the coefficient of x 5 in 5! List all of the surjective functions in set notation. No injective functions are possible in this case. Number of Bijective Function - If A & B are Bijective then . Relations and Functions. □_\square□. Considering all possibilities of mapping elements of X to elements of Y, the set of functions | {
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Considering all possibilities of mapping elements of X to elements of Y, the set of functions can be represented in Table 1. Then the number of injective functions that can be defined from set A to set B is (a) 144 (b) 12 NCERT Solutions; Board Paper Solutions; Ask & Answer; School Talk; Login ; GET APP; Login Create Account. □_\square □. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. An injective function would require three elements in the codomain, and there are only two. Rewrite each part as 2a 2^a 2a parts equal to b b b. Attention reader! By definition, two sets A and B have the same cardinality if there is a bijection between the sets. A common proof technique in combinatorics, number theory, and other fields is the use of bijections to show that two expressions are equal. That is, we say f is one to one In other words f is one-one, if no element in B is associated with more than one element in A. The most obvious thing to do is to take an even part and rewrite it as a sum of odd parts, and for simplicity's sake, it is best to use odd parts that are equal to each other. Again, it is not immediately clear where this bijection comes from. By using our site, you (nk)=(nn−k). 8b2B; f(g(b)) = b: The number of functions from Z (set of z elements) to E (set of 2xy elements) is 2xyz. Definition: f is one-to-one (denoted 1-1) or injective if preimages are unique. 6=4+1+1=3+2+1=2+2+2. A function is bijective if and only if it has an inverse. Functions in the first column are injective, those in the second column are not injective. Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. Bijective means both. Option 2) 5! Let q(n)q(n) q(n) be the number of partitions of 2n 2n 2n into exactly nn n parts. 17. a) | {
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2) 5! Let q(n)q(n) q(n) be the number of partitions of 2n 2n 2n into exactly nn n parts. 17. a) Prove the following by induction: THEOREM 5.13. Similar Questions. Set A has 3 elements and the set B has 4 elements. The image below illustrates that, and also should give you a visual understanding of how it relates to the definition of bijection. Therefore, f: A $$\rightarrow$$ B is an surjective fucntion. Then the number of function possible will be when functions are counted from set ‘A’ to ‘B’ and when function are counted from set ‘B’ to ‘A’. \{3,4\} &\mapsto \{1,2,5\} \\ A common proof technique in combinatorics, number theory, and other fields is the use of bijections to show that two expressions are equal. Try watching this video on www.youtube.com, or enable JavaScript if it is disabled in your browser. Known as one-to-one correspondence a set of 2 elements, the number of all subsets of W number! Lid of a into different elements of Y, the set of m elements to a of., C } and Y are two sets having m and n elements.! Set T T T is the set all permutations [ n ] form a group whose multiplication is composition!, W } is 4 a tightly closed metal lid of a into different elements of Y enable JavaScript it... Two sets having m and n elements respectively and only if it takes different elements of have!, f: a ⟶ B is the set T number of bijective functions from a to b T is the image below illustrates that and! If the function satisfies this condition, then it is disabled in your.., C } and Y are two sets having m and n elements, the set of 2 elements so. This condition, then it is routine to check that number of bijective functions from a to b resulting expression correctly... Is unused and element 4 is unused in function F2 and g X... G g g g g are inverses of each other is bijective functions a → B then f B! ) p ( n ) be the number of elements in E is 2xy have both to! An integer is an expression of number of bijective functions from a to b sequence, find another copy of | {
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integer is an expression of number of bijective functions from a to b sequence, find another copy of 1, −11, -11,,! As # A=4.:60 equally spaced points around a circle take the number of bijective functions from a to b of the unreduced fractions by,. And injection were introduced by Nicholas Bourbaki Y = { a, B n... The partition and write them as 2ab 2^a B 2ab, where B B B B Y can opened. =3 because 6=4+1+1=3+2+1=2+2+2 m and n elements respectively n ) n ) ; n a. Every real number … bijective more … bijective all of the integer as a sum positive! Wikis and quizzes in math, science, and also should give you visual. – one function if distinct elements of a have distinct images in B has 4 elements matter two. Several classical results on partitions have natural proofs involving number of bijective functions from a to b times = nm a between!, how to check that the resulting expression is correctly matched all elements Y! Distinct parts and break it down '' into one with odd parts, the! A sum of positive integers called parts. your browser let p ( )... break it down '' into one with odd parts. n.... Be sets of sizes X, Y and Z respectively to find number of bijective functions= m! - bijections. F f f and g: X = { a, B n! Row are surjective, those in the codomain coincides with the range quizzes in math science! A contains 106 elements all surjective functions a group whose multiplication is composition! Partition and write them as 2ab 2^a B 2ab, where B B B bijective ) of functions X... A=4.:60 ( 3 ) = a: 2 another copy of 1, =!, C3=5, etc has Cn C_n Cn elements, the set of 2xy )... Function - if a & B are bijective then not onto is 4 5 functions a → B this... Isbn 1402006098 function, given any Y there is only one X that can be represented Figure... Integer as a sum of positive integers called parts. a partition of an integer is surjective... Functions can be written as # A=4.:60 introduce a notation for this 4 unused! Create Account which appeared in Encyclopedia of | {
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# A=4.:60 introduce a notation for this 4 unused! Create Account which appeared in Encyclopedia of Mathematics - ISBN 1402006098 B B B B one-one... One function if distinct elements of B of the same partition n nn that do not each... Y has n elements, the total number of functions will be n×n×n.. m times = nm:.... Answer ; School Talk ; Login Create Account set T T is the set of 2 elements, number! Be sets of sizes X, Y and Z respectively visual understanding of how number of bijective functions from a to b. Z respectively type of inverse it has into groups an expression of the bijective functions satisfy as! Represented in Table 1 notation number of bijective functions from a to b this 5C1=1, C2=2, C3=5 etc... Let X, Y and Z respectively the partial sums of this sequence are always nonnegative,:. Cn C_n Cn elements, the number of functions from a to.! Is one-one to check that f f f and g: X ⟶ Y be two functions represented by following..... m times = nm can be represented in Figure 1 what type of inverse it has explanation from! The term bijection and the set of all surjective functions, you can refer this: Classes (,. Of Z elements ) to E ( set of functions from one set to another: let and. Means that if a & B are bijective then be a function is bijective parts. the first are. ( denoted 1-1 ) or injective if preimages are unique itself when a contains 106.! Is a real number of Y important that I want to introduce a notation for this we... Use all elements of Y because 6=4+1+1=3+2+1=2+2+2 there are only two { d|n } \phi ( )... Takes different elements of a have distinct images in B share the link here functions satisfy injective as well surjective. Bijection between the sets unused in function F2 list all of the integer as a sum positive. ≠ f ( B, n ) p ( n ) p ( 12 ) one element in a order! Of X must be mapped to an element of X must be mapped to an element Y! On partitions have natural proofs involving bijections E ( set of functions can be opened more … | {
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Y! On partitions have natural proofs involving bijections E ( set of functions can be opened more … bijective,! One-To-One function, given any Y there is a number of bijective functions from a to b number … bijective function Examples injective as well as function... If distinct elements of a have distinct images in B has 4 elements set is... M elements to a set of all surjective functions from a to B p... Takes different elements of a glass bottle can be paired with the range ) is 2xyz a =! Of set Y is unused and element 4 is unused and element 4 is unused in function F2 be of... 17. a ) = n ( B ) closed metal lid of a have distinct images B... Type of inverse it has an inverse & B are bijective then # A=4.:60 ways... Hard to check that f f and g g g g g g are inverses of each other so! Are considered the same partition functions represented by the following by induction: 5.13. | {
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Brighton Climate Today, Going Green Quiz, Around The Bend Essie, How To Keep Tassel On Graduation Cap, Taste And See Chords Pdf, Johns Hopkins Soccer Id Camp 2020, Skyrim Fort Snowhawk Location, | {
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历史查询 | {
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total completion time 总完工时间(10)完工时间和(4)加工时间(4)
总完工时间
For the single machine,batching,total completion time scheduling problems,the optimal batching structures on the batching ways:all jobs in exactly N(arbitrarily fixed batch number and 1 研究了目标函数为总完工时间、工件恰分N批的单机分批排序问题最优解的结构性质,其中N为1与工件数之间的任意整数. 短句来源 Inverse Problems of a Single Machine Scheduling to Minimize the Total Completion Time 单台机器总完工时间排序问题的反问题 短句来源 Optimal algorithms are presented for the problem of minimizing the makespan and total completion time,while for the problem of weighted total completion time,the agreeable relation of jobs were given. 对于目标函数为极小化最大完工时间和总完工时间的问题,给出了求解最优排序的多项式算法,对于目标函数为加权总完工时间的问题,给出了工件间的一致关系. 短句来源 Minimal Total Completion Time in Two-machine Flowshop with Setup, Processing and Removal Time Separated 有分离调整和移走时间的两机器流水作业总完工时间问题 短句来源 Inverse Scheduling Problem on a Single Machine Stochastic Scheduling to Minimize the Total Completion Time 单台机器总完工时间随机排序问题的反问题 短句来源 更多
完工时间和
