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The integral at the end is identical to the original integral. We can continue by adding the integral to both sides:
$\displaystyle 2\int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx={{e}^{x}}\sin \left( x \right)}+{{e}^{x}}\cos \left( x \right)$
Finally, we divide by 2 and have the antiderivative we were trying to find:
$\displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx=\tfrac{1}{2}{{e}^{x}}\sin \left( x \right)}+\tfrac{1}{2}{{e}^{x}}\cos \left( x \right)+C$
In working this type of problem you must be aware of that the original integrand showing up again can happen and what to do if it does. As long as the coefficient is not +1, we can proceed as above. The same thing happens if we do not use the tabular method. (If the coefficient is +1 then the other terms on the right will add to zero and you need to make different choices for u and dv.)
### Reduction Formulas.
Another use of integration by parts is to produce formulas for integrals involving powers. An integral whose integrand is of less degree than the original, but of the same form results. The formula is then iterated to continually reduce the degree until the final integral can be integrated easily.
Example 3: Find $\displaystyle \int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}$
Let $u={{x}^{n}},\ du=n{{x}^{n-1}}dx,\ dv={{e}^{x}}dx,\ v={{e}^{x}}$
$\displaystyle \int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}={{x}^{n}}{{e}^{x}}-n\int_{{}}^{{}}{{{x}^{n-1}}{{e}^{x}}dx}$
This is a reduction formula; the second integral is the same as the first, but of lower degree. Here is how it is used. At each step the integrand is the same as the original, but one degree lower. So the formula can be applied again, three more times in this example.
Most textbooks have a short selection of reduction formulas.
### Final Thoughts. | {
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Most textbooks have a short selection of reduction formulas.
### Final Thoughts.
Back in the “old days”, BC (before calculators), beginning calculus courses spent a lot of time on the topic of “Techniques of Integration.” This included integration by parts, algebraic techniques, techniques known as trig-substitutions, and others. Mathematicians and engineers had tables of integrals listing over a thousand forms and students were taught how to use the tables and distinguish between similar forms in the tables. (See the photo below from the fourteenth edition of the CRC tables (c) 1965.) Current textbooks often contain such sections still.
Today, none of this is necessary. CAS calculators can find the antiderivatives of almost any integral. Websites such as WolframAlpha are also available to do this work.
I’m not sure why the College Board recently expanded slightly the list of types of antiderivatives tested on the exams. Certainly a few of the basic types should be included in a course, but what students really need to know is how to write the integral appropriate to a problem, and what definite and indefinite integrals mean. This, in my opinion, is far more important than being able to crank out antiderivatives of increasingly complicated expressions: let technology do that – or buy yourself an integral table. Just saying … .
# Trapezoids – Ancient and Modern
The other day, in the course of about 10 minutes, I came across two interesting things about Trapezoidal approximations that I thought I would share with you.
Cuneiform writing | {
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Cuneiform writing
The first was a link to a story about how the ancient Babylonian astronomers sometime between 350 and 50 BCE used trapezoids to, in effect, find the area under a velocity-time graph tracking Jupiter’s motion. This was an NPR story based on a January 2016 Science magazine article in which the author, Mathieu Ossendrijver discusses his work deciphering cuneiform tablets written over 1,400 years before the technique showed up in Europe.
The second was a question asked on the AP Calculus bulletin board. A teacher asked, “Can someone please help me answer this question a student posed the other day. We were comparing left, right and midpoint and trapezoidal approximations. He asked since the trapezoidal calculation is the best estimate what is the use of LRAM and RRAM?” Here is an expanded form of my answer.
There are several things to consider here.
1. First, if all you need is an estimate of the area or integral of a continuous function then a Trapezoid sum is certainly better than the left Riemann sum (left-RΣ) or the right-RΣ. Better, yes, the “best” maybe not: midpoint sums are about as good and parabola sums (Simpson’s Rule) are better.
2. Another reason to do left RΣ and right RΣ with small values of n is simply to give students practice in setting up Riemann sums so that they will be familiar with them when they move on to finding their limits and getting ready to define definite integrals. | {
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3. A RΣ for a function f on a closed interval [a, b] is formed by partitioning the interval into subintervals and taking exactly one function value from each closed subinterval, multiplying the value by the width of that subinterval and adding these results. You may pick the function value any way you want – left end, middle, right end, any place at random in the subintervals and someplace else in the next subinterval. One way is to pick the smallest function value in each subinterval; this gives a RΣ called the lower RΣ. Likewise, you could pick the largest value in each subinterval; this gives the upper RΣ. Now it is true that
lower RΣ ≤ (any/all other RΣs) ≤ upper RΣ
Then as you add more partition points (n approaches infinity, or Δx approaches 0, etc.) the lower sum increases and the upper sum decreases. The series of lower sums is increasing and bounded above (by the upper sum) and therefore converges to its least upper bound. The upper sum decreases forming a decreasing series that is bounded below and therefore converges to its greatest lower bound.
If the lower RΣ and the upper RΣ approach the same value, then ALL the other RΣs approach that same value by the Squeeze theorem. This value is then defined as the definite integral of f from a to b.
In most AP calculus course, the textbooks do not deal with upper and lower sums. Instead, they deal with left RΣ and right RΣ on intervals on which f is only increasing (or only decreasing). In this case the lower RΣ = left RΣ and the upper RΣ = the right RΣ (or the other way around for decreasing functions).
So, this is why you need the left RΣ and right RΣ; not so much to approximate, but to complete the theory leading to the definite integral. | {
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## Section5.4Intersections of General Surfaces
Although CalcPlot3D does not determine the intersection of two surfaces for you, it does make a great tool for visually checking that the parameterization you have worked out as the intersection of two surfaces is indeed correct.
Below you will find two examples to explore this concept. Hopefully this visualization process will help increase your confidence in and your appreciation for the results you have obtained.
###### Exploration5.4.1
Determine the vector-valued function that traces out the intersection of the surfaces defined by the equations below using the parameter $t\text{.}$
State the answer as a vector-valued function. Then visually verify this result, plotting the two surfaces in CalcPlot3D, and then plotting the curve to check that it really does represent the intersection of these two surfaces.
Here we can let $x=t\text{.}$ Then $y=t^2\text{.}$ What does $z$ equal (as a function of $t$)?
Then the vector-valued function we obtain that traces out the intersection of these surfaces is:
\begin{equation*} \vec{\textbf{r}}(t) = t \hat{\textbf{i}} + t^2 \hat{\textbf{j}} + (t^2 + t^4) \hat{\textbf{k}} \end{equation*}
This represents the same curve as specified by the parametric functions:
\begin{align*} x &= t\\ y &=t^2\\ z &=t^2 + t^4 \end{align*}
Now let’s verify this in CalcPlot3D visually!
a. Open the CalcPlot3D app.
b. Once the app is loaded and active, enter the first function listed above ($z =$ x^2 + y^2) in the default function object on the left and press Enter (or click on the Graph button). The surface plot of this paraboloid should appear in the plot window.
c. Now, to enter the second function, go to the Add to graph dropdown menu (just above the default function), and select Function: y = f (x, z). This will set the function to y = 1 by default. Enter x^2 in the textbox and press Enter (or use the Graph button). | {
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d. To extend the second surface farther up the paraboloid, change the range of $z$ (just below y = x^2) to go from -2 to 4. You may want to use the scroll-wheel on the mouse to zoom-out a little. [Alternatively you could click on the Format Axes button located just to the right of the 3D Mode button. Then set $z$-max to 4. This will automatically change the upper $z$-clip value to 8. Let’s change this value to 4 also.]
e. Make the surfaces semi-transparent using the button or by typing the T key to get a clearer view of the intersection of the surfaces. Press the E key to turn off the edges on the surfaces.
f. Next we need to graph the space curve to see how well it fits the intersection of the surfaces. Select Space Curve: r(t) from the Add to graph dropdown menu. A space curve object should appear just below this menu.
Enter the three parametric equations we obtained (each in terms of $t$). Then enter a range of -2 to 2. If you press Enter on the second value, it should produce the curve on the plot. If it does not appear, click the Graph button.
g. I like the look of this one better with a constant color, so select the checkbox titled Use Constant Primary Color, at the bottom of this object.
h. Finally rotate the graph to see if it looks like we found the correct intersection curve. Except for different coloring, this should look like the image in Figure 5.4.1 below.
###### Exploration5.4.2
Determine the vector-valued function that traces out the intersection of the surfaces defined below, assuming that $y = 2\sin t\text{.}$
State the answer as a vector-valued function. Then visually verify the result, plotting the two surfaces in CalcPlot3D, and then plotting the curve to check that it really does represent the intersection of these two surfaces.
Note that we can parameterize the ellipse given by the first equation with
Now that we've parameterized $x$ and $y\text{,}$ the second surface equation makes it easy to determine $z$ in terms of $t\text{.}$ | {
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After simplifying the expression, we obtain, $z = 2 + 2\sin^2 t\text{.}$
Then the vector-valued function we obtain that traces out the intersection of these surfaces is:
\begin{equation*} \vec{\textbf{r}}(t) = \sqrt{2}\cos t \hat{\textbf{i}} + 2\sin t \hat{\textbf{j}} + (2 + 2\sin^2 t) \hat{\textbf{k}} \end{equation*}
This represents the same curve as specified by the parametric functions:
\begin{align*} x &= \sqrt{2}\cos t\\ y &= 2\sin t\\ z &= 2 + 2\sin^2 t \end{align*}
Now let’s verify this in CalcPlot3D visually!
a. Open the CalcPlot3D app.
b. Once the app is loaded and active, enter the second function listed above (z = x2 + y2) in the default function object on the left and press Enter (or click on the Graph button). The surface plot of this paraboloid should appear in the plot window.
c. Now, to enter the first function, which is stated implicitly, go to the Add to graph dropdown menu (just above the default function), and select Implicit Surface. Once this new object definition appears on the left, enter the implicit equation defining the first surface in the textbox and press Enter (or use the Graph button). This surface should look like an elliptic cylinder.
d. To make the intersection more clear, let's first zoom out once using the Zoom-out button, so the $x$- and $y$-axes run from -4 to 4 instead of from -2 to 2. Now click on the Format Axes button located just to the right of the 3D Mode button, and set the upper $z$-clip to 4.
e. Next make the surfaces semi-transparent using the transparency button or by typing the T key to get a clearer view of the intersection of the surfaces. Press the E key to turn off the edges on the surfaces.
f. Next we need to graph the space curve to see how well it fits the intersection of the surfaces. Select Space Curve: r(t) from the Add to graph dropdown menu. A space curve object should appear just below this menu.
Enter the three parametric equations we obtained (each in terms of $t$). | {
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Enter the three parametric equations we obtained (each in terms of $t$).
You will enter them like this:
\begin{align*} x &= \textbf{sqrt(2)cos(t)}\\ y &= \textbf{2sin(t)}\\ z &= \textbf{2 + 2(sin(t))\^{}2} \end{align*}
g. Now enter a range of 0 to $2\pi\text{.}$ If you press Enter on the second value, it should produce the curve on the plot. If it does not appear, click the Graph button.
I also like the look of this one better with a constant color, so select the checkbox titled Use Constant Primary Color, at the bottom of this object.
h. Finally rotate the graph to see if it looks like we found the correct intersection curve. This should look like the image in Figure 5.4.2 below.
Notice that the intersection of these two surfaces may appear a little polygonal rather than smooth like the curve we just graphed. This is because we used an implicit surface with the default resolution. To improve this resolution, you can adjust the # Cubes/axis to a higher number, like 25. See the result in Figure 5.4.3. | {
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# Is there a general procedure for inducing a formula / recognizing a pattern?
For some problems, it's pretty simple to deduce what a formula should look like after you enumerate a few examples, but for some problems, it's not so clear. For these less clear examples, is there some procedure I can follow to find the formula?
For example, just now, I was working on the problem:
Find the expected number of tosses it takes to get $$k$$ 6s in a row.
The recursive formula is
$$E[N_k ] = 6(E[N_{k - 1}] + 1)$$
where $$N_k$$ is the number of tosses it takes to get $$k$$ 6s in a row. The initial condition is $$E[N_1] = 6$$ because $$N_1$$ is a geometric random variable with $$1/6$$ probability of success.
Then I enumerated a few cases:
$$E[N_1] = 6 \\ E[N_2] = 42 \\ E[N_3] = 258 \\ E[N_4] = 1554$$
It's not obvious to me what the formula should be. But I think it should look something like $$E[N_k] = 6^k + (k - 1) * \text{something} + \text{maybe other terms}$$
But I don't know what this "something" and "maybe other terms" should be. If I sit here long enough I could probably figure it out, but is there some kind of general procedure that I can apply to a wide array of problems?
• – John Omielan Apr 17 at 20:29
One good way is to try to convert this to a more standard linear recursion.
