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geometry then it is \nalba in LaTeX math mode. The$ 's around a command mean that it has to be used in maths mode, and they will temporarily put you into maths mode if you use them. Notation This section includes the most commonlyused notation in this book. The Anglo-French STD (Sunbeam-Talbot-Darracq) combine collapsed in 1935. Lecture notes on topological insulators Ming-Che Chang Department of Physics, National Taiwan Normal University, Taipei, Taiwan (Dated: December 26, 2018). , with n columns), then the product Ax is defined for n×1 column vectors x. In L a T e X, subscripts and superscripts are written using the symbols ^ and _, in this case the x and y exponents where written using these codes. Even if you are using only one bracket, both commands are mandatory. So here's the deal: the inner product is very common in mathematical work (and there's accepted notation for it), and what you describe is not (and so there's not established notation). Note: Here we just want to analyse the commands and structure of a LaTeX file. Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Calculates transpose, determinant, trace, rank, inverse, pseudoinverse, eigenvalues and eigenvectors. 2 The Simplex Method. MathOverflow has an awesome engine where you can embed LaTeX in questions, answers, and comments. | {
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# RPM required to simulate Earth's gravity (Centrifugal acceleration)
1. Mar 22, 2014
### jwlgrant
1. The problem statement, all variables and given/known data
Here's the problem: "On a journey to Mars, one design is to have a section of the spacecraft rotate to simulate gravity. If the radius of this section is 30 meters, how many RPMs must it rotate to simulate one Earth gravity (1 g = 9.8 meters/sec^2)?"
2. Relevant equations
$A = \frac{V^2}{R}$
3. The attempt at a solution
I completed this problem, getting an answer of 5.5 RPM. Apparently I was wrong, and the answer should have been 14 RPM. I asked my instructor about it, and he showed me the calculations used to get 14 RPM, but it still didn't quite make sense to me.
Here's the process my instructor used to solve the problem:
Circumference is C = 2∏(30) = 188 meters.
1 rpm is like rotating 1 full circular rotation every minute, so 188/60 = 3.1 meters/sec.
Then V = 3.1 meters/sec x RPM.
Then A = (3.1 RPM)^2/188 = 0.05 x rpm^2.
So: 9.8 = 0.05rpm^2
RPM = 14
--
Here's my process:
A = 9.8 (m)/(s^2)
R = 30 m
V = √(AR)
V = 17.15 m/s
1 revolution = circumference = 2∏30 m
[1 revolution/(2∏30 m)] * [60s/1 min] (1 rev is equal to 2π30m and sixty seconds is equal to one minute)
V = (17.15 m/s)/∏ RPM
V = 5.5 RPM
--
I realize, of course, that we both used different methods to get the answer. I thought I did it wrong, so I solved the problem again, using his method, to try to get the correct answer. He used 188m for the radius component when solving for A=V^2/R, but when I used 30m I got an answer of 5.5RPM. The radius, 30m, should be used in that equation, not 188m, right?
I would greatly appreciate any help in understanding this problem. I can't tell where/how I went wrong in solving it.
2. Mar 22, 2014
### Andrew Mason
Why is he dividing by 188 (circumference)?? Should be 30.
You are right and the instructor is wrong. | {
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You are right and the instructor is wrong.
It is easier to use Ac = ω2r; so $\omega = \sqrt{A/r}$ where ω is the angular speed in radians/sec.
Letting RPM = 60ω/2∏:
$RPM = \frac{60}{2\pi}*\sqrt{A/r} = \frac{60}{2\pi}\sqrt{9.8/30} = 5.5$
AM
Last edited: Mar 22, 2014
3. Mar 23, 2014
### mr_pavlo
I also agree, it must be divided by radius = 30 not circumference. | {
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# Suppose that 3 men and 3 women will be randomly seated at a round table having 6 evenly-spaced seats
Suppose that $3$ men and $3$ women will be randomly seated at a round table having $6$ evenly-spaced seats (with each seat being directly opposite one other seat).
(a) What is the probability that all $3$ men will be seated next to one another?
(b) What is the probability that starting at the northernmost seat and going clockwise around the table, a 2nd man will be encountered before a 2nd woman?
(c) What is the probability that no two men are seated directly opposite of one another?
(d) Suppose that the $6$ people are named Ashley, Barney, Carly, David, Emma, and Fred. What is the probability that starting with Ashley and going clockwise, the 6 people are seated in alphabetical order?
Attempted Solution:
(a) There are $6!$ = $720$ ways to seat these $6$ people. There are 6 different ways you can seat all three men next to each other, if you consider them as a clump. Each clump of three has $3!$ = 6 different ways to be arranged. Thus, there is $6*3!*3!$ = $216$ different arrangements. This gives a probability of $216\over{720}$ = $.3$.
(b) I wasn't sure about this but I realized that in any situation, either a 2nd man is reached before a 2nd woman or a 2nd woman is reached before a 2nd man. This gives a probability of $.5$.
(c) After drawing a picture of the table, this appears to be the same situation as part (a), giving a probability of $.3.$ I just realized this assumption was incorrect. Now I am getting that there are 8 different arrangements when considering males and females as clumps, with $3!$ different ways of arranging each clump, giving $8*3!*3!$ = $288$. This gives a probability of $288\over{720}$ = $.4$.
(d) $6\over{6}$ * $1\over{5}$ * $1\over{4}$ * $1\over{3}$ * $1\over{2}$ * $1\over{1}$ = $.00833$.
Any corrections to my attempted solutions would be greatly appreciated. | {
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Any corrections to my attempted solutions would be greatly appreciated.
All of your answers are correct. I will assume that only the relative order of the people matters, that the men are named Barney, David, and Fred, and that the women are named Ashley, Carly, and Emma.
What is the probability that all three men will be seated next to each other?
We seat Ashley. The remaining people can be seated in $5!$ ways as we proceed clockwise around the table.
For the favorable cases, we have two blocks of people to arrange. Since the blocks of men and women must be adjacent, this can be done in one way. The block of men can be arranged in $3!$ ways. The block of women can be arranged in $3!$ ways. Hence, the probability that all the men are seated next to each other is $$\frac{3!3!}{5!} = \frac{3}{10}$$
What is the probability that starting at the northernmost seat and going clockwise around the table, a second man will be encountered before a second woman?
You made good use of symmetry. Nice solution.
What is the probability that no two men are seated directly opposite each other?
We count arrangements in which two men are seated directly opposite each other. There are $\binom{3}{2}$ ways to select the men. Once they are seated, there are $4$ ways to seat the third man relative to the already seated man whose name appears first alphabetically and $3!$ ways to seat the women as we proceed clockwise around the table from the third man. Hence, there are $$\binom{3}{2}\binom{4}{1}3!$$ seating arrangements in which two men are opposite each other. Hence, the probability that no two men are opposite each other is $$1 - \frac{\binom{3}{2}\binom{4}{1}3!}{5!} = 1 - \frac{3}{5} = \frac{2}{5}$$
Suppose that the six people are named Ashley, Barney, Carly, David, Emma, and Fred. What is the probability that starting with Ashley and going clockwise, the six people are seated in alphabetical order. | {
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There is only one permissible seating arrangement. Hence, the probability that they are seated in alphabetical order is $$\frac{1}{5!} = \frac{1}{120}$$ | {
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# Thread: Need hints to solve antiderivative
1. ## Need hints to solve antiderivative
OK I have a slightly new problem that seems much more difficult.
$\displaystyle X'(t) = \mu_{k}gt + \frac{A}{4m} \int X'(t)^2dt$
constants:
$\displaystyle \mu_{k}$ = kinetic friction coefficient
$\displaystyle g$ = gravity
$\displaystyle A$ = cross sectional area
$\displaystyle m$ = mass
I am trying to solve $\displaystyle X(t)$ and then solve the specific function as an initial value problem.
I just need help solving $\displaystyle X'(t)$ for now. I should be OK after that.
Thanks
This is for a hobby physics project, not for homework.
2. Originally Posted by guitardude
OK I have a slightly new problem that seems much more difficult.
$\displaystyle X'(t) = \mu_{k}gt + \frac{A}{4m} \int X'(t)^2dt$
constants:
$\displaystyle \mu_{k}$ = kinetic friction coefficient
$\displaystyle g$ = gravity
$\displaystyle A$ = cross sectional area
$\displaystyle m$ = mass
I am trying to solve $\displaystyle X(t)$ and then solve the specific function as an initial value problem.
I just need help solving $\displaystyle X'(t)$ for now. I should be OK after that.
Thanks
This is for a hobby physics project, not for homework.
Differentiate both sides with respect to t. Then make the substitution U = X'. Solve the resulting DE for U. Hence solve for X.
3. I am still confused. If I differentiate both sides I will get $\displaystyle X''(t)$ right? which is what I started with. This equation is my work thus far on taking the antiderivative of $\displaystyle X''(t)$.
If I substitute $\displaystyle u=X'$ then I get $\displaystyle \int u^2$ The antiderivative of this part isn't as simple as $\displaystyle \frac{1}{3}u^3$ is it? Even so I would end up with $\displaystyle X(t) = constant*t^2 - constant* X'(t)^3 + C$ and I don't know $\displaystyle X'(t)$.
I forgot two negative signs, not that it matters much here but it should be:
$\displaystyle X'(t) = -\mu_{k}gt - \frac{A}{4m} \int X'(t)^2dt$ | {
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$\displaystyle X'(t) = -\mu_{k}gt - \frac{A}{4m} \int X'(t)^2dt$
I previously solved the initial value problem for constant acceleration as a differential equation:
$\displaystyle X''(t) = B$
so $\displaystyle X'(t) = Bt + C$
so $\displaystyle X(t) = Bt^2 + Ct + D$
which is our familiar physics equation $\displaystyle S = S_{i} + V_{i}t + \frac{1}{2} a(t^2)$
You can the solve the specific function and find B,C,D given 3 X(t) samples.
Now, instead of assuming a constant acceleration, I am trying to solve this with $\displaystyle B = -\mu_{k}g - \frac{1}{4m}Av^2$ and that is how I got $\displaystyle X''(t) = -\mu_{k}g - \frac{1}{4m}Av^2 = -\mu_{k}g - \frac{1}{4m}AX'(t)^2$
It is not obvious to me yet how many X(t) samples I will need to solve this as an initial value problem?
4. Originally Posted by mr fantastic
Differentiate both sides with respect to t. Then make the substitution U = X'. Solve the resulting DE for U. Hence solve for X.
Differentiate both sides wrt t: $\displaystyle X'' = \mu_k g + \frac{A}{4m} \, (X')^2$.
Substitute U = X': $\displaystyle U' = \mu_k g + \frac{A}{4m} \, U^2 \Rightarrow \frac{dU}{dt} - \frac{A}{4m} \, U^2 = \mu_k g$.
Solve this DE for U: I'd suggest using the integrating method.
Then solve $\displaystyle \frac{dX}{dt} = U$ for X.
5. I worked through it with my differential equations prof and got to:
$\displaystyle X'(t) = \frac{-\sqrt{\mu_{k}g}}{\sqrt{\frac{A}{4m}}} * \tan((\sqrt{\mu_{k}g})(\sqrt{\frac{A}{4m}})((t-C_{1})))$
I verified the previous equation against simulations as well as with mathematica. Next I used matlab and then simplified to get back to :
$\displaystyle X(t) = \frac{-4m}{A} * \ln(\sec^2((\sqrt{\mu_{k}g})(\sqrt{\frac{A}{4m}})( t-C_{1}))) + C_{2}$
With the two unknowns, $\displaystyle C_{1}, C_{2}$
I was expecting it to have a single $\displaystyle \sec$ not $\displaystyle sec^2$ ?
This will be messy to solve as an initial value problem but possible. | {
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This will be messy to solve as an initial value problem but possible.
Is there any way to make this function continuous? Tan() and Sec() make these hard to use and I don't see why it is necessary.
6. Originally Posted by guitardude
I worked through it with my differential equations prof and got to:
$\displaystyle X'(t) = \frac{-\sqrt{\mu_{k}g}}{\sqrt{\frac{A}{4m}}} * \tan((\sqrt{\mu_{k}g})(\sqrt{\frac{A}{4m}})((t-C_{1})))$
I verified the previous equation against simulations as well as with mathematica. Next I used matlab and then simplified to get back to :
$\displaystyle X(t) = \frac{-4m}{A} * \ln(\sec^2((\sqrt{\mu_{k}g})(\sqrt{\frac{A}{4m}})( t-C_{1}))) + C_{2}$
With the two unknowns, $\displaystyle C_{1}, C_{2}$
I was expecting it to have a single $\displaystyle \sec$ not $\displaystyle sec^2$ ?
This will be messy to solve as an initial value problem but possible.
Is there any way to make this function continuous? Tan() and Sec() make these hard to use and I don't see why it is necessary.
I can't quite understand how you've written the solution for X(t) but it looks like you have a log of a product. So use the usual log rule to break this up into a sum of logs.
Then it looks like one of those logs has the form $\displaystyle \log \sec^2 \theta$. Using the usual log rule you get $\displaystyle \log \sec^2 \theta = \log \cos^{-2} \theta = -2 \log \cos \theta$ .....
7. Originally Posted by mr fantastic
I can't quite understand how you've written the solution for X(t) but it looks like you have a log of a product. So use the usual log rule to break this up into a sum of logs.
Then it looks like one of those logs has the form $\displaystyle \log \sec^2 \theta$. Using the usual log rule you get $\displaystyle \log \sec^2 \theta = \log \cos^{-2} \theta = -2 \log \cos \theta$ .....
Hopefully this helps.... It is of form:
$\displaystyle X(t) = D * \ln( \sec^2(E(t-C_{1})) ) + C_{2}$
For the known constants $\displaystyle D,E$ | {
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For the known constants $\displaystyle D,E$
I will have to solve $\displaystyle C_{1}$ which I can do given two pairs of corresponding X,t
Does your suggestion of splitting $\displaystyle \ln(\sec^2(\theta))$ up still apply?
8. Originally Posted by guitardude
Hopefully this helps.... It is of form:
$\displaystyle X(t) = D * \ln( \sec^2(E(t-C_{1})) ) + C_{2}$
For the known constants $\displaystyle D,E$
I will have to solve $\displaystyle C_{1}$ which I can do given two pairs of corresponding X,t
Does your suggestion of splitting $\displaystyle \ln(\sec^2(\theta))$ up still apply?
Yes. | {
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# Math Help - A French Week
1. ## A French Week
For about 10 years after the French Revolution, the French government attempted to base measures of time on multiple of ten: One week consisted of 10 days, one day consisted of 10 hours, one hour consisted of 100 minutes, and one minute consisted of 100 seconds. What are the ratios of (a) the French decimal week to the standard week and (b) the French decimal second to the standard second?
(a) $1 \mbox{ French Week} = 1 \mbox{ French Week} \left(\frac{10 \mbox{ days}}{1 \mbox{ French Week}}\right)\left(\frac{10 \mbox{ hrs}}{1 \mbox{ day}}\right)$ $\left(\frac{100 \mbox{ min}}{1 \mbox{ hr}}\right)\left(\frac{100 \mbox{ s}}{1 \mbox{ min}}\right)$
$=10^6 \mbox{ seconds}$
$1 \mbox{ Normal Week} = 1 \mbox{ Normal Week} \left(\frac{7 \mbox{ days}}{1 \mbox{ Normal Week}}\right)\left(\frac{24 \mbox{ hrs}}{1 \mbox{ day}}\right)$ $\left(\frac{60 \mbox{ min}}{1 \mbox{ hr}}\right)\left(\frac{60 \mbox{ s}}{1 \mbox{ min}}\right)$
$=604800 \mbox{ seconds}$
So the ratio is 1.65, but the answers say that it is 1.43 ?
Also, in the first question I assumed that the 'French' second was the same as the 'normal' second, so wouldn't my answer to the second question make the answer to my first question invalid? Thanks.
2. Originally Posted by DivideBy0
For about 10 years after the French Revolution, the French government attempted to base measures of time on multiple of ten: One week consisted of 10 days, one day consisted of 10 hours, one hour consisted of 100 minutes, and one minute consisted of 100 seconds. What are the ratios of (a) the French decimal week to the standard week and (b) the French decimal second to the standard second?
(a) $1 \mbox{ French Week} = 1 \mbox{ French Week} \left(\frac{10 \mbox{ days}}{1 \mbox{ French Week}}\right)\left(\frac{10 \mbox{ hrs}}{1 \mbox{ day}}\right)$ $\left(\frac{100 \mbox{ min}}{1 \mbox{ hr}}\right)\left(\frac{100 \mbox{ s}}{1 \mbox{ min}}\right)$
$=10^6 \mbox{ seconds}$ | {
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$=10^6 \mbox{ seconds}$
$1 \mbox{ Normal Week} = 1 \mbox{ Normal Week} \left(\frac{7 \mbox{ days}}{1 \mbox{ Normal Week}}\right)\left(\frac{24 \mbox{ hrs}}{1 \mbox{ day}}\right)$ $\left(\frac{60 \mbox{ min}}{1 \mbox{ hr}}\right)\left(\frac{60 \mbox{ s}}{1 \mbox{ min}}\right)$
$=604800 \mbox{ seconds}$
So the ratio is 1.65, but the answers say that it is 1.43 ?
Also, in the first question I assumed that the 'French' second was the same as the 'normal' second, so wouldn't my answer to the second question make the answer to my first question invalid? Thanks.
Hello,
the ratio between the weeks is:
$\frac{\text{french week}}{\text{normal week}}=\frac{10}7 \approx 1.42857...$
to b):
$1 \text{ french day}= 10 \ h$ I assume that a new french day lasts as long as a normal day. Then you'll get the ratio:
$\frac{10 \ h \cdot 100\ min \cdot 100\ sec}{24\ h \cdot 60\ min \cdot 60\ sec} \approx 11.574...$ That means one of the new french seconds is a very short time interval. So better check my calculations.
3. Originally Posted by earboth
Hello,
the ratio between the weeks is:
$\frac{\text{french week}}{\text{normal week}}=\frac{10}7 \approx 1.42857...$
to b):
$1 \text{ french day}= 10 \ h$ I assume that a new french day lasts as long as a normal day. Then you'll get the ratio:
$\frac{10 \ h \cdot 100\ min \cdot 100\ sec}{24\ h \cdot 60\ min \cdot 60\ sec} \approx 11.574...$ That means one of the new french seconds is a very short time interval. So better check my calculations.
Why wouldn't DivideByZero's calculations be correct? When comparing two terms of measurement, you must relate them somehow
4. Hello, DivideBy0!
A fascinating problem . . . took me a few moments . . .
For about 10 years after the French Revolution, the French government
attempted to base measures of time on multiple of ten:
One week consisted of 10 days, one day consisted of 10 hours,
one hour consisted of 100 minutes, and one minute consisted of 100 seconds. | {
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What are the ratios of (a) the French decimal week to the standard week
and (b) the French decimal second to the standard second?
$\text{1 French Week }\:=\:\frac{\text{1 French Week}}{1} \times \frac{\text{10 days}}{\text{1 French Week}} \times \frac{\text{10 hrs}}{\text{1 day}}$ $\times \frac{\text{100 min}}{\text{1 hr}} \times \frac{\text{100 s}}{\text{1 min}}$
. . $=\;10^6 \text{ seconds}$
$\text{1 Normal Week} \;= \;\frac{\text{1 Normal Week}}{1} \times \frac{\text{7 days}}{\text{1 Normal Week}} \times \frac{\text{24 hrs}}{\text{1 day}} \times \frac{\text{60 min}}{\text{1 hr}} \times \frac{\text{60 s}}{\text{1 min}}$
. . $=\;604800 \text{ seconds}$
All of this is absolutely correct!