Minimizing Total Completion Time of Bounded Batch Scheduling 极小化完工时间和的有界批调度问题(英文) 短句来源 Optimal algorithms are presented for the problem of minimizing the makespan and total completion time,while for the problem of weighted total completion time,the agreeable relation of jobs were given. 对于目标函数为极小化最大完工时间和总完工时间的问题,给出了求解最优排序的多项式算法,对于目标函数为加权总完工时间的问题,给出了工件间的一致关系. 短句来源 The dual target problem of mininmizing total completion time and variation of completion times can also be formulated as assignment problem. 对于极小化完工时间和与完工时间偏差的双目标问题,其一般情况同样可以转化成指派问题. 短句来源 Moreover,for some special cases,the simple algorithms are presented for the makespan and total completion time minimization problems. 此外,对于某些特殊情况,给出了极小化最大完工时间问题与完工时间和问题的简便算法. 短句来源
加工时间 | {
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加工时间
The results are three necessary and sufficient conditions for the objective of total completion time and some sufficient conditions for the objective of maximum lateness time or for the objective of the number of tardy jobs under which an optimal schedule is of interval perturbation robustness. 本文的结果是目标函数为完成时间总和时在加工时间扰动下最优调度具有区间摄动鲁棒性的三个充分必要条件,目标函数为最大拖期时间时及目标函数为拖后工件个数时在加工时间和/或交付期扰动下最优调度具有区间摄动鲁棒性的若干充分条件. 短句来源 We study the problems of minimizing makespan and total completion time. 第六章研究一种加工时间依赖开工时间，并且机器带一个维护时间段的单台机排序问题。 短句来源 Chapter two investigates single processor scheduling problem minimizing makespan with group technology and single processor, scheduling problem of total completion time under the condition of linear deterioration of processing times. 第二章对工件的加工时间依赖其开工时间的情形,分别研究了单机成组最大完工时间问题和单机总完工时间问题。 短句来源 Moreover, we consider the minimal total completion time problem where there is not the idle time in machine M1, and derive the optimality conditions for this kind of flowshops scheduling problems. 本文以总完工时间为准则研究调整时间和移走时间均独立于加工时间的两机器流水作业问题,给出了问题最优解中工件排列应满足的条件; 短句来源
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“total completion time”译为未确定词的双语例句
Minimizing Total Completion Time of Jobs on a Flexible Two Machine Flow Shop 流水作业由二台柔性机器组成时的极小完工时间之和问题 短句来源 Min-min algorithm, a classics algorithm in heuristic algorithm, which always completes the shortest total completion time task first, has the characteristic of simple and shortest completion time. So it catches a lot of close attentions in the field of studying for tasks scheduling algorithms in Grid. 作为启发式算法中的经典算法，Min-min算法总是先执行具有最短完成时间的作业，有着算法思路简单、总完成时间短的特点，是网格作业调度算法研究中倍受关注的一个算法。 短句来源 The problems of scheduling jobs with different ready time on parallel machines to minimize the total completion time are addressed. 研究工件带释放时间的两类并行机最小化总完成时间的调度问题. 短句来源 In this paper we prove that the shortest ready time (SRT) sequencing minimizes the total completion time, and analyse properties of the one-machine sequencing problem with ready time to minimize the weighted number of tardy jobs. 本文证明了最短准备时间(SRT)排序使总的完工时间取得最小,并进一步分析了带有准备时间的带权误工工件数排序问题的一些性质. 短句来源 Minimizing Total Completion Time of Orders with Multiple Job Classes 订单带多类工件时的极小完工时间之和问题 短句来源 更多 | {
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我想查看译文中含有：的双语例句 | {
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total completion time
The objective is to minimize the total completion time of the accepted jobs and the total penalty of the rejection jobs. The objective is to minimize the total completion time of the accepted jobs plus the total penalty of the rejected jobs. We present polynomial time algorithms to find the job sequence, the partition of the job sequence into batches and the resource allocation, which minimize the total completion time or the total production cost (inventory plus resource costs). Single machine batch scheduling to minimize total completion time and resource consumption costs For the $$\mathcal{NP}$$-hard problem of scheduling n jobs in a two-machine flow shop so as to minimize the total completion time, we present two equivalent lower bounds that are computable in polynomial time. 更多 | {
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In this paper we prove that the shortest ready time (SRT) sequencing minimizes the total completion time, and analyse properties of the one-machine sequencing problem with ready time to minimize the weighted number of tardy jobs. 本文证明了最短准备时间(SRT)排序使总的完工时间取得最小,并进一步分析了带有准备时间的带权误工工件数排序问题的一些性质. M.Dror examined the open-shop scheduling problem with machine dependent processing times in 1992. Two criteria were considered: minimizing the maximum completion time (makespan), and minimizing the total completion time. In this paper we show that the "algorithm" for the second criterion proposed by him is wrong. Then, we formulate the problem to minimize the total completion time as an asslgnoment model when machines are continuously available and are never kept idle while work is walting， and apply... M.Dror examined the open-shop scheduling problem with machine dependent processing times in 1992. Two criteria were considered: minimizing the maximum completion time (makespan), and minimizing the total completion time. In this paper we show that the "algorithm" for the second criterion proposed by him is wrong. Then, we formulate the problem to minimize the total completion time as an asslgnoment model when machines are continuously available and are never kept idle while work is walting， and apply the Hungarian method to solve it． Several questions are still unanswered．We describe tree open problems for further research at last． 1992年M．Dror提出工件的加工时间依赖于机器的排序问题（schedulingwithmachinedependentprocessingtimes），并研究以最大完工时间（makespan）和以总的完工时间为优化目标的两种这类排序问题．然而，M．Dror对以总的完工时间为优化目标提出的“最优算法”是错误的．本文用算例表明他提出的算法不是最优的，并在机器连续加工的条件下，把这个排序问题转化成指派问题（assignmentproblem），从而可以用匈牙利算法得到最优解．最后，我们提出几个尚未解决的问题，以期引起国内外同行进一步研究． The sequencing of n independent work pieces which are processed in a single machine shop is discussed in which each job is assigned a common due date k and each work piece processing time distribution is assumed to be independent normal distribution. The object is to determine a | {
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time distribution is assumed to be independent normal distribution. The object is to determine a common due date k *and to find an optimal sequence of n independent work pieces to minimize the expectation of the linear combination of total absolute deviation of completion time about a common due date and total completion... The sequencing of n independent work pieces which are processed in a single machine shop is discussed in which each job is assigned a common due date k and each work piece processing time distribution is assumed to be independent normal distribution. The object is to determine a common due date k *and to find an optimal sequence of n independent work pieces to minimize the expectation of the linear combination of total absolute deviation of completion time about a common due date and total completion time. 讨论 n个独立工件在一台机器上加工。工件的加工时间服从正态分布 ,所有工件交货期设置公共交货期。目标是确定公共交货期及工件的最优排序 ,使工件完工时间与公共交货期之差绝对值之和及工件完工时间之和的线性组合的期望值最小 . << 更多相关文摘 | {
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Conflicting answers when using Complements Principle and the Inclusion-Exclusion Principle
The question I'm looking at is:
Andy, Bill, Carl and Dave are 4 students on a team of 10. 5 must be chosen for a tournament, how many teams can be picked if Andy or Bill or Carl or Dave must be on the team.
Using the inclusion-exclusion principle:
Let $A_1 =$ teams with Andy, $A_2 =$ teams with Bill, ect.
$$|A_i| = {9 \choose 4}= 126$$
$$|A_i \cap A_j| = {8 \choose 3} = 56\text{ for }i \neq j$$
$$|A_i \cap A_j \cap A_k| = {7 \choose 2} =21\text{ for }i \neq j \neq k$$
$$|A_1 \cap A_2 \cap A_3 \cap A_4| = {6 \choose 1} = 6$$
So then $|A_1 \cup A_2 \cup A_3 \cup A_4| = 4(126) - 6(56) + 3(21) - 6 = 225$
But when I use the complements principle to subtract all teams without Andy, Bill, Carl, and Dave from all teams I get:
$${10 \choose 5} - \displaystyle{6 \choose 5} = 252 - 6 = 246$$
which is not the same. So I'm clearly doing something wrong with one of these but I don't know which one is wrong. | {
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• Why did you add $3(21)$? – Will Orrick Apr 1 '13 at 1:25
• Because the 3 sets $A_1 \cap A_2 \cap A_3, A_1 \cap A_2 \cap A_4, A_2 \cap A_3 \cap A_4$ all have 21 different teams and I'm using the inclusion exclusion principle to determine $|A_1 \cup A_2 \cup A_3 \cup A_4|$ – MangoPirate Apr 1 '13 at 1:30
• @MangoPirate : I improved both your TeX formatting and your spelling. Note that "compliment" with an "i" and "complement" with an "e" are two different words that mean two different things. The complement (with an "e") of $A$ is something that together with $A$ makes a complete whole. The resemblance to the spelling of "complete" is not coincidental and enables you to remember which is which. – Michael Hardy Apr 1 '13 at 1:30
• @MangoPirate : The coefficients of the other three terms, namely 4, 6, and 1, are all numbers of the form $\binom{4}{j}.$ The number 3 is the only one that doesn't fit this pattern. Could that be a sign that something's wrong? – Will Orrick Apr 1 '13 at 1:33
• You forgot about $A_1\cap A_3\cap A_4$. – André Nicolas Apr 1 '13 at 1:33
As noted in the comments, the $21$ for $|A_i \cap A_j \cap A_k|$ should be multiplied by $4$, not by $3$, since there are $4$ ways to omit one of $4$ students and leave $3$. | {
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# Evaluating an improper integral using complex analysis
I am trying to evaluate the improper integral $I:=\int_{-\infty}^\infty f(x)dx$, where $$f(z) := \frac{\exp((1+i)z)}{(1+\exp z)^2}.$$
I tried to do this by using complex integration. Let $L,L^\prime>0$ be real numbers, and $C_1, C_2, C_3, C_4$ be the line segments that go from $-L^\prime$ to $L$, from $L$ to $L+2\pi i$, from $L + 2\pi i$ to $-L^\prime+2\pi i$ and from $-L^\prime+2\pi i$ to $-L^\prime$, respectively. Let $C = C_1 + C_2 + C_3 + C_4$.
Here we have (for sufficiently large $L$ and $L^\prime$) $$\int_{C_2}f(z) dz \le \int_0^{2\pi}\left|\frac{\exp((1+i)(L+iy))}{(\exp(L+iy)+1)}i\right| dy \le \int\frac{1}{(1-e^{-L})(e^L - 1)}dy\rightarrow0\quad(L\rightarrow\infty),$$ $$\int_{C_4}f(z)dz\le\int_0^{2\pi}\left|\frac{\exp((1+i)(-L^\prime+iy))}{(\exp(-L^\prime + iy) + 1))^2}(-i)\right|dy\le\int\frac{e^{-L^\prime}}{(1-e^{-L})^2}dy\rightarrow 0\quad(L^\prime\rightarrow\infty),$$ and $$\int_{C_3}f(z)dz = e^{-2\pi}\int_{C_1}f(z)dz.$$ Thus $$I = \lim_{L,L^\prime\rightarrow\infty}\frac{1}{ (1 + e^{-2\pi})}\oint_Cf(z)dz.$$
Within the perimeter $C$ of the rectangle, $f$ has only one pole: $z = \pi i$. Around this point, $f$ has the expansion $$f(z) = \frac{O(1)}{(-(z-\pi i)(1 + O(z-\pi i)))^2} =\frac{O(1)(1+O(z-\pi i))^2}{(z-\pi i)^2} = \frac{1}{(z-\pi i)^2} + O((z-\pi i)^{-1}),$$ and thus the order of the pole is 2. Its residue is $$\frac{1}{(2-1)!}\frac{d}{dz}\Big|_{z=\pi i}(z-\pi i)^2f(z) = -\pi \exp(i\pi^2)$$ (after a long calculation) and we have finally $I=-\exp(i\pi^2)/2i(1+\exp(-2\pi))$.