Letting $$a_n=6(a_{n-1}+1)=6a_{n-1}+6$$
We rewrite this as $$6=a_n-6a_{n-1}=a_{n-1}-6a_{n-2}$$
Whence $$a_n=6a_{n-1}+a_{n-1}-6a_{n-2}=7a_{n-1}-6a_{n-2}$$
and this can now be solved by standard means, yielding $$\boxed{a_n=\frac 65\times \left(6^n-1\right)}$$ | {
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• What do you mean by 'standard means' here? – roulette01 Apr 17 at 20:22
• It's a linear recursion, so you look at the characteristic polynomial $x^2-7x+6=(x-1)(x-6)$. Thus the solution is of the form $a_n=A6^n+B$ and then solve for $A,B$ using initial values. – lulu Apr 17 at 20:23
• Ah I see. I've never formally learned recursion. My understanding is there are a lot of analogies between how recursive formulas are solved vs. how ODEs are solved, and this general solution seems to be analogous to formulas in second order ODEs. – roulette01 Apr 17 at 20:25
• If you were to solve this problem without recursion, and by just inspecting the first several enumerations, is there a general approach to guess what that formula should be? – roulette01 Apr 17 at 20:26
• I think guessing is hard. You could guess that $a_n=A6^n+p(n)$ where $p(n)$ is a low degree polynomial...but I don't know how reliable that sort of guesswork is. Works in this case, though. – lulu Apr 17 at 20:28
If you call $$a_k=E[N_k]$$ then you end up with a linear induction relation: $$a_k-6a_{k-1}=6$$
Since it is linear, it solves as the sum of general solution of homegenous equation $$H_k-6H_{k-1}=0$$, which is $$H_k=6^kH_0$$ and one particular solution of the full equation.
The theory says that if the RHS is of the form $$P(k)r^k$$ then you can find a particular solution of the form $$Q(k)r^k$$ with $$\deg(Q)=\deg(P)+m$$ where $$m$$ is the multiplicity of the root of the homogenous equation.
In this case $$RHS = 6\times 1^k$$ and the single root of homogenous equation is $$r=6$$. So the multiplicity of $$1$$ is just $$m=0$$ (i.e. not a root) this means $$\deg(Q)=0+0=0$$.
All that to say that our particular solution is simply a constant...
Therefore let search for $$S_k=c$$ then $$c-6c=6\iff c=-\frac 65$$
Our general solution is then $$a_k=H_k+S_k=6^kH_0-\frac 65$$
We now solve for initial conditions:
$$a_1=6=6H_0-\frac 65\iff H_0=\frac 65$$ and we get $$E[N_k]=\frac 65(6^k-1)$$ | {
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$$a_1=6=6H_0-\frac 65\iff H_0=\frac 65$$ and we get $$E[N_k]=\frac 65(6^k-1)$$
is there some kind of general procedure that I can apply to a wide array of problems?
Around 1960 there was some work on a General Problem Solver which had very limited capabilities but much progress has been made since 1960. Still, there is no way to solve all problems and probably never will be.
However, the solution of linear recurrences is known for a long time. For example, $$a_n=6(a_{n-1}+1)=6a_{n-1}+6$$ is easily solved using the theory.
For this particular difference equation with initial value $$\,a_1\!=\!6\,$$ a lookup of $$\,6,42,258,1554\,$$ in OEIS returns OEIS sequence A105281 "a(0)=0; a(n)=6*a(n-1)+6." This entry has a program
(PARI) a(n)=if(n<0, 0, (6^n-1)*6/5)
which gives a formula for the solution.
The Mathematica program has a function RSolve which solves recurrences and also has limited capabilities
You can use the Mathematica (or WolframAlpha) command
RSolve[a[n] == 6 a[n - 1] + 6 && a[1] == 6, a[n], n]
to get the particular solution $$\,a_n = 6 (6^n-1)/5.\,$$ | {
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# Math Help - coin tossing
1. ## coin tossing
2 gamblers bet $1 each on the successive tosses of a coin. each as$6. what is the probability that
a. they break even after 6 tosses?
b. one player wins all the money on the tenth toss?
my working:
a. since there are 6 tosses and equal probability of a head and tail,
the ans should be 6 x (0.5)^6
b. in order to win all the money o the other player, he needs to win 8 games and loss 2 games, where the 10th toss have to be a win.
so, (9 choose 7) x ( 0.5)^10
but i didnt get the answer
2. Originally Posted by alexandrabel90
2 gamblers bet $1 each on the successive tosses of a coin. each as$6. what is the probability that
a. they break even after 6 tosses?
b. one player wins all the money on the tenth toss?
my working:
a. since there are 6 tosses and equal probability of a head and tail,
the ans should be 6 x (0.5)^6
b. in order to win all the money o the other player, he needs to win 8 games and loss 2 games, where the 10th toss have to be a win.
so, (9 choose 7) x ( 0.5)^10
but i didnt get the answer
hi alexandrabel90,
for the first one, they both must win 3 times each,
so that's the probability of either winning exactly 3 times and losing exactly 3 times.
for the second one, you must add the probability for the 2 players.
this is different from the first one because one or other could win the game on the 10th go, having won 7 of the preceding 9. | {
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As I read the question we have two players each having $6 to play with. Each player either wins$1 or looses $1 on each toss of the coin. So it is possible for one player of loose all in six tosses- loses six straight. So for both to breakeven in six tosses each would wins three and lose three. Is that a misreading of the question? If not, the answer to part a) is $\frac{\binom{6}{3}}{2^6}$. If it is a misreading, please clarify the question. 4. it is not a misreading of the question. your answer is correct for part (a)..i never think that we would need to count then chances of wining since there is only a 0.5 chance of a win and lose. for part (b),Archie, what do you mean by adding the probability of the 2 players? does it mean (9 choose 7) x ( 0.5)^10 x 2? 5. Hello, alexandrabel90! Two gamblers bet$1 each on the successive tosses of a coin. | {
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Each has $6. .What is the probability that: a. they break even after 6 tosses? They each win 3 games and lose 3 games. There are $_6C_3 \:=\:\frac{6!}{3!\,3!} \:=\:20$ ways this can happen. Answer: . $(20)\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\ri ght)^3 \:=\20)\,\frac{1}{64} \:=\:\frac{5}{16}" alt="(20)\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\ri ght)^3 \:=\20)\,\frac{1}{64} \:=\:\frac{5}{16}" /> b. one player wins all the money on the tenth toss? my working: In order to win all the money of the other player, he needs to win 8 games and loss 2 games, where the 10th toss has to be a win. So: . $_9C_7\left(\tfrac{1}{2}\right)^{10}$ . . . . correct, so far $_9C_7\left(\tfrac{1}{2}\right)^{10} \:=\:\tfrac{36}{1024} \:=\:\frac{9}{256}$ This is the probability that $A$ wins all of $B$'s money on the 10th game. It's also the probability that $B$ wins all of $A$'s money on the 10th game. Therefore: . $P(one\text{ player wins all the money}) \:=\:\frac{9}{256} + \frac{9}{256} \;=\;\frac{9}{128}$ 6. Originally Posted by alexandrabel90 for part (b),Archie, what do you mean by adding the probability of the 2 players? does it mean (9 choose 7) x ( 0.5)^10 x 2? Yes, that was it, because you calculated the probability of only one out of the two winning on the tenth throw. Soroban has shown that really nicely. Plato has made an interesting point also. If we interpret "break-even" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even on one's own throws while the other has more wins than losses or vice versa on their throws. In this situation, one could break even on one's own throws but make a profit if the other player does badly, or make a loss if the other player does well, "if" losing a toss loses a dollar and winning a toss wins a dollar. Or, if the two players lost every time, "player A" wins "player B"'s 6 dollars and "player B" wins "player A"'s 6 dollars, so they break even. They could also both win 1 and lose 5 | {
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"player B" wins "player A"'s 6 dollars, so they break even. They could also both win 1 and lose 5 each, win 2 and lose 4 each, win 4 and lose 2 each, win 5 and lose 1 each, win 6 and lose 0 each. That's the thing with probability questions, quite often open to alternative interpretations. The answer book will distinguish given it's answers. We've been interpreting this as..... If one player loses on a throw, then the second player wins the first player's dollar. In this case then, both would have to win 3 times and lose 3 times to end up with the 6 dollars they began with. It is probably safest to assume there is just 6 tosses rather than 6 each. 7. Originally Posted by alexandrabel90 2 gamblers bet$1 each on the successive tosses of a coin. each as $6. what is the probability that b. one player wins all the money on the tenth toss? Originally Posted by Archie Meade Yes, that was it, Plato has made an interesting point also. If we interpret "break-even" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even on one's own throws while the other has more wins than losses or vice versa on their throws. Look at the way part b) is put. It seems to me that in this game on any toss player A gets a dollar from player B or gives a dollar to player B. That is, A wins or looses on any toss. Moreover, the game is over when one player is out of money. If player A wins in each of the first six tosses she has all of the money in six tosses. If B wins five of the first six tosses he has 11 dollars and she has 1. So the question becomes “how can the game be over in exactly ten tosses?” 8. Brilliant analysis, Plato! it's not over yet. 9. I reckon that, in order for one player to win 8 to 2, the score had to stand at 5-1, at which point one player has$10 and the other has \$2 (win 5 and lose 1... 6+5-1=10, win 1 and lose 5.... 6+1-5=2). | {
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Then the score can go to 6-1 or 5-2.
From 6-1 the score must go to 6-2, since 7-1 would end the game.
From there to 7-2 and then to 8-2.
From 5-2 the score goes to 6-2, to 7-2, to 8-2.
Beginning at 5-1, there are two ways for the leading player to win on the 10th throw.
So, we count the number of ways of getting to a 5-1 lead and double that.
$2\binom{6}{5}=2\binom{6}{1}$
$P=\frac{2(6)}{2^{10}}=\frac{3}{2^8}=\frac{3}{256}$
Then double this to get the result for either player winning by 8 to 2. | {
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## ParthKohli 4 years ago $\Huge \mathsf{\text{Primitive Pythagorean Triplets.}}$
1. ParthKohli
Give it a start, Kingy.
2. KingGeorge
Well, they're triplets of the form $$(a,b,c)$$ such that $$a^2+b^2=c^2$$ and $$\gcd(a,b,c)=1\(. Unknown to many high school students, there are in fact several formulas for generating an infinite number of these triplets. 3. ParthKohli $2n,n^2 - 1, n^2 + 1$? 4. ParthKohli I know one. \(3,4,5$$.
5. KingGeorge
I just can't type that correctly can I? Well, they're triplets of the form $$(a,b,c)$$ such that $$a^2+b^2=c^2$$ and $$\gcd(a,b,c)=1$$. Unknown to many high school students, there are in fact several formulas for generating an infinite number of these triplets.
6. ParthKohli
Another one! $$5,12,13$$!
7. KingGeorge
Correct, (3,4,5) is a triplet as well as (5,12,13) and (7,24,25) among infinite others.
8. ParthKohli
Can you tell me the formula(s) for it, please?
9. KingGeorge
The most well known one is $2mn, \quad m^2-n^2, \quad m^2+n^2$Given any $$m,n$$ such that $$m>n$$.
10. ParthKohli
I see :) Are there any more?
11. KingGeorge
What you gave above, only works for $$m$$ even. So it generates an infinite number, but not all of them.
12. ParthKohli
Can this formula generate all such triplets, or you have to use some other formulae to generate more?
13. anonymous
another one $(x,y,z)=(2k+1,2k^2+2k,2k^2+2k+1)$
14. ParthKohli
Oh wait. It does generate all.
15. KingGeorge
There is also $(a,b,c)=\left(mn,\;\;\frac{m^2-n^2}{2},\;\;\frac{m^2+n^2}{2}\right)$This is a different formula that what I gave above since it only works with $$(m,n)$$ such that $$n>m\ge1$$ and $$\gcd(m,n)=1$$. This does not generate all of them.
16. KingGeorge
Ccorrection, $$m>n\ge1$$.
17. anonymous
if you want them "primitive" makes sure $$m,n$$ are opposite parity, and no common factors
18. ParthKohli
Well. I'd just use the first one :)
19. KingGeorge | {
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18. ParthKohli
Well. I'd just use the first one :)
19. KingGeorge
Satellite is correct. The first formula I wrote only gives primitive triplets for $$m,n$$ such that $$m-n$$ is odd. And $$\gcd(m,n)=1$$. However, any two positive integers will give you a triplet.
20. anonymous
algebra shows you that $$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$$ if you want a proof that every such triple is generated by such an $$m$$ and $$n$$ try working through the attachment
21. ParthKohli
If $$\gcd(m,n) = 1$$, then it automatically means that $$m - n$$ is odd.
22. anonymous
no, try 5 and 3
23. ParthKohli
Oh grr.
24. anonymous
really, if you have time, work through the worksheet i sent you you will see why all triples come from this formula
25. ParthKohli
Yes, I am looking at it.
26. anonymous
it requires no more than algebra
27. ParthKohli
All set. I got that formula well.
28. KingGeorge
There is also a multitude of fascinating properties of primitive triplets. For example: $$c$$ is always odd. 2,3,4 divide exactly one of $$a$$ or $$b$$, 5 divides exactly one of $$a,b$$ or $$c$$. $$a+b+c$$ is always even among others. These are some of the simpler facts.
29. ParthKohli
30. anonymous
have fun. there are several steps to the proof, but it is all there
31. anonymous
what do you mean by primitive triplets?