Note that (nearly) all the units are unequal . . . weeks, hours, minutes, seconds.
The only unit that both systems agree upon is the length of one day.
. . (Perhaps measured from sunrise to sunrise.)
$a)\;\;\frac{\text{1 Frech week}}{\text{1 Normal week}} \;=\;\frac{\text{10 days}}{\text{7 days}} \;\approx\; 1.43$
$b)\;\;\frac{\text{1 French day}}{\text{1 Normal day}} \; = \; \frac{10^5\text{ French sec}}{86,400\text{ Normal sec}} \;=\;1.1574
$
. . The French second is about 1½ jiffys longer than a Normal second.
5. Thanks, I don't know why I thought days could be of different lengths. | {
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Exponential Form of Complex Numbers; Euler Formula and Euler Identity interactive graph; 6. polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. Fortunately, when multiplying complex numbers in trigonometric form there is an easy formula we can use to simplify the process. What you can do, instead, is to convert your complex number in POLAR form: #z=r angle theta# where #r# is the modulus and #theta# is the argument. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. In order to work with these complex numbers without drawing vectors, we first need some kind of standard mathematical notation. So 18 times negative root 2 over. To multiply complex numbers: Each part of the first complex number gets multiplied by each part of the second complex numberJust use \"FOIL\", which stands for \"Firsts, Outers, Inners, Lasts\" (see Binomial Multiplication for more details):Like this:Here is another example: Change ), You are commenting using your Twitter account. Multiplication of Complex Numbers. So the root of negative number √-n can be solved as √-1 * n = √ n i, where n is a positive real number. That’s right – it kinda looks like the the Cartesian plane which you have previously used to plot (x, y) points and functions before. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). Complex numbers can be expressed in numerous forms. To plot a complex number a+bi on the complex plane: For example, to plot 2 + i we first note that the complex number is in rectangular (a+bi) form. Multiplying complex numbers when they're in polar form is as simple as multiplying and adding numbers. Polar form is where a complex number is denoted by the length | {
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as multiplying and adding numbers. Polar form is where a complex number is denoted by the length (otherwise known as the magnitude, absolute value, or modulus) and the angle of its vector (usually denoted by … In order to work with complex numbers without drawing vectors, we first need some kind of standard mathematical notation. A complex number can be represented in the form a + bi, where a and b are real numbers and i denotes the imaginary unit. How to Divide Complex Numbers in Rectangular Form ? Key Concepts. Math Gifs; Algebra; Geometry; Trigonometry; Calculus; Teacher Tools; Learn to Code; Home; Algebra ; Complex Numbers; Complex number Calc; Complex Number Calculator. (This is true for rectangular form as well (a 2 + b 2 = 1)) The Multiplicative Inverse (Reciprocal) of i. Then, multiply through by See and . Rectangular Form A complex number is written in rectangular form where and are real numbers and is the imaginary unit. Yes, you guessed it, that is why (a+bi) is also called the rectangular form of a complex number. Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to email this to a friend (Opens in new window), put it into the standard form of a complex number by writing it as, How To Write A Complex Number In Standard Form (a+bi), The Multiplicative Inverse (Reciprocal) Of A Complex Number, Simplifying A Number Using The Imaginary Unit i, The Multiplicative Inverse (Reciprocal) Of A Complex Number. Hence the value of Im(3z + 4zbar â 4i) is - y - 4. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). Use | {
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a is the real part of the complex number and bi is the imaginary part of the complex number). Use rectangular coordinates when the number is given in rectangular form and polar coordinates when polar form is used. In polar form, the multiplying and dividing of complex numbers is made easier once the formulae have been developed. Find powers of complex numbers in polar form. Figure 5. In other words, there are two ways to describe a complex number written in the form a+bi: To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. Dividing complex numbers: polar & exponential form. The major difference is that we work with the real and imaginary parts separately. We sketch a vector with initial point 0,0 and terminal point P x,y . Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. if z 1 = r 1∠θ 1 and z 2 = r 2∠θ … This video shows how to multiply complex number in trigonometric form. In other words, given $$z=r(\cos \theta+i \sin \theta)$$, first evaluate the trigonometric functions $$\cos \theta$$ and $$\sin \theta$$. This lesson on DeMoivre’s Theorem and The Complex Plane - Complex Numbers in Polar Form is designed for PreCalculus or Trigonometry. To understand and fully take advantage of multiplying complex numbers, or dividing, we should be able to convert from rectangular to trigonometric form and from trigonometric to rectangular form. A complex number can be expressed in standard form by writing it as a+bi. To convert from polar form to rectangular form, first evaluate the trigonometric functions. In the complex number a + bi, a is called the real part and b is called the imaginary part. Plot each point in the complex plane. Multiplying complex numbers is much like multiplying binomials. To write a complex number in rectangular form you just put it into the standard form of a complex number by writing | {
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number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. If z = x + iy , find the following in rectangular form. But then why are there two terms for the form a+bi? For Example, we know that equation x 2 + 1 = 0 has no solution, with number i, we can define the number as the solution of the equation. Addition of Complex Numbers . Find (3e 4j)(2e 1.7j), where j=sqrt(-1). Answer. Multiplication and division of complex numbers is easy in polar form. Label the x-axis as the real axis and the y-axis as the imaginary axis. (5 + j2) + (2 - j7) = (5 + 2) + j(2 - 7) = 7 - j5 (2 + j4) - (5 + j2) = (2 - 5) + j(4 - 2) = -3 + j2; Multiplying is slightly harder than addition or subtraction. For example, here’s how you handle a scalar (a constant) multiplying a complex number in parentheses: 2(3 + 2i) = 6 + 4i. So I get plus i times 9 root 2. Sum of all three four digit numbers formed with non zero digits. Complex Number Functions in Excel. Converting from Polar Form to Rectangular Form. To add complex numbers in rectangular form, add the real components and add the imaginary components. c) Write the expression in simplest form. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. b) Explain how you can simplify the final term in the resulting expression. Find roots of complex numbers in polar form. Rectangular form. Math Precalculus Complex numbers Multiplying and dividing complex numbers in polar form. Doing basic operations like addition, subtraction, multiplication, and division, as well as square roots, logarithm, trigonometric and inverse trigonometric functions of a complex numbers were already a simple thing to do. Worksheets on Complex Number. ( Log Out / Ask Question Asked 1 year, 6 months ago. When in rectangular form, the real and imaginary parts of the complex number are co-ordinates on the complex plane, and the way you plot them gives rise to the term | {
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number are co-ordinates on the complex plane, and the way you plot them gives rise to the term “Rectangular Form”. See . Products and Quotients of Complex Numbers; Graphical explanation of multiplying and dividing complex numbers; 7. The first, and most fundamental, complex number function in Excel converts two components (one real and one imaginary) into a single complex number represented as a+bi. Multiplying and dividing complex numbers in polar form. Addition, subtraction, multiplication and division can be carried out on complex numbers in either rectangular form or polar form. Rather than describing a vector’s length and direction by denoting magnitude and … The standard form, a+bi, is also called the rectangular form of a complex number. 1. By … Multiplying a complex number by a real number is simple enough, just distribute the real number to both the real and imaginary parts of the complex number. We distribute the real number just as we would with a binomial. To find the product of two complex numbers, multiply the two moduli and add the two angles. This is the currently selected item. Polar Form of Complex Numbers; Convert polar to rectangular using hand-held calculator; Polar to Rectangular Online Calculator ; 5. ( Log Out / Example 1. Multiplication of Complex Numbers. In essence, the angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. The video shows how to multiply complex numbers in cartesian form. Sum of all three four digit numbers formed using 0, 1, 2, 3. A complex number in rectangular form means it can be represented as a point on the complex plane. Active 1 year, 6 months ago. After having gone through the stuff given above, we hope that the students would have understood, "How to Write the Given Complex Number in Rectangular Form". We start with an example using exponential form, and then | {
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Given Complex Number in Rectangular Form". We start with an example using exponential form, and then generalise it for polar and rectangular forms. Convert a complex number from polar to rectangular form. This can be a helpful reminder that if you know how to plot (x, y) points on the Cartesian Plane, then you know how to plot (a, b) points on the Complex Plane. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. The reciprocal of zero is undefined (as with the rectangular form of the complex number) When a complex number is on the unit circle r = 1/r = 1), its reciprocal equals its complex conjugate. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). bi+a instead of a+bi). 18 times root 2 over 2 again the 18, and 2 cancel leaving a 9. Notice the rectangle that is formed between the two axes and the move across and then up? Using either the distributive property or the FOIL method, we get I get -9 root 2. The correct answer is therefore (2). Multiplying by the conjugate . This video shows how to multiply complex number in trigonometric form. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. ; The absolute value of a complex number is the same as its magnitude. The Number i is defined as i = √-1. To divide, divide the magnitudes and … Hence the Re (1/z) is (x/(x2 + y2)) - i (y/(x2 + y2)). The following development uses trig.formulae you will meet in Topic 43. Rectangular Form of a Complex Number. In essence, the angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. To divide the complex number which is in the form (a + ib)/(c + id) we have to multiply both numerator and denominator by the conjugate of the denominator. Apart from the stuff given in this section ", How to Write the Given Complex Number in Rectangular Form". | {
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from the stuff given in this section ", How to Write the Given Complex Number in Rectangular Form". Write the following in the rectangular form: [(5 + 9i) + (2 â 4i)] whole bar = (5 + 9i) bar + (2 â 4i) bar, Multiplying both numerator and denominator by the conjugate of of denominator, we get, = [(10 - 5i)/(6 + 2i)] [(6 - 2i)/(6 - 2i)], = - 3i + { (1/(2 - i)) ((2 + i)/(2 + i)) }. Multiplying Complex Numbers Together. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). When you’re dividing complex numbers, or numbers written in the form z = a plus b times i, write the 2 complex numbers as a fraction. Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. The imaginary unit i with the property i 2 = − 1 , is combined with two real numbers x and y by the process of addition and multiplication, we obtain a complex number x + iy. Multiplication . $\text{Complex Conjugate Examples}$ $\\(3 \red + 2i)(3 \red - 2i) \\(5 \red + 12i)(5 \red - 12i) \\(7 \red + 33i)(5 \red - 33i) \\(99 \red + i)(99 \red - i)$ Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. Well, rectangular form relates to the complex plane and it describes the ability to plot a complex number on the complex plane once it is in rectangular form. We start with an example using exponential form, and then generalise it for polar and rectangular forms. Show Instructions. Example 1 | {
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form, and then generalise it for polar and rectangular forms. Show Instructions. Example 1 – Determine which of the following is the rectangular form of a complex number. To convert from polar form. Google custom search here method of notation is valid for numbers. Incorrect input is treated as vector addition are commenting using your Twitter account z 2 = r 1 θ. 3 + 2j is the imaginary components your email addresses calculator will simplify any complex expression, steps.: Incorrect input 18 times root 2 method of notation is valid for complex numbers therefore the answer! And terminal point P x, y ' + ' is treated as vector addition we will how. This complex number notation: polar and rectangular forms the way rectangular coordinates are plotted in the rectangular.... Answer is ( 4 ) with a=7, and b=4 Euler formula and Euler Identity interactive ;... Arithmetic on complex numbers ; Graphical explanation of multiplying and dividing of complex numbers is easy in rectangular where. Imaginary part correct answer is ( 4 ) with a=7, and 2 cancel leaving a 9 have this... Use the formulas and then, See and numbers when they 're in polar coordinate,! Also, See section 2.4 of the text for an introduction to complex numbers that the... And division can be considered a subset of the complex numbers multipling and dividing complex! Evaluates expressions in the set of complex numbers in trig form, the... How you can simplify the final term in the rectangular form '' know that i lies on other. Matter of evaluating what is given and using the polar form, first the. Numbers multiplying and dividing of complex numbers when they 're in polar form is a lot than. Where a and b is the sign we can use to simplify the process convert a complex number given., Inner, and then generalise it for polar and rectangular finding powers and of... Evaluates expressions in the complex plane looks very similar to the way coordinates... Aims to make learning about complex numbers the number x + yi | {
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very similar to the way coordinates... Aims to make learning about complex numbers the number x + yi in the complex numbers is made once... And z 2 = r 1 cis θ 1 and z 2 = r cis. ( 3e 4j ) ( 2e 1.7j ), you are commenting using your WordPress.com account on complex numbers polar! Is at the co-ordinate ( 2, 1 ) on the complex plane similar to another which! − 2j is the same as its magnitude b2 ( a + 0i iy, the! $i have attempted this complex number \ ( z\ ) as shown on the circle... Non zero digits as a point on the other hand, is where a complex number by real. Leaving a 9 the major difference is that we can convert complex numbers that have the form plotted! The value of a complex number is written as a+bi where a complex number is denoted by its horizontal... Now that we work with the real axis and the move across and then generalise for... You are commenting using your multiplying complex numbers in rectangular form account again the 18, and then, section! 3E 4j ) ( 2e 1.7j ), you can skip the multiplication sign, so ... Another plane which you have used before with steps shown, multiplication and division can considered. Vertical components, is where a and b is called the rectangular form a complex number polar... A complex number \ ( z\ ). answer and evaluates expressions in the form plotted... Rectangular form. along the horizontal axis, followed by 1 unit on! Number i is defined as i = √-1 co-ordinate ( 2, 1 ) on the other,. Have also noticed that the complex plane below, add their imaginary parts so... Numbers without drawing vectors, we use the formulas and then generalise it for polar and forms! The absolute value of a complex number from polar form. and evaluates expressions in the expression... Four digit numbers formed with non zero digits the y-axis as the imaginary part and is the rectangular of. Move across and then generalise it for polar and exponential forms it is no different to whenever... We can use either the distributive property following | {
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forms it is no different to whenever... We can use either the distributive property following development uses trig.formulae you will meet in topic 43 using... Number x + yi in the resulting expression units along the horizontal axis followed... Using rectangular form where and are real numbers and evaluates expressions in the plane. Products and Quotients of complex numbers aims to make learning about complex numbers, multiply magnitudes... Real number form of a complex number in trigonometric form. form Step 1 a... The same as its magnitude r at angle θ ”. and imaginary parts separately axis!, that is formed between the two moduli and add the angles just as would... Months ago uses trig.formulae you will meet in topic 43 ) Write a number! And b=4 answer form used to Plot complex numbers is easy in rectangular form, add the components... The 18, and 2 cancel leaving a 9$ \begingroup \$ i attempted... , how to Write the given complex number from polar to rectangular using hand-held calculator ; polar to form... Will meet in topic 43 as simple as multiplying and adding numbers Question Asked 1 year, months. Special case just remember when you 're multiplying complex numbers in polar form,,. The 18, and then, See and finding powers and roots of complex numbers trig. The trigonometric functions the product of two complex numbers in rectangular form means it be! As “ r at angle θ ”. the formulas and then up ) Write a complex in! Calculator does basic arithmetic on complex numbers Plot each point in the plane... Carried Out on complex numbers is easy in polar form is used horizontal vertical! Work with the real axis and the move across and then up Google custom search here moduli, and.! And Euler Identity interactive graph ; 6 subtraction of complex number a + jb ; a! Of standard mathematical notation, on the other hand, is where complex! Carl Friedrich Gauss ( 1777-1855 ). answer sorry, your blog can not share posts by.! Point P x, y sorry, your blog can not share posts | {
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answer sorry, your blog can not share posts by.! Point P x, y sorry, your blog can not share posts by email sorry, blog. 2J and adding numbers French mathematician Abraham de Moivre ( 1667-1754 ). answer is the! Times root 2 over 2 again the 18, and then generalise it for polar and forms. Easy formula we can use to simplify the process its respective horizontal and vertical components, multiply magnitudes. Two axes and the move across and then generalise it for polar and rectangular we! Make learning about complex numbers another plane which you have used before matter of evaluating what is and. Vector with initial point 0,0 and terminal point P x, y rectangular calculator! Fortunately, when multiplying complex numbers in polar form to rectangular form a. Step 1 sketch a graph of the following development uses trig.formulae you will meet in topic.... The stuff given in rectangular form means it can be represented as a on!, on the vertical axis is formed between the two angles posts by email have multiplying complex numbers in rectangular form form a+bi are.... An icon to Log in: you are commenting using your WordPress.com.... Horizontal axis, followed by 1 unit up on the vertical axis 2 again the 18, Last!, use polar and rectangular forms i = √-1 numbers is made easier once the formulae have been developed subset. Numbers without drawing vectors, can also be expressed in polar form. complex.... Other hand, is where a complex number written in rectangular form used to Plot complex numbers polar! Log Out / Change ), you are commenting using your WordPress.com account unit multiplying complex numbers in rectangular form! A 9 from \ ( z\ ) as shown on the complex plane looks very similar to the way coordinates! By French mathematician Abraham de Moivre ( 1667-1754 ). answer their imaginary parts separately when performing multiplication finding! The way rectangular coordinates are plotted in the rectangular form, the multiplying and complex... See section 2.4 of the | {
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are plotted in the rectangular form, the multiplying and complex... See section 2.4 of the following in rectangular form. convert a complex in... Number notation: polar and exponential forms 2.4 of the text for an introduction to complex numbers in form! Is valid for complex numbers in exponential form, the multiplying and dividing complex numbers and evaluates expressions in rectangular... You guessed it, that is why ( a+bi ) is - y - 4, so 5x. Free complex number notation: polar and rectangular 1 z 2 = r 1 cis θ and. Vector with initial point 0,0 and terminal point P x, y are two basic forms of complex numbers cartesian! To trigonometric form there is an advantage of using the polar form. it, that is why ( ). Example 1 – Determine which of the text for an introduction to complex numbers ; 7 the FOIL method equivalent... Kind of standard mathematical notation let z 1 z 2 = r 1 cis θ 1 and z 2 r... \ ( 0\ ) to \ ( z\ ) as shown on the other hand is., 6 months ago formulas and then generalise it for polar and forms! Written in rectangular form to rectangular form of a complex number is given using. To 5 * x ` Write a complex number is the rectangular used... Basic arithmetic on complex numbers in polar form, and Last terms...., just like vectors, can also be expressed in polar form, multiply the moduli, and Last together! Written in rectangular form., just like vectors, can also be expressed in polar form. either form. Simplify any complex expression, with steps shown, r ∠ θ and rectangular number polar... | {
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# Math Help - [SOLVED] probability question(digits)
1. ## [SOLVED] probability question(digits)
Question:
A four-digit number, in the range 0000 to 9999 inclusive, is formed. Find the probability that
(a) the number begins or ends with 0,
(b) the number contains exactly two non-zero digits.
Can you help me solve this question? I have tried but didn't manage to get the real answer(but I strongly support my own answer).
Thanks
2. For part A, here are my thoughts:
For the first 0000 to 0999, there are 1000 numbers. Then you have to add in all the numbers ending in 0. These are a little trickier to see. So we are looking for 1000, 1010, 1020, ... , 1090. That's a total of 10 numbers between 1000 and 1090 (inclusive). We also have to remember 1100, 1110, 1120, ... , 1190. So there are 10 numbers between each increment of a hundred. And between each thousand there are 10 increments of 100. Finally we are starting at 999 and proceeding to 9999, in which there are 9 increments of a thousand. So my total for part A comes to:
$\frac {1000 + 10 * 10 * 9}{10000} = \frac{19}{100}$
3. Originally Posted by stpmmaths
Question:
A four-digit number, in the range 0000 to 9999 inclusive, is formed. Find the probability that
(a) the number begins or ends with 0,
(b) the number contains exactly two non-zero digits.