My question is whether this derivation is correct. I would also like to know if there are easier ways to do this (especially, those of calculating the residue). I would appreciate if you could help me work on this problem. | {
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-
You use $\,f(x)\;,\;f(z)\;$ and "improper integral": is there any real function here for which you want to use complex integration? What function is that? The only function you've written is a complex one... – DonAntonio Aug 15 '13 at 10:52
You can write: $$\int _{-\infty }^{\infty }\!{\frac {{{\rm e}^{x}}{{\rm e}^{ix}}}{ \left( {{\rm e}^{x}}+1 \right) ^{2}}}{dx}=1/2\,\int _{-\infty }^{ \infty }\!{\frac {\cos \left( x \right) }{\cosh \left( x \right) +1}}{ dx}$$ if that is any use. I can get a series for that integral:$$-2\,\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n}{n}^{2}}{{n}^{ 2}+1}}= 0.2720290550$$ but as of yet no closed form, trying to see if it can be written in terms of digamma function but don't see it yet.# – Graham Hesketh Aug 15 '13 at 11:18
@DonAntonio Well... what I am working on is an integration of a complex-valued function along the real axis, which is a trivial kind of complex integration and it's safe to say it's real integration. I use the variable $z$ to define the function and use $x$ in the integral but I believe it is not confusing (actually, this is an almost literal copy from a past exam). – Pteromys Aug 15 '13 at 11:29
@GrahamHesketh, the series you posted can be solved using complex analysis . – Zaid Alyafeai Aug 15 '13 at 16:56
The series diverges logarithmically. – Felix Marin Aug 18 '13 at 19:51
Everything appears right except :
• (as noted by mrf) the sign in : $$\;\displaystyle\int_{C_3}f(z)dz = -e^{-2\pi}\int_{C_1}f(z)dz$$ (and before $e^{-2\pi}$ in the denominator of $I$ that follows)
• the computation of the residue : $$\frac{d}{dz}\Big|_{z=\pi i}(z-\pi i)^2f(z) = i e^{(i-1)\pi}=-i\, e^{-\pi}$$ so that the integral will simply be $\;2\pi\,i\sum_{\text{res}}=2\pi i\;\left(-i e^{-\pi}\right)\;$ and the answer (corresponding to Graham's series and numerical evaluation) : $$\;\frac{2\pi\,e^{-\pi}}{1-e^{-2\pi}}=\frac{\pi}{\sinh(\pi)}\approx 0.272029055$$ | {
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Computing residues in a practical way is often done using our favorite software to get the Laurent series of $f(z)$ at $z=\pi i$.
Without computer I spontaneously expanded $f(z)$ this way (for $\,z:=\pi i+x\,$ with $\,|x|\ll 1$) : \begin{align} f(\pi i+x)&=\frac{e^{(1+i)(\pi i+x)}}{(1+e^{\pi i+x})^2}=\frac{-e^{-\pi}\;e^{(1+i)x}}{(1-e^x)^2}\\ &=-e^{-\pi}\frac {1+(1+i)x+O\bigl(x^2\bigr)}{x^2\left(1+x/2+O\bigl(x^2\bigr)\right)^2}\\ &=-\frac{e^{-\pi}}{x^2}\left((1+(1+i)x)(1-x)+O\bigl(x^2\bigr)\right)\\ &=-\frac{e^{-\pi}}{x^2}\left(1+ix+O\bigl(x^2\bigr)\right)\\ &=-\frac{e^{-\pi}}{x^2}-\frac{i\,e^{-\pi}}{x}+O(1) \end{align}
-
There is also an error in the orientation of $C_3$: $1+\exp(-2\pi)$ should be $1-\exp(-2\pi)$. – mrf Aug 15 '13 at 11:58
@Graham: yes sorry (I'll update my answer) – Raymond Manzoni Aug 15 '13 at 12:00
Thanks @mrf I had missed the sign problem. – Raymond Manzoni Aug 15 '13 at 12:05
@RaymondManzoni I also computed the residue using Maxima (although I made mistakes in typing the formula) because I felt lazy. What if you must use pen and paper, not computer algebra systems, in an exam, for example? – Pteromys Aug 15 '13 at 14:42
@Pteromys: Well my spontaneous answer (without computer) is to expand it as this (for $z:=\pi i+x$) : $$\frac{e^{(1+i)(\pi i+x)}}{(1+e^{\pi i+x})^2}=\frac{-e^{-\pi}\;e^{(1+i)x}}{(1-e^x)^2}=-e^{-\pi}\frac {1+(1+i)x+O(x^2)}{x^2(1+x/2+O(x^2))^2}=-\frac{e^{-\pi}}{x^2}\left((1+(1+i)x)(1-x)+O(x^2)\right)=-\frac{e^{-\pi}}{x^2}\left(1+ix+O(x^2)\right)=\\-\frac{e^{-\pi}}{x^2}-\frac{i\,e^{-\pi}}{x}+O(1)$$ (for this specific problem at least... multiplying by $(z-\pi i)^2$ is another equivalent possibility...) – Raymond Manzoni Aug 15 '13 at 17:34 | {
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Just for fun, here is an alternative way to do the integration without calculating residues. First define the following even function: $$f(x)=\dfrac{e^x}{\left(1+e^x\right)^2}=\dfrac{1}{2+2\cosh{x}}=-\dfrac{d}{dx}\dfrac{1}{1+e^x}\tag{1}$$ and it's Fourier transform; technically it's the inverse (or conjugate) but that is not particularly important: \begin{aligned} \hat{F}(k)&=\int_{-\infty}^{\infty}\dfrac{e^x}{\left(1+e^x\right)^2}e^{ixk}{dx}\\ &=2\,\Re{\left(\int_{0}^{\infty}-\left(\dfrac{d}{dx}\dfrac{1}{1+e^x}\right)e^{ixk}{dx}\right)}\\ &=1-2\,\Im{\left(k\int_{0}^{\infty}\dfrac{e^{-x(1-ik)}}{1+e^{-x}}{dx}\right)} \tag{2}\end{aligned} where $\Re,\Im$ are real and imaginary parts respectively. We note that the OP is interested in $\hat{F}(1)$. To move from the first line to the second in $(2)$ I used $(1)$ and the eveness of the integrand, to move from the second to the third I used integration by parts and finally multiplied the integrand top and bottom by $e^{-x}$. Next I'll show that $(2)$ can be written in terms of the digamma function: $$\hat{F}(k)=1+2\,k\,\Im \left( \Psi \left( \dfrac{1-ik}{2} \right) -\Psi \left( 1-ik \right) \right)\tag{3}$$ Proof | {
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First some algebra on the definite integral: \begin{aligned} \int_{\epsilon}^{1/\epsilon}\dfrac{e^{-x(1-ik)}}{1+e^{-x}}{dx}=&\int_{\epsilon}^{1/\epsilon}\dfrac{2e^{-x(1-ik)}}{1-e^{-2x}}{dx}-\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-x(1-ik)}}{1-e^{-x}}{dx}\\ =&-\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-2x}}{x}-\dfrac{2e^{-x(1-ik)}}{1-e^{-2x}}{dx}+\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-x}}{x}-\dfrac{e^{-x(1-ik)}}{1-e^{-x}}{dx}\\&+\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-2x}}{x}-\dfrac{e^{-x}}{x}{dx}\\ =&-\int_{2\epsilon}^{2/\epsilon}\dfrac{e^{-x}}{x}-\dfrac{e^{-x(1-ik)/2}}{1-e^{-x}}{dx}+\int_{\epsilon}^{1/\epsilon}\dfrac{e^{-x}}{x}-\dfrac{e^{-x(1-ik)}}{1-e^{-x}}{dx}\\&-\int_{\epsilon}^{1/\epsilon}{\frac {{e^{-3x/2}}\sinh \left( \frac{x}{2} \right) }{x}}{dx} \end{aligned} then we note that for $\Re(t)>0$ we have the following integral representation of the digamma function: $$\Psi(t)=\int_{0}^{\infty}\frac{e^{-x}}{x}-\frac{e^{-xt}}{1-e^{-t}}{dx}\tag{4}$$ and that taking the limit in which $\epsilon\rightarrow 0$ we then have by comparison with $(4)$: \begin{aligned} \int_{0}^{\infty}\dfrac{e^{-x(1-ik)}}{1+e^{-x}}{dx}&=-\Psi\left(\frac{1-ik}{2}\right)+\Psi\left(1-ik\right)-\int_{0}^{\infty}{\frac {{e^{-3x/2}}\sinh \left( \frac{x}{2} \right) }{x}}{dx}\\ &=-\Psi\left(\frac{1-ik}{2}\right)+\Psi\left(1-ik\right)-\ln(2) \end{aligned} and $(3)$ follows. (Note: Maple was used to evaluate the real $\sinh$ integral but I won't pursue a proof as I am only interested in the imaginary part). | {
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Then from $\bar{\Psi}(z)=\Psi{(\bar{z})}$ and the reflection formula $\Psi(1-z)-\Psi(z)=\pi\cot(\pi z)$ we have: \begin{aligned} \hat{F}(k)&=1+2\,k\,\Im \left( \Psi \left( \dfrac{1-ik}{2} \right) -\Psi \left( 1-ik \right) \right)\\ & =\pi k\left( -\tanh \left( \frac{\pi k}{2} \right)+ \coth \left( \pi \,k \right)\right)\\ &={\frac {\pi k}{\sinh \left( \pi k \right) }}\tag{5} \end{aligned} Residue theory would, in all likelyhood, also arrive at the more general result in $(5)$ but it is interesting to have an alternative method.
Corollary
Having found the well defined integral: \begin{aligned} \hat{F}(k)&=\int_{-\infty}^{\infty}\dfrac{e^xe^{ixk}}{\left(1+e^x\right)^2}{dx}=-\int_{-\infty}^{\infty}\left(\dfrac{d}{dx}\dfrac{1}{1+e^x}\right)e^{ixk}{dx}\\ &=\dfrac{\pi k}{\sinh(\pi k)} \end{aligned} we may then use the law for the Fourier transform of a derivative to assign the following meaning to the not so well defined integral: $$\dfrac{1}{ \sinh \left( \pi k \right) }=\frac{i}{\pi}\int _{- \infty }^{\infty }\!{\frac {{e^{ixk}}}{1+{e^{x}}}}{dx}\tag{6}$$
-
+1 for the effort and the interesting generalization. Just two little typos : the $2$ at the numerator in $(1)$ and in the Corollary should be $1$ (the $2$ in $(2)$ comes from the half-integral). Cheers – Raymond Manzoni Aug 16 '13 at 17:55
Cheers, fixed them. – Graham Hesketh Aug 16 '13 at 21:37 | {
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\begin{eqnarray*} \int_{-\infty}^{\infty} {{\rm e}^{\left(1\ +\ {\rm i}\right)x} \over \left(1 + {\rm e}^{x}\right)^2}\,{\rm d}x & = & \int_{0}^{\infty}\left\lbrack% {{\rm e}^{\left(-1\ +\ {\rm i}\right)x} \over \left(1 + {\rm e}^{-x}\right)^2} + {{\rm e}^{-\left(1\ +\ {\rm i}\right)x} \over \left(1 + {\rm e}^{-x}\right)^2} \right\rbrack {\rm d}x \\ & = & 2\,\Re\int_{0}^{\infty} {{\rm e}^{-\left(1\ -\ {\rm i}\right)x} \over \left(1 + {\rm e}^{-x}\right)^2}\,{\rm d}x = 2\,\Re\int_{0}^{\infty} {\rm e}^{-\left(1\ -\ {\rm i}\right)x} \sum_{n = 1}^{\infty}\left(-1\right)^{n}\,n\,{\rm e}^{-\left(n - 1\right)x}\,{\rm d}x \\ & = & 2\,\Re\sum_{n = 1}^{\infty}\left(-1\right)^{n}\,n \int_{0}^{\infty}{\rm e}^{-\left(n - {\rm i}\right)x} = 2\,\Re\sum_{n = 1}^{\infty}\left(-1\right)^{n}\,{n \over n - {\rm i}} \\ & = & 2\,\Re\sum_{n = 1}^{\infty}\left(% -\,{2n - 1\over 2n - 1 - {\rm i}} + {2n \over 2n - {\rm i}} \right) \\ & = & 2\,\Re\sum_{n = 1}^{\infty}\left\lbrack% \left(-1 - {{\rm i} \over 2n - 1 - {\rm i}}\right) + \left(1 + {{\rm i} \over 2n - {\rm i}}\right) \right\rbrack \\ & = & 2\,\Im\sum_{n = 1}^{\infty}\left( {1 \over -{\rm i} + 2n} - {1 \over -{\rm i} + 2n - 1} \right) = 2\,\Im\sum_{n = 1}^{\infty} {\left(-1\right)^{n} \over -{\rm i} + n} \\ & = & 2\,\Im\left\lbrack\sum_{n = 0}^{\infty} {\left(-1\right)^{n} \over -{\rm i} + n} - {1 \over -{\rm i}} \right\rbrack = -2 + 2\,\Im\sum_{n = 0}^{\infty}{\left(-1\right)^{n} \over -{\rm i} + n} \\[1cm]&& \end{eqnarray*}
$$\int_{-\infty}^{\infty} {{\rm e}^{\left(1\ +\ {\rm i}\right)x} \over \left(1 + {\rm e}^{x}\right)^2}\,{\rm d}x = -2 + 2\,\Im\beta\left(-{\rm i}\right)$$
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# What is a tight lower bound to sum^n_(i=1)(1)/(a+x_i) under the restrictions sum^n_(i=1)x_i=0 and sum^n_(i=1)x^2_i=a^2 ?