32. KingGeorge
$$\gcd(a,b,c)=1$$.
33. anonymous
"primitive" no common factors
34. anonymous
oh
35. anonymous
so 3, 4, 5 but not 6, 8, 10
36. ParthKohli
Whoa! Primitive triplets are $$a,b,c$$ such that $$\gcd(a,b,c) = 1$$. Other words, they all are co-prime.
37. anonymous
@ParthKohli method of obtaining the formula i gave u...we will talk about later if u want... http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee
38. ParthKohli
@mukushla I'd see that tutorial tomorrow; tired somewhat at the moment. Looks impressive by looking at it though.
39. anonymous
np...:)
40. ParthKohli
Haha$\Huge \ddot \smile$
41. anonymous | {
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39. anonymous
np...:)
40. ParthKohli
Haha$\Huge \ddot \smile$
41. anonymous
why do we need to have opposite parity? and what's with m-n being odd?
42. ParthKohli
Well. Isn't that the same thing?
43. KingGeorge
$$m-n$$ is odd if and only if they have opposite parity. So either $$m$$ or $$n$$ is odd, but not both.
44. anonymous
i understand that. but is it necessary for m-n to be odd?
45. KingGeorge
If you want a primitive triplet, it is.
46. KingGeorge
If they are both even or both odd, $$\gcd(a,b,c)\ge2$$
47. anonymous
how do i prove it?
48. KingGeorge
Note that $$2mn$$ is always even. Also notice that if $$m,n$$ are both odd, then $$m^2$$ and $$n^2$$ are both odd. If they are both even, $$m^2$$ and $$n^2$$ are both even. Now, $$m^2-n^2$$ and $$m^2+n^2$$ will both be even as well since even-even=even and odd-odd=odd. Hence, $$a,b,c$$ are all even, and their gcd must be divisible by 2.
49. anonymous
that was so easy and so stupid of me to not get it. i am sorry for troubling you on this. if i am to prove 5 exactly divides one of the triplet, is assuming all of them to be non divisible by 5 the right way?
50. KingGeorge
When I first did it, I assumed $$a$$ was not divisible by 5, and deduced that either $$b$$ or $$c$$ was divisible by 5. Then assume $$5$$ divides $$a$$ and show that it doesn't divide $$b$$ or $$c$$. This probably isn't the fastest way to do it thought.
51. anonymous
you can work by cases assume $$m$$ and $$n$$ are not divisible by 5 then prove that either $$m^2-n^2$$ or $$m^2+n^2$$ must be for example, if $$m\equiv n$$ mod 5, then $$m^2-n^2\equiv 0$$ mod 5 | {
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# Evaluate $\int \sqrt{1-x^2}\,dx$
I have a question to calculate the indefinite integral: $$\int \sqrt{1-x^2} dx$$ using trigonometric substitution.
Using the substitution $u=\sin x$ and $du =\cos x\,dx$, the integral becomes: $$\int \sqrt{\cos^2 u} \, \cos u \,du = \int \|{\cos u}\| \cos u\, du$$
Q: (part a) At what point (if at all) is it safe to say that this is the equivalent of ? $$\int \cos^2 u\, du = \int \frac {1 + \cos 2u} {2} du$$ (this is easy to solve, btw).
In lectures, it was made abundantly clear that over certain intervals (eg $0 \le u \le \pi/2$) that $cos u$ is +ve and is safe to do so, but in the indefinite form, the same argument cannot be made (eg $\pi/2 \le u \le n\pi$).
Q: (part b) Is it safe to declare it ok due to the nature of the original integral, which, using a sqrt() must return a +ve number? It could then be argued that it was the substitution which artificially added a -ve aspect...
Any suggestions on how to proceed?
PS: This is a 1st year calculus course and am revising for exams ;) | {
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PS: This is a 1st year calculus course and am revising for exams ;)
• The original integrand allows for $x$ to be in the interval $[-1,1]$ which means that $\cos$ can be negative. In my experience ignoring this is almost never a problem. Rather than worrying about the details of when $\cos$ is positive or negative, recall that you can always prove that your final answer is correct by showing that its derivative equals the original integrand. – Spencer Jun 22 '14 at 5:21
• The lecturer has been pounding into us the need to think correctly on fundamental concerns, which is why I raise it here. Although I understand I can show equivalency by derivation, it is not a proof (as far as I am aware) – cmroanirgo Jun 22 '14 at 5:37
• The definition of an indefinite integral is that it is the anti-derivative of the integrand (up to an additive constant). There is no more legitimate proof. – Spencer Jun 22 '14 at 5:39
• You have it backwards. To do this with trig substitution, $x=\sin u$, not $u=\sin x$. – alex.jordan Jun 22 '14 at 5:40
• @Spencer I don't think that's right. $x$ is in $[-1,1]$, and since $x=\sin(u)$, well, see my answer below. – alex.jordan Jun 22 '14 at 5:44
Since $x$ ranges from $-1$ to $1$, and you are using the substitution $x=\sin(u)$, you can make this substitution with $u\in[-\pi/2,\pi/2]$, and then $\cos(u)$ is unambiguously positive.
• It was the logic of discussion which had me confused. This clarifies the why. Recognising the limiting domain (to real numbers) from -1 to 1 is the key ingredient. – cmroanirgo Jun 22 '14 at 6:13
You have received good answer which all conclude that $\sqrt{\cos^2 x} = \cos x$. | {
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You have received good answer which all conclude that $\sqrt{\cos^2 x} = \cos x$.
But, for the time being, let us assume you still ignore if it is safe or not. So, let us write $$\int \sqrt{1-x^2} dx=\pm \int \cos^2 u\, du =\pm \int \frac {1 + \cos 2u} {2} du=\pm \Big(\frac{u}{2}+\frac{1}{4} \sin (2 u)\Big)$$ But now, let us consider the definite integral $$\int_0^1 \sqrt{1-x^2} dx=\pm \int_0^{\frac{\pi}{2}} \cos^2 u\, du =\pm \int_0^{\frac{\pi}{2}} \frac {1 + \cos 2u} {2} du=\pm \frac{\pi}{4}$$ But now, consider now the area between the curve $y=\sqrt{1-x^2}$ and the $x$ axis. All the curve is above the axis and the area is then positive. So, $\pm$ should be just replaced by $+$.
Is this making things clearer ?
Note that there is a difference between substitution as one initially learns it (let $u=g(x)$ and what in the OP is called "trigonometric substitution." The latter process, when one is feeling pedantic, is actually called inverse trigonometric substitution.
In the example we are discussing, the substitution "really is" let $u=\arcsin x$. So $u$ naturally lives in the interval $[-\pi/2,\pi/2]$, and therefore $\cos u$ is non-negative.
Hint: first integrate by parts to get,
$$\int\sqrt{1-x^2}\,\mathrm{d}x=x\sqrt{1-x^2}+\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x.$$
Now attempt the trigonometric substitution.
To answer your question: yes, it is safe and doesn't really matter. When you do trig. substitution in your first year calculus course, you are always assuming that $\cos$ is positive as a result you can do:
$$\sqrt{\cos^2 x} = \cos x$$
and not have any problems. Also, take into account what @spencer said; whatever your final answer, you can just find it's derivative and prove yourself right or wrong. | {
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• Absolutely right but for first year, as far as I know, the integral is calculated with keeping in mind that $\cos$ is in between that intervals which you specified, otherwise $\sqrt{\cos^2 x} \rightarrow \cos x$ doesn't make sense. Depending on which textbook you use, go to the trig. substitution chapter and from an example with $\cos x$ it will say specifically that $\cos x$ is bounded between $0 \le x \le \pi/2$ – Jeel Shah Jun 22 '14 at 5:40
• The bounds for the integral and the bounds for the function are not the same. I believe that is where you are confused. All we are saying is that we want to consider $\cos x$ when it is positive and therefore, bound it between those bounds. This doesn't imply that the entire integral is bounded between those. – Jeel Shah Jun 22 '14 at 5:42 | {
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# simplify $(a_1 + a_2 +a_3+… +a_n)^m$
How to simplify this best $(a_1 + a_2 +a_3+... +a_n)^m$ for $m=n, m<n, m>n$
I could only get $\sum_{i=0}^{m}\binom{m}{i}a_i^i\sum_{j=0}^{m-i}\binom{m-i}{j}a_j ...$
-
It is already as simple as it gets if you don't have other information on the $a_i$s. – Phira May 30 '12 at 15:06
they are just variable ... let's say like a and b in binomial expansion – Santosh Linkha May 30 '12 at 15:07
Note that Pascal's Tetraedron, then its $4th$ dimensional analogue, then the $5th$, is a graphical way to visualise the multinomial theorem. – Alyosha May 5 '13 at 16:59
The simplification for this type of expansion is done through the multinomial theorem. The multinomial theorem is a generalization of the binomial case to any arbitrary number of terms in the sum to be exponentiated.
The multinomial theorem is written as follows:
$$(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}x_{t}^{k_{t}}\$$
Where the multinomial co-efficient is defined as:
$${n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}$$
It may also be useful to you to note that the multinomial co-efficient is always expressible as products of binomial co-efficients [Graham, Knuth, Patashnik, Concrete Mathematics (2nd edition)]:
$${n \choose k_1, k_2, \ldots, k_m} = {x_1+x_2+\cdots+x_m \choose x_2+\cdots+x_m}\cdots{x_{m-1}+x_m \choose x_m}$$
A fuller explanation can be found on Wikipedia, or Wolfram MathWorld
-
is there any relation between $k_1, k_2, ...$ isn't $k_1+k_2+ ...+k_m=n$ again going to be combination of k's?? – Santosh Linkha May 30 '12 at 15:18
The only relation is that the sum of all $k_i$ is equal to n. I do not believe there is an equality relationship such as $k_1>k_2$ or anything like that. – Shaktal May 30 '12 at 15:20
Oh ... thank you both!! – Santosh Linkha May 30 '12 at 15:21 | {
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This is called the Multinomial theorem: $$(a_1 + a_2 +a_3+... +a_n)^m=\sum_{k_1+k_2+...+k_n=m}\frac{m!}{k_1!\cdot...\cdot k_n!}a_1^{k_1}\cdot...\cdot a_n^{k_n}$$
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# Prove that $\frac{a_1}{a_2}+ \frac{a_2}{a_3}+\frac{a_3}{a_4}+…+\frac{a_{n-1}}{a_n} \le \frac{n}{2}$
Let $n \ge 2$ be a positive integer and let $a_1, a_2, ... a_n$ be positive numbers such that $$a_1\le a_2, a_1+a_2\le a_3, a_1+a_2+a_3\le a_4, ... ,a_1+a_2+...+a_{n-1}\le a_n$$ prove that $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{n-1}}{a_n} \le \dfrac{n}{2} \hspace{2cm} (1)$$ When does the equality holds?
Solution: I proceed as follows using Mathematical Induction.
For $n=2, \frac{a_1}{a_2} \le 1$. Let the (1) be true for $n=k$ i.e $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k} \le \dfrac{k}{2} \hspace{2cm} (2)$$ We need to prove $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k+1}{2}$$ Consider $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k}+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k}{2} +\dfrac{a_{k}}{a_{k+1}} \hspace{2cm} (3)$$ Since $$a_1+a_2+...+a_{k-1}+a_{k}\le a_{k+1} \hspace{2cm} (4)$$ also $$a_1+a_2+...+a_{k-1}\le a_{k} \hspace{2cm} (5)$$ Using 5 in 4, we get $$a_{k}+a_{k}\le a_{k+1}$$ $$\dfrac{a_{k}}{a_{k+1}} \le \dfrac{1}{2}$$ using in (3), we get $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k}+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k+1}{2}$$ Is the procedure is correct. And when the equality holds... Thanks for any assistance | {
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• Your proof looks good. As to the question about when equality holds, the clues are right there in your proof. How big can $\frac{a_1}{a_2}$ be? Can you force equality? What about $\frac{a_2}{a_3}, \frac{a_3}{a_4},\ldots$ ? – quasi Jan 7 '17 at 18:36
• How do you get $a_k+a_k \leq a_{k+1}$ from (4) and (5)? – rtybase Jan 7 '17 at 18:42
• Oops -- I fell in the same trap that yasir fell in -- ignore my first comment. Instead, look at rtybase's objection. – quasi Jan 7 '17 at 18:46
• @rtybase Subtracting (5) from (4) – MattG88 Jan 7 '17 at 18:49
• @rtybase -- you can't subtract inequalites that are in the same direction. You can add them, but you can't subtract them. A common illusion (which I fell victim to as well, even though I know better). – quasi Jan 7 '17 at 19:13 | {
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The idea is "local adjustments". If $k$ is the smallest integer such that $a_1+\cdots+a_{k-1} < a_k$, we adjust up $a_1$, $\cdots$, $a_{k-1}$ with a scale factor of $s=\frac{\sum_{i=1}^{k}a_i}{2\sum_{i=1}^{k-1}a_i}$ for all, and adjust down $a_k$ by a scale factor of $t=\frac{\sum_{i=1}^{k}a_i}{2a_k}$, and achieve a large value for LFS, and make $\sum_{i=1}{k}a_i=a_k$. Note that since $\sum_{i=1}^{k}a_i$ is not changed, and $a_1$, $\cdots$, $a_{k-1}$ are scaled up by the same factor, all conditions are still satisfied. We just need to prove that really the left hand side gets larger or equal.