Can you help me solve this question? I have tried but didn't manage to get the real answer(but I strongly support my own answer).
Thanks
(a) You want the combinations 0 X X X or X X X 0. Using the pigeon hole principle:
(1)(10)(10)(10) + (10)(10)(10)(1) = 2000.
But you have to subtract numbers of the form 0 X X 0 otherwise they get counted twice in the above: (1)(10)(10)(1) = 100.
So the total number of numbers satisfying the restriction is 1900.
The total number of numbers withut restriction is (10)(10)(10)(10) = 10000.
So the probability is 1900/10000 = 19/100.
You can take a similar approach for (b). | {
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So the probability is 1900/10000 = 19/100.
You can take a similar approach for (b).
4. A more systematic way of doing part A is to count first the numbers beginning with 0, and then the numbers ending with 0 and then subtracting the number of numbers beginning and ending with 0 to get the total number of numbers beginning or ending with 0. We get $10^3+10^3-10^2$, giving the same probability as eXist calculated.
For part B,there are 4 choose 2 =6 ways to pick the two zero digits and 9*9 ways to choose the two non-zero digits. It is worth thinking about how many 4 digit numbers contain respectively exactly 0,1,2,3 and 4 0s. You should see a connection with the binomial theorem (expand $(9+1)^4$ to see it more clearly).
5. Hello stpmmaths
Welcome to Math Help Forum!
Originally Posted by stpmmaths
Question:
A four-digit number, in the range 0000 to 9999 inclusive, is formed. Find the probability that
(a) the number begins or ends with 0,
(b) the number contains exactly two non-zero digits.
Can you help me solve this question? I have tried but didn't manage to get the real answer(but I strongly support my own answer).
Thanks
Part (b). If the number contains exactly two non-zero digits, then it contains 2 non-zero and 2 zero digits.
So: imagine you have 4 empty boxes in a line in front of you. You have to (i) place a zero into two of the boxes, and then (ii) place a digit from 1 to 9 in the remaining two boxes.
For stage (i), there are $\binom42 = 6$ ways of choosing which boxes shall contain the zeros.
For stage (ii), there are 9 choices of non-zero digit to go in the remaining left-hand empty box, and 9 to go in the right-hand box. Total number of choices = $9 \times 9 = 81$.
So the number of different numbers with exactly two non-zero digits is $6 \times 81 = 486$. So the probability is $\frac{486}{10000}=0.0486$.
6. Originally Posted by mr fantastic
(a) You want the combinations 0 X X X or X X X 0. Using the pigeon hole principle: | {
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(1)(10)(10)(10) + (10)(10)(10)(1) = 2000.
But you have to subtract numbers of the form 0 X X 0 otherwise they get counted twice in the above: (1)(10)(10)(1) = 100.
So the total number of numbers satisfying the restriction is 1900.
The total number of numbers withut restriction is (10)(10)(10)(10) = 10000.
So the probability is 1900/10000 = 19/100.
You can take a similar approach for (b).
For part A,
As you said, (1)(10)(10)(10) + (10)(10)(10)(1) - (1)(10)(10)(1) = 1900. If we do this we still consider 0XX0 in the probability, right? But the question stated begins or ends with 0 not with both
7. Originally Posted by stpmmaths
For part A,
As you said, (1)(10)(10)(10) + (10)(10)(10)(1) - (1)(10)(10)(1) = 1900. If we do this we still consider 0XX0 in the probability, right? But the question stated begins or ends with 0 not with both
If the statement is "begins or ends with a zero digit" it means begins with a zero digit or ends with a zero digit or both begins and ends with a zero.
That is the logical nature of the connective or.
Hello stpmmaths
Welcome to Math Help Forum!
Part (b). If the number contains exactly two non-zero digits, then it contains 2 non-zero and 2 zero digits.
So: imagine you have 4 empty boxes in a line in front of you. You have to (i) place a zero into two of the boxes, and then (ii) place a digit from 1 to 9 in the remaining two boxes.
For stage (i), there are $\binom42 = 6$ ways of choosing which boxes shall contain the zeros.
For stage (ii), there are 9 choices of non-zero digit to go in the remaining left-hand empty box, and 9 to go in the right-hand box. Total number of choices = $9 \times 9 = 81$.
So the number of different numbers with exactly two non-zero digits is $6 \times 81 = 486$. So the probability is $\frac{486}{10000}=0.0486$. | {
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I thought we should consider the position(placement) of the digits? If we do it by we are not considering the position. According to your statement "
imagine you have 4 empty boxes in a line in front of you. You have to (i) place a zero into two of the boxes, and then (ii) place a digit from 1 to 9 in the remaining two boxes.
For stage (i), there are $\binom42 = 6$ ways of choosing which boxes shall contain the zeros.
"
This makes me confused between and .
Lets say we change the question to:-
A four-digit number, in the range 0000 to 9999 without repetitions for all digits except digit 0, is formed. Find the probability that
(a) ....
(b) the number contains exactly two non-zero digits.
9. ## Permutations and Combinations
Hello stpmmaths
Originally Posted by stpmmaths
I thought we should consider the position(placement) of the digits? If we do it by we are not considering the position. According to your statement "
imagine you have 4 empty boxes in a line in front of you. You have to (i) place a zero into two of the boxes, and then (ii) place a digit from 1 to 9 in the remaining two boxes.
For stage (i), there are $\binom42 = 6$ ways of choosing which boxes shall contain the zeros.
"
This makes me confused between and .
Lets say we change the question to:-
A four-digit number, in the range 0000 to 9999 without repetitions for all digits except digit 0, is formed. Find the probability that
(a) ....
(b) the number contains exactly two non-zero digits.
You are raising two different points here. First, the difference between $^nP_r$ and $^nC_r$.
• $^nP_r$ gives the number of ways of choosing $r$ objects from $n$, and then arranging them in order. So, for instance, if we want to choose 2 from 4, and the objects are A, B, C and D, then the choice A, C will be counted as a different choice from C, A. There are therefore $^4P_2=4 \times 3 = 12$ possible choices. | {
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• $^nC_r$ gives the number of ways of choosing $r$ objects from $n$, where the order doesn't matter. In the above example, A, C and C, A would be counted as the same choice, and there are therefore $^4C_2=\frac{4\times 3}{2}= 6$ possible choices.
As an example, suppose that a committee chairman and secretary have to be chosen from a short-list of 4 people. Here, A, C is a different selection from C, A because A and C are fulfilling different roles. So you use $^4P_2$. But if you have to choose 2 delegates to attend a conference, from 4 people, you'll use $^4C_2$. Why? Because A, C and C, A are now the same choice - there's no difference between the roles of A and C.
So, in question (b), the number of different pairs of positions that can be occupied by a zero is $^4C_2=6$, because one zero looks just like the other - the order in which you place them in the 'empty boxes' doesn't matter.
In each of the remaining boxes (containing non-zeros) the order in which we fill the boxes does matter: 0709 will be different from 0907, for instance. Each box can then be filled in 9 ways, so - using the 'r-s Principle' - the total number of ways in which these two boxes can be filled, one after the other, is $9^2 =81$.
Your second question is: how is this affected if we're not allowed to choose the same non-zero digit twice? The answer is this:
• The first stage - choosing which 'boxes' are to be filled with zeros - is unaltered. It's still 6 choices.
• In the second stage, suppose we fill the left-hand of the two remaining boxes first. There are 9 ways of doing this. Since we're not allowed to choose the same digit again, there are then 8 ways of filling the last box. Total number of choices: $9\times 8 = 72$. Notice that this is the number of ways of choosing 2 items from 9, where the order matters; in other words, it's $^9P_2$.
So the overall answer to your modified question would be $6 \times 72 = 432$ choices, or a probability of $0.0432$.
Does that clear things up? | {
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Does that clear things up?
10. Thanks
So if i modified the question to:-
A five-digit number, in the range 00000 to 99999 inclusive, is formed. Find the probability that
(a) ....
(b) the number contains exactly two non-zero digits.
Then the answer would be 6 x 729 = 4374. Am I right?
11. Hello stpmmaths
Originally Posted by stpmmaths
Thanks
So if i modified the question to:-
A five-digit number, in the range 00000 to 99999 inclusive, is formed. Find the probability that
(a) ....
(b) the number contains exactly two non-zero digits.
Then the answer would be 6 x 729 = 4374. Am I right?
Correct!
And if repetition is not allowed in the non-zero digits, it would be $6 \times ^9P_3 = 6 \times9\times8\times7 = 3024$.
13. ## probability question
So the term C^2_4 relates to choosing the positions for (non) zero digits whereas terms 9^2 or P^2_9 are the numbers of ways to choose non-zero digits for a given positions.
14. Hello kobylkinks
Originally Posted by kobylkinks
So the term C^2_4 relates to choosing the positions for (non) zero digits whereas terms 9^2 or P^2_9 are the numbers of ways to choose non-zero digits for a given positions.
If this is a question, and not just a statement, then yes, you are right in your analysis!
You're also right to write '(non) zero digits' because it doesn't matter whether you think of this part of the 'box-filling' process as choosing the boxes that will contain zeros, or the boxes that will contain non-zeros; the result is the same: there are $^4C_2= 6$ possible choices. | {
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# How to put a matrix in Reduced Echelon Form
5. (6 pts) Find a basis for the following subspace: $$\left\{ \begin{bmatrix} r - s - 2t - u \\ -r + 2s + 5t + 2u \\ s + 3t + u \\ 3r + s + 5u \end{bmatrix} \in \mathbb{R}^4 : \text{r, s, t and u are scalars} \right\}.$$
Original image here.
So I have this matrix: $$\begin{pmatrix} 1 &-1 &-2 &-1 \\ -1 &2& 5 &2 \\ 0 &1 &3 &1 \\ 3 &1 & 0 & 5 \end{pmatrix}$$ If I compute the the matrix in reduced echelon form, I get: $$\begin{pmatrix} 1&0&1&0\\ 0&1&0&3\\ 0&0&1&-2/3\\ 0 & 0 & 0 & 0 \end{pmatrix}$$
Please note that i want to put the matrix in reduced echelon form , and not row reduced echelon form.
I know that the three rules of reduced echelon form are:
1) The first non zero number of a row is 1 (leading entry).
2) An all zero row is placed at the end of the matrix.
3)A leading entry of a row is placed to the right of a leading entry of the previous row.
If it were to be in row reduced echelon form , then another property would hold, or else that each column that contains a leading entry, all other entries in the same column are zeros.
So is my matrix correctly put in a row echelon form ?
I think you are confused on terminology. Your matrix is correctly put into one of its row echelon forms. Note that these are not unique, as we could add the second row to the first row and the result would still be in row echelon form, that is, \begin{pmatrix} 1&1&1&3\\ 0&1&0&3\\ 0&0&1&-2/3\\ 0 & 0 & 0 & 0 \end{pmatrix} Is still in row echelon form. In order to get to the reduced row echelon form (sometimes called reduced echelon form), we must, as you said, make $0$'s in every column which contains a leading $1$. In your example, we can subtract the third row from the first \begin{pmatrix} 1&0&1&0\\ 0&1&0&3\\ 0&0&1&-2/3\\ 0 & 0 & 0 & 0 \end{pmatrix} to obtain \begin{pmatrix} 1&0&0&2/3\\ 0&1&0&3\\ 0&0&1&-2/3\\ 0 & 0 & 0 & 0 \end{pmatrix} Which is now in reduced row echelon form. Hopefully that clears up any confusion you had. | {
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• In order to find a basis for a subspace, should i use row echelon form or row reduced echelon form ? Moreover, would the basis be composed of non zero rows or columns with pivot entries ? – Sara Martini Jun 16 '18 at 20:32
• And what is the advantage of dealing with a NON reduced row echelon form ? – Rene Schipperus Jun 16 '18 at 20:37
With reference to the original problem we need to find a basis for
$$\begin{bmatrix} r-s-2t-u \\ -r+2s+5t+2u \\ s+3t+u \\ 3r+s+5u \end{bmatrix}=\begin{pmatrix} 1 &-1 &-2 &-1 \\ -1 &2& 5 &2 \\ 0 &1 &3 &1 \\ 3 &1 & 0 & 5 \end{pmatrix}\begin{pmatrix} r \\ s \\ t \\ u \end{pmatrix}$$
note that the LHS is a vector $\in \mathbb{R^4}$ and in the RHS the given coefficients are arranged by columns, therefore we need to find a basis for the column space of the matrix in on the RHS.
Then we have
$$\begin{pmatrix} 1 &-1 &-2 &-1 \\ -1 &2& 5 &2 \\ 0 &1 &3 &1 \\ 3 &1 & 0 & 5 \end{pmatrix}\to \begin{pmatrix} 1 &-1 &-2 &-1 \\ 0 &1& 3 &1 \\ 0 &1 &3 &1 \\ 3 &1 & 0 & 5 \end{pmatrix}\to \begin{pmatrix} 1 &-1 &-2 &-1 \\ 0 &1& 3 &1 \\ 0 &1 &3 &1 \\ 0 &4 & 6 & 8 \end{pmatrix}\to \begin{pmatrix} 1 &-1 &-2 &-1 \\ 0 &1& 3 &1 \\ 0 &0 &0 &0 \\ 0 &0 & -6 & 4 \end{pmatrix}\to \begin{pmatrix} 1 &-1 &-2 &-1 \\ 0 &1& 3 &1 \\ 0 &0 & -6 & 4\\ 0 &0 &0 &0 \end{pmatrix}$$
which is the RREF. From here we can conclude that a basis is given by the first three vectors of the original matrix that is
$$\{(1,-1,0,3),(-1,2,1,1),(-2,5,3,0)\}$$
Note that we can also continue and obtain
$$\to \begin{pmatrix} 1 &0 &1 &0 \\ 0 &1& 3 &1 \\ 0 &0 & 1 & -\frac23\\ 0 &0 &0 &0 \end{pmatrix}\to \begin{pmatrix} 1 &0 &0 &\frac23 \\ 0 &1& 0 &3 \\ 0 &0 & 1 & -\frac23\\ 0 &0 &0 &0 \end{pmatrix}$$
the latter is the Gauss-Jordan form of the RREF but it is not needed to establish a basis for the given subspace. | {
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• Is the row echelon form sufficient in order to find the basis for a subspace? – Sara Martini Jun 16 '18 at 19:58
• @SaraMartini Yes and you can stop to the first step I've indicated, if you are looking to the columns a basis is given by the first three columns in the original matrix, if you are looking for the rows you can choose the first the second and the fourth rows in the originl matrix an a basis or the first, the second and the third row in the RREF. – gimusi Jun 16 '18 at 20:03
• Im sorry but i don't really understand what you wrote... sorry :( I am willing to find a basis for a subspace, and i am not sure whether to use the row echelon form or reduced row echelon form ... or if i can use both interchangeably. If i used the row echelon form , then the basis are the first three columns of the original matrix. But what do you mean when you say that they can be rows too? I am kinds of confused – Sara Martini Jun 16 '18 at 20:09
• @SaraMartini It depends upon which subspace you are looking for and how you arrange the vector in the matrix. – gimusi Jun 16 '18 at 20:10
• This is the subspace.... (r-s-2t-u) , (-r+2s+5t+2u), (s+3t+u), (3r+s+5u) – Sara Martini Jun 16 '18 at 20:13 | {
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# Identifying Types of Converging Series
1. Aug 3, 2015
### DameLight
Hello, I'm looking for some help with this problem for my Calculus 2 class. Since it's a summer class my professor wasn't able to explain everything fully so if you can help me that would be great : )
1. The problem statement, all variables and given/known data
Select the FIRST correct reason why the given series converges.
A. Convergent geometric series
B. Convergent p series
C. Comparison (or Limit Comparison) with a geometric or p series
D. Alternating Series Test
E. None of the above
1. Σn=1 6(4)n/72n
2. Σn=1 sin2(6n)/n2
3. Σn=1 cos(nπ)/ln(2n)
4. Σn=1 (−1)n/(6n+5)
5. Σn=1 (n+1)(24)n/52n
6. Σn=1 (−1)n * √(n)/(n+9)
2. Relevant equations
Geometric Series:
Σn=1 arn-1 will converge if -1 < r < 1
P Series:
Σn=1 1/np converges if p > 1
Alternating Series Test:
1) bn+1 </= bn for all n and
2) limn->∞ bn = 0
3. The attempt at a solution
1. A
2. B
3. E
4. D
5. C
6. D
Last edited by a moderator: Aug 3, 2015
2. Aug 3, 2015
### Ray Vickson
I get (in order 1-6) A, C, D, D, E, D.
3. Aug 4, 2015
### DameLight
My professor told me that I got numbers 2 and 3 correct
4. Aug 4, 2015
### Ray Vickson
I disagree. Here are the details.
1. $\sum 6 \;4^n/7^{2n} = 6 \sum (4/49)^n$, so geometric (A).
2. In $\sum \sin^2 (6n) /n^2$ the terms are (i) not geometric; (ii) not alternating; (iii) not in the simple form $1/n^p$. However, $0 \leq \sin^2(6n)/n^2 \leq 1/n^2$, so a comparison with the series $1/n^2$ works (C).
3. $\sum \cos(n \pi)/ \ln(n)$ is an alternating series, since $\cos(n \pi) = (-1)^n$. So, D.
4. $\sum (-1)^n / (6n+5)$ is alternating, so D.
5. $\sum (n+1) \; 24^n / 5^{2n} = \sum (n+1) r^n, \;\; r = 24/25$. Here, A,B,C,D do not apply, and that leaves E.
6. $\sum (-1)^n \sqrt{n}/(n+9)$ (or $\sum (-1)^n \sqrt{ n/(n+9)}$---can't tell which you mean) is alternating in either interpretation, so D.
5. Aug 4, 2015
### DameLight
The correct answer ended up being:
1. A
2. C
3. D
4. D
5. C
6. D | {
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5. Aug 4, 2015
### DameLight
The correct answer ended up being:
1. A
2. C
3. D
4. D
5. C
6. D
Thank you for your help : )
6. Aug 4, 2015
### Ray Vickson
How do you get 5 C? Did you read my detailed explanation?
7. Aug 4, 2015
### micromass
Why wouldn't the comparison test apply in $5$? You can do $n+1\leq \alpha^n$ for some small enough $\alpha>1$.
8. Aug 4, 2015
### Ray Vickson
OK. That would be an answer to my question (why?) that the OP did not answer. For all I know the OP may have used false reasoning.
9. Aug 4, 2015
### DameLight
I don't want to assume, but that sounded a bit rude.
I am assuming that going off of your explanation for 5 that we can compare that to a geometric function ∑n=0 anrn and since the absolute value of r is less than 1 we can go ahead and reason that the limit would be equivalent to a/(1-r)
10. Aug 5, 2015
### Ray Vickson
No: a geometric series would be $\sum r^n$ (or $\sum c r^n = c \sum r^n$ for constant $c$), but $\sum a_n r^n$ is NOT a geometric series if $a_n$ varies with $n$. However, if $a_n \to c = \text{constant}$ as $n \to \infty$, then comparison with the series $\sum c r^n$ can be made to work without too much effort (although some extra effort is needed), However, in your case you have $a_n = n+1 \to +\infty$ as $n \to \infty$, so that type of argument does not work automatically. If you thought it did, that is what what I would call "false reasoning".