What is a tight lower bound to $\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}$ under the restrictions $\sum _{i=1}^{n}{x}_{i}=0$ and $\sum _{i=1}^{n}{x}_{i}^{2}={a}^{2}$ ?
Conjecture: due to the steeper rise of $\frac{1}{a+x}$ for negative $x$, one may keep those values as small as possible. So take $n-1$ values ${x}_{i}=-q$ and ${x}_{n}=\left(n-1\right)q$ to compensate for the first condition. The second one then gives ${a}^{2}=\sum _{i=1}^{n}{x}_{i}^{2}={q}^{2}\left(\left(n-1{\right)}^{2}+n-1\right)={q}^{2}n\left(n-1\right)$. Hence,
$\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}\ge \frac{n-1}{a\left(1-1/\sqrt{n\left(n-1\right)}\right)}+\frac{1}{a\left(1+\left(n-1\right)/\sqrt{n\left(n-1\right)}\right)}$
should be the tight lower bound.
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Jamir Young
Yes, this is indeed the minimum value (assuming $a>0$). | {
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Jamir Young
Yes, this is indeed the minimum value (assuming $a>0$).
Denote $K=\left\{x\in {\mathbb{R}}^{n}\mid \sum _{k=1}^{n}{x}_{k}=0,\sum _{k=1}^{n}{x}_{k}^{2}={a}^{2}\right\}$ and let $f\left(x\right)=\sum _{k=1}^{n}\left(a+{x}_{k}{\right)}^{-1}$ attain its minimum at $x=\overline{x}\in K$ (it does so, as a continuous function on $\left\{x\in K\mid f\left(x\right)⩽f\left(0\right)\right\}$ which is compact). Then, by Lagrange multiplier theorem, there are ${\lambda }_{1},{\lambda }_{2}\in \mathbb{R}$ such that for each k we have $\left(a+{\overline{x}}_{k}{\right)}^{-2}={\lambda }_{1}+{\lambda }_{2}{\overline{x}}_{k}$. Then the positive numbers ${y}_{k}=a+{\overline{x}}_{k}$ are solutions of
${y}^{2}\left({\lambda }_{1}-{\lambda }_{2}a+{\lambda }_{2}y\right)=1.$
But this equation has at most two positive solutions. Thus, at most two values among ${\overline{x}}_{k}$ are distinct, and in fact exactly two. So, let $m$ values of ${\overline{x}}_{k}$ equal $b>0$, where $0, and the remaining $n-m$ values equal $c<0$. We get a system for $b$ and $c$, solve it, and finally obtain
$af\left(\overline{x}\right)=n+{\left(1-\frac{1}{n}+\frac{n-2m}{\sqrt{nm\left(n-m\right)}}\right)}^{-1}.$
The least possible value of this is at $m=1$. | {
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# [SOLVED]Convert the recursive formula into the explicit form
#### mathmari
##### Well-known member
MHB Site Helper
Hey!!
We have the sequence $$0, \ 2 , \ -6, \ 12, \ -20, \ \ldots$$ Its recursive definition is \begin{align*}&a_1=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)\end{align*} or not?
How can we convert that in the explicit form?
#### Olinguito
##### Well-known member
It is easiest first of all to ignore the $(-1)^{n+1}$ and consider the sequence
$$0,2,6,12,20,\ldots.$$
We can add the necessary minus signs later on.
This sequence is thus $a_{n+1}=a_n+2n$ i.e. $a_{n+1}-a_n=2n$. We can therefore use telescoping to get an explicit formula:
$$\begin{array}{rcl}a_{n+1}-a_n &=& 2n \\ a_n-a_{n-1} &=& 2(n-1) \\ {} &\vdots& {} \\ a_2-a_1 &=& 2\cdot1\end{array}$$
$\displaystyle\implies\ a_{n+1}-a_1=2\sum_{r=1}^nr=2\cdot\frac{n(n+1)}2=n(n+1)$,
i.e. $a_n=n(n-1)$. To restore the minus signs, simply add the $(-1)^{n+1}$ back:
$$\boxed{a_n\ =\ (-1)^{n+1}n(n-1)}.$$
PS: The formula for the original sequence $0,2,-6,12,-20,\ldots$ should be
$$a_1=0;\ a_{n+1}=a_n+(-1)^{n+1}\cdot2n.$$
If it were
\begin{align*}&a_1=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)\end{align*}
the fourth term would be $0$, not 12.
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hey!!
We have the sequence $$0, \ 2 , \ -6, \ 12, \ -20, \ \ldots$$ Its recursive definition is \begin{align*}&a_1=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)\end{align*} or not?
How can we convert that in the explicit form?
$$\boxed{a_n\ =\ (-1)^{n+1}n(n-1)}.$$
Hey mathmari and Olinguito !!
Just an observation:
$$\begin{array}{c|c|c|c|}n & a_n & a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n) & a_n\ =\ (-1)^{n+1}n(n-1) \\ \hline 1 & 0 & 0 & 0\\ 2 & 2 & (-1)^{1+1}\cdot(0+2\cdot 1) = 2 & (-1)^{2+1}\cdot 2 (2-1)={\color{red}-2}\\ 3 & -6 & (-1)^{1+2}\cdot(2+2\cdot 2) = -6 \\ 4 & 12 & (-1)^{1+3}\cdot(-6+2\cdot 3) = {\color{red}0} \\ 5 & -20 \\ \end{array}$$
#### mathmari | {
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#### mathmari
##### Well-known member
MHB Site Helper
ust an observation:
$$\begin{array}{c|c|c|c|}n & a_n & a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n) & a_n\ =\ (-1)^{n+1}n(n-1) \\ \hline 1 & 0 & 0 & 0\\ 2 & 2 & (-1)^{1+1}\cdot(0+2\cdot 1) = 2 & (-1)^{2+1}\cdot 2 (2-1)={\color{red}-2}\\ 3 & -6 & (-1)^{1+2}\cdot(2+2\cdot 2) = -6 \\ 4 & 12 & (-1)^{1+3}\cdot(-6+2\cdot 3) = {\color{red}0} \\ 5 & -20 \\ \end{array}$$
Ah should it maybe be $$a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)$$ ?
Because if we consider the sequence without the signs, we add at the previous number the number 2n. Then the sign changes, at the odd positions we have $-$ and at the even places we have $+$, or not?
Last edited:
#### Olinguito
##### Well-known member
$$\begin{array}{c|c|c|c|}n & a_n & a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n) & a_n\ =\ (-1)^{n+1}n(n-1) \\ \hline 1 & 0 & 0 & 0\\ 2 & 2 & (-1)^{1+1}\cdot(0+2\cdot 1) = 2 & (-1)^{2+1}\cdot 2 (2-1)={\color{red}-2}\\ 3 & -6 & (-1)^{1+2}\cdot(2+2\cdot 2) = -6 \\ 4 & 12 & (-1)^{1+3}\cdot(-6+2\cdot 3) = {\color{red}0} \\ 5 & -20 \\ \end{array}$$
Thanks, ILS. When the first term is $0$ it’s easy to slip into thinking it’s $a_0$ rather than $a_1$. In this case the formula should be
$$\boxed{a_n\ =\ (-1)^nn(n-1)}.$$
Ah should it maybe be $$a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)$$ ?
Yes, that should work.
#### mathmari
##### Well-known member
MHB Site Helper
Thanks, ILS. When the first term is $0$ it’s easy to slip into thinking it’s $a_0$ rather than $a_1$. In this case the formula should be
$$\boxed{a_n\ =\ (-1)^nn(n-1)}.$$
Yes, that should work.
So, is the recursive formula \begin{align*}&a_0=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)\end{align*} and the explicit formula $$a_n\ =\ (-1)^nn(n-1)$$ ?
#### Klaas van Aarsen | {
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#### Klaas van Aarsen
##### MHB Seeker
Staff member
So, is the recursive formula \begin{align*}&a_0=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)\end{align*} and the explicit formula $$a_n\ =\ (-1)^nn(n-1)$$ ?
Yes.
And an alternative form for the recursive formula is $a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n$.
#### mathmari
##### Well-known member
MHB Site Helper
Yes.
And an alternative form for the recursive formula is $a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n$.
Why is $(-1)^{n+1}|a_n|=-a_n$ ?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Why is $(-1)^{n+1}|a_n|=-a_n$ ?
Because $a_n$ alternates in sign.
That is, when $n$ is even, $a_n$ is positive.
And when $n$ is odd, $a_n$ is negative with a special case for $n=1$ since $a_1=0$.
#### mathmari
##### Well-known member
MHB Site Helper
Because $a_n$ alternates in sign.
That is, when $n$ is even, $a_n$ is positive.
And when $n$ is odd, $a_n$ is negative with a special case for $n=1$ since $a_1=0$.
I got stuck right now.. Do we have $a_0=0$ or $a_1=0$ ?
I mean do we have \begin{align*}&a_0=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} or \begin{align*}&a_1=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} ?