Really there is just one case, but let's do three cases: 1) $k=2$, and 2) $k=n$; and 3) $2<k<n$.
1) $k=2$, which means $a_1<a_2$. Now The only values changed are $\frac{a_{1}}{a_2}$ and $\frac{a_{2}}{a_{3}}$, and we want to show that $\frac{a_{1}}{a_2} + \frac{a_{2}}{a_{3}} \le \frac{sa_{1}}{ta_2} + \frac{ta_{2}}{a_{3}}$. This is easy as $s=\frac{a_1+a_2}{2a_1}$ and $t=\frac{a_1+a_2}{2a_2}$, and $sa_1=ta_2=\frac{a_1+a_2}{2}$: $$\frac{a_{1}}{a_2} + \frac{a_{2}}{a_{3}} \le \frac{a_{1}}{a_1} + \frac{a_{1}}{a_{3}} = \frac{sa_{1}}{ta_2} + \frac{a_{1}}{a_{3}} < \frac{sa_{1}}{ta_2} + \frac{ta_{2}}{a_{3}},$$ where the first inequality requires a simple check.
2) $k=n$. This is easy as The only value changed is $\frac{a_{n-1}}{a_n}$ but we scale up $a_{n-1}$ and down $a_n$. | {
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3) $2<k<n$. Now The only values changed are $\frac{a_{k-1}}{a_k}$ and $\frac{a_{k}}{a_{k+1}}$, and we want to show that $\frac{a_{k-1}}{a_k} + \frac{a_{k}}{a_{k+1}} \le \frac{sa_{k-1}}{ta_k} + \frac{ta_{k}}{a_{k+1}}$. Note that since the equality holds for $k-1$, we have $2a_{k-1}=\sum_{i=1}{k-1}a_i$. So $sa_{k-1}= \frac{2a_{k-1}+a_k}{4}$ and $ta_k=\frac{2a_{k-1}+a_k}{2}=2sa_{k-1}$. So we want to prove that $$\frac{a_{k-1}}{a_k} + \frac{a_{k}}{a_{k+1}} \le \frac{1}{2} + \frac{2a_{k-1}+a_k}{2a_{k+1}},$$ which is equivalent to the following after multiplying $2a_ka_{k+1}$: $$2a_{k-1}a_{k+1}+2a_k^2 \le a_ka_{k+1}+(2a_{k-1}+a_k)a_k,$$ which is equivalent to $$(a_{k+1}-a_k)(a_k-2a_{k-1}) \ge 0,$$ which is true since $a_{k+1}>a_k$ and $a_k \ge \sum_{i=1}^{k-1}a_i = 2a_{k-1}$.
So if we adjust up/down all $a_i$ and make all equalities in the condition hold, we achieve the largest value for $\frac{a_1}{a_2}+\cdots+\frac{a_{n-1}}{a_n}$, which happens to be $\frac{n}{2}$. | {
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Class Notes (838,022)
Mathematics (1,919)
MATH 136 (168)
Lecture 20
Lecture 20.pdf
4 Pages
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School
Department
Mathematics
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MATH 136
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Robert Sproule
Semester
Winter | {
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Wednesday, February 26 − Lecture 20: Linear independence Concepts: 1. Linearly independent set. 2. Recognize that subsets of linearly independent sets are linearly independent. 3. Characterize a linearly independent set as one being a set where no vector isa linear combination of the others. 20.1 Definition – Generalization of the definition of linear independence. Let v , v ...., v 1 2, k be k vectors in a vector space V. The vectors v , v ..., v are said to be linearly 1 2, k independent if and only if the only way that α 1 1 α v 2 2.. + α v = 0k k can hold true is if α ,1α ,2....., α kre all zeroes. If v1, v2,.., vkare not linearly independent (i.e. there existsα , α , .1...2 α not alk zero such that α v1 1α v +2 2... + α v = k k then they are said to be linearly dependent. x 20.1.1 Example – We know that the two functions e and sinx are vectors in the set F described above. Then S = Span{e , sinx} = { αe + βsinx : α, β belong to ℝ } is a x subspace of F. Show that the set {e , sinx} is linearly independent. Solution: Let us first recall that two functions f and g are equal on their domain Dif and only if f(x) = g(x) for all x in D. Suppxse αe + βsinx = 0(x) for all x (where 0(x) denotes the zero function; it maps every x in the domain to 0.). x - Suppose there exists α ≠ 0 such that αe + βsinx = 0(x). x - Then e = (–β/α)sinx for all x. This includes the value x = π. - But 0 ≠ e = (–β/α)sinπ = 0, a contradiction. - Then α must be 0. x x - Suppose β ≠ 0 is such that, 0e + βsinx = 0(x). Then (–0/β)e = sinx for all x. - Since sin(π/2) =1 ≠ 0, we have a contradiction. - So β must also be 0. - So {e , sinx} is linearly independent. 20.1.2 Proposition − The vectors M = {v , v , ..1, v2} with k , 2, are linearly independent if and only if no vector in M is a linear combination of the other vectors in M. Proof : This is similar to the previously given proof for the case of V = ℝ . n (⇐) - Suppose none of these vectors is a linear combination of the others. - Suppose M is | {
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V = ℝ . n (⇐) - Suppose none of these vectors is a linear combination of the others. - Suppose M is not linearly independent. - Then there exists α , α 1...2., α not kll zeroes such that α v + α v +1 1..+ α2 2= k k 0. - Suppose α is not zero. Rearrange the order of {v , v , ...., v } so that p = 1, .i.e., p 1 2 k α 1s not zero. - Then v = 1 (α /α )2v +1− (2 /α ) v +3, 1.., 3 − (α /α )v . k 1 k - Hence v is a linear combination of the others. Contradiction. 1 (⇒) - Suppose {v , v1, .2.., v } ks linearly independent. - Suppose one vector of {v , v , .1..,2v } is k linear combination of the others. Say it is v . 1 - Then there exists α ....2, α suck that such that v = α v + 1.... 2 2 v . k k - Then v − 1 v − 2 v2+ , ..3, 3 α v . = 0. k k - Since the coefficient of v is n1t zero this contradicts our hypothesis. - Then no vector of {v , v ,1...2, v } iska linear combination of the others. If a vector in U is a linear combination of the others we will refer to it as being “redundant”. It doesn't contribute anything to its span. 20.2 Definition – Let {v 1, v2,..., vk} be a spanning family for a vector space V. That is, Span{v , v ..., v } = V. Then for every vector v in V there exists scalars α , α , ....., α 1 2, k 1 2 k such that v = α v1 1α v +2 2... + α v . k k If for each vectorv this set of scalars α , α , ..., α satisfying v = α v + α v + ..... + α v 1 2 k 1 1 2 2 k k is unique then we say that Span{v 1, v2,.., v k satisfies the unique representation property with respect to its spanning family. 20.2.1 Example – The spanning family Span{(1, 0), (0,1)} = Span{ e , e } = ℝ 1 2 2 satisfies the unique representation property since for an arbitrary vector v= (a, b) in ℝ , a and b are the unique scalars associated to v so that v = ae + be . 1 2 20.2.2 Theorem − Let S = {v , v , .1..,2v } be a k | {
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More Less | {
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# How many ten-digit numbers are there in which every digit is 2 or 3, and no two 3s are adjacent?
How many ten-digit numbers are there in which every digit is 2 or 3, and no two 3s are adjacent?
Taken from the 2008 IMC https://chiuchang.org/wp-content/uploads/sites/2/2018/02/2008-IWYMIC-Individual.x17381.pdf
my attempt
The number of ten digit numbers in which the digits are either 2 or 3 is $$2^{10}$$ and the numbers of ten digit numbers where there is no pair of adjacent numbers that are the same is $$2$$ E.g($$2323232323$$ & $$3232323232$$) and we know that the number of ten digit numbers consisting of pairs of adjacent $$2$$s and adjacent $$3$$s is the same due to symmetry therefore the answer would be $$\frac{2^{10}-2}{2}=511$$ however this doesn't taken into account the possibilities of having adjacent pairs of $$3$$s and $$2$$s in the same number.
• This is the same as this question
– lulu
Sep 24 '19 at 18:45
• And similar to this, more recent, one. Sep 25 '19 at 5:47
Suppose you have r number of 3 in a line. Now we know 3 can't be adjacent so I have to put at least 1 in between them. For example-r=3 |-|-|(| denotes 3 and - denotes 2). Now we have x=10-(r+r-1) 2's remaining. Now we can put these 2's in r+1 slots(in the example we can put in 4 slots slot1|slot2|slot3|slot4). Now this becomes a standard problem Stars and bars of x1+x2+x3+..xl=p (where l denotes the number of slots and where every xi can be 0 and p denotes number of 2 we want to distribute which is x in our case). Number of solution of above equation is given by:
$$\binom {p+l-1}{l-1}$$
here p=10-(r+r-1) and l=r+1(slots) putting these value in above formula we get:
$$\binom {11-r}{r}$$
Now you can vary r from 0-10 and just add them you will get 144 as the answer. Here you can see you do not need to vary from 6 -10 because when we place 6 or more number we don't have enough 2's to put in between them so they will contribute 0. | {
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Suppose you have $$k$$ $$3$$s and it does not end with a $$3$$.
That the same thing as saying you have $$k$$ characters $$32$$ and $$10-k-k=10-2k$$ twos. So of the $$10-2k + k=10-k$$ characters you must choose $$k$$ spaces for the $$k$$ $$32$$ characters. There $${10-k \choose k}$$ ways to do that.
Now suppose you have $$k$$ $$3$$s and it does end with a $$3$$.
If we just ignore the last place and put the $$3$$ in it, that is the same thing as saying you have $$k-1$$ characters $$32$$ to place and $$10-k-(k-1)=11-2k$$ $$2$$s to place. There are $${10-k\choose k-1}$$ ways to do that.
So there are $${10-k \choose k}+ {10-k\choose k-1}$$ ways to place $$k$$ threes.
Now we can have at most $$5$$ threes. (Any more and we won't have enough $$2$$s to go between all $$3$$s.)
So there are
$$\sum_{k=0}^5 {10-k \choose k}+ {10-k\choose k-1}$$ ways. (Assume $${10\choose -1} = 0$$.... after all.... this would be the number of ways to choose $$0$$ threes and have a $$3$$ at the end which isn't possible.)
So $$({10\choose 0}) + ({9\choose 1}+ {9\choose 0}) +({8\choose 2}+{8\choose 1}) + ({7\choose 3}+{7\choose 2}) + ({6\choose 4}+{6\choose3}) + ({5\choose 5} + {5\choose 4})=$$
$$1 + (9+1) + (\frac {8*7}2 + 8) + (\frac {7*6*5}6 -\frac {7*6}2) + (\frac {6*5*4*3}{24}+\frac{6*5*4}6) + (1 +5)=$$
$$1 + 10 +(28+8) + (35+21) + (15+20) + 6=$$
$$11+36+56+35+5=144$$ | {
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# Finding the inverse of a number under a certain modulus
How does one get the inverse of 7 modulo 11?
I know the answer is supposed to be 8, but have no idea how to reach or calculate that figure.
Likewise, I have the same problem finding the inverse of 3 modulo 13, which is 9.
The inverses can be computed using the extended euclidean algorithm. As long as $\gcd(x,n) = 1$ the inverse $x^{-1} \bmod n$ exists and is $y$ from the extended euclidean algorithm where $$xy + kn = 1 = \gcd(x,n)$$ For example we get $$7\cdot 8 - 5\cdot 11 = 1\\ 9 \cdot 3 - 2\cdot 13 = 1$$
Note that you can also obtain negative numbers, for example $$7\cdot (-3) + 2\cdot 11 = 1$$ In this case you can use congruences $\bmod n$ to obtain the canonical inverse: $-3 \equiv 8 \pmod{11}$
• Thanks but I'm really struggling with this, could you provide me with a sample, i be stuck for two days trying to figure it out – Dan W May 4 '15 at 12:07
• @DanW What part of this do you struggle with? – AlexR May 4 '15 at 12:09
• Basically, i want to know the inverse of 7 within mod 11, i need a figure!!, i understand the gcd and what it is doing/proving. – Dan W May 4 '15 at 12:20
• @DanW Can you show me what the extended euclidean algorithm gives you wen used for $\gcd(7,11)$? – AlexR May 4 '15 at 12:23
• A B Q R 11 7 1 4 7 4 1 3 4 3 1 1 – Dan W May 4 '15 at 12:28
To find the inverse of $7$, $\pmod{11}$, you must find a solution to $7x\equiv 1 \pmod{11}$. (Then $x$ satisfies the definition of inverse.)
As suggested in other answers, one way to do this is with the extended Euclidean algorithm, and in fact this is the best general purpose algorithm for this type of problem.
But for small values, you can also try 'adding the modulus':
$7x\equiv 1\equiv 12\equiv 23\equiv 34\equiv 45\equiv 56 \pmod{11}$.