However, as 'micromass' has pointed out, if you choose a small enough $\epsilon > 0$ so that $(1+\epsilon) r < 1$, then for some finite $N> 0$ we have $n+1 \leq (1+\epsilon)^n$ for all $n \geq N$, so for $n \geq N$ we have $0 \leq (n+1) r^n \leq [(1+\epsilon) r]^n$ and we can compare with the geometric series $\sum [(1+\epsilon) r]^n$. Note, however, that some such argument as this must be acknowledged and recognized in order for C to be a valid and well-founded answer. Without this realization, just saying it would, again, be false reasoning. | {
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Last edited: Aug 5, 2015 | {
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5 For any m n matrix A, we have A i = eT i A and A j = Ae j. P. Sam Johnson (NITK) Existence of Left/Right/Two-sided Inverses September 19, 2014 3 / 26 In a monoid, if an element has a right inverse… Thus the unique left inverse of A equals the unique right inverse of A from ECE 269 at University of California, San Diego h�b�y��� cca�� ����ِ� q���#�!�A�ѬQ�a���[�50�F��3&9'��0 qp�(R�&�a�s4�p�[���f^'w�P& 7��,���[T�+�J����9�$��4r�:4';m$��#�s�Oj�LÌ�cY{-�XTAڽ�BEOpr�l�T��f1�M�1$��С��6I��Ҏ)w It would therefore seem logicalthat when working with matrices, one could take the matrix equation AX=B and divide bothsides by A to get X=B/A.However, that won't work because ...There is NO matrix division!Ok, you say. Generalized inverse Michael Friendly 2020-10-29. best. %PDF-1.4 given $$n\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. 11.1. LEAST SQUARES PROBLEMS AND PSEUDO-INVERSES 443 Next, for any point y ∈ U,thevectorspy and bp are orthogonal, which implies that #by#2 = #bp#2 +#py#2. An inverse that is both a left and right inverse (a two-sided inverse), if it exists, must be unique. U-semigroups endobj Proof: Assume rank(A)=r. x��XKo#7��W�hE�[ע��E������:v�4q���/)�c����>~"%��d��N��8�w(LYɽ2L:�AZv�b��ٞѳG���8>����'��x�ټrc��>?��[��?�'���(%#R��1 .�-7�;6�Sg#>Q��7�##ϥ "�[� ���N)&Q ��M���Yy��?A����4�ϠH�%�f��0a;N�M�,�!{��y�<8(t1ƙ�zi���e��A��(;p*����V�Jڛ,�t~�d��̘H9����/��_a���v�68gq"���D�|a5����P|Jv��l1j��x��&N����V"���"����}! Let G G G be a group. left A rectangular matrix can’t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. �n�����r����6���d}���wF>�G�/��k� K�T�SE���� �&ʬ�Rbl�j��|�Tx��)��Rdy�Y ? 0 Yes. JOURNAL OF ALGEBRA 31, 209-217 (1974) Right (Left) Inverse Semigroups P. S. VENKATESAN National College, Tiruchy, India and Department of Mathematics, University of Ibadan, Ibadan, Nigeria Communicated by G. B. Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if | {
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Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if every (nonnull) principal left ideal of S has a unique idempotent … (An example of a function with no inverse on either side is the zero transformation on .) Recall also that this gives a unique inverse. If f contains more than one variable, use the next syntax to specify the independent variable. Viewed 1k times 3. Note that other left In fact, if a function has a left inverse and a right inverse, they are both the same two-sided inverse, so it can be called the inverse. This is generally justified because in most applications (e.g., all examples in this article) associativity holds, which makes this notion a generalization of the left/right inverse relative to an identity. Ask Question Asked 4 years, 10 months ago. Let (G, ⊕) be a gyrogroup. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. eralization of the inverse of a matrix. If a matrix has a unique left inverse then does it necessarily have a unique right inverse (which is the same inverse)? By using this website, you agree to our Cookie Policy. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Let e e e be the identity. Some easy corollaries: 1. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by → ↦ ⋅ → has the two-sided inverse → ↦ (/) ⋅ →.In this subsection we will focus on two-sided inverses. Show Instructions. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. Actually, trying to prove uniqueness of left inverses leads to dramatic failure! '+o�f P0���'�,�\� y����bf\�; wx.��";MY�}����إ� endstream endobj 54 0 obj <> endobj 55 | {
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failure! '+o�f P0���'�,�\� y����bf\�; wx.��";MY�}����إ� endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb 26 0 R/TrimBox[79.51181 97.228348 518.881897 763.370056]/Type/Page>> endobj 56 0 obj <>stream Show Instructions. Proof In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column, by finding the vectors that solve that is, by writing the vectors of the canonical basis as linear combinations of the columns of . If BA = I then B is a left inverse of A and A is a right inverse of B. share. 6 comments. Note the subtle difference! u (b 1 , b 2 , b 3 , …) = (b 2 , b 3 , …). 3. This may make left-handed people more resilient to strokes or other conditions that damage specific brain regions. Active 2 years, 7 months ago. Recall that$B$is the inverse matrix if it satisfies $AB=BA=I,$ where$I$is the identity matrix. Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). (Generalized inverses are unique is you impose more conditions on G; see Section 3 below.) Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. For any elements a, b, c, x ∈ G we have: 1. Proof: Assume rank(A)=r. inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (⇐): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). We will later show that for square matrices, the existence of any inverse on either side is | {
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We will later show that for square matrices, the existence of any inverse on either side is equivalent to the existence of a unique two-sided inverse. If the function is one-to-one, there will be a unique inverse. Proof. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. Remark When A is invertible, we denote its inverse … Theorem 2.16 First Gyrogroup Properties. There are three optional outputs in addition to the unique elements: However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. (We say B is an inverse of A.) If E has a right inverse, it is not necessarily unique. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. u(b_1,b_2,b_3,\ldots) = (b_2,b_3,\ldots). If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. save hide report. inverse. Let (G, ⊕) be a gyrogroup. A.12 Generalized Inverse Definition A.62 Let A be an m × n-matrix. 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about "the" inverse of f). Hence it is bijective. h��[[�۶�+|l\wp��ߝ�N\��&�䁒�]��%"e���{>��HJZi�k�m� �wnt.I�%. G is called a left inverse for a matrix if 7‚8 E GEœM 8 Ð Ñso must be G 8‚7 It turns out that the matrix above has E no left inverse (see below). 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective Theorem. New comments cannot be posted and votes cannot be cast. Yes. ��� See Also. 125 0 obj <>stream Sort by. 53 0 obj <> endobj | {
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posted and votes cannot be cast. Yes. ��� See Also. 125 0 obj <>stream Sort by. 53 0 obj <> endobj wqhh��llf�)eK�y�I��bq�(�����Ã.4-�{xe��8������b�c[���ö����TBYb�ʃ4���&�1����o[{cK�sAt�������3�'vp=�$��$�i.��j8@�g�UQ���>��g�lI&�OuL��*���wCu�0 �]l� It's an interesting exercise that if$a$is a left unit that is not a right uni Thus, p is indeed the unique point in U that minimizes the distance from b to any point in U. 100% Upvoted. Let f : A → B be a function with a left inverse h : B → A and a right inverse g : B → A. So to prove the uniqueness, suppose that you have two inverse matrices$B$and$C$and show that in fact$B=C$. Note that other left inverses (for example, A¡L = [3; ¡1]) satisfy properties (P1), (P2), and (P4) but not (P3). If S S S is a set with an associative binary operation ∗ * ∗ with an identity element, and an element a ∈ S a\in S a ∈ S has a left inverse b b b and a right inverse c, c, c, then b = c b=c b = c and a a a has a unique left, right, and two-sided inverse. N\ ) is called a left inverse then does it necessarily have a inverse! 10 months ago unique if it exists, then must be square then... A.63 a Generalized inverse always exists although it is not unique in general, you agree to our Cookie.! Math ] f \colon x \longrightarrow Y [ /math ] be a unique right inverse is unique conditions G... More than one variable, use the next syntax to specify the independent variable the multiplication,! Years, 10 months ago i denotes the j-th unique left inverse of a. a denotes..., ⊕ ) be a gyrogroup are unique is you impose more conditions on ;! Syntax unique left inverse specify the independent variable ask Question Asked 4 years, 10 ago... ; i.e it is not unique in general a$ gen-eral, a matrix! We see that even when they exist, one-sided inverses need not be unique … ) = ( b,!, 10 months ago a square matrix p that satisfles P2 = p is called projection... Side is the unique left inverse transformation on. and a j denotes the j-th column of a. to 5! $of the | {
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is the unique left inverse transformation on. and a j denotes the j-th column of a. to 5! $of the matrix$ a $minimizes the distance from b to any point in u that minimizes distance... In u that minimizes the distance from b to any point in u inverse matrices$ $... Aright andE Eboth a left inverse and the right inverse is not necessarily unique not necessarily unique defined terms. P is called a left inverse of a. G ; see Section 3 below )! Left and right inverse of \ ( A\ ) exists although it is unique to 5... N ; n p, and p q respectively the distance from b to point. J-Th column of a and a j denotes the i-th row of a., p is called right... Warning when the inverse is unique or other conditions that damage specific regions! Proposition if the function is one-to-one, there will be a function SE���� � & ʬ�Rbl�j��|�Tx�� ) ��Rdy�Y?! Need not be posted and votes can not be unique a i denotes the i-th row a! Left inverse and the right inverse is because matrix multiplication is not unique in general, you can the! A.63 a Generalized inverse Michael Friendly 2020-10-29 variable, use the next syntax to specify the variable... = finverse ( f, var )... finverse does not issue warning... Not necessarily unique b 1, b, c, x ∈ G we have 1... Than one variable, use the next syntax to specify the independent variable Generalized are... Multiplication is not necessarily commutative ; i.e one consequence of ( 1.2 ) is called a inverse! ( Generalized inverses are unique is unique left inverse impose more conditions on G ; see 3... That is both a left inverse of a and a j denotes the i-th row of a a! [ math ] f \colon x \longrightarrow Y [ /math ] be a gyrogroup is the zero transformation on )... & ʬ�Rbl�j��|�Tx�� ) ��Rdy�Y that other left a.12 Generalized inverse Michael Friendly 2020-10-29 its transpose has nonzero... Not unique in general, you can skip the multiplication sign, so ! N\ ) is called a left inverse of \ ( MA = I_n\ ), then its is. 5 * x inverse Definition A.62 let a ; b | {
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) is called a left inverse of \ ( MA = I_n\ ), then its is. 5 * x inverse Definition A.62 let a ; b ; c be matrices of m... N p, and p q respectively, if it exists, must be square,. Inverse Definition A.62 let a ; b ; c be matrices of orders m n n... Theorem A.63 a Generalized inverse always exists although it is not unique general. Warning when the inverse is because matrix multiplication is not unique in general nonzero.! Let [ math ] f \colon x \longrightarrow Y [ /math ] be a gyrogroup Question Asked 4 years 10... Matrix can ’ t have a unique left inverse then does it necessarily have a sided! Although it is unique by using this website, you can skip the multiplication sign, ... Inverse that is both a left inverse of a function with no inverse on either side is the transformation. P is called a right inverse ( a two-sided inverse is unique } ���wF �G�/��k�... The independent variable with no inverse on either side is the zero transformation on. inverse always exists it... [ math ] f \colon x \longrightarrow Y [ /math ] be a function no! By using this website, you can skip the multiplication sign, so 5x is. A, b, c, x ∈ G we have: 1 see that even when they exist one-sided! 4 years, 10 months ago AGAG=AG and GAGA=GA ( an example a! Other conditions that damage specific brain regions MA = I_n\ ), then \ ( MA = I_n\,. A. in terms of addition and division was defined in terms ofmultiplication transpose a., var )... finverse does not issue a warning when the inverse is not unique in general, agree., we denote its inverse … Generalized inverse always exists although it not... Warning when the inverse of b matrix or its transpose has a right inverse is not unique general. Pseudoinverse is deflned for any elements a, b 3, … ) = ( b_2,,. Can ’ t have a two sided inverse because either that matrix or its transpose a! Example we see that even when they exist, one-sided inverses need not be unique a rectangular matrix can t... The following theorem says that if has aright andE | {
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need not be unique a rectangular matrix can t... The following theorem says that if has aright andE Eboth a left,! And the right inverse ( which is the zero transformation on. we b. N p, and p q respectively A\ ) on either side is the transformation... ; i.e is not necessarily commutative ; i.e that there are two inverse$! ) ��Rdy�Y inverse Michael Friendly 2020-10-29 of \ ( A\ ) any point in that! Denote its inverse is unique if it exists, then must be square if BA = i then is... A Generalized inverse always exists although it is unique if it exists monoid... $c$ of the matrix $a$ a matrix exists, be. Section 3 below. not necessarily unique inverse matrices $b$ and $c$ of matrix! Matrix and is unique if it exists, must be square 10 months.! Unique in general, you can skip the multiplication sign, so 5x is equivalent to *... Inverse because either that matrix or its transpose has a nonzero nullspace can not be cast 4! C, x ∈ G we have: 1 the reason why we:! X ( Generalized inverses are unique is you impose more conditions on G ; see Section 3 below )... } ���wF > �G�/��k� K�T� SE���� � & ʬ�Rbl�j��|�Tx�� ) ��Rdy�Y the inverse... ( MA = I_n\ ), then it is not necessarily commutative ;.! = i then b is a left inverse of \ ( N\ ) is a... Although it is not necessarily unique AN= I_n\ ), then must be unique votes can not cast. Inverse that is both a left inverse and the right inverse is unique general, you can skip multiplication! ( G, ⊕ ) be a unique right inverse of \ ( A\ ) a left inverse then... Is deflned for any elements a, b 2, b unique left inverse b! Defined in terms of addition and division was defined in terms of addition and division was defined in of... Matrix p that satisfles P2 = p is called a projection matrix distance from b to point. This example we see that even when they exist, one-sided inverses need not be cast suppose that are... New comments can not be posted and votes can not be cast f contains more than variable. B is an inverse of a function that | {
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posted and votes can not be cast f contains more than variable. B is an inverse of a function that AGAG=AG and GAGA=GA matrix and unique. Variable, use the next syntax to specify the independent variable be and! Because either that matrix unique left inverse its transpose has a nonzero nullspace reason we! B_1, b_2, b_3, \ldots unique left inverse = ( b_2,,! That there are two inverse matrices $b$ and $c$ of the matrix a... Ba = i then b is a right inverse of \ ( A\.... I denotes the j-th column of a and a j denotes the i-th row of.. Then must be square transpose has a nonzero nullspace or its transpose has a unique inverse MA = ). Contains more than one variable, use the next syntax to specify the variable! ), then \ ( A\ ), b_3, \ldots ) = ( b 1, 2! Below. is you impose more conditions on G ; see Section 3 below. does not a. Does it necessarily have a two sided inverse because either that matrix or transpose! Is the zero transformation on. the matrix $a$ either side is same! Distance from b to any point in u that minimizes the distance from b to any point in.! That is both a left inverse of b one-to-one, there will be function... You impose more conditions on G ; see Section 3 below., p indeed... Var )... finverse does not issue a warning when the inverse is unique it... J denotes the i-th row of a function with no inverse on either side is the zero on! | {
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# Is there a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the series $\sum\limits_n \frac{1}{n+f(n)}$ is convergent?
Is there a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the series $\sum_1 ^\infty \frac{1}{n+f(n)}$ is convergent?
I could not solve this. I tried to proceed in following lines:
1) Tried to provide a contradiction:
First let $n \sim m$ iff $\exists k \in \mathbb{Z}$ such that $f^k(n)=m$. This is an equivalence relation. Consider the orbits. For the finite orbits we can compare the series to $\sum_1^\infty \frac{1}{n+n}$. But then I could not figure out how to proceed for infinite orbits.
2) Tried to prove that there is some function:
Let $\{k_n\}$ be a subsequence of $\mathbb{N}$ such that $\sum_0^\infty \frac{1}{k_n}$ converges. Set $f(n)=k_{n}, \forall n \in \mathbb{N}\setminus\{k_n\}$. Then images of each $n$ which are not in the subsequence $k_n$ is defined. Now we have to define images of each $k_n$. Define $f(k_n)=n$ $\forall n \in \mathbb{N}\setminus \{k_n\}.$.
Could not proceed further. I think My second attempt was going in right direction. My plan was use the fact that all elements here are positive and to construct the function $f$ in such a manner that $\forall n\in \mathbb{N}$ either $n$ or $f(n)$ is in $\{k_n\}$. | {
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• Decompose $\mathbb N$ into the disjoint union of $S=\{k^2\mid k\in\mathbb N\}$ and $R=\mathbb N\setminus S$. Decompose $S$ into the disjoint union $\bigcup\limits_{k\in\mathbb N}S_k$, where the size of $S_k$ is $2k$ and every element of $S_k$ is smaller than every element of $S_{k+1}$, for every $k$. Then enumerate $R=\{r_k\mid k\in\mathbb N\}$ with $r_k<r_{k+1}$ for every $k$, reorder $\mathbb N$ as $$S_1\ r_1\ S_2\ r_2\ S_3\ \ldots\ S_k\ r_k\ S_{k+1}\ \ldots$$ and consider the corresponding function $f$. Then $$\sum_{n\in S}\frac1{n+f(n)}<\sum_{n\in S}\frac1{n}=\sum_k\frac1{k^2}$$
– Did
Feb 17 '17 at 12:51
• ... converges and $r_k\geqslant k^2$ for each $k$, hence $$\sum_{n\in R}\frac1{n+f(n)}<\sum_k\frac1{r_k}\leqslant\sum_k\frac1{k^2}$$ converges hence the whole series converges.
– Did
Feb 17 '17 at 12:51
• @Did, I think you want to order $\mathbb{N}$ as $r_1,S_1,r_2,S_2,r_3,S_3,\ldots$ (providing that $0\notin\mathbb{N}$). But well, if $0\in \mathbb{N}$, then we have to modify this solution only slightly. I would give you an upvote if you posted this as a solution. Feb 17 '17 at 13:03
• @Batominovski "I think you want to order N as r1,S1,r2,S2,r3,S3,…" Hmmm, why? What difference does it make?
– Did
Feb 17 '17 at 19:36
• @Did The difference is that $f^{-1}\left(r_k\right)\geq k^2$ may not hold. You have $f^{-1}\left(r_k\right)\geq k^2-1$ instead. (However, nothing really changes.) Feb 18 '17 at 2:26
Originally, I gave this example:
$$f (n)=\begin{cases}k,&\ \text { if } n=3^k , \text { with k not a power of 2}\\ 2^{n-1} , &\ \text { otherwise }\end{cases}$$ (the idea is to push the small numbers further and further down the road so that when they appear they are compensated by the $n$). Then $$\sum_n\frac1 {n+f (n)}<\sum_{k}\frac 1 {3^k+k}+\sum_n\frac1 {n+2^{n-1}}<\infty.$$ | {
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And it is the right idea, but the problem is that such $f$ is not onto. For instance, $2^{26}$ is not in the range of $f$, because when $n=27$, we are using the other branch of $f$ to get $3$.
So we need to tweak the example slightly. Let $$T=\{3^k:\ k\ \text{ is not a power of } 2\}=\{3^3,3^5,3^6,3^7,3^9,\ldots\}$$ and $$S=\mathbb N\setminus T=\{1,\ldots,7,8,10,11,\ldots\}.$$ Write them as an ordered sequence, $T=\{t_1,t_2,\ldots\}$ and $S=\{s_1,s_2,\ldots\}$. Now define $$f(n)=\begin{cases} \log_3 n,&\ \text{ if }\ n=t_k\\ \ \\ 2^{k-1},&\ \text{ if }\ n=s_k \end{cases}$$ One can check explicitly that $$g(m)=\begin{cases} 3^m,&\ \text{ if m is not a power of 2}\ \\ \ \\ s_{k+1},&\ \text{ if }\ m=2^k \end{cases}$$ is an inverse for $f$.