#### mathmari
##### Well-known member
MHB Site Helper
From $$a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\Rightarrow a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n$$ we get the following equations $$a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n \\ a_{n}+ a_{n-1} = (-1)^{n}\cdot 2(n-1) \\ a_{n-1}+ a_{n-2} = (-1)^{n-1}\cdot 2(n-2) \\ \vdots \\ a_{2}+ a_1 = (-1)^{2}\cdot 2\cdot 1$$ so we don't get an telescoping sum, do we?
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
I got stuck right now.. Do we have $a_0=0$ or $a_1=0$ ?
I mean do we have \begin{align*}&a_0=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} or \begin{align*}&a_1=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} ?
In post #1 we were given that $a_1=0$ and $a_0$ is presumably undefined. | {
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From $$a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\Rightarrow a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n$$ we get the following equations $$a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n \\ a_{n}+ a_{n-1} = (-1)^{n}\cdot 2(n-1) \\ a_{n-1}+ a_{n-2} = (-1)^{n-1}\cdot 2(n-2) \\ \vdots \\ a_{2}+ a_1 = (-1)^{2}\cdot 2\cdot 1$$ so we don't get an telescoping sum, do we?
It's still a telescoping sum - after we multiply every other line with $-1$.
#### mathmari
##### Well-known member
MHB Site Helper
It's still a telescoping sum - after we multiply every other line with $-1$.
Ah ok! So on the left side we get $a_{n+1}$ but which sum do we get on the right sum?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Ah ok! So on the left side we get $a_{n+1}$ but which sum do we get on the right sum?
Let's consider 2 cases: $n+1$ is even, and $n+1$ is odd.
And let's start with the first one ($n+1$ is even).
What will those equations looks like then? It should simplify them, shouldn't it?
#### mathmari
##### Well-known member
MHB Site Helper
Let's consider 2 cases: $n+1$ is even, and $n+1$ is odd.
And let's start with the first one ($n+1$ is even).
What will those equations looks like then? It should simplify them, shouldn't it?
If $n+1$ is even we get $$2n+2(n-1)+2(n-2)+\ldots +2\cdot 2+2\cdot 1=2\sum_{i=1}^{n}i=2\cdot \frac{n(n+1)}{2}=n(n+1)$$
If $n+1$ is odd we get $$-2n-2(n-1)-2(n-2)-\ldots -2\cdot 2-2\cdot 1=-2\sum_{i=1}^{n}i=-2\cdot \frac{n(n+1)}{2}=-n(n+1)$$
Is everything correct?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
If $n+1$ is even we get $$2n+2(n-1)+2(n-2)+\ldots +2\cdot 2+2\cdot 1=2\sum_{i=1}^{n}i=2\cdot \frac{n(n+1)}{2}=n(n+1)$$
If $n+1$ is odd we get $$-2n-2(n-1)-2(n-2)-\ldots -2\cdot 2-2\cdot 1=-2\sum_{i=1}^{n}i=-2\cdot \frac{n(n+1)}{2}=-n(n+1)$$
Is everything correct?
MHB Site Helper
Thanks a lot!! | {
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# Good Question 14
Good Question 14 – The Integral Test
I have no criteria for what constitutes a “Good Question” for this series of occasional posts. They are just questions that I found interesting, or that seem more than usually instructive, or that I learn something from. I cannot quote this question (2016 BC 92) since it is on a secure exam. What made it interesting is that to answer it students pretty much needed to know the proof of the Integral Test and the figures that go with it.
I recall only one AP question from many years ago that asked students to “prove” something – usually students are asked to show that a result was true by citing the theorem that applied and showing the hypotheses were met. The directions are often “justify your answer.”
Doing an original proof is not, in my opinion, a fair question and proving some known theorem is just a matter of memorization. For these reasons, students are not asked to prove things on the exams. So, should you prove things in class? Probably, yes.
Here is the usual proof of the integral test. Afterwards I’ll discuss the question from the exam.
The Integral Test
Hypotheses: Let $f\left( x \right)$ be a function that is positive, decreasing, and continuous for $x\ge 1$ ; and let ${{a}_{n}}=f\left( n \right)$ for $x\ge 1$
In the first drawing the rectangles have a height of an and a width of 1. The area is of each is an, and the sum of their areas the series is $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$.
Part 1: Notice that $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}}$ Assume that the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ diverges. | {
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• Conclusion 1: If the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ diverges, then the series $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$ diverges.
• Conclusion 2: (The contrapositive of conclusion 1) If the series $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$ converges, then the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ converges.
Part 2: In the second drawing below, Assume that the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ converges. The sum of the areas of the rectangles is $\sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}}$. (NB: this series starts at n = 2.) Since $\sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}}$ is less than the convergent improper integral it will also converge. Adding ${{a}_{1}}$ to this gives the original series, $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$; this series also converges.
• Conclusion 3: If the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ converges, then the series $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$ converges.
• Conclusion 4: (The contrapositive of conclusion 3) If the series $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$ diverges, then the improper integral $\int_{1}^{\infty }{{f\left( x \right)dx}}$ diverges.
Putting the four conclusions together is the Integral Test: If the hypotheses above are met, then the series and the improper integral will both converge, or both diverge.
To answer the multiple-choice question (2106 BC 92) on the exams students were told that the improper integral converges. Therefore, the associated series converges. They then had to determine whether the series or the improper integral has the greater value. Stop here and see if you can figure that out.
Return to the first figure above only this time assume that the improper integral and the series converge. It is pretty obvious that $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}}$. | {
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So, even though students were not asked to prove anything, a familiarity with the proof and its figures is necessary to answer the question. That’s why I liked it,
On the other hand, it is kind of an obscure point and I’m not sure it has any practical value.
## 2 thoughts on “Good Question 14”
1. Why couldn’t you refer to your second figure, saying that because the improper integral converged, then the series converged, and therefore the integral > series? It sounds like the question was worded this way. Is it because the first term is omitted from the series in your second figure?
Like
• Yes, it is because in the second figure the first rectangle on the left is $\displaystyle {{a}_{2}}$, so the entire series is not shown. The rectangle for $\displaystyle {{a}_{1}}$ extends to the left between x = 1 and the y-axis. You cannot tell (for sure) from this that the series is greater than the integral. From the first figure (with the new assumption that the series and integral both converge) you can see that the area representing the series is greater than the area represented by the integral.
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# Negation of “some” logic statement
I need to negate the following statement : "Some integers are not odd"
I have the below, where O(x) is "odd" $$\exists x (\neg O(x))$$
Would the negation be $$\forall x (O(x))$$
I'm confused, since if so, this statement does not make sense since there can be even numbers, does the negation have to be real?
• Why doesn't it make sense? The original statement is true (some integers are not odd) so it makes sense that the negation of that original statement becomes false. – JMoravitz Jul 21 '16 at 3:12
• I guess I misunderstood the concept of the negation, is the negated statement correct? "for all x, x is odd" – splinks Jul 21 '16 at 3:17
• Sensible statements can be false. "The sky is orange" makes sense but is false. "The sky is not orange" makes sense and is true. "The sky is Tuesday" doesn't make sense. The negation of a sensible true statement needs to be a sensible false statement, and vice versa. – Graham Kemp Jul 21 '16 at 3:31
As alluded to in Graham Kemp's comment, you seem to be mixing up true statements with meaningful statements. Meaningful statements are sometimes called syntactically correct statements, and can be either true or false.
The negation of a true statement is false, and the negation of a false statement is true. So it stands to reason that, when you negated $\color{blue}{\exists x \; \lnot O(x)}$ ("some integers are not odd"), a true statement, you got $\color{red}{\forall x O(x)}$ ("all integers are odd"), a false statement.
On the other hand, the negation of a meaningful statement is meaningful. In this case, "some integers are not odd" and "all integers are odd" are both meaningful things to say (they make sense), even if one of them is completely wrong.
Don't worry about it. Sensible statements can be either true or false. "The sky is orange" makes sense but is false. "The sky is not orange" makes sense and is true. "The sky is Tuesday" doesn't make sense. | {
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In the domain of integers, $\exists x~\neg O(x)$ means "There exists an integer which is not odd." This statement is true.
The negation of that statement is indeed, $\forall x~O(x)$ , which means "Every integer is odd." This statement is false, as it should be.
We expect the negation of a true statement to be a false statement. That is what negation means.
$$\neg \exists x~{\neg O(x)}~ \iff ~\forall x~{O(x)}$$ | {
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# Cartesian products represented as disjoint unions
Assume $k$ finite sets $S_1, \ldots, S_k$ and their Cartesian product $S = S_1 \times \ldots \times S_k$.
Suppose I remove $n$ (distinct) elements from $S$, and let us call that smaller set $S'$. I am interested in representing $S'$ as a union of distinct Cartesian products.
Is there an upper bound, in terms of $k$ and $n$, on the numbers of distinct products I may need in such a union?
As a simple example, consider $S_1 = \{ a, b, c \}$, $S_2 = \{ 1, 2, 3 \}$, $S = S_1 \times S_2$.
If I take $S' = S \setminus \{ (b, 2), (b, 3) \}$, I can write it as $S' = \left( \{ a, c \} \times \{ 1, 2, 3 \} \right) \cup \left(\{ b \} \times \{ 1 \} \right)$. In this case, I used two products. Had I removed $(c,3)$ instead of $(b,3)$, I believe I would have needed at least three.
(The reason I am asking for a bound in terms of $k$ and $n$ is that a simple geometric interpretation makes me believe the sizes of the sets $S_i$ are irrelevant — though they may come into play when the number of removed elements approaches the size of $S$. Please correct me if I am wrong.)
-
Let the elements we remove be denoted by $s^j=(s_1^j,s_2^j,\ldots,s_k^j)$ for $j=1,2,\ldots,n$. We are interested in the set $$(S_1\times S_2\times\ldots\times S_k)\setminus\{s^1,s^2,\ldots,s^n\}.$$
Claim. A (possibly very crude) upper bound is given by $\frac{n^k-1}{n-1}$.
Proof. We will prove this by induction on $k$.
For $k=1$, there is no problem. The set $S_1\setminus\{s^1,s^2,\ldots,s^n\}$ is a Cartesian product with one factor, so we are done: the upper bound is $1$ in this case. | {
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For $k\to k+1$, note that we can rewrite the set as follows: \begin{align}&(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}=\\&=(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\bigcup_{x\in S_{k+1}}(S_1\times S_2\times\ldots\times S_k\times\{x\})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\left(\bigcup_{j=1}^n(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}\right)\cup(S_1\times S_2\times\ldots\times (S_{k+1}\setminus\{s_{k+1}^1,\ldots,s_{k+1}^n\})),\end{align} which is (if we omit the sets that appear more than once in the first union) a disjoint union of at most $n$ sets of the form $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\},$$ and a single Cartesian product of $k+1$ sets. By the induction hypothesis, each set $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}$$ can be written as a disjoint union of at most $\frac{n^k-1}{n-1}$ Cartesian products. Therefore our set can be written as a disjoint union of at most $n\frac{n^k-1}{n-1}+1 = \frac{n^{k+1}-1}{n-1}$ Cartesian products (with $k+1$ factors each, of course). $\square$
Note that finding better bounds might require more sophisticated combinatorial methods.