Then from $7x\equiv 56\pmod{11}$, we can cancel $7$, obtaining $x\equiv 8 \pmod{11}$. | {
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Then from $7x\equiv 56\pmod{11}$, we can cancel $7$, obtaining $x\equiv 8 \pmod{11}$.
Here's an illustration of finding the multiplicative inverse of $37 \bmod 100$ using the extended Euclidean algorithm. (I used bigger numbers for this example so that the relationships are a little clearer).
On each line, $n=100s+37t$. We start the table with two lines giving $n=100$ and $n=37$ in the obvious way. Then $q$ gives the rounded-down ratio of the current and previous value of $n$. Using this, the next line is calculated by subtracting $q$ copies of the current line entries from the previous line entries.
$$\begin{array}{c|c|c|c} n & s & t & q \\ \hline 100 & 1 & 0 & \\ 37 & 0 & 1 & 2 \\ 26 & 1 & -2 & 1 \\ 11 & -1 & 3 & 2 \\ 4 & 3 & -8 & 2 \\ 3 & -7 & 19 & 1 \\ 1 & 10 & \color{red}{-27} & 3 \\ \end{array}$$
On the last line, $1 = 10\times 100 + (-27)\times 37$, so $\color{red}{-27}\times 37 \equiv 1 \bmod 100$
Bringing the result positive, $-27 \equiv \color{red}{73} \bmod 100$. And $37 \times 73 = 2701 \equiv 1 \bmod 100 \quad \checkmark$.
As you can perhaps see, you don't actually need to calculate the $s$ values at all to get the modular inverse. I left them in to help understand the table.
The corresponding tables for your particular questions: $$\begin{array}{c|c|c|c} n & s & t & q \\ \hline 11 & 1 & 0 & \\ 7 & 0 & 1 & 1 \\ 4 & 1 & -1 & 1 \\ 3 & -1 & 2 & 1 \\ 1 & 2 & \color{red}{-3} & 3 \\ \end{array}$$ and $7^{-1} \equiv -3 \equiv \color{red}8 \bmod 11$. $\quad 7\times 8 = 56 = 55+1\quad \checkmark$
$$\begin{array}{c|c|c|c} n & s & t & q \\ \hline 13 & 1 & 0 & \\ 3 & 0 & 1 & 4 \\ 1 & 1 & \color{red}{-4} & 3 \\ \end{array}$$ and $3^{-1} \equiv -4 \equiv \color{red}9 \bmod 13$. $\quad 3\times 9 = 27 = 26+1\quad \checkmark$
${\rm mod}\ 11\!:\,\ \dfrac{1}7\equiv \dfrac{12}{-4}\equiv -3\$ (see Gauss's algorithm. for an algorithmic version of this). | {
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Or, compute the Bezout identity $\,\gcd(11,7) = 2(11)-3(7) = 1\,$ by the Extended Euclidean Algorithm (see here for a convenient version). Thus $\ {-}3(7)\equiv 1\pmod{11}$
${\rm mod}\ 13\!:\,\ \dfrac{1}3\equiv \dfrac{-12}{3}\equiv -4\$
Generally inverting $\,a\,$ mod $\,m\,$ is easy if $\,a\mid m\pm1,\$ i.e. $\ m = ab\pm 1$
${\rm mod}\ ab-1 \!:\qquad ab\equiv 1\,\Rightarrow\, a^{-1}\equiv b$
${\rm mod}\ ab+1 \!:\,\ a(-b)\equiv 1\,\Rightarrow\, a^{-1}\equiv -b$
Above are special cases: $\,3\mid 13\!-\!1\$ and $\ 7\equiv -4\mid 11\!+\!1$
To find the inverse of $7$ modulo $11$, we must solve the equivalence $7x \equiv 1 \pmod{11}$. To do this, we use the Extended Euclidean Algorithm to express $1$ as a linear combination of $7$ and $11$. The coefficient of $7$ will be the inverse modulo $11$. By the Euclidean Algorithm, \begin{align*} 11 & = 1 \cdot 7 + 4\\ 7 & = 1 \cdot 4 + 3\\ 4 & = 1 \cdot 3 + 1\\ 3 & = 3 \cdot 1 \end{align*} We now take the equation $4 = 1 \cdot 3 + 1 = 3 + 1$, solve for $1$, then work backwards until we obtain $1$ as a linear combination of $7$ and $11$. \begin{align*} 1 & = 4 - 3\\ & = 4 - (7 - 4)\\ & = 2 \cdot 4 - 7\\ & = 2(11 - 7) - 7\\ & = 2 \cdot 11 - 3 \cdot 7 \end{align*} Since $2 \cdot 11 - 3 \cdot 7 = 1$, $$-3 \cdot 7 = 1 - 2 \cdot 11 \Longrightarrow -3 \cdot 7 \equiv 1 \pmod{11}$$ Hence, $-3$ is the inverse of $7 \pmod{11}$. To express the inverse as one of the residues $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, we add $11$ to $-3$ to obtain $-3 + 11 \equiv 8 \pmod{11}$. Hence, $7^{-1} \equiv 8 \pmod{11}$.
Check: $7 \cdot 8 \equiv 56 \equiv 1 + 5 \cdot 11 \equiv 1 \pmod{11}$.
To verify you understand the algorithm, try to find the inverse of $3$ modulo $13$.
As said by AlexR, you can find $x,y\in\mathbb Z$ with $7x+11y=1$ using Extended Euclidean algorithm (see this answer for how to best use it).
You can also use elementary modular arithmetic 'tricks':
$\bmod{11}\!:\ 7x\equiv 1\equiv -21\stackrel{:7}\iff x\equiv -3\equiv 8$. | {
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$\bmod{11}\!:\ 7x\equiv 1\equiv -21\stackrel{:7}\iff x\equiv -3\equiv 8$.
$\bmod{11}\!:\ 7x\equiv -4x\equiv 1\equiv 12\stackrel{:(-4)}\iff x\equiv -3\equiv 8$.
We could divide by $7$ and $-4$ because $(7,11)=(-4,11)=1$. | {
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# Number of Lattice Points in a Triangle
Problem
Let the co-ordinates of the vertices of the $\triangle OAB$ be $O(1,1)$, $A(\frac{a+1}{2},1)$ and $B(\frac{a+1}{2},\frac{b+1}{2})$ where $a$ and $b$ are mutually prime odd integers, each greater than $1$. Then find the number of lattice points inside $\triangle OAB$, i.e., not on the borders of $\triangle OAB$. How does the answer change if the restriction $\operatorname{gcd}(a,b)=1$ is removed?
Solution
Let $L(a,b)$ be the number of lattice points inside $\triangle OAB$. The linear transformation $T(x,y)=(x-1,y-1)$ on the triangle $OAB$ do not have any effect on $L(a,b)$. Hence $L(a,b)$ is equal to the number of lattice points inside the triangle $O'A'B'$, where $O'=(0,0)$, ${\textstyle A'=(\frac{a-1}{2},0)}$ and ${\textstyle B'=(\frac{a-1}{2},\frac{b-1}{2})}$.
Set ${C'=(0,\frac{b-1}{2})}$. By symmetry the number of lattice points inside the rectangle $O'B'C'$ is $L(a,b)$. The number of lattice point inside the rectangle $O'A'B'C'$ is ${\textstyle \frac{a-3}{2} \cdot \frac{b-3}{2}}$. Consequently
$$2L(a,b) = \frac{a-3}{2} \cdot \frac{b-3}{2} - K$$
where $K$ are the number of lattice points on the straight line between $O'$ and $B'$. This line is given by $y= \frac{b-1}{a-1}x$, yielding
$$K = |\{ {\textstyle 0 < x < \frac{a-1}{2} \mid \frac{b-1}{a-1}x \in \mathbb{N} \}| }$$
By letting $d=\text{gcd}(a-1,b-1)$, we obtain $a-1=rd$ and $b-1=sd$ for two positive coprime integers $r$ and $s$. Therefore
$$\frac{b-1}{a-1}x = \frac{sx}{r} \in \mathbb{N} \;\; \Leftrightarrow \;\; \frac{x}{r} \in \mathbb{N}$$
Now ${\textstyle 0 < x < \frac{a-1}{2} = \frac{rd}{2}}$, implying ${\textstyle 0 < \frac{x}{r} = \frac{d}{2}}$. Since $d$ is even (since $d=\gcd(a-1,b-1)$ and $a$ and $b$ are both odd positive integers) give us ${\textstyle K = \frac{d}{2}-1}$. Thus by (1)
$${\textstyle L(a,b) = \dfrac{(a-3)(b-3)}{8} - \dfrac{\text{gcd}(a-1,b-1)}{4} + \dfrac{1}{2}}$$
But I think that by Pick's Theorem the answer should be, | {
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But I think that by Pick's Theorem the answer should be,
$${\textstyle L(a,b) = \dfrac{(a-1)(b-1)}{8} - \dfrac{\text{gcd}(a-1,b-1)}{4} + \dfrac{1}{2}}$$
Which one is correct?
• You can easily figure out which one is correct by putting in some small values for $a$ and $b$. – Ted Sep 14 '14 at 6:06
• Try $a = 1$ and $b = 10$. This yields a degenerate triangle with $0$ lattice points strictly inside it. Which answer gives you $0$? – JimmyK4542 Sep 14 '14 at 6:17
• @JimmyK4542: Both $a$ and $b$ are odd and mutually prime and each is greater than $1$. – user170039 Sep 14 '14 at 6:26
• Fine, try a less trivial case like $a = 3$ and $b = 5$. The formula in the gray box gives $L(3,5) = 0$ while the other formula gives $L(3,5) = 1$. Which one is right? – JimmyK4542 Sep 14 '14 at 6:27
• There are $0$ lattice points inside the triangle with vertices $O(1,1)$, $A(2,1)$, $B(2,3)$. So the first formula is correct (in that case). I have no idea where the second formula goes wrong because you have left out the details of how you got that. – JimmyK4542 Sep 14 '14 at 6:34
The area of $\Delta OAB$ is $K = \dfrac{1}{2} \cdot \dfrac{a-1}{2} \cdot \dfrac{b-1}{2} = \dfrac{(a-1)(b-1)}{8}$.
The number of points on the boundary (line segments $OA$, $AB$, and $BO$) is
$\underbrace{\dfrac{a-1}{2}}_{OA}+\underbrace{\dfrac{b-1}{2}}_{OB}+\underbrace{\text{gcd}\left(\dfrac{a-1}{2},\dfrac{b-1}{2}\right)}_{BO}$ (you missed the first two terms of this).
So, by Pick's Theorem, the area is
$K = I+\dfrac{1}{2}B-1 = I+\dfrac{1}{2}\left[\dfrac{a-1}{2}+\dfrac{b-1}{2}+\text{gcd}\left(\dfrac{a-1}{2},\dfrac{b-1}{2}\right)\right]-1$
$= I + \dfrac{a-1}{4} + \dfrac{b-1}{4}+\dfrac{\text{gcd}(a-1,b-1)}{4}-1$.
Hence, the number of points in the interior is
$I = \dfrac{(a-1)(b-1)}{8} - \dfrac{a-1}{4} - \dfrac{b-1}{4} - \dfrac{\text{gcd}(a-1,b-1)}{4} + 1$
$= \dfrac{(a-3)(b-3)}{8} - \dfrac{\text{gcd}(a-1,b-1)}{4} + \dfrac{1}{2}$ (after a bit of simplification). | {
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# What is the combinatorial interpretation of the product of binomial coefficients?
Full disclosure: This question is relating to a homework question. It's not a homework question itself, but rather a clarifying question to help myself get a handle on the actual question.
Suppose I had three sets $X$, $Y$, and $Z$ with cardinalities $|X| = x$, $|Y| = y$, and $|Z| = z$, with $z \leq y \leq x$
I know that $x \choose y$ is the number of ways of selecting $y$ elements from $X$.
However, I'm unclear of the combinatorial interpretation of the following:
$${x \choose y} {y \choose z}$$
So my question is: in general, is there an intuitive interpretation for the product of the number of choices?
• If $x\le y\le z$ then both of those coefficients are $0$ – Adam Hughes Sep 19 '14 at 1:50
• You're right, sorry; I'm updating my question now – Jason Baker Sep 19 '14 at 1:51
• There's two interpretations. 1) How many ways are there to first pick $y$ elements out of $x$, and then pick $z$ elements out of these $y$? 2) How many ways are there to divide $x$ into sets of size $z$, $y-z,$ and $x-y$? – Semiclassical Sep 19 '14 at 1:54
• @Semiclassical That makes sense; thanks for the confirmation – Jason Baker Sep 19 '14 at 1:57
The answer to your question is affirmative: Yes, there is an intuitive interpretation for the number of choices.
We know at least two interpretations of the binomial coefficient $\binom{n}{k}$ which are useful for combinatorial proofs. These are
• Set theoretical view: $\binom{n}{k}$ as number of $k$-element subsets of an $n$-element set
• Geometrical view: $\binom{n}{k}$ as number of lattice pathes of length $n$ containing $k$ horizontal $(1,0)$-steps and $n-k$ vertical $(0,1)$-steps
The geometrical view is valid because there are $\binom{n}{k}$ choices to select $k$ steps in horizontal direction leaving the remaining $n-k$ steps for the vertical direction.