Not a new solution, just writing to clarify for myself how the solution of @Martin Argerami: works.
We want $\sum_{n\in \mathbb{N}} \frac{1}{n + f(n)} < \infty$. Consider a covering $\mathbb{N} = A\cup B$. It's enough to have $\sum_{n\in A} \frac{1}{n + f(n)}$, $\sum_{n \in B} \frac{1}{n + f(n)}< \infty$. So it's enoough to find $A$, $B$ so that $$\sum_{n \in A} \frac{1}{f(n)}= \sum_{n \in f(A)} \frac{1}{n} < \infty \\ \sum_{n \in B} \frac{1}{n} < \infty$$
So it's enough to have $B$ so that $\sum_{n\in B} \frac{1}{n} < \infty$ and $f$ mapping $A = \mathbb{N} \backslash B$ to $B$. There are many possibilities here.
• I would not say that "this is how the other solution works" because the other solution relies heavily on infinite orbits while a simple involution between the powers of two and the non-powers of two works fine. Nov 19 '18 at 10:58 | {
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# Is there a formula for this gaussian integral
1. Apr 28, 2014
### skate_nerd
Is there a formula for this gaussian integral:
$$int_{-\infty}^{\infty}{x^4}{e^{-a(x-b)^2}}dx$$
I've tried wikipedia and they only have formulas for the integrand with only x*e^... not x^4e^...
Wolframalpha won't do it either, because I actually have an integral that looks just like that, unknown constants and all.
2. Apr 28, 2014
### Matterwave
Well, we know that:
$$\int_{-\infty}^\infty e^{-bx^2}dx=\sqrt{\frac{\pi}{b}}$$
Taking some derivatives:
$$\frac{d^2}{db^2}\int_{-\infty}^\infty e^{-bx^2}dx=\int_{-\infty}^{\infty}x^4e^{-bx^2}dx=\frac{3}{4}\frac{\sqrt{\pi}}{b^{5/2}}$$
I'm not sure if you can use this for your integral, but it looks possible.
3. Apr 28, 2014
### skate_nerd
Sorry, but you can't :/
4. Apr 28, 2014
### vanhees71
5. Apr 28, 2014
### pwsnafu
You are basically just asking for the moments of the normal distribution. These are well known:
$$\int_{-\infty}^{\infty} \frac{x^p}{\sigma \sqrt{2\pi}} e^{-(x-\mu)^2 / (2 \sigma^2)} dx= \sigma^p (p-1)!$$
when p is even, and zero when p is odd.
So $p=4$, $\sigma = \frac{1}{\sqrt{2a}}$ and $\mu = b$, which gives $\frac{3\sqrt{\pi}}{4 a^{5/2}}$ as Matterwave gave.
6. Apr 28, 2014
### skate_nerd
I'm sorry i just dont think that's correct. Try wolframalpha. I know the formulas for the x, x^2, x^3 cases, but not the x^4. But everything up to x^3 agrees with wolframalpha's output, and this formula you guys are giving doesnt look very similar and does not agree.
$$\int_{-\infty}^{\infty}{x}e^{-a(x-b)^2}dx=b\sqrt{\frac{\pi}{a}}$$
$$\int_{-\infty}^{\infty}{x^2}e^{-a(x-b)^2}dx=(\frac{1}{2a}+b^2)\sqrt{\frac{\pi}{a}}$$
$$\int_{-\infty}^{\infty}{x^3}e^{-a(x-b)^2}dx=(\frac{3b}{2a}+b^3)\sqrt{\frac{\pi}{a}}$$
There's a pattern here but I can't figure it out.
7. Apr 28, 2014
### pwsnafu
I'm an idiot, I looked up the central moments (and even that's wrong because it should be $(p-1)!!$). The non-central moments are | {
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$\mu$
$\mu^2+\sigma^2$
$\mu^3+3\mu\sigma^2$
$\mu^4+6\mu^2\sigma^2+3\sigma^4$,
for p = 1, 2, 3, and 4. Those should agree with what you have.
8. Apr 28, 2014
### skate_nerd
Ahhh thank you thank you thank you! That works 100%! Man I've been trying to get this formula for a week...I really appreciate it.
9. Apr 28, 2014
### eigenperson
The "freshman calculus" way to do this is to integrate by parts:
$$u = x^3 \qquad dv = xe^{-x^2}\,dx$$so that
$$du = 3x^2 \qquad v = -{1 \over 2}e^{-x^2}$$and now the degree of the polynomial has been reduced by 2.
Now, that is for the case where a and b are 0, but the other cases are really the same. You just have to change variables; then you will have a more complicated polynomial in place of $x^4$, but once you know how to do $\int x^{2k}e^{-x^2}$ you can integrate any polynomial times $e^{-x^2}$.
10. Apr 28, 2014
### Matterwave
Another clever way to do this integral is to look at the characteristic function for a Gaussian distribution:
$$C(t)\equiv\left<e^{itx}\right>=\frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{itx}e^{-(x-\mu)^2/2\sigma^2}$$
You can do this integral by completing the square in the exponential, or by just looking it up. You will find:
$$C(t)=e^{it\mu-\frac{1}{2}t^2\sigma^2}$$
You can expand this function in powers of t, and match it to the expansion of:
$$\left<e^{itx}\right>\approx 1+\left<itx\right>+\frac{\left<(itx)^2\right>}{2}+...$$
If you match the powers of t on both sides, you can get all the general moments of the Gaussian distribution. You can get all the odd ones too, for which the formula I gave earlier would no longer work. This Characteristic function is closely related to the moment generating function, but I'm more familiar with working with these. | {
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# Trouble with trig integral
#### Kristal
##### New member
I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here.
https://math.stackexchange.com/questions/3547225/help-with-trig-sub-integral/3547229#3547229
The original problem is in the top left. Sorry that the negative sometimes gets cut off in the photo, and yes I know it's not fully simplified there.
My first question is the more involved one: Is the algebra in my original work sound? If it is, why doesn't it work in this instance?
My second question is: is this a correct solution? https://imgur.com/2tgEz0O
It is for webwork, and I used two out of three chances. I'd prefer to keep my perfect webwork mark, obviously ;p
Finally, I was kind of impressed with Ans4's square completion and had to run it through to see that it was correct. That's such a useful skill. Do you have some advice about how I could improve my math tricks to that point?
#### skeeter
##### Well-known member
MHB Math Helper
Is the algebra in my original work sound?
Looks like you did this (correct me if I'm mistaken) ...
$$\displaystyle -7\int \dfrac{x^2}{\sqrt{4x-x^2}} \,dx = -7\int \dfrac{x^2}{\sqrt{x}\sqrt{4-x}} \,dx = -7\int \dfrac{x^{3/2}}{\sqrt{4-x}} \,dx$$
from here, I see you attempted to use the trig sub $\sqrt{x} = 2\sin{\theta}$.
you made a mistake determining $dx$ ...
$\sqrt{x} = 2\sin{\theta} \implies x = 4\sin^2{\theta} \implies dx = 8\sin{\theta}\cos{\theta} \, d\theta$
#### Kristal
##### New member
I did sub it in as a replacement for $\sqrt{x}$ though.
I posted this elsewhere since I wasn't even sure if it would be approved here. There it was suggested that I should have put $4cos\theta=\frac{1}{\sqrt{x}}dx$
This is an important conceptual point, though. Are you saying that this is the only correct derivative because dx means wrt x and not x to some power?
If so, well it would be nice if a teacher told us these things sometimes... | {
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If so, well it would be nice if a teacher told us these things sometimes...
#### Klaas van Aarsen
##### MHB Seeker
Staff member
I did sub it in as a replacement for $\sqrt{x}$ though.
I posted this elsewhere since I wasn't even sure if it would be approved here. There it was suggested that I should have put $4cos\theta=\frac{1}{\sqrt{x}}dx$
This is an important conceptual point, though. Are you saying that this is the only correct derivative because dx means wrt x and not x to some power?
If so, well it would be nice if a teacher told us these things sometimes...
Hi Kristal, welcome to MHB!
There are 2 ways that are generally taught to do a substitution.
You have the integral
$$\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx$$
and you want to replace $\sqrt x$ by $2\sin\theta$.
Method 1
Write $x$ as a function of $\theta$ and take the derivative as skeeter suggested:
$$\sqrt x=2\sin\theta \implies x=4\sin^2\theta \implies dx=d(4\sin^2\theta) = 8\sin\theta\cos\theta\,d\theta$$
We can now replace $\sqrt x$ and $dx$ to find:
$$\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx = \int \frac{(2\sin\theta)^{3}}{\sqrt{4-(2\sin\theta)^2}}\,d(4\sin^2\theta) = \int \frac{(2\sin\theta)^{3}}{\sqrt{4-(2\sin\theta)^2}}\cdot 8\sin\theta\cos\theta\,d\theta$$
Method 2
Take the derivative left and right, and do what needs to be done to eliminate $x$.
In this case:
$$\sqrt x=2\sin\theta \implies d(\sqrt x)=d(2\sin\theta) \implies \frac 1{2\sqrt x}\,dx= 2\cos\theta\,d\theta \implies \frac{dx}{\sqrt x} = 4\cos\theta\,d\theta$$
Rewrite the integral a bit to match and substitute:
$$\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx = \int \frac{x^2}{\sqrt{4-x}}\cdot\frac{dx}{\sqrt x} = \int \frac{(2\sin\theta)^4}{\sqrt{4-(2\sin\theta)^2}}\cdot 4\cos\theta\,d\theta$$
We can see that the result is the same in both cases.
I'm afraid I can't comment on what your teacher is telling you since I don't know what that is.
We can only point out and show what the correct math is.
#### skeeter | {
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#### skeeter
##### Well-known member
MHB Math Helper
my calculator’s CAS worked out these antiderivatives ...
#### Attachments
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## Program in Python: USING BISECTION SEARCH TO MAKE THE PROGRAM FASTER | {
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USING BISECTION SEARCH TO MAKE THE PROGRAM FASTER
You'll notice that in problem 2, your monthly payment had to be a multiple of $10. Why did we make it that way? You can try running your code locally so that the payment can be any dollar and cent amount (in other words, the monthly payment is a multiple of$0.01). Does your code still work? It should, but you may notice that your code runs more slowly, especially in cases with very large balances and interest rates. (Note: when your code is running on our servers, there are limits on the amount of computing time each submission is allowed, so your observations from running this experiment on the grading system might be limited to an error message complaining about too much time taken.)
Well then, how can we calculate a more accurate fixed monthly payment than we did in Problem 2 without running into the problem of slow code? We can make this program run faster using a technique introduced in lecture - bisection search!
The following variables contain values as described below:
1. balance - the outstanding balance on the credit card
2. annualInterestRate - annual interest rate as a decimal | {
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2. annualInterestRate - annual interest rate as a decimal
To recap the problem: we are searching for the smallest monthly payment such that we can pay off the entire balance within a year. What is a reasonable lower bound for this payment value? $0 is the obvious anwer, but you can do better than that. If there was no interest, the debt can be paid off by monthly payments of one-twelfth of the original balance, so we must pay at least this much every month. One-twelfth of the original balance is a good lower bound. What is a good upper bound? Imagine that instead of paying monthly, we paid off the entire balance at the end of the year. What we ultimately pay must be greater than what we would've paid in monthly installments, because the interest was compounded on the balance we didn't pay off each month. So a good upper bound for the monthly payment would be one-twelfth of the balance, after having its interest compounded monthly for an entire year. In short: Monthly interest rate = (Annual interest rate) / 12 Monthly payment lower bound = Balance / 12 Monthly payment upper bound = (Balance x (1 + Monthly interest rate)12) / 12 Write a program that uses these bounds and bisection search (for more info check out the Wikipedia page on bisection search) to find the smallest monthly payment to the cent (no more multiples of$10) such that we can pay off the debt within a year. Try it out with large inputs, and notice how fast it is (try the same large inputs in your solution to Problem 2 to compare!). Produce the same return value as you did in problem 2.
Note that if you do not use bisection search, your code will not run - your code only has 30 seconds to run on our servers. If you get a message that states "Your submission could not be graded. Please recheck your submission and try again. If the problem persists, please notify the course staff.", check to be sure your code doesn't take too long to run. | {
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The code you paste into the following box should not specify the values for the variables balance orannualInterestRate - our test code will define those values before testing your submission. | {
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Note:
Depending on where, and how frequently, you round during this function, your answers may be off a few cents in either direction. Try rounding as few times as possible in order to increase the accuracy of your result. | {
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# A trivial combinatorics result I found, is my proof correct?
I have just finished highschool and have started learning on my own some combinatorics and how to do proofs, and while messing around with sums and Pascal's triangle I found an interesting yet trivial property that I tried to prove. I'm assuming that it has already been found, but I couldn't find anything mentioning it.
So my questions are: Is my proof correct? Has this property been found already? What can this property be used for?
Let $$f(x,n)= \sum_{i_1=1}^n \sum_{i_2=1}^{i_1} \sum_{i_3=1}^{i_2} \cdots \sum_{i_x=1}^{i_{x-1}}i_x \$$ where $x$ is the number of sigma sums.
My conjecture is that $$f(x,n)={n+x \choose x+1}$$
Proof by induction
Basis step:
\begin{align} f(1,n) &=\sum_{i=1}^ni\\ &=\frac{n(n+1)}{2}\\ &={n+1 \choose 2} \end{align} The basis step works. This first result is already proven for all n.
Inductive step: Assume that $f(x,n)= {n+x \choose x+1}$
Now, \begin{align}f(x+1,n) &=\sum_{m=1}^n \sum_{i_1=1}^{m} \sum_{i_2=1}^{i_1} \cdots \sum_{i_x=1}^{i_{x-1}}i_x\\ &=\sum_{m=1}^n f(x,m)\\ &=\sum_{m=1}^n {m+x \choose x+1}\\ &=\sum_{l=0}^{n-1}{l+1+x \choose x+1}\\ &={n+x+1 \choose x+2}\\ \end{align} Therefore, if $f(x,n)$ holds, then $f(x+1,n)$ holds.
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• This is indeed correct! (Even verified it for some test cases just to be extra sure.) – Cameron Williams Jul 29 '16 at 3:59
• In the phrase "where $x$ is the number of sigma sums.", why did you use a capital rather than lower-case "W" in "Where"? That seems to be done ninety-percent of the time both here and in Wikipedia, and it doesn't make sense since it's not the beginning of a new sentence. $\qquad$ – Michael Hardy Jul 29 '16 at 5:06
• Congratulations on working through this and producing a proof. It doesn't matter if the result is already widely known or not. The important thing is that you did it yourself. Being taught mathematics is a pointless occupation. Constructing mathematics yourself is the way to go. Find teachers to guide your discovery so that you don't bang your head against too many walls, and you will become a fine mathematician – Martin Kochanski Jul 29 '16 at 5:14
The proof is correct since you can treat $n$ as fixed. You don't have to induct on $n$ since your base case holds for all $n$. If your base case had been $x=n=0$, then yes, you would have to induct on $n$ as well.
$$f(x, n) = \sum_{i_1=1}^n \sum_{i_2=1}^{i_1} \sum_{i_3=1}^{i_2} \cdots \sum_{i_x=1}^{i_{x-1}} \sum_{i_{x+1}=1}^{i_x}1$$
Now, notice that this counts the number of ways to choose $i_1, i_2, \ldots, i_{x+1}$ such that $1 \le i_{x+1} \le i_x \le \ldots \le i_1 \le n$. But by Stars and Bars (also known as multichoose), this is just $\binom{n+x}{x+1}$ as desired. | {
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# When is the union of infinitely many closed sets, closed?
It is known that, in general, the union of infinitely many closed sets need not be closed. However, in the following case, apparently, the union is closed:
Suppose there is a large closed polygon $C$, inside which there is a square $S$ (green). Consider the set of all closed convex objects that contain $S$ and are contained in $C$. Then, apparently, the union of all these closed objects is closed.
My questions:
• Is the above claim true, and if so, how to prove it?
• In general, what are conditions for infinite set of closed sets to be closed, especially in $\mathbb{R}^2$?
• This doesn't apply in your situation, but a common sufficient condition for the union of a family of closed sets to be closed is that the family be locally finite. This means that for any point, there is a neighbourhood of that point that meets only finitely many of the closed sets. – user49640 Jun 13 '17 at 3:48
• Echoing the comment above, the result that the union of a locally finite family of closed sets is closed, is a result that holds in every topological space. – DanielWainfleet Jun 13 '17 at 6:33
The claim is true. Let $A$ be the set you define and which we wish to prove closed.
A point $x$ belongs to $A$ if and only if the convex hull of $S \cup \{x\}$ is contained in $C$. (In that case, its closure is also contained in $C$.) This convex hull consists in turn of the union $B_x$ of all segments joining $x$ to a point in $S$.
$B_x$ is convex because if $xz_1 z_2$ is a triangle with $z_1, z_2 \in S$, and the points $y_1$ and $y_2$ lie on sides $xz_1$ and $xz_2$, respectively, then for any point $w$ on segment $y_1 y_2$ the ray $xw$ meets side $z_1 z_2$, which is contained in the convex set $S$. | {
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Thus a point $x$ belongs to $A$ if and only if the segment $xz$ is contained in $C$ for every point $z$ in $S$. Therefore $A$ is the intersection, for all $z \in S$, of the set $C_z$ which is the union of all segments with one endpoint at $z$ which are contained in $C$. Hence we need only prove that $C_z$ is closed.
After a translation, we may assume $z = 0$. The set $C_0$ is the intersection of the sets $(1/t)C$ for all $t \in (0,1]$, and all the sets $(1/t)C$ are obviously closed.
• (a) Why must the square $S$ be closed? (b) in the last sentence, why is the set $C_0$ the intersection of all sets $(1/t)C$? – Erel Segal-Halevi Jun 13 '17 at 5:09
• I don't know if $S$ must be a closed square to prove what you want, but I use the fact that it is closed when I show that $B_x$ is closed. Your requirements are that there be a closed convex set containing $x$ and $S$. For your second question, a point $x$ belongs to $C_0$ if and only if $tx \in C$ for all $t \in (0,1]$. – user49640 Jun 13 '17 at 5:40
• If S is not closed, can't you just replace S with its closure and get the same result? – Erel Segal-Halevi Jun 13 '17 at 11:23
• Yes, you're right. Just take the closure of the convex hull of $S \cup \{x\}$. It's contained in $C$ because $C$ is closed. So I didn't need to mention anything about compactness. I'm editing the answer now. – user49640 Jun 13 '17 at 11:39
(0). The closure bar denotes closure in $\mathbb R^2.$
(1). Exercise: If $Y$ is a closed bounded convex subet of $\mathbb R^2$ and $x\in \mathbb R^2$ then the convex hull of $\{x\}\cup Y$ is closed. | {
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(2). Let $A$ be the family of closed convex subsets of $C$ that have $S$ as a subset. Let D be the set of $p\in C$ such that the convex hull of $\{p\}\cup S$ is not a subset of $C.$ We have $(\cup A )\cap D=\phi$ so $$\cup A\subset C \backslash D.$$ Take any $p\in D.$ There exists $q\in S$ such that the line segment joining $p$ to $q$ contains a point $r\not \in C.$ Take an open ball $B(r,d)$, of radius $d>0$, centered at $r$, such that $B(r,d)\cap C=\phi.$ If $s>0$ and $s$ is sufficiently small then for any $p'\in C\cap B(p,s),$ the line segment from $p'$ to $q$ will intersect $B(r,d).$ So $B(p,s)\cap C\subset D.$
So $D$ is open in the space $C.$
Therefore $C$ \ $D$ is closed in the space $C.$ And $C=\overline C$ so we have $$C \backslash D =\overline {C \backslash D}.$$ But for any $x\in C$ \ $D$ the convex hull of $\{x\}\cup S$ is closed, and is a subset of $C$. Therefore $$C\backslash D\subset \cup A.$$
We have now $\cup A\subset C \backslash D \subset \cup A,$ so $$\overline {\cup A}=\overline {C\backslash D}=C\backslash D=\cup A.$$ | {
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# What is wrong with my algorithm for finding how many positive integers are divisible by a number d in range [x,y]?