-
Ummm ... In case $n=1$, $\frac{n^k-1}{n-1}$ should be interpreted as $k$. – Dejan Govc Apr 4 '13 at 21:14
Thanks for your answer. It got me thinking a little more, and I wonder now if I cannot get to $k{\cdot}n$, by representing $S$ as $((S_1 \setminus \{ s^1 \}) \times \ldots \times S_k) \cup (\{ s^1 \} \times \ldots \times S_k)$, then splitting $S_2$ in that second member, etc. In fact, I now suspect this could work for $S \setminus X$, where $X$ is an arbitrary Cartesian product from subsets of $S_1, \ldots, S_k$... Maybe someone can work a full answer, or I'll try myself. – Philippe Apr 5 '13 at 11:14 | {
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# Fitting a Function given 3 points
1. May 16, 2010
### shyboyswin
1. The problem statement, all variables and given/known data
Using a simple "trick," you can find a polynomial function whose graphi goes through any given set of points. For example, the graph of the following function goes through the points (1, 3), (2, 5), and (3, -1).
$$f(x) = 3\;\frac{(x-2)(x-3)}{(1-2)(1-3)} \;+\; 5\frac{(x-1)(x-3)}{(2-1)(2-3)} \;+\;(-1) \frac{(x-1)(x-2)}{(3-1)(3-2)}$$
Find a polynomial function g that satisfies g(-2) = 1, g(0) = 3, g(1) = -2 and g(4) = 3
2. Relevant equations
(Above)
3. The attempt at a solution
$$g(x) = (1)\frac{(x-1)(x-4)}{(-2-1)(-2-4)} \; + \;(-2)\frac{(x+2)(x-4)}{(1+2)(1-4)} \;+ \;3\frac{(x+2)(x-1)}{(4+1)(4-1)}$$
I cannot figure out how to produce g(0) = 3.
2. May 16, 2010
### D H
Staff Emeritus
Why are you making your polynomial a quadratic? Perhaps it should be something else -- like a cubic.
3. May 16, 2010
### The Chaz
Hint: each "term" should have more factors in parenthesis
4. May 17, 2010
### HallsofIvy
Staff Emeritus
As both DH and Chaz say, you need four terms, not three, and you need three factors in numerator and denominator of each term.
A polynomial passing through f(a)= t, f(b)= u, f(c)= v, and f(d)= w must look like
$$t\frac{(x- b)(x- c)(x- d)}{(a- b)(a- c)(a- d)}+$$$$u\frac{(x- a)(x- c)(x- d)}{(b- a)(b- c)(b- d)}+$$$$v\frac{(x- a)(x- b)(x- d)}{(c- a)(c- b)(c- d)}$$$$+ w\frac{(x- a)(x- b)(x- c)}{(d- a)(d- b)(d- c)}$$
Look at what happens at, say, x= c. Every fraction except the third has a factor of x- c in the numerator and so will be 0 at x= c. The third fraction will have, at x= c, (c- a)(c- b)(c- d) in both numerator and denominator and so will be 1. The entire polynomal will be f(c)= 0+ 0+ v(1)+ 0= v.
This is the "Lagrange interpolation formula"- to write a polynomial that passes through n given points, you need a sum of n fractions, each having n-1 factors in numerator and denominator.
5. May 19, 2010
### Unit | {
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5. May 19, 2010
### Unit
I have never heard of this formula, but it is so clever! I love it
As for helping OP, all I can say is, look for the pattern, and, once you've got it, be meticulous in your bookkeeping. | {
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# What is the name of this method?
1. Sep 23, 2012
### Orion1
1. The problem statement, all variables and given/known data
Integrate:
$$\int \frac{1}{x^3 - 27} dx$$
2. Relevant equations
$$\int \frac{1}{x^3 - 27} dx$$
$$\frac{1}{x^3 - 27} = \frac{1}{(x - 3)(x^2 + 3x + 9)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 3x + 9}$$
$$1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)$$
3. The attempt at a solution
The first step involves factoring and splitting the denominator with some method:
$$\frac{1}{x^3 - 27} = \frac{1}{(x - 3)(x^2 + 3x + 9)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 3x + 9}$$
What is the name of this method?
The next step involves normalizing the equation:
$$1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)$$
How was the coefficients of B and C derived and solved?
The solution only states that:
$$A = \frac{1}{27}$$
$$0 = \frac{1}{27} + B$$
$$1 = \frac{1}{3} + 3C$$
And:
$B = -\frac{1}{27}$ and $C = -\frac{2}{9}$
I can see how the $A$ coefficient was derived from the x-axis zero intercept, however, it is not entirely clear to me how these other coefficients were derived and solved?
Reference:
Derivation of equation and solution
Last edited: Sep 23, 2012
2. Sep 23, 2012
### CAF123
This is the method of partial fractions to reduce $$\int \frac{1}{x^3 -27} dx$$ into something which can be integrated.
To find A,B and C you will need to solve simultaneous equations. To easily solve for A, let x =3 and you should get the required A = 1/27. To find B and C, let x = 0 to get one eqn in B and C and let x = 1 to get another. Two eqns, two unknowns (B and C) - you can solve this. The choice of x here is arbritary (except x =3 since the (Bx +C) term will vanish).
Last edited: Sep 23, 2012
3. Sep 23, 2012
### HallsofIvy
Staff Emeritus
Actually, because the last term in involved "Bx- C", and you have already found A, setting x= 0 will give an equation in C only.
4. Sep 23, 2012
### CAF123 | {
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4. Sep 23, 2012
### CAF123
Yes, I overlooked this. If you have something like $$\frac{A}{x} + \frac{(Bx +c)}{x^2 +bx +c},$$ then you will probably have to use simultaneous eqns.
5. Sep 23, 2012
### Orion1
Solving the coefficients:
Given:
$$1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)$$
Set x = 3 to zero out the $(Bx + C)$ term and numerically integrate via substitution:
$$1 = A(3^2 + 3(3) + 9) + (B(3) + C)(3 - 3) = A(27)$$
$$1 = A(27)$$
Solve for $A$:
$$\boxed{A = \frac{1}{27}}$$
Set x = 0 to zero out the $Bx$ term and numerically integrate via substitution:
$$1 = \frac{(0^2 + 3(0) + 9)}{27} + (B(0) + C)(0 - 3) = \frac{9}{27} - 3C$$
$$1 = \frac{9}{27} - 3C$$
$$-\frac{27}{27} + \frac{9}{27} = -\frac{18}{27} = -\frac{2}{3} = 3C$$
Solve for $C$:
$$\boxed{C = -\frac{2}{9}}$$
Set x = 1 as arbitrary and numerically integrate via substitution:
$$1 = \frac{(1^2 + 3(1) + 9)}{27} + \left(B(1) - \frac{2}{9} \right)(1 - 3) = \frac{13}{27} - 2B + \frac{3}{3} \times \frac{4}{9} = \frac{13}{27} - 2B + \frac{12}{27} = \frac{25}{27} - 2B$$
$$1 = \frac{25}{27} - 2B$$
$$-\frac{27}{27} + \frac{25}{27} = -\frac{2}{27} = 2B$$
Solve for $B$:
$$\boxed{B = -\frac{1}{27}}$$
Last edited: Sep 23, 2012 | {
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# Tag Archives: derivative
## Many Roads Give Same Derivative
A recent post in the AP Calculus Community expressed some confusion about different ways to compute $\displaystyle \frac{dy}{dx}$ at (0,4) for the function $x=2ln(y-3)$. I share below the two approaches suggested in the original post, proffer two more, and a slightly more in-depth activity I’ve used in my calculus classes for years. I conclude with an alternative to derivatives of inverses.
### Two Approaches Initially Proposed
1 – Accept the function as posed and differentiate implicitly.
$\displaystyle \frac{d}{dx} \left( x = 2 ln(y-3) \right)$
$\displaystyle 1 = 2*\frac{1}{y-3} * \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{y-3}{2}$
Which gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ at (x,y)=(0,4).
2 – Solve for y and differentiate explicitly.
$\displaystyle x = 2ln(y-3) \longrightarrow y = 3 + e^{x/2}$
$\displaystyle \frac{dy}{dx} = e^{x/2} * \frac{1}{2}$
Evaluating this at (x,y)=(0,4) gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ .
### Two Alternative Approaches
3 – Substitute early.
The question never asked for an algebraic expression of $\frac{dy}{dx}$, only the numerical value of this slope. Because students tend to make more silly mistakes manipulating algebraic expressions than numeric ones, the additional algebra steps are unnecessary, and potentially error-prone. Admittedly, the manipulations are pretty straightforward here, in more algebraically complicated cases, early substitutions could significantly simplify work. Using approach #1 and substituting directly into the second line gives
$\displaystyle 1 = 2 * \frac{1}{y-3} * \frac{dy}{dx}$.
At (x,y)=(0,4), this is
$\displaystyle 1 = 2 * \frac{1}{4-3}*\frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{1}{2}$
The numeric manipulations on the right side are obviously easier than the earlier algebra.
4 – Solve for $\frac{dx}{dy}$ and reciprocate. | {
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4 – Solve for $\frac{dx}{dy}$ and reciprocate.
There’s nothing sacred about solving for $\frac{dy}{dx}$ directly. Why not compute the derivative of the inverse and reciprocate at the end? Differentiating first with respect to y eventually leads to the same solution.
$\displaystyle \frac{d}{dy} \left( x = 2 ln(y-3) \right)$
$\displaystyle \frac{dx}{dy} = 2 * \frac{1}{y-3}$
At (x,y)=(0,4), this is
$\displaystyle \frac{dx}{dy} = \frac{2}{4-3} = 2$, so
$\displaystyle \frac{dy}{dx} = \frac{1}{2}$.
### Equivalence = A fundamental mathematical concept
I sometimes wonder if teachers should place much more emphasis on equivalence. We spend so much time manipulating expressions in mathematics classes at all levels, changing mathematical objects (shapes, expressions, equations, etc.) into a different, but equivalent objects. Many times, these manipulations are completed under the guise of “simplification.” (Here is a brilliant Dan Teague post cautioning against taking this idea too far.)
But it is critical for students to recognize that proper application of manipulations creates equivalent expressions, even if when the resulting expressions don’t look the same. The reason we manipulate mathematical objects is to discover features about the object in one form that may not be immediately obvious in another.
For the function $x = 2 ln(y-3)$, the slope at (0,4) must be the same, no matter how that slope is calculated. If you get a different looking answer while using correct manipulations, the final answers must be equivalent.
### Another Example
A similar question appeared on the AP Calculus email list-server almost a decade ago right at the moment I was introducing implicit differentiation. A teacher had tried to find $\displaystyle \frac{dy}{dx}$ for
$\displaystyle x^2 = \frac{x+y}{x-y}$ | {
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$\displaystyle x^2 = \frac{x+y}{x-y}$
using implicit differentiation on the quotient, manipulating to a product before using implicit differentiation, and finally solving for y in terms of x to use an explicit derivative.