Let's apply the second approach to | {
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Let's apply the second approach to
\begin{align*} \binom{x}{y}\binom{y}{z}\qquad z \leq y \leq x\tag{1} \end{align*}
When looking at $\binom{y}{z}$ you may think on all pathes of length $y$ starting at $A=(0,0)$ and ending in $B=(z,y-z)$ thereby containing exactly $z$ horizontal $(1,0)$-steps and $y-z$ vertical $(0,1)$-steps. All these pathes are enclosed within a rectangle of length $z$ and height $y-z$.
Similarly we consider $\binom{x}{y}$ as the number of lattice pathes of length $x$ containing $y$ $(1,0)$-steps and $x-y$ $(0,1)$-steps. But here we let all these pathes start in $B=(z,y-z)$ and so they end up in $C=(z+y,x-z)$.
We observe following
Combinatorial interpretation of $$\binom{x}{y}\binom{y}{z}$$
This is the number of lattice pathes of length $x+y$ starting in $A=(0,0)$ ending up in $C=(z+y,x-z)$ and passing through the point $B=(z,y-z)$.
The pathes contain only horizontal $(1,0)$-steps and vertical $(0,1)$-steps.
A possible interpretation is pairing sets $X$ and $Y$ with no elements of $Y$ left so that would require selecting $y$ elements from $X$ or $\displaystyle \binom xy$, then making triplets with no elemts of $Z$ left that would require selecting $z$ pairs from these $y$ pairs or $\displaystyle \binom yz$. | {
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# Number of points of accumulation of a sequence
Can a sequence have infinitely many points of accumulation i.e. we can extract infinitely many subsequences from it s.t. they all converge to their respective point of accumulation? I have the feeling it would mean that the period of repetition of something could be infinitely big.
• Since there are countably many rational numbers, there is a sequence that includes them all. The accumulation points of this sequence is the entire set of real numbers. Apr 29 '16 at 17:29
• I like your example a lot ! Apr 29 '16 at 18:15
• There's a classic theorem that if you walk around a circle in discrete steps of $a$ radians, where $a/\pi$ is irrational, then the set of points that you visit is dense in the circle. This implies that $(\sin(an))_{n\in\mathbb N}$ is dense in $[-1, 1]$, and you can take $a=1$ for a cute example of a sequence with infinitely many accumulation points. Apr 29 '16 at 23:16
• See also this question. Some of the posts linked there might be of interest, too. Apr 30 '16 at 8:23
Start with $0,1$. Then travel backwards to $0$ in steps of $1/2$, so $1/2,0$. Then travel forwards to $1$ in steps of $1/4$, so $1/4,2/4,3/4,1$. Then travel backwards to $0$ in steps of $1/8$, so $7/8$, $6/8$, $5/8$, and so on. Continue.
Every real between $0$ and $1$ is an accumulation point of our sequence. | {
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Every real between $0$ and $1$ is an accumulation point of our sequence.
• Your sequence is bounded AND has infinitely many points of accumulation. Thank you for answering :) Apr 29 '16 at 16:28
• @MehdiSlimani: You are welcome. If you prefer we could use $0,1,1/2,0,1/3,2/3,1,3/4,2/4,1/4,0,1/5,2/5,\dots$ and then we can quote the result that the rationals in $[0,1]$ are dense in $[0,1]$. Apr 29 '16 at 16:29
• I'm a bit confused. How can subsequences of your sequence converge if each member is in a different space? Maybe I don't have the necessary knowledge yet Apr 29 '16 at 16:36
• For every real $x$ between $0$ and $1$, there is a subsequence of our sequence that has limit $x$. For in our back and forth travels, we get within $1$ of $x$, then within $1/2$ of $x$, then within $1/4$ of $x$, and so on. With the modified version in which we take steps $1$, then $1/2$, then $1/3$, then $1/4$, then $1/5$, and so on, our travels bring us to every rational between $0$ and $1$, so again there is a subsequence that converges to $x$. Apr 29 '16 at 16:39
• Ok, I hadn't understood it well. Thank you again Apr 29 '16 at 16:43
Yes, this is possible. For example consider the sequence $a_n$ for $n \ge 2$ defined as the smallest divisor of $n$ greater than $1$.
The accumulation points are all the prime numbers. Subsequences witnessing them are for instance the $p$-th powers for each prime $p$.
• Simple and nice example, thank you a lot :) Apr 29 '16 at 16:24
The rationals are a countable set. We can define a 1-to-1 function $f:N\to Q$ with $Q=\{f(n):n\in N\}.$ Consider the sequence $S=(f(n))_{n\in N}.$ Every real number is a limit point of a subsequence of $S.$
All the answers have uncountably many accumulation points. If you only want countably many, consider the sequence:
$1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,\cdots$
Every positive integer is an accumulation point, and nothing else.
If you further want it to be bounded: | {
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If you further want it to be bounded:
$\frac11,\frac11,\frac12,\frac11,\frac12,\frac13,\frac11,\frac12,\frac13,\frac14,\cdots$
Suppose each row of the infinite matrix below converges to a different real number
\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & \dots \\ a_{21} & a_{22} & a_{23} & a_{24} & \dots \\ a_{31} & a_{32} & a_{33} & a_{34} & \dots \\ a_{41} & a_{42} & a_{43} & a_{44} & \dots \\ \vdots & \vdots & \vdots & \vdots & \end{bmatrix}
Then the sequence
$a_{11}, a_{21}, a_{12}, a_{31}, a_{22}, a_{13}, a_{41},a_{32}, a_{23}, a_{14}, a_{51}, \dots$
contains an infinite number of convergent subsequences.
• Some authors define accumulation point not so strictly, namely that it is enough that a subsequence converges to it. Apr 30 '16 at 9:37 | {
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# How many random selections needed to be 90%, 95%, 99% confident that a specific selection occurred?
For example, let's say that I have a bag of 100 marbles labeled 1 to 100. A "selection" is randomly picking a marble from the bag and then placing it back into the bag. The marble selected is not known.
How many random selections do I need to make to be X% confidence that a specific marble was one of the marbles I selected, say marble #42?
Is there a formula or a distribution that helps me solve for X (90%, 95%, 99% confidence)?
Bernoulli or Binomial distributions seem close, but I can not quite make them solve for what I want.
Each time you grab a marble you have a 1 out of 100 chance of selecting the correct one. What you are asking is what are the odds that at least one of these will be correct which would be an inclusive "or" statement
$$P_{selected} = p \cup p \cup p \cup p \cup p... \cup p$$
However this is not trivial to calculate since you could select your marble multiple times. Thus it's better to use the negative of this and calculate when it was not selected.
$$P_{selected}^\sim = (1-p)\cap(1-p)\cap(1-p)(1-p)\cap(1-p)...(1-p)\cap(1-p)$$
which can be easily calculated to N times by
$$P_{selected}^\sim = (1-p)^N$$
Where you can then show
$$P_{selected} = 1- P_{selected}^\sim = 1-(1-p)^N$$
and after some algebra
$$N = \frac{\ln(1-P_{selected})}{\ln(1-p)}$$
so 90% confidence is N = 229.
95% confidence is N = 298
99% confidence is N = 458.
If $X$ has $k$ distinct values, each appearing with equal probability $1/k$, then it follows discrete uniform distribution. Probability of drawing a specific $x$ in a single draw is
$$\Pr(X=x) = \frac{1}{k}$$
probability of drawing value other than $x$ is
$$\Pr(X \ne x) = \frac{k-1}{k} = 1-\frac{1}{k}$$
so if you make $n$ draws, probability that in neither of the draws $x$ is drawn is
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$$\left(1-\frac{1}{k} \right)^n$$
from here you can easily find that probability of drawing at least one $x$ in $n$ draws is
$$1-\left(1-\frac{1}{k} \right)^n$$
Now all you need to know is find such $n$ that leads to prespecified probability.
This can be easily checked by simulation.
set.seed(123)
f <- function(n, k = 100) 1-((1-1/k)^n)
g <- function() which(sample(100, 1e4, replace = TRUE) == 42)[1]
sim <- replicate(5e4, g())
that returns
> quantile(sim, c(0.9, 0.95, 0.99))
90% 95% 99%
230.00 301.00 460.01
> f(c(230, 301, 460))
[1] 0.9008952 0.9514495 0.9901782
(Notice that example above concerns not with confidence intervals but with exact probabilities.)
This shows us that we need to sample $n$ values until $x$ occurs, so we are talking about negative binomial distributed variable with parameters $p=1/k$ and $r=1$ (observe $n$ values until $X=x$ occurs once, where probability of observing $x$ is $1/k$). Results obtained using negative binomial distribution agree with the simulation.
> pnbinom(c(230, 301, 460)-1, 1, 1/k)
[1] 0.9008952 0.9514495 0.9901782
So you intuition about binomial distribution was close, as it follows negative binomial distribution. Negative binomial with $r=1$ simplifies to geometric distribution. This is how we ended up with distribution that is traditionally used for modeling discrete waiting times.
> pgeom(c(230, 301, 460)-1, 1/k)
[1] 0.9008952 0.9514495 0.9901782
In fact, if you recall, cumulative distribution function for geometric distribution is
$$F(n) = 1 - (1-p)^n$$
where $p$ is a probability of observing $x$ in a single draw. So geometric distribution can be used to obtain confidence intervals for the number of samples that need to be taken.
# 95% CI using quantile function of geometric distribution
> qgeom(c(0.025, 0.975), 1/k)
[1] 2 367
Let $M=100$. Let's say you make $N$ selections and you are looking for the occurrence of marble $m$. | {
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Let $M=100$. Let's say you make $N$ selections and you are looking for the occurrence of marble $m$.
Let $E_n$ be the event that marble $m$ is drawn at draw $n$. Let $F_n=E_n^c$ be the event that marble $m$ is NOT drawn.
Note that $P(E_n)=1/M$. This means, $P(F_n)=1-1/M$.
Let $F$ be the failure event that marble $m$ is not drawn for any $N$. Success is then $F^c$.
$P(F^c)=1-P(F)$
Assuming independent draws, we have:
$P(F)=P(F_1 \cap F_2 \cap F_3 \cap \ldots \cap F_N)=\prod_{n=1}^N P(F_n)=(1-1/M)^N$
Therefore, $P(F^c)=1-(1-1/M)^N$.
For the $c \in [0,1]$ (e.g. $c=0.9 \equiv 90\%$ confidence) confidence, set the LHS to $c$ and solve for $N$.
$\Rightarrow N=\dfrac{\ln(1-c)}{\ln(1-1/M)}$ | {
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# Linear Algebra: Preserving the null space
What does it mean when a book says that row operations preserve the null space? And why should that be true? I have read that row operations are equivalent to multiplying a vector on the left by an invertible elementary matrix. And I think I understand that the nullspace is the set of all vectors from $u \in U$ which get mapped to the zero vector in $V$ if $T:U\rightarrow V$ is linear. But I'm still not sure what this means.
-
It means that performing an elementary row operation on a matrix does not change the null space of the matrix. That is, if $A$ is a matrix, and $E$ is an elementary matrix of the appropriate size, then the matrix $EA$ has the same null space as $A$.
To see why this is true, suppose first that $x$ is in the null space of $A$. This means that $Ax=\vec 0$. Multiplying both sides of this equation by $E$, we see that $(EA)x=E\vec 0 =\vec 0$, meaning that $x$ is also in the null space of $EA$. Now suppose that $x$ is in the null space of $EA$, so that $(EA)x=\vec 0$. As you mentioned, $E$ is invertible, so we can multiply this equation by $E^{-1}$:
$$Ax=IAx=(E^{-1}E)Ax=E^{-1}(EA)x=E^{-1}\vec 0=\vec 0\;,$$
showing that $x$ is in the null space of $A$. In other words, a vector is in the null space of $EA$ if and only if it is in the null space of $A$, and $EA$ and $A$ have the same null space.
-
Thanks a lot, the if and only if explanation helps me connect these facts together now! – mcrocker Feb 12 '12 at 4:32
Say we have $V, W$ vector spaces and a linear transformation $T : V \to W$. The null space of $T$ is defined $N(T) = \{x \in V : T(x) = 0\}$.
When talking about row operations, we are talking about the matrix representation of the linear transformation. So if $\alpha = \{ x_1, ..., x_n \}$ is a basis of $V$ and $\beta = \{ y_1, ..., y_n \}$ is a basis of $W$ we have a matrix, say $A$, such that $A = [T]_\alpha^\beta$. | {
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Say $T(x_i) = a_{i1}y_1 + ... + a_{in}y_n$. Then the matrix A will be the following $$A = \left[\begin{array}{ccc} a_{11} & ... & a_{1n} \\ ... & ... & ... \\ a_{n1} & ... & a_{nn} \end{array}\right].$$
Row operations are as follows
(i) Swap two rows.
(ii) Multiply row by non zero constant.
(iii) Add multiple of one row to a different row.