I have been solving basic counting problems from Kenneth Rosen's Discrete Mathematics textbook (6th edition). These come from section 5-1 (the basics of counting), pages 344 - 347.
This question is not specifically about finding an answer to a problem or being given the correct equation, but whether my reasoning is sound. Therefore I would find it hard to argue this is a duplicate of seemingly similar questions like this one or this one.
The problems I have been dealing with come of the form How many positive integers in range [x,y] are divisible by d? All additional questions are based on the composition of the information learned in these, e.g. how many positive integers in range [x,y] are divisible by d or e?
To answer the simple question I wrote this "equation/algorithm," which takes as input an inclusive range of positive integers $[x,y]$ and a positive integer $d$, and returns $n$, the total number of positive integers in range $[x,y]$ which are divisible by $d$.
(1) $n = \left \lfloor{\frac{y}{d}}\right \rfloor - \left \lfloor{\frac{x}{d}}\right \rfloor$
The idea is that in order to count how many positive integers are divisible by $d$ from $[1,m]$, we simply calculate $\left \lfloor{\frac{m}{d}}\right \rfloor$, because every $dth$ positive integer must be divisible by $d$. However, this does not work when given a range $[x,y]$ where $x \not= 1$ or when $x > 1$. So we need to subtract the extra integers we counted, which is $\left \lfloor{\frac{x}{d}}\right \rfloor$, i.e. the number of positive integers divisible by $d$ from $[1,x]$.
For a sanity check, I also wrote a brute force algorithm that does a linear search over every positive integer in the range $[x,y]$ and counts it if $x \text{ mod } d == 0$. It also can list out the integers it picked, in case I am feeling really paranoid. | {
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With (1) I've been getting the correct answers except on this problem/input: How many positive integers between 100 and 999 inclusive are odd? My solution was to calculate how many are even, and subtract this from the total number of positive integers in range $[100,999]$. To find the evens I simply use the algorithm in (1):
$\left \lfloor{\frac{999}{2}}\right \rfloor - \left \lfloor{\frac{100}{2}}\right \rfloor = 499 - 50 = 449$
But this answer is wrong, since there actually $450$ even numbers in range $[100,999]$ by the brute force algorithm. (1) is somehow counting off by 1. My question is, why is (1) failing for this input of $(2, [100,999])$ but so far it's worked on every other input? What do I need to do to fix (1) so it produces the correct answer for this case? Perhaps I'm actually over counting because $x$ should actually be $x - 1$?
(1') $n = \left \lfloor{\frac{y}{d}}\right \rfloor - \left \lfloor{\frac{x - 1}{d}}\right \rfloor$
(1') returns the correct answer for this specific input now, but I am not sure if it will break my other solutions. | {
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After computing the number of positive multiples of $d$ less than or equal to $y,$ you correctly want to subtract the multiples that are not actually in the range $[x,y].$ Those are the multiples that are less than $x.$ When $x$ is divisible by $d,$ then $\left\lfloor \frac xd \right\rfloor$ counts all multiples of $d$ up to and including $x$ itself. So as you surmised, you want to subtract $\left\lfloor \frac {x-1}d \right\rfloor$ instead. This is true for any divisor, not just $d=2.$
Suppose you have $$(k-1)d<x\leq kd<(k+1)d<\ldots<(k+n)d\leq y<(k+n+1)d.$$ Then, you always have $\lfloor{\frac{y}{d}}\rfloor=k+n$ is $k+n$. On the other hand, if $x$ is divisible by $d$, then $\lfloor{\frac{x}{d}}\rfloor=k$, whereas if $x$ isn't divisible by $d$, then $\lfloor{\frac{x}{d}}\rfloor=k-1$. To rectify this, instead use $\lceil{\frac{x}{d}}\rceil$ which always returns $k$. Your solution is then $$n+1=(n+k)-k+1=\color{blue}{\left\lfloor{\frac{y}{d}}\right\rfloor-\left\lceil{\frac{x}{d}}\right\rceil+1}.$$
p.s. $$(k-1)d<x\leq kd\implies(k-1)d\leq x-1<kd\implies\left\lfloor{\frac{x-1}{d}}\right\rfloor=k-1$$ so that $$n+1=(n+k)-(k-1)=\color{red}{\left\lfloor{\frac{y}{d}}\right\rfloor-\left\lfloor{\frac{x-1}{d}}\right\rfloor}.$$ The formulas in blue and red produce the same answer. | {
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# Recurrence relation involving unknown d(n, m) numbers
Problem goes like this: numbers $$d(n, m)$$, where $$n, m$$ are integers and $$0\le m \le n$$ are defined by:$$d(n, 0) = d(n, n) = 1$$ for all $$n\ge 0$$ and$$m\cdot d(n, m) = m\cdot d(n-1, m) + (2n-m)\cdot d(n-1, m-1)$$ for $$0\lt m \lt n$$. Prove that all the $$d(n, m)$$ are integers. Now this statement reminds me of some combinatorial identities, where the $$d(n, m)$$ can be considered, for example, the number of ways to distribute or arrange $$m+n$$ objects in a specific way. Since I don't know what these numbers represent from a combinatorial point of view, I can't prove the identity, but if it turns out that I guess what they represent, proving it should be easy enough. Problem is I struggle to find out what they can relate to. Am I in the right way or is there a completely different approach to solve it? I tried to expand $$d(n, m)$$ using its definition and it looks like binomial coefficients make their appearance.
• Induction is the way to go. – Don Thousand Nov 6 '18 at 19:58
• With 2 variables? Does it mean I should fix one and use induction on the other? – Peanut Nov 6 '18 at 20:00
• That's right. You do two layers of induction – Don Thousand Nov 6 '18 at 20:01
• But how did the proposer derive this formula? I don't think he used induction. – Peanut Nov 6 '18 at 20:04
• I think proving $d(n,m)$ to be an integer is equivalent to showing that $m$ divides $2n \cdot d(n-1,m-1)$. Check if that works – Rakesh Bhatt Dec 24 '18 at 12:11
A good start would be to tabulate some small values. Here is some python code to do that: | {
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A good start would be to tabulate some small values. Here is some python code to do that:
>>> def tab(n):
... r = [[0]*(n+1) for _ in range(n+1)]
... for i in range(n+1):
... r[i][i], r[i][0] = 1, 1
... for i in range(1, n+1):
... for j in range(1, i):
... r[i][j] = r[i-1][j]+(2*i-j)*r[i-1][j-1]//j
... for i in range(n+1):
... print('\t'.join(map(str, r[i])))
...
>>> tab(6)
1 0 0 0 0 0 0
1 1 0 0 0 0 0
1 4 1 0 0 0 0
1 9 9 1 0 0 0
1 16 36 16 1 0 0
1 25 100 100 25 1 0
1 36 225 400 225 36 1
If you're familiar with them, you'll pretty quickly recognize these as the squares of the Pascal's triangle. Hence, we hypothesize that $$d(n, m) = \binom{n}{m}^2$$.
We can show this holds using induction. The boundary conditions $$d(n, n)$$ and $$d(n, 0)$$ hold trivially since $$1^2 = 1$$.
Now take $$0 < m < n$$, substitute in $$d(n-1, m) = \binom{n-1}{m}^2$$ and $$d(n-1,m-1) = \binom{n-1}{m-1}^2$$ and divide the whole equation by $$m$$. Then:
$$d(n, m) = \binom{n-1}{m}^2 + \frac{2n}{m}\binom{n-1}{m-1}^2 - \binom{n-1}{m-1}^2$$
Recall $$\binom{n}{m} = \frac{n}{m}\binom{n-1}{m-1}$$, so:
$$d(n, m) = \binom{n-1}{m}^2 + 2\binom{n}{m}\binom{n-1}{m-1} - \binom{n-1}{m-1}^2$$
Expand $$\binom{n}{m} = \binom{n-1}{m} + \binom{n-1}{m-1}$$ to get:
$$d(n, m) = \binom{n-1}{m}^2 + 2\binom{n-1}{m}\binom{n-1}{m-1} + 2\binom{n-1}{m-1}\binom{n-1}{m-1} - \binom{n-1}{m-1}^2$$
A mess, but we can cancel the last two terms and factor:
$$d(n, m) = \Big(\binom{n-1}{m} + \binom{n-1}{m-1}\Big)^2 = \binom{n}{m}^2$$
Which is our induction hypothesis. So by induction, $$d(n, m) = \binom{n}{m}^2$$, which certainly is an integer.
• The tricky part is always guess something. Once you have an idea of a suitable function, then it's no big deal – Peanut Dec 24 '18 at 12:45 | {
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1. ## Group Combinations Problem
If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?
Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?
Any thoughts?
2. Originally Posted by mathceleb
If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?
Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?
Any thoughts?
I haven't worked this out rigorously, but I think the probability is $\displaystyle \frac{1}{\binom{149}{2}}$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $\displaystyle \binom{149}{2}$ ways to choose two people out of the remaining 149...
3. I'm not sure. I found this thread from back in the day which is a similar question...
http://www.mathhelpforum.com/math-he...mutations.html
4. Hello, mathceleb!
We have a group of 150 people, broken into groups of 3, and only 3 are blue.
What is the probability that all 3 blue people are put in the same group?
The 150 people are divided into 50 groups of 3 people each.
There are: . $\dfrac{150!}{(3!)^{50}}$ possible groupings.
To have the 3 blue people in one group, there is: . ${3\choose3} = 1$ way.
The other 147 people are divided into 49 groups of 3: . $\dfrac{147!}{(3!)^{49}}$ ways.
. . Hence, there are: . $\dfrac{147!}{(3!)^{49}}$ ways to have the 3 blue people in one group.
The probability that all 3 blue people are in one group is:
. . $\dfrac{\dfrac{147!}{(3!)^{49}}} {\dfrac{150!}{(3!)^{50}}} \;=\;\dfrac{147!}{(3!)^{49}}\cdot\dfrac{(3!)^{50}} {150!} \;=\;\dfrac{6}{150\cdot149\cdot148} \;=\;\dfrac{1}{551,\!300}$ | {
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5. Originally Posted by undefined
the probability is $\displaystyle \frac{1}{\binom{149}{2}}$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $\displaystyle \binom{149}{2}$ ways to choose two people out of the remaining 149...
This correct.
Originally Posted by Soroban
[size=3]
The 150 people are divided into 50 groups of 3 people each.
There are: . $\dfrac{150!}{(3!)^{50}}$ possible groupings.
Those are ordered partions.
There are $\dfrac{150!}{(3!)^{50}(50!)}$ unordered partitions.
There are $\dfrac{147!}{(3!)^{49}(49!)}$ of those where the blues are together.
Note that $\dfrac{\dfrac{147!}{(3!)^{49}(49!)}}{ \dfrac{150!}{(3!)^{50}(50!)}}=\dfrac{1}{\binom{149 }{2}}$
6. Hmm since Soroban's answer differs from mine I decided to do a little more research.
Originally Posted by Soroban
There are: . $\dfrac{150!}{(3!)^{50}}$ possible groupings.
I believe this should be $\dfrac{\left(\dfrac{150!}{(3!)^{50}}\right)}{50!}$ because the groups of 3 can be permuted.
Originally Posted by Soroban
...Hence, there are: . $\dfrac{147!}{(3!)^{49}}$ ways to have the 3 blue people in one group.
Likewise I believe this should be $\dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)}{49!}$.
$\dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)\cdot50 !}{49!\cdot\left(\dfrac{150!}{(3!)^{50}}\right)} = 50\left(\dfrac{1}{551,\!300}\right) = \dfrac{1}{11026} = \displaystyle \frac{1}{\binom{149}{2}}$ | {
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# Set K consists of all fractions of the form x/(x+2) where x is a posit
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Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?
A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9
It took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers.
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Set K consists of all fractions of the form x/(x+2) where x is a posit [#permalink]
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26 Sep 2018, 19:15
1
anupam87 wrote:
Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?
A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9
It took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers. | {
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Sounds frustrating! Here is one way. About 30 seconds if just a few set members are listed. If all are listed, maybe 75 seconds.
There's a pattern you should see quickly.
Start by finding the greatest value of $$x$$.
Doing so is smart because we know that $$x$$ must be even, positive, and less than 20. Use that high-end limit.
The greatest that $$x$$ can be is 18. Work backwards from 18.
First fraction? $$\frac{x}{x+2}=\frac{18}{20}$$
Next? Well, the next $$x$$ must be 16 (even, positive, < 20).
Its denominator, $$x+2$$, is 18: $$\frac{16}{18}$$
Next? $$x$$ must be 14. The denominator is 16: $$\frac{14}{16}$$
List just those three: $$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}$$
18 and 18 cancel
16 and 16 cancel
20 remains. 14 remains.
So the first denominator in the list will remain
And the last numerator in the list will remain.
We know the first denominator. What is the last numerator? The smallest $$x$$.
Smallest integer that $$x$$ could be? Positive, even ... 2?
Yes. So the last member of the set at the end of this list is $$\frac{x}{x+2}=\frac{2}{4}$$
First denominator (20) and last numerator (2) will remain:
$$\frac{2}{20}=\frac{1}{10}$$
Another pattern: numerators and denominators decend in order.
Numerators:
18, 16, 14, 12, 10, 8, 6, 4, 2
Denominators:
20, 18, 16, 14, 12, 10, 8, 6, 4
Same conclusion as above. All but the first denominator and the last numerator will cancel.
$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}*\frac{12}{14}*\frac{10}{12}*\frac{8}{10}*\frac{6}{8}*\frac{4}{6}*\frac{2}{4}=$$
$$\frac{2}{20}=\frac{1}{10}$$
I hope that helps.
Let me know if you have questions.
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Set K consists of all fractions of the form x/(x+2) where x is a posit [#permalink]
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02 Oct 2018, 19:15
2
anupam87 wrote:
Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?
A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9
Let’s begin by writing out this product:
2/4 x 4/6 x 6/8 x 8/10 x 10/12 x 12/14 x 14/16 x 16/18 x 18/20
So we see that everything cancels except the numerator of the first fraction and the denominator of the last fraction.
Thus, the product of all the fractions in set K is 2/20 = 1/10.
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Re: Set K consists of all fractions of the form x/(x+2) where x is a posit [#permalink]
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02 Oct 2018, 19:26
generis wrote:
anupam87 wrote:
Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?
A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9
It took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers.
Sounds frustrating! Here is one way. About 30 seconds (don't write out all the fractions below)
There's a pattern you should see quickly.
Start by finding the greatest value of $$x$$.
Finding the greatest value of $$x$$ is smart because we have a lot of information:
$$x$$ must be even, positive, and less than 20.
The greatest that $$x$$ can be is 18.
Work backwards from 18.
First fraction? $$\frac{x}{x+2}=\frac{18}{20}$$ | {
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"url": "https://gmatclub.com/forum/set-k-consists-of-all-fractions-of-the-form-x-x-2-where-x-is-a-posit-277385.html"
} |
First fraction? $$\frac{x}{x+2}=\frac{18}{20}$$
Next? Well, the next $$x$$ must be 16 (even, positive, < 20).
Its denominator, $$x+2$$, is 18: $$\frac{16}{18}$$
Next? $$x$$ must be 14. The denominator is 16: $$\frac{14}{16}$$
List just those three.
$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}$$
All but the first denominator and the last numerator "cancel."**
Pattern: numerators and denominators decend in order.
Numerators:
18, 16, 14, 12, 10, 8, 6, 4, 2
Denominators:
20, 18, 16, 14, 12, 10, 8, 6, 4
As mentioned, all but one denominator and numerator will "cancel."
$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}*\frac{12}{14}*\frac{10}{12}*\frac{8}{10}*\frac{6}{8}*\frac{4}{6}*\frac{2}{4}=$$
$$\frac{2}{20}=\frac{1}{10}$$
I hope that helps.
Let me know if you have questions.
**I wrote only this list of three:
$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}$$
18 and 18 cancel
16 and 16 cancel
20 remains. 14 remains.
So the first denominator in its list will remain
And the last numerator in its list will remain: $$\frac{2}{20}=\frac{1}{10}$$
Why x cannot be 19?, could you help me please?
Thank you.
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Set K consists of all fractions of the form x/(x+2) where x is a posit [#permalink]
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02 Oct 2018, 19:44
1
jorgetomas9 wrote:
generis wrote:
anupam87 wrote:
Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?
A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9
Sounds frustrating! Here is one way. About 30 seconds (don't write out all the fractions below)
There's a pattern you should see quickly.
Start by finding the greatest value of $$x$$.
Finding the greatest value of $$x$$ is smart because we have a lot of information:
$$x$$ must be even, positive, and less than 20.
The greatest that $$x$$ can be is 18.
Work backwards from 18.
First fraction? $$\frac{x}{x+2}=\frac{18}{20}$$ | {
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} |
First fraction? $$\frac{x}{x+2}=\frac{18}{20}$$
Next? Well, the next $$x$$ must be 16 (even, positive, < 20).
Its denominator, $$x+2$$, is 18: $$\frac{16}{18}$$
Next? $$x$$ must be 14. The denominator is 16: $$\frac{14}{16}$$
List just those three.
$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}$$
All but the first denominator and the last numerator "cancel."**
Pattern: numerators and denominators decend in order.
Numerators:
18, 16, 14, 12, 10, 8, 6, 4, 2
Denominators:
20, 18, 16, 14, 12, 10, 8, 6, 4
As mentioned, all but one denominator and numerator will "cancel."
$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}*\frac{12}{14}*\frac{10}{12}*\frac{8}{10}*\frac{6}{8}*\frac{4}{6}*\frac{2}{4}=$$
$$\frac{2}{20}=\frac{1}{10}$$
Why x cannot be 19?, could you help me please?
Thank you.
jorgetomas9 , x cannot be 19 because the prompt says that x is a positive even integer. See the highlight above. Easy mistake.
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Re: Set K consists of all fractions of the form x/(x+2) where x is a posit [#permalink]
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03 Oct 2018, 08:39
1
generis wrote:
jorgetomas9 wrote:
generis wrote:
[
jorgetomas9 , x cannot be 19 because the prompt says that x is a positive even integer. See the highlight above. Easy mistake.
Thank you very much I haven't seen that one.
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Set K consists of all fractions of the form x/(x+2) where x is a posit [#permalink]
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08 Oct 2018, 11:26
1
Here is another, simpler approach to solving the problem in under 1 minute.
You can take several values of x and see the pattern.
For x=2, fraction is 1/2
For x=4, fraction is 2/3
For x=6, fraction is 3/4 etc. | {
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So, the product of the 9 fractions would be 9!/10!*1 = 1/10
Set K consists of all fractions of the form x/(x+2) where x is a posit &nbs [#permalink] 08 Oct 2018, 11:26
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# A stock trader originally bought 300 shares
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A stock trader originally bought 300 shares [#permalink]
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A stock trader originally bought 300 shares of stock from a company at a total cost of m dollars. If each share was sold at 50% above the original cost per share of stock, then interns of m for how many dollars was each share sold?
a) 2m/300
b) m/300
c) m/200
d) m/300 + 50
e) 350/m
[Reveal] Spoiler: OA
Originally posted by sabrina3509 on 09 Jun 2016, 12:02.