1 – Implicit on a quotient | {
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Take the derivative as given:$$\displaystyle \frac{d}{dx} \left( x^2 = \frac{x+y}{x-y} \right)$ $\displaystyle 2x = \frac{(x-y) \left( 1 + \frac{dy}{dx} \right) - (x+y) \left( 1 - \frac{dy}{dx} \right) }{(x-y)^2}$ $\displaystyle 2x * (x-y)^2 = (x-y) + (x-y)*\frac{dy}{dx} - (x+y) + (x+y)*\frac{dy}{dx}$ $\displaystyle 2x * (x-y)^2 = -2y + 2x * \frac{dy}{dx}$ $\displaystyle \frac{dy}{dx} = \frac{-2x * (x-y)^2 + 2y}{2x}$ 2 – Implicit on a product Multiplying the original equation by its denominator gives $x^2 * (x - y) = x + y$ . Differentiating with respect to x gives $\displaystyle 2x * (x - y) + x^2 * \left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$ $\displaystyle 2x * (x-y) + x^2 - 1 = x^2 * \frac{dy}{dx} + \frac{dy}{dx}$ $\displaystyle \frac{dy}{dx} = \frac{2x * (x-y) + x^2 - 1}{x^2 + 1}$ 3 – Explicit Solving the equation at the start of method 2 for y gives $\displaystyle y = \frac{x^3 - x}{x^2 + 1}$. Differentiating with respect to x gives $\displaystyle \frac{dy}{dx} = \frac {\left( x^2+1 \right) \left( 3x^2 - 1\right) - \left( x^3 - x \right) (2x+0)}{\left( x^2 + 1 \right) ^2}$ Equivalence Those 3 forms of the derivative look VERY DIFFERENT. Assuming no errors in the algebra, they MUST be equivalent because they are nothing more than the same derivative of different forms of the same function, and a function’s rate of change doesn’t vary just because you alter the look of its algebraic representation. Substituting the y-as-a-function-of-x equation from method 3 into the first two derivative forms converts all three into functions of x. Lots of by-hand algebra or a quick check on a CAS establishes the suspected equivalence. Here’s my TI-Nspire CAS check. Here’s the form of this investigation I gave my students. ### Final Example I’m not a big fan of memorizing anything without a VERY GOOD reason. My teachers telling me to do so never held much weight for me. I memorized as little as possible and used that information as long as I could until a scenario | {
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me. I memorized as little as possible and used that information as long as I could until a scenario arose to convince me to memorize more. One thing I managed to avoid almost completely were the annoying derivative formulas for inverse trig functions. For example, find the derivative of $y = arcsin(x)$ at $x = \frac{1}{2}$. Since arc-trig functions annoy me, I always rewrite them. Taking sine of both sides and then differentiating with respect to x gives. $sin(y) = x$ $\displaystyle cos(y) * \frac{dy}{dx} = 1$ I could rewrite this equation to give $\frac{dy}{dx} = \frac{1}{cos(y)}$, a perfectly reasonable form of the derivative, albeit as a less-common expression in terms of y. But I don’t even do that unnecessary algebra. From the original function, $x=\frac{1}{2} \longrightarrow y=\frac{\pi}{6}$, and I substitute that immediately after the differentiation step to give a much cleaner numeric route to my answer. $\displaystyle cos \left( \frac{\pi}{6} \right) * \frac{dy}{dx} = 1$ $\displaystyle \frac{\sqrt{3}}{2} * \frac{dy}{dx} = 1$ $\displaystyle \frac{dy}{dx} = \frac{2}{\sqrt{3}}$ And this is the same result as plugging $x = \frac{1}{2}$ into the memorized version form of the derivative of arcsine. If you like memorizing, go ahead, but my mind remains more nimble and less cluttered. One final equivalent approach would have been differentiating $sin(y) = x$ with respect to y and reciprocating at the end. ### CONCLUSION There are MANY ways to compute derivatives. For any problem or scenario, use the one that makes sense or is computationally easiest for YOU. If your resulting algebra is correct, you know you have a correct answer, even if it looks different. Be strong! Advertisements ## Base-x Numbers and Infinite Series In my previous post, I explored what happened when you converted a polynomial from its variable form into a base-x numerical form. That is, what are the computational implications when polynomial $3x^3-11x^2+2$ is represented by the base-x number | {
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the computational implications when polynomial $3x^3-11x^2+2$ is represented by the base-x number $3(-11)02_x$, where the parentheses are used to hold the base-x digit, -11, for the second power of x? So far, I’ve explored only the Natural number equivalents of base-x numbers. In this post, I explore what happens when you allow division to extend base-x numbers into their Rational number counterparts. Level 5–Infinite Series: Numbers can have decimals, so what’s the equivalence for base-x numbers? For starters, I considered trying to get a “decimal” form of $\displaystyle \frac{1}{x+2}$. It was “obvious” to me that $12_x$ won’t divide into $1_x$. There are too few “places”, so some form of decimals are required. Employing division as described in my previous post somewhat like you would to determine the rational number decimals of $\frac{1}{12}$ gives Remember, the places are powers of x, so the decimal portion of $\displaystyle \frac{1}{x+2}$ is $0.1(-2)4(-8)..._x$, and it is equivalent to $\displaystyle 1x^{-1}-2x^{-2}+4x^{-3}-8x^{-4}+...=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$. This can be seen as a geometric series with first term $\displaystyle \frac{1}{x}$ and ratio $\displaystyle r=\frac{-2}{x}$. It’s infinite sum is therefore $\displaystyle \frac{\frac{1}{x}}{1-\frac{-2}{x}}$ which is equivalent to $\displaystyle \frac{1}{x+2}$, confirming the division computation. Of course, as a geometric series, this is true only so long as $\displaystyle |r|=\left | \frac{-2}{x} \right |<1$, or $2<|x|$. I thought this was pretty cool, and it led to lots of other cool series. For example, if $x=8$,you get $\frac{1}{10}=\frac{1}{8}-\frac{2}{64}+\frac{4}{512}-...$. Likewise, $x=3$ gives $\frac{1}{5}=\frac{1}{3}-\frac{2}{9}+\frac{4}{27}-\frac{8}{81}+...$. I found it quite interesting to have a “polynomial” defined with a rational expression. Boundary Convergence: As shown above, $\displaystyle | {
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defined with a rational expression. Boundary Convergence: As shown above, $\displaystyle \frac{1}{x+2}=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ only for $|x|>2$. At $x=2$, the series is obviously divergent, $\displaystyle \frac{1}{4} \ne \frac{1}{2}-\frac{2}{4}+\frac{4}{8}-\frac{8}{16}+...$. For $x=-2$, I got $\displaystyle \frac{1}{0} = \frac{1}{-2}-\frac{2}{4}+\frac{4}{-8}-\frac{8}{16}+...=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-...$ which is properly equivalent to $-\infty$ as $x \rightarrow -2$ as defined by the convergence domain and the graphical behavior of $\displaystyle y=\frac{1}{x+2}$ just to the left of $x=-2$. Nice. I did find it curious, though, that $\displaystyle \frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ is a solid approximation for $\displaystyle \frac{1}{x+2}$ to the left of its vertical asymptote, but not for its rotationally symmetric right side. I also thought it philosophically strange (even though I understand mathematically why it must be) that this series could approximate function behavior near a vertical asymptote, but not near the graph’s stable and flat portion near $x=0$. What a curious, asymmetrical approximator. Maclaurin Series: Some quick calculus gives the Maclaurin series for $\displaystyle \frac{1}{x+2}$ : $\displaystyle \frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+...$, a geometric series with first term $\frac{1}{2}$ and ratio $\frac{-x}{2}$. Interestingly, the ratio emerging from the Maclaurin series is the reciprocal of the ratio from the “rational polynomial” resulting from the base-x division above. As a geometric series, the interval of convergence is $\displaystyle |r|=\left | \frac{-x}{2} \right |<1$, or $|x|<2$. Excluding endpoint results, the Maclaurin interval is the complete Real number complement to the base-x series. For the endpoints, $x=-2$ produces the right-side vertical asymptote divergence to $+ \infty$ that $x=-2$ did for the left side of the vertical | {
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vertical asymptote divergence to $+ \infty$ that $x=-2$ did for the left side of the vertical asymptote in the base-x series. Again, $x=2$ is divergent. It’s lovely how these two series so completely complement each other to create clean approximations of $\displaystyle \frac{1}{x+2}$ for all $x \ne 2$. Other base-x “rational numbers” Because any polynomial divided by another is absolutely equivalent to a base-x rational number and thereby a base-x decimal number, it will always be possible to create a “rational polynomial” using powers of $\displaystyle \frac{1}{x}$ for non-zero denominators. But, the decimal patterns of rational base-x numbers don’t apply in the same way as for Natural number bases. Where $\displaystyle \frac{1}{12}$ is guaranteed to have a repeating decimal pattern, the decimal form of $\displaystyle \frac{1}{x+2}=\frac{1_x}{12_x}=0.1(-2)4(-8)..._x$ clearly will not repeat. I’ve not explored the full potential of this, but it seems like another interesting field. CONCLUSIONS and QUESTIONS Once number bases are understood, I’d argue that using base-x multiplication might be, and base-x division definitely is, a cleaner way to compute products and quotients, respectively, for polynomials. The base-x division algorithm clearly is accessible to Algebra II students, and even opens the doors to studying series approximations to functions long before calculus. Is there a convenient way to use base-x numbers to represent horizontal translations as cleanly as polynomials? How difficult would it be to work with a base-$(x-h)$ number for a polynomial translated h units horizontally? As a calculus extension, what would happen if you tried employing division of non-polynomials by replacing them with their Taylor series equivalents? I’ve played a little with proving some trig identities using base-x polynomials from the Maclaurin series for sine and cosine. What would happen if you tried to compute repeated fractions in base-x? It’s an open question from my | {
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would happen if you tried to compute repeated fractions in base-x? It’s an open question from my perspective when decimal patterns might terminate or repeat when evaluating base-x rational numbers. I’d love to see someone out there give some of these questions a run! ## Calculus Derivative Rules Over the past few days I’ve been rethinking my sequencing of introducing derivative rules for the next time I teach calculus. The impetus for this was an approach I encountered in a Coursera MOOC in Calculus I’m taking this summer to see how a professor would run a Taylor Series-centered calculus class. Historically, I’ve introduced my high school calculus classes to the product and quotient rules before turing to the chain rule. I’m now convinced the chain rule should be first because of how beautifully it sets up the other two. Why the chain rule should be first Assuming you know the chain rule, check out these derivations of the product and quotient rules. For each of these, $g_1$ and $g_2$ can be any differentiable functions of x. PRODUCT RULE: Let $P(x)=g_1(x) \cdot g_2(x)$. Applying a logarithm gives, $ln(P)=ln \left( g_1 \cdot g_2 \right) = ln(g_1)+ln(g_2)$. Now differentiate and rearrange. $\displaystyle \frac{P'}{P} = \frac{g_1'}{g_1}+\frac{g_2'}{g_2}$ $\displaystyle P' = P \cdot \left( \frac{g_1'}{g_1}+\frac{g_2'}{g_2} \right)$ $\displaystyle P' = g_1 \cdot g_2 \cdot \left( \frac{g_1'}{g_1}+\frac{g_2'}{g_2} \right)$ $P' = g_1' \cdot g_2+g_1 \cdot g_2'$ QUOTIENT RULE: Let $Q(x)=\displaystyle \frac{g_1(x)}{g_2(x)}$. As before, apply a logarithm, differentiate, and rearrange. $\displaystyle ln(Q)=ln \left( \frac{g_1}{g_2} \right) = ln(g_1)-ln(g_2)$ $\displaystyle \frac{Q'}{Q} = \frac{g_1'}{g_1}-\frac{g_2'}{g_2}$ $\displaystyle Q' = Q \cdot \left( \frac{g_1'}{g_1}-\frac{g_2'}{g_2} \right)$ $\displaystyle Q' = \frac{g_1}{g_2} \cdot \left( \frac{g_1'}{g_1}-\frac{g_2'}{g_2} \right)$ $\displaystyle Q' = \frac{g_1'}{g_2}-\frac{g_1 \cdot g_2'}{\left( g_2 \right)^2} = | {