Say we swap the $i$th and $j$th row and call the new matrix $A'$. This new matrix defines a new linear transformation $T':V \to W$ such that $T'(x) = A'x$. We want to show that $N(T) = N(T')$. For any $x \in V$, say $$T(x) = b_1y_1 + ... + b_iy_i + ... + b_jy_j + ... + b_ny_n.$$ Then since $T'$ is just rows swapped $$T'(x) = b_1y_1 + ... +b_jy_j + ... + b_iy_i + ... b_ny_n.$$ Thus if $T(x) = 0$, $T'(x) = 0$ and vice versa. So $N(T) = N(T')$.
For (ii), think about if we multiply the $i$th row by $\beta$, then if the $i$th scalar for $T(x)$ is $c_i$, then the $i$th scalar for $T'(i)$ is $\beta c_i$. So if $T(x) = 0$, then $c_i = 0$, so $\beta c_i = 0$. All other scalars stay the same, thus $T'(x) = 0$. And if $T'(x) = 0$, $\beta^{-1}c_i = 0$, so $T(x) = 0$. Once again $N(T) = N(T')$.
I leave (iii) to you.
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For what it's worth, I think the following way of thinking about why row operations do not change the null space is worthwhile:
An important observation to make:
One may observe that multiplication of a matrix $A$ by a column vector $\bf x$ amounts to taking a linear combination of the columns of $A$. The coefficients of the particular linear combination are the coordinates of $\bf x$. | {
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For example: $$\tag{1} \underbrace{\left[ \matrix{a_{11}&a_{12}&a_{13} \cr a_{21}&a_{22}&a_{23}\cr a_{31}&a_{32}&a_{33} }\right]}_{A} \left[\matrix{x\cr y\cr z}\right] =\left[\matrix{x a_{11} +ya_{12}+ z a_{13} \cr x a_{21} +ya_{22}+ z a_{23} \cr x a_{31} +ya_{32}+ z a_{33}} \right] = x\underbrace{\left[\matrix{\color{maroon}{a_{11}}\cr\color{darkgreen}{ a_{21}}\cr \color{darkblue}{a_{31}}}\right]}_{\bf a_1} +y\underbrace{\left[\matrix{\color{maroon}{a_{12}}\cr\color{darkgreen}{ a_{22}}\cr \color{darkblue}{a_{32}}}\right]}_{\bf a_2} +z\underbrace{\left[\matrix{\color{maroon}{a_{13}}\cr\color{darkgreen}{ a_{23}}\cr\color{darkblue}{ a_{33}}}\right]}_{\bf a_3}$$
A second important observation to make:
$\bf x$ is in the null space of $A$ if and only if $A{\bf x}=\bf0$. This means that the linear combination of the columns of $A$ whose coefficients are the coordinates of ${\bf x}=\Bigl[{\textstyle{x\atop y}\atop\scriptstyle z }\Bigr]$ is the zero vector:
$$\tag{2} x\underbrace{\left[\matrix{\color{maroon}{a_{11}}\cr\color{darkgreen}{ a_{21}}\cr \color{darkblue}{a_{31}}}\right]}_{\bf a_1} +y\underbrace{\left[\matrix{\color{maroon}{a_{12}}\cr\color{darkgreen}{ a_{22}}\cr \color{darkblue}{a_{32}}}\right]}_{\bf a_2} +z\underbrace{\left[\matrix{\color{maroon}{a_{13}}\cr\color{darkgreen}{ a_{23}}\cr\color{darkblue}{ a_{33}}}\right]}_{\bf a_3} =\bf0$$
Now on to row operations:
Here, we make our third and final observation:
If one performs a row operation on $A$, then the corresponding right hand side of $(1)$ is obtained by performing the same row operation to each of ${\bf a_1}$, ${\bf a_2}$, and ${\bf a_3}$. | {
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It follows from all of this that multiplying a row of $A$ by a non-zero number does not affect the null space. For example if row 1 of $A$ were multiplied by 2 then the right hand side of (1) would be: $$x\underbrace{\left[\matrix{\color{maroon}{2a_{11}}\cr\color{darkgreen}{ a_{21}}\cr \color{darkblue}{a_{31}}}\right]}_{\bf a_1} +y\underbrace{\left[\matrix{\color{maroon}{2a_{12}}\cr\color{darkgreen}{ a_{22}}\cr \color{darkblue}{a_{32}}}\right]}_{\bf a_2} +z\underbrace{\left[\matrix{\color{maroon}{2a_{13}}\cr\color{darkgreen}{ a_{23}}\cr\color{darkblue}{ a_{33}}}\right]}_{\bf a_3} =\bf z$$ If (2) holds, it is easily seen that ${\bf z}=\bf0$.
It should be fairly obvious that interchanging two rows of $A$ does not affect the null space. For example if rows 1 and 3 of $A$ were interchanged, the right hand side of (1) would become $$x\left[\matrix{\color{darkblue}{ a_{31}}\cr \color{darkgreen}{ a_{21}}\cr \color{maroon}{ a_{11}}}\right] +y\left[\matrix{\color{darkblue}{ a_{32}}\cr \color{darkgreen}{ a_{22}}\cr \color{maroon}{ a_{12}}}\right] +z\left[\matrix{\color{darkblue}{ a_{33}}\cr\color{darkgreen}{ a_{23}}\cr\color{maroon}{ a_{13}}}\right] =\bf z$$ If (2) holds, then $\bf z$ would be $\bf 0$.
Finally, if a multiple of a row of $A$ were added to another row of $A$, the null space would be unchanged. For example if twice row 1 of $A$ were added to to row 2, then the right hand side of $(1)$ would become:
$$\tag{3} x\left[\matrix{\color{maroon}{ a_{11}}\cr \color{darkgreen}{ a_{21}}+2\color{maroon}{ a_{11}}\cr\color{darkblue} a_{31}}\right] +y\left[\matrix{\color{maroon}{ a_{12}}\cr\color{darkgreen}{ a_{22}}+2 \color{maroon}{a_{12}}\cr\color{darkblue} a_{32}}\right] +z\left[\matrix{\color{maroon}{ a_{13}}\cr\color{darkgreen}{ a_{23}}+2 \color{maroon}{a_{13}}\cr\color{darkblue} a_{33}}\right] =\bf z$$ | {
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If $(2)$ holds, then $\bf z=0$. In particular, looking at the second component of the left hand side of $(3)$: \eqalign{ x(\color{darkgreen}{a_{21}}+2\color{maroon}{a_{11}})+ y(\color{darkgreen}{a_{22}}+2\color{maroon}{a_{12}})+ z(\color{darkgreen}{a_{23}}+2\color{maroon}{a_{13}}) &= (x \color{darkgreen}{a_{21}}+ y \color{darkgreen}{a_{22}}+ z \color{darkgreen}{a_{23}} )\cr &\ \ \ \ \ \ \ +2( x\color{maroon}{a_{11}} + y\color{maroon}{a_{12}}+ z\color{maroon}{a_{13}} )\cr&=0+2\cdot0\cr&=0}.
-
I think the most natural way to think about this is to note that the kernel of a matrix is also the orthogonal complement to the column-space of it's transpose. (this is the fundamental theorem of linear algebra)
So when you do row operations on a matrix, this is the same as doing linear combinations of the columns in it's transpose. This doesn't change the overall space spanned by those columns in the transpose, so it doesn't change the kernel of the original matrix.
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# Is it more practice and intuition or rather algorithmic to solve third degree polynomials of this type?
Consider
$x^3 - 6x^2 + 11x - 6 = 0$
I can not reasonably factor this intuitively in any short amount of time with my skill level. Is this the only hope to solving such equations by hand? What tools can I use to help ease the pain or make it more clear to myself?
I try to:
$(x + ?)(x^2 - 6x)$ but it does not seem promising, in my eyes
I cant do anything, that I know about, with:
$x(x^2 - 6x + 11) = 6$
I don't want to give up, but I do anyway and ask my calculator, there is three solutions, alas I have no idea how that helps me (with my knowledge, it does not)!
Should I just practice with smaller polynomials more, and this one might look easier?
Any suggestions are appreciated, I will tag this as homework because it should be.
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Note that trying to factorize anything as $x(x^2-6x+11)=6$ is almost useless, since $6$ need not factor nicely. The rare instance when such a factorization is useful, is for example $(x+1)(x+2)(x+3)=6$, in which case we can observe that $x=0$ is a root. – Calvin Lin Jan 14 '13 at 20:21
Do you mean $6x^2$ instead of $6x$? – Git Gud Jan 14 '13 at 20:22
Yes, sorry forgot to square that one. – Leonardo Jan 14 '13 at 20:23
You should also know that there are closed formulas to get the roots of polynomials of degree up to 4. – Git Gud Jan 14 '13 at 20:28
If you have a polynomial with integer coefficients of the form $$p(x) = x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + a_0$$ and if you have any hope of factorizing this "nicely", you first need to check if the divisors of the constant term, $a_0$, are factors. In your case, I assume the correct polynomial you are interested in is $p(x) = x^3 - 6x^2+11x-6$. The divisors of $-6$ are $\{\pm1,\pm2, \pm3, \pm6\}$. A quick check reveals that $x^*=1,2,3$ satisfy $p(x^*) = 0$. Hence, we have $$x^3 - 6x^2 + 11x - 6 =(x-1)(x-2)(x-3)$$ | {
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Luckily this is a nice equation, thank you I have at least one extra tool in my belt and I will need to read up more on the rational root theorem among other things. Your answer is very clear and concise. – Leonardo Jan 14 '13 at 20:27
Your equation is $$x^3-6x^2+11x-6=0$$ The coefficients of the equation are $1,-6,+11,-6$ and the summation of them is $$1-6+11-6=0$$ Whenever you face to this fact, you always have $x=1$ as one solution. In fact the equation has a factor as form $x-1$. It means that $$x^3-6x^2+11x-6=(x-1)(ax^2+bx+c)$$ for some proper available constants $a,b,c$.
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I kind of get what you and Will Jagy are saying, but it will take time to absorb. The only thing that is immediately obvious from your last equations is that c = 6. I appreciate your helps. – Leonardo Jan 14 '13 at 20:22
@Leonardo: Look at Marvis's. I noted some points just for starting. He gave you the answer in detailed. – Babak S. Jan 14 '13 at 20:25
$x=1$ is an evident solution. Therefore we can write $$x^3 - 6 x^2 + 11 x - 6 = (x-1)(a x^2 + b x + c)$$ and solve for $a,b,c$ carefully. Then the rest is the quadratic formula.
I suppose the main advice is to check for integer roots and rational roots before factoring, when you have no feeling for what is going on in the problem. There is a Rational Roots Theorem that applies here.
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$$x^3 - 6x^2 + 11x - 6 =0$$ $$x^3 -1-( 6x^2 - 11x + 6)+1 =0$$ $$x^3 -1-( 6x^2 - 11x + 5) =0$$ $$x^3 -1-( 6x^2 - 6x-5x + 5) =0$$ $$(x-1)(x^2+x+1)-( 6x(x-1)-5(x-1) =0$$ $$(x-1)(x^2+x+1)-(x-1)(6x-5) =0$$ $$(x-1)(x^2+x+1-6x+5) =0$$ $$(x-1)(x^2-5x+6) =0$$ $$(x-1)(x^2-2x-3x+6=0$$ $$(x-1)(x(x-2)-3(x-2)=0$$ $$(x-1)(x-2)(x-3)=0$$ $$x_1=1,x_2=2,x_3=3$$
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Nice Adi,nice. ;-) – Babak S. Jan 14 '13 at 20:26
My brain does not work like yours, but I hope it starts to soon. – Leonardo Jan 14 '13 at 20:29
thank you Babak – Adi Dani Jan 14 '13 at 23:31 | {
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You need to know a few things. For example if $p(x)$ is a polynomial of odd degree with rational coefficients it will necessarily have a real root.
Another useful fact is that if $a$ is a root of $p(x)=x^3+px^2+qx+r=0$ then $(x-a)$ is a factor of $p(x)$. This works because: $x^3-a^3=(x-a)(x^2+ax+a^2)$ and $x^2-a^2=(x-a)(x+a)$ so if $p(a)=0$ we can write:$$p(x)=p(x)-p(a)=[x^3-a^3]+p[x^2-a^2]+q[x-a]=(x-a)([x^2+ax+a^2]+p[x+a]+q)$$
We also note that if $p,q,r,a \in \mathbb Z$, then if $p(a)=a(a^2+pa+q)+r=0$ then $a$ must be a factor of $r$.
[There are various generalisations and extensions of these comments].
So one thing we do when faced with a cubic with integer coefficients to factorise is to try the factors ($\pm$) of the constant term as "easy candidates" - and once we have one factor we can divide through and solve the remaining quadratic.
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# Which Grows Faster: Factorial or Double Exponentiation
Which of the functions among $$2^{3^n}$$ or $$n!$$ grows faster?
I know that $$n^n$$ grows faster than $$n!$$ and $$n!$$ grows faster than $$c^n$$ where $$c$$ is a constant, but what is it in my case?