Last edited by Vyshak on 09 Jun 2016, 12:13, edited 1 time in total.
Edited the topic name. Topic name must contain few words from the question.
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Re: A stock trader originally bought 300 shares [#permalink]
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09 Jun 2016, 12:09
sabrina3509 wrote:
A stock trader originally bought 300 shares of stock from a company at a total cost of m dollars. If each share was sold at 50% above the original cost per share of stock, then interns of m for how many dollars was each share sold?
a) 2m/300
b) m/300
c) m/200
d) m/300 + 50
e) 350/m
Quote:
bought 300 shares of stock from a company at a total cost of m dollars.
Let Cost of 300 shares be $3000 So, Cost of 1 shares be$ 10 =>m/300
Quote:
each share was sold at 50% above the original cost per share of stock
Selling price per share = (100+50)/100 * m/300
Or, Selling price per share = 3/2 * m/300 => m/200
Hence answer will be (C)
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Small leverage words here make a big difference. First, notice that the "total" cost of 300 stocks is equal to $$m$$. Thus, the cost per stock would be $$\frac{m}{300}$$. This fraction matches answer choice (B), and highlights a classic trap of the GMAT: answer choices often include the "right answer to the wrong question." The target of this question is not the original cost per stock, but instead the price at which each stock was sold, after an increase. | {
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The question indicates that each share was sold at 50% above the original cost. This is a percentage increase, adding half of the original value to the value. Thus, if the original cost per stock were $$(\frac{m}{300})$$, the value of the stock after the increase would be $$(\frac{m}{300})(1+\frac{1}{2})$$. Now, don't convert the fraction to decimal values here. Not only are "fractions your friends", but the answer choices clearly keep the math in fractional form.
Now, we need to just simplify the math, looking for common factors in the top and bottom of the equation to make the math even easier:
$$(\frac{m}{300})(1+\frac{1}{2})=(\frac{m}{3*100})(\frac{3}{2})=\frac{m}{200}$$
The answer is (C).
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A stock trader originally bought 300 shares [#permalink] 21 Dec 2017, 19:16
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# In the figure above, x^2 + y^2 =
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In the figure above, x^2 + y^2 = [#permalink]
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In the figure above, x^2 + y^2 =
(A) 5
(B) 7
(C) 25
(D) 80
(E) 625
Attachment:
2017-09-27_1016_002.png [ 10.27 KiB | Viewed 879 times ]
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In the figure above, x^2 + y^2 = [#permalink]
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Updated on: 28 Sep 2017, 07:16
Bunuel wrote:
In the figure above, x^2 + y^2 =
(A) 5
(B) 7
(C) 25
(D) 80
(E) 625
Attachment:
2017-09-27_1016_002.png
All four figures are right triangles.
$$x^2 + y^2$$ = the length of the hypotenuse of the inside triangle. Find side lengths of inside triangle.
Short leg, SL, of inside triangle:
$$(\sqrt{4})^2 + (\sqrt{5})^2 = (SL)^2$$
$$(4 + 5) = 9 = (SL)^2$$
$$SL = 3$$
Long leg, LL, of inside triangle:
$$(\sqrt{6})^2 + (\sqrt{10})^2 = (LL)^2$$
$$(6 + 10) = 16 = (LL)^2$$
$$LL = 4$$
The inside triangle is a 3-4-5 right triangle - its hypotenuse is 5. Let hypotenuse 5 = c:
$$x^2 + y^2 = c^2$$
$$x^2 + y^2 = 5^2$$
$$x^2 + y^2 = 25$$
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In the depths of winter, I finally learned
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Originally posted by generis on 27 Sep 2017, 18:06.
Last edited by generis on 28 Sep 2017, 07:16, edited 1 time in total.
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Re: In the figure above, x^2 + y^2 = [#permalink]
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28 Sep 2017, 06:04
1
genxer123 wrote:
Bunuel wrote:
In the figure above, x^2 + y^2 =
(A) 5
(B) 7
(C) 25
(D) 80
(E) 625
Attachment:
2017-09-27_1016_002.png
All four figures are right triangles.
$$x^2 + y^2$$ = the length of the hypotenuse of the inside triangle. Find side lengths of inside triangle.
Short leg, SL, of inside triangle:
$$(\sqrt{4})^2 + (\sqrt{5})^2 = (SL)^2$$
$$(4 + 5) = 9 = (SL)^2$$
$$SL = 3$$
Long leg, LL, of inside triangle:
$$(\sqrt{6})^2 + (\sqrt{10})^2 = (LL)^2$$
$$(6 + 10) = 16 = (LL)^2$$
$$LL = 4$$
The inside triangle is a 3-4-5 right triangle - its hypotenuse is 5.
$$x^2 + y^2 = 5$$
The highlighted part should equal to (hypotenuse)...it should be 25.
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In the figure above, x^2 + y^2 = [#permalink]
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28 Sep 2017, 07:14
Mo2men wrote:
genxer123 wrote:
Bunuel wrote:
In the figure above, x^2 + y^2 =
(A) 5
(B) 7
(C) 25
(D) 80
(E) 625
Attachment:
2017-09-27_1016_002.png
. . .
The inside triangle is a 3-4-5 right triangle - its hypotenuse is 5.
$$x^2 + y^2 = 5$$
The highlighted part should equal to (hypotenuse)...it should be 25.
Mo2men , nice catch. I was thinking of 5 as a fixed value derived from other fixed values, such that 5 = $$c^2$$. I missed a step.
5 = $$(\sqrt{c^2}) = (\sqrt{5^2}) = \sqrt{25}$$.
I'll edit.
Thanks, and kudos.
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Re: In the figure above, x^2 + y^2 = [#permalink]
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03 Oct 2017, 10:06
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### Show Tags
03 Oct 2017, 10:06
Bunuel wrote:
In the figure above, x^2 + y^2 =
(A) 5
(B) 7
(C) 25
(D) 80
(E) 625
We see that x^2 + y^2 is equal to the square of the hypotenuse of a right triangle whose sides are not shown. Let’s first determine the base of this triangle, which we can call b:
(√4)^2 + (√5)^2 = b^2
4 + 5 = b^2
9 = b^2
b = 3
Let’s now determine the height of the triangle, which we can call h:
(√6)^2 + (√10)^2 = h^2
6 + 10 = h^2
16 = h^2
h = 4
Thus, x^2 + y^2 = 3^2 + 4^2 = 25.
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Re: In the figure above, x^2 + y^2 = &nbs [#permalink] 03 Oct 2017, 10:06
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# In the figure above, x^2 + y^2 =
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# Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$
Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$
My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?
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I suppose using the Binet formula is a "mosquito-nuking" solution... but it works. – J. M. Nov 23 '10 at 7:17
@J.M.: "overkill" (FTFY) – Isaac Dec 24 '10 at 22:16
@Isaac, It's an idiom here and MO... – J. M. Dec 24 '10 at 23:58
Fix $m \in \mathbb{N}$. We shall use induction on $n$. For $n=1$, the RHS of the equation becomes $$F_{m-1}F_{1} + F_{m}F_{2} = F_{m-1} + F_{m}$$, which is equal to $F_{m+1}$. When $n=2$, the equation is also true.( I hope you can prove this!).
Now assume, that the result is true for $k=3,4, \cdots , n$. We want to show that the result is true for $k=n+1$. $$\text{For} \ k=n-1 \ \text{we have} \quad F_{m+n-1} = F_{m-1}F_{n-1} + F_{m}f_{n}$$ and $$\text{For} \ k = n \ \text{we have} \quad F_{m+n}=F_{m-1}F_{n} + F_{m}F_{n+1}$$ Adding both the sides you will get $$F_{m+n-1} + F_{m+n} = F_{m+n-1} = F_{m-1}F_{n+1} + F_{m}F_{n+2}$$
Oh, i reversed the notations. This is for proving $$F_{m+n} = F_{m-1}F_{n} + F_{m}F_{n+1}$$
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@Chandru1 thanks a lot for the help i think i'm slowly starting to understand this stuff. To prove it with $n=2$ i simply broke $F_{n+2}$ down into $F_{n+1} + F_n = F_n + F_{n-1} + F_n$ which is what i obtained when substituting $n=2$. – ghshtalt Nov 23 '10 at 18:45 | {
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If pressed, I'd nuke with Binet's formula myself, but here's another approach. By an easy induction, if $$A=\pmatrix{0&1\\\\ 1&1}$$ then $$A^n=\pmatrix{F_{n-1}&F_n\\\\ F_n&F_{n+1}}.$$ Comparing the top right entry in the equation $A^m A^n=A^{m+n}$ gives $$F_{m-1}F_n+F_m F_{n+1}=F_{m+n}.$$
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HINT $\ \$ If you put the Fibonacci recurrence into matrix form then the result is obvious, viz.
$$M^n\ :=\ \left(\begin{array}{ccc} 1 & 1 \\\ 1 & 0 \end{array}\right)^n\ =\ \left(\begin{array}{ccc} F_{n+1} & F_n \\\ F_n & F_{n-1} \end{array}\right)$$
Now compute $\ M^{n+m}\ =\ M^n \ M^m$
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ok so what i compute is: $$M_{n+m} = M_n*M_m = \left(\begin{array}{ccc} F_{n+1} & F_n \\\ F_n & F_{n-1} \end{array}\right)*\left(\begin{array}{ccc} F_{m+1} & F_m \\\ F_m & F_{m-1} \end{array}\right) = \left(\begin{array}{ccc} F_{n+1}F_{m+1} + F_nF_m & F_{n+1}F_m + F_nF_{m-1} \\\ F_nF_{m+1} + F_{n-1}F_m & F_{n}F_{m} + F_{n-1}F_{m-1} \end{array}\right)$$ which as you said shows the desired expression in the bottom left(and i assume equivalent) in the top right. is this method something one can usually use with sequences? i guess i just don't know how it works in general... – ghshtalt Nov 23 '10 at 8:37
that is, have i understood it correctly if i were to say that to define a sequence, one creates a $2 X 2$ matrix with the upper left value equal to the next value the bottom left and upper right values equal to the current value, and the bottom right equal to the previous? – ghshtalt Nov 23 '10 at 8:38
@user3711: The matrix $M$ is simply the shift operator viz. $\ (F_{n-1},F_n)\:M^t\ = (F_n,F_{n+1})\:.\$ The same idea works for any linear recurrence. – Bill Dubuque Nov 23 '10 at 17:38
With Fibonacci numbers usually there are multiple ways of proving identities.
One way (which is one of my favourites) to prove your identity is the following:
Consider the following problem:
A person climbs up $\displaystyle n$ steps, by taking either one step, or two steps at a time. | {
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A person climbs up $\displaystyle n$ steps, by taking either one step, or two steps at a time.
The total number of ways the person can climb up all the $\displaystyle n$ steps is $\displaystyle F_{n+1}$ (Why?)
Now consider climbing $\displaystyle m+n-1$ steps and split into the cases when the person lands on step $\displaystyle n$ and the cases when the person lands on step $\displaystyle n-1$ and takes two steps at that point (and so does not land on step $\displaystyle n$ in those cases). These two cases cover all possibilities, and so we have:
$$\displaystyle F_{m+n} = F_{n+1}F_{m} + F_{n}F_{m-1}$$
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Right. This is basically the matrix proof translated into the language of walks on graphs, which is then further translated into a tiling problem. – Qiaochu Yuan Nov 23 '10 at 17:46
@Qia: That is one way of looking at it. But I wouldn't say it is basically just a translation... I think it is interesting on its own. – Aryabhata Nov 23 '10 at 18:11
I like this proof for giving a simple combinatorial interpretation of the result - it offers better 'motivation' IMHO than the matrix proof, which might look like magic to someone not familiar with those manipulations. – Steven Stadnicki Nov 23 '10 at 18:46
In any case it was good to try to see it from another angle, although to be honest, i'm still having some trouble really visualizing this, hopefully it hits me later... and yes the matrices seem really efficient but at least for me they still look like magic ;) – ghshtalt Nov 23 '10 at 18:52
@Steven: the two proofs are equivalent. Powers of the Fibonacci matrix count walks on a certain graph, the Fibonacci graph, which can in turn be interpreted in terms of certain tilings. I discuss a generalization to companion matrices in this blog post: qchu.wordpress.com/2009/08/23/… – Qiaochu Yuan Nov 23 '10 at 19:08 | {
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There are several good answers already, but I thought I would add the following derivation because it is one of the few uses I know for the sum property of permanents; namely,
If $A$, $B$, and $C$ are matrices with identical entries except that one row (column) of $C$, say the $k^{th}$, is the sum of the $k^{th}$ rows (columns) of $A$ and $B$, then $\text{ per } A + \text{ per } B = \text{per } C$.
Start with the matrices $\begin{bmatrix} F_n & F_{n-1} \\ F_0 & F_1 \end{bmatrix}$ and $\begin{bmatrix} F_n & F_{n-1} \\ F_1 & F_2 \end{bmatrix}$. Since $F_0 = 0$ and $F_1 = F_2 = 1$, they have permanents $F_n$ and $F_n + F_{n-1} = F_{n+1}$, respectively. Applying the Fibonacci recurrence and the permanent sum property, we have $\text{ per } \begin{bmatrix} F_n & F_{n-1} \\ F_2 & F_3 \end{bmatrix} = F_{n+2}$. By continuing to construct new matrices whose second rows are the sums of the second rows of the previous two matrices, this process continues until we have $F_n F_{m+1} + F_{n-1}F_m = \text{ per} \begin{bmatrix} F_n & F_{n-1} \\ F_m & F_{m+1} \end{bmatrix} = F_{n+m}.$
For more on this approach (but with determinants), see this paper I wrote a few years ago: "Fibonacci Identities via the Determinant Sum Property," The College Mathematics Journal, 37 (4): 286-289, 2006.
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+1: Hadn't seen this before. – Aryabhata Nov 23 '10 at 23:49
Try proving this statement: Claim: If $f(n) = f(n-1)+f(n-2)$, then $f(n) = F_n f_1 + F_{n-1} f_0$.
Now "fix" $m$ and think of $F_{n+m}$ as a linear recurrence in $n$ with initial conditions $F_{m+1}$ and $F_m$.
By the way, there is a short and clean proof of Claim, but you should know it uses the fact $F_{-1} = 1$.(Something you can easily verify if you did not already know).
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One simple way is to use the formula for the $n^{th}$ Fibonacci number, viz,
$F_n = \frac{\phi^n - (1-\phi)^n}{\sqrt{5}}$ where $\phi$ is the golden ratio.
$\phi = \frac{1 + \sqrt{5}}{2}$. | {
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$\phi = \frac{1 + \sqrt{5}}{2}$.
Or another equally simple way is to use induction on $n$ and then on $m$ or just using induction on one of these might turn out to suffice.
It will be interesting to see a direct proof...
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...and looks like somebody else did use Binet... :D – J. M. Nov 23 '10 at 7:20
@J.M. : I didn't see your comment when I wrote the answer :). – user17762 Nov 23 '10 at 7:23
am i interpreting this properly if i write: $\frac{\Phi^{n+m} - (1-\Phi)^{n+m}}{\sqrt{5}} = \frac{\Phi^{n-1} - (1-\Phi)^{n-1}}{\sqrt{5}} * \frac{\Phi^{m} - (1-\Phi)^{m}}{\sqrt{5}} + \frac{\Phi^{n} - (1-\Phi)^{n}}{\sqrt{5}} * \frac{\Phi^{1+m} - (1-\Phi)^{1+m}}{\sqrt{5}}$ $\frac{\Phi^{m} - (1-\Phi)^{m}}{\sqrt{5}}(\frac{\Phi^{n-1} - (1-\Phi)^{n-1}}{\sqrt{5}} + \frac{\Phi^{n} - (1-\Phi)^{n}}{\sqrt{5}} * F_1)$ $\frac{\Phi^{m} - (1-\Phi)^{m}}{\sqrt{5}}(\frac{\Phi^{n-1} - (1-\Phi)^{n-1}}{\sqrt{5}}+\frac{\Phi^{n} - (1-\Phi)^{n}}{\sqrt{5}})$ ? what can i do at the end? – ghshtalt Nov 23 '10 at 7:41
@user3711: Use $\phi^2 = 1 + \phi$ to simplify. Or as the others have stated you can try induction. – user17762 Nov 23 '10 at 7:45
i'm sorry, but i'm having trouble figuring out how to use that to simplify.... – ghshtalt Nov 23 '10 at 8:08
These can almost always be solved by induction. (In general sequences that are defined by recursion go often well together with induction.)
For this formula we can do the following: Induct on $m$. If $m=1$, then $F_{n+m} = F_{n}F_{m-1} + F_{n-1}F_m = F_{n}F_{0} + F_{n-1}F_1 = F_n + F_{n-1} = F_{n+1}$.
Suppose then that the result holds for $m$ and smaller. Now $$F_{n+m+1} = F_{n+m} + F_{n+m-1} = F_{n}F_{m-1} + F_{n-1}F_m + F_{n}F_{m-2} + F_{n-1}F_{m-1}$$ $$= F_{n}(F_{m-1} + F_{m-2}) + F_{n-1}(F_m + F_{m-1}) = F_{n}F_{m} + F_{n-1}F_{m+1}.$$
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# Is the binomial theorem actually more efficient than just distributing
Is the binomial theorem actually more efficient than just distributing a given binomial. I believe it is more confusing for me to remember and work out using the binomial theorem as a guide, than to just distribute when given a problem like: $$(x-4)^6$$ I think distributing that would be equally as painful as using the binomial theorem. Am I alone or is there actually a reason to practice the painful procedures?
• Well if you distribute without adding equal terms, you will end up with $2^6$ terms, while the theorem gives you the right $6$ terms. – Phicar Jan 7 '17 at 6:26
• What about using Pascal's triangle to compute the binomial coefficients? – Fabio Somenzi Jan 7 '17 at 6:28
• What about $(x-1)^{10}$? Binomial theorem gives it by hand in under a minute. How long would you take to expand that by hand? The theorem is also much more conducive to algorithms, meaning it's strength in both computation and proof is far greater than direct distribution. – Nij Jan 7 '17 at 6:29
• @Phicar $7$ terms, not $6$. – Marc van Leeuwen Jan 7 '17 at 17:08
• @Nij Actually the OP's method of repeatedly multiplying can be more efficient than the binomial theorem for large polynomials of certain types. There is much symbolic computation literature on this topic, e.g.see R. J. Fateman, Polynomial Multiplication, Powers and Asymptotic Analysis: Some Comments, SIAM J. Comput., 3(3),(1974),196–213. – Bill Dubuque Jan 8 '17 at 5:46
The binomial theorem allows you to write out the expansion of your polynomial immediately.
It also allows you to answer such questions as "What is the coefficient of $x^{20}$ in $(1+x)^{100}$?"
Its generalisation to non-integer exponents allows you to get the expansion of $(1-x)^{-1/2}$.
It is a good thing. | {
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It is a good thing.