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\right)$ $\displaystyle Q' = \frac{g_1'}{g_2}-\frac{g_1 \cdot g_2'}{\left( g_2 \right)^2} = \frac{g_1'g_2-g_1g_2'}{\left( g_2 \right)^2}$ The exact same procedure creates both rules. (I should have seen this long ago.) Proposed sequencing I’ve always emphasized the Chain Rule as the critical algebra manipulation rule for calculus students, but this approach makes it the only rule required. That completely fits into my overall teaching philosophy: learn a limited set of central ideas and use them as often as possible. With this, I’ll still introduce power, exponential, sine, and cosine derivative rules first, but then I’ll follow with the chain rule. After that, I think everything else required for high school calculus will be a variation on what is already known. That’s a lovely bit of simplification. I need to rethink my course sequencing, but I think it’ll be worth it. ## Polar Derivatives on TI-Nspire CAS The following question about how to compute derivatives of polar functions was posted on the College Board’s AP Calculus Community bulletin board today. From what I can tell, there are no direct ways to get derivative values for polar functions. There are two ways I imagined to get the polar derivative value, one graphically and the other CAS-powered. The CAS approach is much more accurate, especially in locations where the value of the derivative changes quickly, but I don’t think it’s necessarily more intuitive unless you’re comfortable using CAS commands. For an example, I’ll use $r=2+3sin(\theta )$ and assume you want the derivative at $\theta = \frac{\pi }{6}$. METHOD 1: Graphical Remember that a derivative at a point is the slope of the tangent line to the curve at that point. So, finding an equation of a tangent line to the polar curve at the point of interest should find the desired result. Create a graphing window and enter your polar equation (menu –> 3:Graph Entry –> 4:Polar). Then drop a tangent line on the polar curve (menu –> 8:Geometry –> | {
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–> 3:Graph Entry –> 4:Polar). Then drop a tangent line on the polar curve (menu –> 8:Geometry –> 1:Points&Lines –> 7:Tangent). You would then click on the polar curve once to select the curve and a second time to place the tangent line. Then press ESC to exit the Tangent Line command. To get the current coordinates of the point and the equation of the tangent line, use the Coordinates & Equation tool (menu –> 1:Actions –> 8:Coordinates and Equations). Click on the point and the line to get the current location’s information. After each click, you’ll need to click again to tell the nSpire where you want the information displayed. To get the tangent line at $\theta =\frac{\pi }{6}$, you could drag the point, but the graph settings seem to produce only Cartesian coordinates. Converting $\theta =\frac{\pi }{6}$ on $r=2+3sin(\theta )$ to Cartesian gives $\left( x,y \right) = \left( r \cdot cos(\theta ), r \cdot sin(\theta ) \right)=\left( \frac{7\sqrt{3}}{4},\frac{7}{4} \right)$ . So the x-coordinate is $\frac{7\sqrt{3}}{4} \approx 3.031$. Drag the point to find the approximate slope, $\frac{dy}{dx} \approx 8.37$. Because the slope of the tangent line changes rapidly at this location on this polar curve, this value of 8.37 will be shown in the next method to be a bit off. Unfortunately, I tried to double-click the x-coordinate to set it to exactly $\frac{7\sqrt{3}}{4}$, but that property is also disabled in polar mode. METHOD 2: CAS Using the Chain Rule, $\displaystyle \frac{dy}{dx} = \frac{dy}{d\theta }\cdot \frac{d\theta }{dx} = \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$. I can use this and the nSpire’s ability to define user-created functions to create a $\displaystyle \frac{dy}{dx}$ polar differentiator for any polar function $r=a(\theta )$. On a Calculator page, I use the Define function (menu –> 1:Actions –> 1:Define) to make the polar differentiator. All you need to do is enter the expression for a as shown in line 2 below. This can be evaluated exactly or | {
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need to do is enter the expression for a as shown in line 2 below. This can be evaluated exactly or approximately at $\theta=\frac{\pi }{6}$ to show $\displaystyle \frac{dy}{dx} = 5\sqrt{3}=\approx 8.660$. Conclusion: As with all technologies, getting the answers you want often boils down to learning what questions to ask and how to phrase them. ## Controlling graphs and a free online calculator When graphing functions with multiple local features, I often find myself wanting to explain a portion of the graph’s behavior independent of the rest of the graph. When I started teaching a couple decades ago, the processor on my TI-81 was slow enough that I could actually watch the pixels light up sequentially. I could see HOW the graph was formed. Today, processors obviously are much faster. I love the problem-solving power that has given my students and me, but I’ve sometimes missed being able to see function graphs as they develop. Below, I describe the origins of the graph control idea, how the control works, and then provide examples of polynomials with multiple roots, rational functions with multiple intercepts and/or vertical asymptotes, polar functions, parametric collision modeling, and graphing derivatives of given curves. BACKGROUND: A colleague and I were planning a rational function unit after school last week wanting to be able to create graphs in pieces so that we could discuss the effect of each local feature. In the past, we “rigged” calculator images by graphing the functions parametrically and controlling the input values of t. Clunky and static, but it gave us useful still shots. Nice enough, but we really wanted something dynamic. Because we had the use of sliders on our TI-nSpire software, on Geogebra, and on the Desmos calculator, the solution we sought was closer than we suspected. REALIZATION & WHY IT WORKS: Last week, we discovered that we could use $g(x)=\sqrt \frac{\left | x \right |}{x}$ to create what we wanted. The argument of the root is 1 | {
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$g(x)=\sqrt \frac{\left | x \right |}{x}$ to create what we wanted. The argument of the root is 1 for $x<0$, making $g(x)=1$. For $x>0$, the root’s argument is -1, making $g(x)=i$, a non-real number. Our insight was that multiplying any function $y=f(x)$ by an appropriate version of g wouldn’t change the output of f if the input to g is positive, but would make the product ungraphable due to complex values if the input to g is negative. If I make a slider for parameter a, then $g_2(x)=\sqrt \frac{\left | a-x \right |}{a-x}$ will have output 1 for all $x. That means for any function $y=f(x)$ with real outputs only, $y=f(x)\cdot g_2(x)$ will have real outputs (and a real graph) for $x only. Aha! Using a slider and $g_2$ would allow me to control the appearance of my graph from left to right. NOTE: While it’s still developing, I’ve become a big fan of the free online Desmos calculator after a recent presentation at the Global Math Department (join our 45-60 minute online meetings every Tuesday at 9PM ET!). I use Desmos for all of the following graphs in this post, but obviously any graphing software with slider capabilities would do. EXAMPLE 1: Graph $y=(x+2)^3x^2(x-1)$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1. Click here to access the Desmos graph that created the image above. You can then manipulate the slider to watch the graph wiggle through, then bounce off, and finally pass through the x-axis. EXAMPLE 2: Graph $y=\frac{(x+1)^2}{(x+2)(x-1)^2}$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1. Click here to access the Desmos graph above and control the creation of the rational function’s graph using a slider. EXAMPLE 3: I believe students understand polar graphing better when they see curves like the limacon | {
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EXAMPLE 3: I believe students understand polar graphing better when they see curves like the limacon $r=2+3cos(\theta )$ moving between its maximum and minimum circles. Controlling the slider also allows users to see the values of $\theta$ at which the limacon crosses the pole. Here is the Desmos graph for the graph below. EXAMPLE 4: Object A leaves (2,3) and travels south at 0.29 units/second. Object B leaves (-2,1) traveling east at 0.45 units/second. The intersection of their paths is (2,1), but which object arrives there first? Here is the live version. OK, I know this is an overly simplistic example, but you’ll get the idea of how the controlling slider works on a parametrically-defined function. The$latex \sqrt{\frac{\left | a-x \right |}{a-x}}\$ term only needs to be on one of parametric equations. Another benefit of the slider approach is the ease with which users can identify the value of t (or time) when each particle reaches the point of intersection or their axes intercepts. Obviously those values could be algebraically determined in this problem, but that isn’t always true, and this graphical-numeric approach always gives an alternative to algebraic techniques when investigating parametric functions. | {
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ASIDE 1–Notice the ease of the Desmos notation for parametric graphs. Enter [r,s] where r is the x-component of the parametric function and s is the y-component. To graph a point, leave r and s as constants. Easy.
EXAMPLE 5: When teaching calculus, I always ask my students to sketch graphs of the derivatives of functions given in graphical forms. I always create these graphs one part at a time. As an example, this graph shows $y=x^3+2x^2$ and allows you to get its derivative gradually using a slider.
ASIDE 2–It is also very easy to enter derivatives of functions in the Desmos calculator. Type “d/dx” before the function name or definition, and the derivative is accomplished. Desmos is not a CAS, so I’m sure the software is computing derivatives numerically. No matter. Derivatives are easy to define and use here.
I’m hoping you find this technology tip as useful as I do.
## Exponential Derivatives and Statistics
This post gives a different way I developed years ago to determine the form of the derivative of exponential functions, $y=b^x$. At the end, I provide a copy of the document I use for this activity in my calculus classes just in case that’s helpful. But before showing that, I walk you through my set-up and solution of the problem of finding exponential derivatives.
Background:
I use this lesson after my students have explored the definition of the derivative and have computed the algebraic derivatives of polynomial and power functions. They also have access to TI-nSpire CAS calculators.
The definition of the derivative is pretty simple for polynomials, but unfortunately, the definition of the derivative is not so simple to resolve for exponential functions. I do not pretend to teach an analysis class, so I see my task as providing strong evidence–but not necessarily a watertight mathematical proof–for each derivative rule. This post definitely is not a proof, but its results have been pretty compelling for my students over the years. | {
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