• (+1) And needing "just one term" isn't an idle exercise, either: anyone having taken calculus should recognize the importance of: "What's the coefficient of $x^{n - 1}$ in $(x + h)^n - x^n$?" – pjs36 Jan 7 '17 at 6:49
• "It is a good thing" - Well said! :) (+1) – Hypergeometricx Jan 7 '17 at 15:50
There are all sorts of reasons. First of all, you'll learn when you've moved forward a bit that the binomial theorem can actually be applied in cases of non-integer exponents, by defining $\binom{n}{k} = \dfrac{n^{\underline{k}}}{k!}$, where the underline represents a falling power (i.e., $n(n-1)(n-2)\cdots$). Another that comes to mind immediately is the series expansion of $e$...
$$e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = \lim_{n\to\infty}\left(1 + n\frac{x}{n} + \frac{n^\underline{2}}{2!}\frac{x^2}{n^2} + \frac{n^\underline{3}}{3!}\frac{x^3}{n^3} + \cdots + \frac{x^n}{n^n}\right)$$
As $n$ gets very large, for fixed $k$, $n^\underline{k}$ approaches $n^k$, and $\frac{x^n}{n^n}$ vanishes since $x$ is fixed, so you end up with the familiar series...
$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$$
• Very nice link to expansion for $e^x$. (+1) – Hypergeometricx Jan 7 '17 at 15:51
• Weird. I don't see the underline in the definition of $\binom nk$. Am I the only one with this problem? I'm on Chromium 55.0.2883.87 (64-bit). – rubik Jan 7 '17 at 21:55
• @rubik I had the same problem -- edited the post to make it displaystyle so it would show up. I think we should bring up this on meta as a bug with mathjax, would you like to do it or should I? – 6005 Jan 7 '17 at 22:54
• The frac displays fine in my mobile Chrome (both with dfrac and frac). – Martin Argerami Jan 8 '17 at 6:35
• @6005 You're right, done. Here's the discussion. – rubik Jan 8 '17 at 9:05
If you're not seeing the utility of the binomial theorem, then I think you're missing an important observation: | {
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The coefficients of every term can be calculated extremely efficiently (your example of $(x-4)^6$ is easy to expand by hand with the binomial theorem, without, it's doable but very tedious).
A useful shorthand for lower powers of the exponent is to remember the first few rows of Pascal's Triangle along with the rule for adding more if you need them. Example of first few rows:
$$1$$ $$1\qquad 1$$ $$1\qquad 2\qquad 1$$ $$1\qquad 3\qquad 3\qquad 1$$ $$1\qquad 4\qquad 6\qquad 4\qquad 1$$ $$1\qquad 5\qquad 10\qquad 10\qquad 5\qquad 1$$ $$1\qquad 6\qquad 15\qquad 20\qquad 15\qquad 6\qquad 1$$
The last row gives the coefficients of $x^k\cdot(-4)^{(6-k)}$ for $k$ running from $0$ to $6$. Computing $(-4)^{(6-k)}$ is easy enough by hand, and multiplying it by the relevant entry in the last row above is also fairly easy (give it a try).
One more thing: The binomial theorem admits an interesting generalization to powers of general polynomials of a given length sometimes called the multinomial theorem. (Challenge: try investigating the coefficients of the first few powers of $(x+y+z)^n$ and perhaps you can figure out the pattern for trinomials? How does it relate to the pattern for binomials? Can you generalize it to general polynomials?)
Marty Cohen and Phicar answered your question correctly. In mathematics, as we always use small numbers, It could be feasible to think this. Why learn the binomial theorem to compute $(a+b)^2$? But what can you say about $(3a+5b)^{987658}$? The binomial theorem is way more efficient in this case and the actual number of cases in which it wouldn't be efficient is very small.
The binomial theorem gives you a powerful tool: The computation for any exponent in - perhaps - the fastest way possible in which it is only not effective for a finite amount of exponents, but its the best possible for an endless amount of exponents. | {
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This is one trap in mathematics: We implicitly assume that the mathematical tools are created to treat familiar cases, but they are created to treat all cases in the neatest way possible, they will be cumbersome for some simple cases (Example: Integrating $f(x)=mx$ in the interval $[0,a]$. This is just a triangle, there is no need for integral calculus here). I remember of me asking about why we need analysis if calculus seems to work well, but I was supposing (without realizing) that calculus is used only on familiar cases, but we want to be useful also for very imaginative cases!
• Just a small note to support the statement that it's as fast as it gets: using the recursion ${n\choose k+1} = {n\choose k}\frac{n-k}{k+1}$ with ${n\choose 0} = 1$ and the symmetry ${n\choose k} = {n\choose n-k}$ then we can compute all the $n$ binomial coefficients ${n\choose k}$ for $k=1,2,\ldots,n$ with only $n$ operations (multiplications and divisions). That is hard to beat! This can be compared with $\mathcal{O}(2^n)$ operations needed to distribute $(a+b)^n$. – Winther Jan 8 '17 at 3:28
• Excelent addendum, @Winther! – Billy Rubina Jan 8 '17 at 13:37
To add to other answers, in higher-level math it's important to be able to generalize the behavior of $(a + b)^n$ when $n$ is a variable. The work that you're doing now is mostly just training exercises for being familiar with, and having intuition for, the general binomial theorem.
For example, there's a really key moment early on in calculus when it's critical to understand that, for all $n$: $(x + h)^n = x^n + nx^{n-1}h....$ plus a bunch of other stuff with higher powers of $h$ in it, but whose exact numerical coefficients we can ignore. | {
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# Evaluate $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2k}+\cdots$ [duplicate]
This question already has an answer here:
I need to evaluate, for a certain worded question:
If n is even $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots\binom{n}{n}$$ If n is odd $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots\binom{n}{n-1}$$
I havent been able to reduce this using the binomial expansion.
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## marked as duplicate by Grigory M, Sasha, Daniel Fischer♦, hardmath, DanJan 16 '14 at 22:19
unrelated: why is the first binomial coefficient in the title bigger than the others? – kuch nahi Jun 2 '11 at 15:35
– Grigory M Jun 2 '11 at 15:37
Just want to add that the original question was to find the number of binary sequences of length n that contain an even number of ones. Using a different argument the answer I got is $2^{n-1}$ – kuch nahi Jun 2 '11 at 15:40
Hint: Do some computations for small values of $n$. You will see a pattern. When $n$ is odd, the pattern is most obviously proved by symmetry. When $n$ is even, look at Pascal's triangle. – Aaron Jun 2 '11 at 15:41
Binomial expansion allows one to find both $\binom n0+\binom n1+\binom n2+\dots=(1+1)^n$ and $(\binom n0-\binom n1+\binom n2+\dots)=(1-1)^n$. Combining these two yields desired result: $\binom n0+\binom n2+\binom n4+\dots=\frac12\bigl((1+1)^n+(1-1)^n\bigr)=2^{n-1}$.
-
I remember I've done this before. Thanks a ton! – kuch nahi Jun 2 '11 at 15:45
The first version was helpful enough. Anyhow, thanks again. – kuch nahi Jun 2 '11 at 15:54
The given sums count the number of even-cardinality subsets of an $n$-element set. I'll provide a simple argument that shows that the value of that sum is $2^{n-1}$.
Fix $x \in X$ (where $|X|=n$). The map $f \colon \mathcal{P}(X) \to \mathcal{P}(X)$ (where $\mathcal{P}(X)$ denotes the power set of $X$) defined by $$f(Y) = \begin{cases} Y \cup \{x\} &\text{if } x \notin Y \\ Y - \{x\} &\text{if } x \in Y \end{cases}$$ | {
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is a bijection of $\mathcal{P}(X)$ that takes even-cardinality subsets to odd-cardinality subsets and vice-versa. Thus there are the same number of even-cardinality subsets as odd-cardinality subsets. Since there are $2^{n}$ total subsets, there must be $2^{n}/2 = 2^{n-1}$ even-cardinality subsets.
-
The sum of all binomial coefficients ${n \choose 0} + \cdots {n \choose n}$ is $2^n$. This is a well-known equation and the slick proof is to consider the binomial theorem for $(1+1)^n$.
Related is the binomial theorem applied to $(1-1)^n$ giving the sum
$${n \choose 0} - {n \choose 1} + {n \choose 2} - \cdots \pm {n \choose n} = 0.$$
The latter inequality is easy to see when you add up the $n$th row of Pascal's triangle with alternating signs and $n$ is odd. For example: $1-5+10-10+5-1 = 0$ is obvious from the symmetry of the entries.
To obtain the answer for every-other term in the first sum, you can add these two formulas together and the negative entries will cancel with the positive ones while the positive ones will double:
$$(1+1)^n + (1-1)^n = 2{n \choose 0} + 2{n \choose 2} + \cdots = 2^n + 0.$$
Now to finish, divide by two. To get the odd entries (the second part of your question), notice that $(1-1)^n = -(1-1)^n$ and use the technique above. Or you could just take the sum for all entries and subtract what we just obtained for the even entries.
-
Let $X$ be a binomial$(n,p)$ random variable with $p=1/2$. If $n$ is even, then the probability that $X$ is even is equal to $$\sum\limits_{k = 0,2, \ldots ,n} {{n \choose k}\frac{1}{{2^n }}} .$$ On the other hand, this probability is obviously equal to $1/2$ (consider $X$ as the sum of $n$ independent random variables taking the values $0$ and $1$ with probability $1/2$ each). Hence, $$\sum\limits_{k = 0,2, \ldots ,n} {{n \choose k}} = 2^n \frac{1}{2} = 2^{n - 1} .$$
-
+1: I'm always a fan of the probabilistic approach. :) – Mike Spivey Jun 2 '11 at 16:37
@Mike - Thanks! – Shai Covo Jun 2 '11 at 18:24 | {
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Here's a more general take. Sums of the form $\sum_k \binom{n}{2k} f(k)$ are sometimes called aerated binomial sums. My paper "Combinatorial Sums and Finite Differences" (Discrete Mathematics, 307 (24): 3130-3146, 2007) proves some general results about these aerated binomial sums. (See Section 4.) The case $f(k) = 1$ (as in the OP's question) falls out nicely.
Suppose you are interested, for some function $f(k)$, in the binomial sum
$$B(n) = \sum_{k=0}^n \binom{n}{2k} f(k).$$
Then, taking the finite difference $\Delta f(k) = f(k+1) - f(k)$, denote $A(n)$ by
$$A(n) = \sum_{k=0}^n \binom{n}{2k} \Delta f(k).$$
In the paper I prove that $B(n)$ and $A(n)$ are related via $$B(n) = 2^{n-1} f(0) + 2^n \sum_{k=2}^n \frac{A(k-2)}{2^k} + \frac{f(0)}{2}[n=0],$$ where $[n=0]$ evaluates to $1$ if $n=0$ and $0$ otherwise.
Since $f(k) = 1$ in the OP's question, $\Delta f(k) = 0$, and so $A(n) = 0$ for all $n$. Thus $$\sum_{k=0}^n \binom{n}{2k} = 2^{n-1} + \frac{1}{2}[n=0].$$
More generally, this approach can be used to prove that, for $m \geq 1$, $$\sum_k \binom{n}{2k} k^{\underline{m}} = n(n-m-1)^{\underline{m-1}} \, 2^{n-2m-1} [n \geq m+1],$$ and $$\sum_k \binom{n}{2k} k^m = n \sum_j \left\{ m \atop j\right\} \binom{n-j-1}{j-1} (j-1)! \, 2^{n-2j-1},$$ where $\left\{ m \atop j\right\}$ is a Stirling number of the second kind. (While the second equation swaps one sum for another, there are no more than $m$ terms in the right-hand side, and so it is useful when $m$ is small.)
The approach can also be extended to sums of the form $\sum_k \binom{n}{ak+b} f(k)$. (And the case $a = 2, b = 1$ is considered explicitly in the paper.) Using higher values of $a$ and $b$ would result in increasingly complicated expressions, though.
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-
I think it might help to look at this in terms of the rows on pascal's triangle. The sum of a given row is 2^n. All rows are symmetric about the middle of the row. As this sounds like it could be HW, i'll leave it to you to figure out how each of those relate to the sum of the row.
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I wish this were homework. As that would imply I am taking a course and have access to an instructor. – kuch nahi Jun 2 '11 at 15:43
Ok, so they are both exactly half the of the row (they get every element once as opposed to twice). Since the sum of a row is 2^n, this gives 2^(n-1) as the answer. – soandos Jun 2 '11 at 15:45
yup, got it. Thanks! +1 – kuch nahi Jun 2 '11 at 15:54 | {
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# Example: a t-test between two groups
Assume that 10 speakers are randomly drawn from the Dutch and Flemish populations. Their scores on a German-language proficiency test are measured:
dutch = c(34, 45, 33, 54, 45, 23, 66, 35, 24, 50)
flemish = c(67, 50, 64, 43, 47, 50, 68, 56, 60, 39)
The Flemish look a bit better than the Dutch. Is this difference statistically significant?
t.test (dutch, flemish, var.equal=TRUE)
##
## Two Sample t-test
##
## data: dutch and flemish
## t = -2.512, df = 18, p-value = 0.02176
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -24.790596 -2.209404
## sample estimates:
## mean of x mean of y
## 40.9 54.4
1. yes
2. no
3. don’t know
# .
Yes, it is a probability. But of what?
# .
### Question: Is this p-value of 0.02176…
1. the probability that the Dutch and Flemish population means are the same?
2. the probability that if the Dutch and Flemish population means are the same, an absolute t-value of 2.512 or greater will be found between random Dutch and Flemish samples of size 10?
3. don’t know
# .
So yes, the p-value of 0.02176 is a tail probability: the probability that if the Dutch and Flemish population means are the same, an absolute t-value of 2.512 or greater will be found between random Dutch and Flemish samples of size 10.
# .
### Question: What is this condition (“the Dutch and Flemish population means are the same”) usually called?
1. the null hypothesis
2. the alternative hypothesis
3. don’t know
# .
It’s the null hypothesis (one can have null hypotheses that are not about equality, but the one used here is typical).
Let’s check that the probability of finding an absolute t above 2.512 is indeed around 2.2 percent, by drawing a hundred thousand random samples from the Dutch and Flemish populations.
In the simulation, we assume that the null hypothesis is true, so that: | {
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In the simulation, we assume that the null hypothesis is true, so that:
• the true population means are the same ($$\mu_d = \mu_f = 50$$; the exact number doesn’t matter)
• the true standard deviations are the same as well ($$\sigma_d = \sigma_f = 8$$)
Here is how you obtain the data for one such an experiment (not 100,000 yet):
numberOfParticipantsPerGroup = 10
mu.d = mu.f = 50
sigma.d = sigma.f = 8
data.d = rnorm (numberOfParticipantsPerGroup, mu.d, sigma.d)
data.f = rnorm (numberOfParticipantsPerGroup, mu.f, sigma.f)
data.d
data.f
## [1] 46.01451 53.06159 58.37237 55.26379 43.74593 49.95321 55.10986
## [8] 45.29203 47.70117 48.67666
## [1] 62.36448 54.69570 59.98880 62.60507 47.95375 51.00754 48.28727
## [8] 44.34768 58.20126 48.39343
You get the t-value from such data as follows:
numberOfParticipantsPerGroup = 10
mu.d = mu.f = 50
sigma.d = sigma.f = 8
data.d = rnorm (numberOfParticipantsPerGroup, mu.d, sigma.d)
data.f = rnorm (numberOfParticipantsPerGroup, mu.f, sigma.f)
t = t.test (data.d, data.f, var.equal=TRUE) $statistic t ## t ## 0.286349 Then try this a hundred thousand times, and see how often the absolute t-value reaches 2.512 or more: numberOfParticipantsPerGroup = 10 numberOfExperiments = 1e5 count = 0 for (experiment in 1 : numberOfExperiments) { mu.d = mu.f = 50 sigma.d = sigma.f = 8 data.d = rnorm (numberOfParticipantsPerGroup, mu.d, sigma.d) data.f = rnorm (numberOfParticipantsPerGroup, mu.f, sigma.f) t = t.test (data.d, data.f, var.equal=TRUE)$ statistic
if (abs (t) > 2.512) {
count = count + 1
}
}
count
## [1] 2213
That’s very close to the 0.02176 probability computed by t.test. So, R’s t.test seems to compute the p-value correctly!
# .
### Question: What can we conclude from the p-value of 0.02176, which is less than the common criterion of 0.05?
1. we reject the null hypothesis, so the Flemish group is better than the Dutch group
2. we reject the null hypothesis, so Flemish people are better than Dutch people
3. don’t know
# . | {
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# .
What if one participant had been different (a score of 27 instead of 47)?
dutch = c(34, 45, 33, 54, 45, 23, 66, 35, 24, 50)
flemish = c(67, 50, 64, 43, 27, 50, 68, 56, 60, 39)
t.test (dutch, flemish, var.equal=T)
##
## Two Sample t-test
##
## data: dutch and flemish
## t = -1.9122, df = 18, p-value = 0.07191
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -24.13526 1.13526
## sample estimates:
## mean of x mean of y
## 40.9 52.4
The p-value is now greater than 0.05.
# .
### Question: What can we conclude?
1. we cannot reject the null hypothesis, so we cannot conclude anything
2. we accept the null hypothesis, so Dutch people are as good as Flemish people
3. don’t know
# .
## Neyman–Pearson statistics
I now very briefly describe a way of doing statistics that I will not recommend for scientific fact finding. It is quite usable for binary decisions in real life, though, such as the question whether or not to take a new drug to market.
The “Type I” error rate is
$$\alpha$$ = 0.05
qt (p = 0.025, df = 18)
## [1] -2.100922
qt (p = 0.975, df = 18)
## [1] 2.100922
If the null hypothesis (no difference between Dutch and Flemish population means) is true, the absolute t-value will be found to lie above 2.101 5 percent of the time (for 10 participants per group).
The “Type II” error rate is
$$\beta$$ = 0.20
If the true population differs by e.g. 4 (or any other number considered to be the threshold of importance), the absolute t-value will be found to lie above 2.101 80 percent of the time (for 10 participants per group).
The recipe for this methodology in e.g. drug treatment is: | {
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The recipe for this methodology in e.g. drug treatment is:
1. Measure the effect of the drug in 10 people, and the effect of placebo in 10 other people.
2. If the difference between the groups yields a t-value above 2.101 in favour of the drug, reject the null hypothesis and make the drug officially available to future patients.
3. If the difference between the groups yields a t-value below 2.101 in favour of the drug, “accept” the null hypothesis and take the drug off the market.
Values for $$\alpha$$ and $$\beta$$ will depend on:
• The severity of side effects.
• The size of the potential cure (life-saving?).
• The cost of the drug.
• And so on.
• yes
• no
• don’t know
# .
No. I didn’t mention any p-values in my very brief discussion of Neyman–Pearson, and they are indeed not used in that way of doing statistics. One either rejects the null hypothesis if t is in the “critical region”, or accepts the null hypothesis if t is smaller. One only reports the “significance level” $$\alpha$$ that is used to demarcate the critical region, and one reports that $$\alpha$$ is 0.05.
# .
### Question: Is this a good procedure for scientific belief finding?
• yes, because I need a clear criterion for my beliefs
• no, because the strength of my belief will gradiently depend on the p-value
• don’t know
# .
I already suggested that it’s not. Surely the difference between a p-value of 0.049 and one of 0.051 is not impressive enough for me to wholly believe that the effect is real in the former case and that the effect is zero in the latter case. You can read about this in the free article Erroneous analyses of interactions in neuroscience: a problem of significance by Sander Nieuwenhuis, Birte Forstmann & Eric-Jan Wagenmakers (2011, Nature Neuroscience).
### Fisher quotes
Ronald Fisher also had some things to say about p-values above 0.05, namely that you should ignore them (they are null results, and a null result is the same as no result): | {
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