• It isn't a proof, but evaluate the expressions with $n$ in the range 1 to 10 and see just how phenomenally fast $2^{3^n}$ grows compared to $n!$ – John Coleman Nov 26 '19 at 0:20
• Write down the values for 1 ≤ n ≤ 10. You can't? Why can't you? – gnasher729 Nov 26 '19 at 13:18
• Why is this in CS instead of Mathematics? – Barmar Nov 26 '19 at 21:08
• @Barmar because it's related to CS in terms of time complexity. Indeed, in this question, there is not a hard border between math and CS. – OmG Nov 27 '19 at 15:14
• – Will Ness Nov 27 '19 at 16:36
You can find the result by taking a $$\log$$. Hence:
$$\log(2^{3^n}) = 3^n$$ $$\log(n!) \leqslant \log(n^n) = n\log n$$
(In the latter equation, we have used the fact that $$n! \leqslant n^n$$, as you note in the question.)
Of course $$3^n$$ grows faster than $$n \log n$$. As $$\log$$ is an increasing function, we can say $$2^{3^{n}}$$ grows faster than $$n^n$$, and also $$n!$$.
• Beware, however, that $\log(f) = O(\log(g))$ does not imply $f = O(g)$, in general. For instance, take $f(n)=n^2$ and $g(n)=n$. – chi Nov 26 '19 at 19:07
• @chi You are right. But, it is true if $\log(f) = o(\log(g))$. – OmG Nov 26 '19 at 19:56
• Another question related with taking log and solving and its danger is given in this link--> cs.stackexchange.com/questions/117584/… – Turing101 Nov 28 '19 at 15:19
Another way to directly compare the two expressions is to take the ratio of consecutive terms:
$$\frac{2^{3^{n+1}}}{2^{3^n}}=2^{2\cdot 3^n}\gg 3^n\gg n+1=\frac{(n+1)!}{n!}$$
(for positive integers $$n$$), and clearly also $$2^{3^1}=8>1!$$, so $$2^{3^n}$$ indeed grows more rapidly than $$n!$$. | {
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• The discussion of $2^{3^1}$ is unnecessary. – leftaroundabout Nov 27 '19 at 1:56
• @leftaroundabout, I guess that's true if you assume the >> signs are significantly greater than, but that's not very rigorous (and if they are only > signs it would then depend on your definition of "grows more rapidly") – boboquack Nov 27 '19 at 7:43
• Well, they only need to be “significantly” greater in a very weak sense, namely the ratio needs to be $>1+\varepsilon$. That is clearly given in this case, specifically $\frac{3^n}{n+1} > 1.5\quad \forall n>0$. – leftaroundabout Nov 27 '19 at 12:05
You noted in your question that $$n^n$$ grows faster than $$n!$$, and that’s a great starting point for comparing the growth of $$2^{3^n}$$ and $$n!$$. Specifically, let’s ask - of $$n^n$$ and $$2^{3^n}$$, which grows faster?
To answer that, let’s try rewriting $$n^n$$ so that it has the same exponential base as $$2^{3^n}$$. Since $$n = 2^{\log_2 n}$$, we have that
$$n^n = (2^{\log_2 n})^n = 2^{n \log_2 n}.$$
Now, is it easier to see how $$n^n$$ and $$2^{3^n}$$ relate?
As a note, this approach is similar to taking the base-2 logs of both expressions. I thought it would be good to include this here because it gives a slightly different perspective on how to arrive at the answer given your initial observation.
Induction.
Base case: $$n=1$$ gives $$2^{3^1}=8$$ and $$1!=1$$ which clearly holds.
Hypothesis: suppose that $$2^{3^k}>k!$$ for some $$k\in\Bbb N$$.
Consider $$n=k+1$$. Then $$2^{3^{k+1}}=2^{3^k\cdot3}>(k!)^3=(k+1)!\cdot\frac{k!(k-1)!}{1+\frac1k}>(k+1)!$$ which is true $$\forall k>1$$ and thus the result follows. | {
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• I tend to interpret "which grows faster?" questions as "whose big-Oh set is bigger?". With that interpretation, this post answers something different. (E.g. $x \mapsto x$ and $x \mapsto (x+1)$ are in the same big-Oh set, but in the spirit of this post the latter grows faster.) – ComFreek Nov 26 '19 at 12:02
• @ComFreek I would really have to question that interpretation. Which grows faster, $f(x) = x$ or $g(x) = 2x$? I think by any reasonable meaning of the words "grow" and "faster", $2x$ grows faster than $x$. My definition of grows faster would be $f(x)$ grows faster than $g(x)$ if there exists an $x_0$ beyond which $f'(x)$ is always greater than $g'(x)$. – Apollys supports Monica Nov 27 '19 at 23:36
• That being said, this proof certainly doesn't prove anything about growth rates. Plug in the functions $f(x) = -1/x$ and $g(x) = 0$ (on the domain $x > 0$) into your proof and you conclude that $g(x)$ grows faster than $f(x)$. This is obviously wrong since $g(x)$ never grows and $f(x)$ never doesn't grow! – Apollys supports Monica Nov 27 '19 at 23:42 | {
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# Bounding an expression involving digamma function
Let $\psi$ be the digamma function. I have a conjecture that
$$\frac ax > \log(x) - \psi(x)$$
holds for all $x > 0$ if (and only if) $a \ge 1$. I do not know how to prove it. Please help.
-
Let
$$f(x) = \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x}.$$
Then for $x > 0$, we have
$$f'(x) = \frac{1}{x^2} - \frac{1}{4\sinh^2(x/2)} = \frac{1}{x^2}\left(1 - \left(\frac{x/2}{\sinh (x/2)} \right)^2 \right) > 0.$$
This shows that $f(x)$ is increasing. Also, it is easy to find that $f(0+) = 0$ and $\lim_{x\to\infty} f(x) = \frac{1}{2}$. Thus we find that $0 < f(x) < \frac{1}{2}$ for $0 < x < \infty$ and hence the Laplace transform $\mathcal{L}f(s) = \int_{0}^{\infty} f(x) \, e^{-sx} \; dx$ of $f(x)$ satisfies
$$0 < \mathcal{L}f(s) < \int_{0}^{\infty} \frac{1}{2} \, e^{-sx} \; dx = \frac{1}{2s}.$$
But we also have
\begin{align*} \mathcal{L}f(s) &= \int_{0}^{\infty} \left( \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x} \right) e^{-sx} \; dx\\ &= \left[ \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \right]_{0}^{\infty} \\ &\qquad + s \int_{0}^{\infty} \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \; dx \\ &= s \int_{0}^{\infty} \left( \frac{x}{2} - \log (sx) + \log s + \log \left(1 - e^{-x} \right) \right) e^{-sx} \; dx \\ &= \frac{1}{2s} + \gamma + \log s - s \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-(n+s)x}}{n} \; dx \\ &= \frac{1}{2s} + \gamma + \log s - \sum_{n=1}^{\infty} \frac{s}{n(n+s)} \\ &= \frac{1}{2s} + \log s - \psi_0(1+s), \end{align*}
where we have used the fact that $\int_{0}^{\infty} e^{-x} \log x \; dx = -\gamma$ and
$$\psi_0(1+x) = -\gamma + \sum_{n=1}^{\infty} \frac{x}{n(n+x)}.$$
Finally, since $\psi_0(1+s) = \frac{1}{s} + \psi_0 (s)$, we have
$$\frac{1}{2s} < \log s - \psi_0(s) < \frac{1}{s}.$$
Therefore the inequality holds if $a \geq 1$ and only if $a > \frac{1}{2}$. | {
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Therefore the inequality holds if $a \geq 1$ and only if $a > \frac{1}{2}$.
Now we prove that $a = 1$ is the sharp bound. We are to calculate the limit
$$\alpha = \lim_{x\to 0} x(\log x - \psi_0(x)).$$
But by our previous calculations,
\begin{align*}\alpha &= \lim_{s\to 0} s(\log s - \psi_0(s)) \\ &= \lim_{s\to 0} s\left( \frac{1}{2s} + \mathcal{L}f(s) \right) \\ &= \frac{1}{2} + \lim_{s\to 0} s \int_{0}^{\infty} f(x) \, e^{-sx} \; dx \\ &= \frac{1}{2} + \lim_{s\to 0} \int_{0}^{\infty} f(u/s) \, e^{-u} \; du \qquad (u = sx) \\ &= \frac{1}{2} + \int_{0}^{\infty} \lim_{s\to 0} f(u/s) \, e^{-u} \; du \\ &= \frac{1}{2} + \int_{0}^{\infty} \frac{1}{2} \, e^{-u} \; du \\ &= 1. \end{align*}
Thus if $\frac{a}{x} > \log x - \psi_0 (x)$ is true for all $x > 0$, then we must have $a \geq \alpha = 1$, as desired.
-
This is a cool technique. (There's a typo after you switch summation and integral.) – Tunococ Aug 16 '12 at 18:25
@Tunococ, Thanks. Also I fixed the typo. Even I cannot believe how I made such a mistake! :) – sos440 Aug 17 '12 at 3:36
Recently I happened to find out that Qi et al (J. Math. Anal. Appl. 310 (2005) 303–308) proved the following inequality, among others, for digamma function $\psi(x)$ when $x>0$:
$$\frac{1}{2x} < \log x - \psi(x) < \frac{1}{2x} + \frac{1}{12x^2}$$
This inequality is sharper than the one given by @sos440 above when x>1/6.
- | {
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# Each year for 4 years, a farmer increased the number of trees in a
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Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
A. 1250
B. 1563
C. 2250
D. 2560
E. 2752
[Reveal] Spoiler:
Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.
Thanks
[Reveal] Spoiler: OA
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Thanks
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imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
A. 1250
B. 1563
C. 2250
D. 2560
E. 2752
Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.
Thanks
Say the number of trees at the beginning of the 4 year period was x, then:
At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$;
At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$;
At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$;
At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$;
At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$;
So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. | {
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If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question.
Hope it's clear.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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08 Jul 2012, 22:32
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imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
A. 1250
B. 1563
C. 2250
D. 2560
E. 2752
Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.
Thanks
The number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population.
If x increases by 25%, how we denote it? (5/4)*x
If next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x
and so on...
For more on this, check: http://www.veritasprep.com/blog/2011/02 ... e-changes/
So if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250 | {
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So if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250
As for your next question, the numbers given would be such that the calculation will not be tough.
Say, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.
Something like 2^8 * x = 2560
and if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", note that you still need to work with
(5/4)^4 * x = 6250
since you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by $$5^{11}$$ which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by $$5^{11}$$ but you wouldn't really need to bother about it.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1838 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 25 Oct 2013, 00:26 4 This post received KUDOS 2 This post was BOOKMARKED Trees increase 1/4th every year, which means 100 trees become 125 after 1 yr & so on So, 125/100 = 5/4 is the resultant (after adding 1/4 as the growth) for 1 year So, for 4 yrs is 5^4/(4^4) From the condition given, inital trees were = 6250 x 4^4 / 5^4 = 2560 _________________ Kindly press "+1 Kudos" to appreciate Manager Joined: 12 Jan 2013 Posts: 212 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 27 Dec 2013, 09:31 Bunuel wrote: imhimanshu wrote: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period. A. 1250 B. 1563 C. 2250 D. 2560 E. 2752 Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be | {
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of trees would be $$(\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$; So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x.. Is there a straightforward "word translation" way in knowing how to interpret wordings like this? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7996 Location: Pune, India Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 01 Jan 2014, 23:34 Expert's post 2 This post was BOOKMARKED aeglorre wrote: Bunuel wrote: imhimanshu wrote: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period. A. 1250 B. 1563 C. 2250 D. 2560 E. 2752 Can someone walk me through the logic behind this | {
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period. A. 1250 B. 1563 C. 2250 D. 2560 E. 2752 Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$; So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x.. Is there a straightforward "word translation" way in knowing how to interpret wordings like this? Actually, it is not ambiguous. Read the | {
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way in knowing how to interpret wordings like this? Actually, it is not ambiguous. Read the statement: Each year a farmer increased the number of trees by 1/4. He did this for 4 years. (In GMAT Verbal and Quant are integrated. You need Verbal skills (slash and burn) in Quant and Quant skills (Data Interpretation) in Verbal. So in the first year, he increased it by 1/4 The next year, he again increased it by 1/4 (of preceding year) Next year, again the same. Next year, again the same. So he did it for a total of 4 years. So if initially the number of trees was x, in the first year he made them (5/4)x _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199 | {
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07 Feb 2014, 18:28
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it can easily solved by compound interest formula
R = 25%
P = original number of trees
N = no of times of compounding within a year = 1
T = total number of years = 4
P+I at the end of compounding years = 6250
hence,
6250 = (1 + 0.25/1)^(4*1) * P
5^4*10 = 1.25^4 * P
(5/1/25)^4 *10 = P
4^4 * 10 = P
2560 = P
Another way to think is to pick ans choice which is exactly divisible by 4. A, B and C are not. Only D and E are.
Pick D and increase it by 25% for 4 times, you should get 6250.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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27 Jul 2014, 10:08
Hi Karishma,
I am not able to solve it by below method, where am I wrong:
6250*(3/4)*(3/4)*(3/4)
Regards,
Ravi
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27 Jul 2014, 22:08
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email2vm wrote:
Hi Karishma,
I am not able to solve it by below method, where am I wrong:
6250*(3/4)*(3/4)*(3/4)
Regards,
Ravi
If I increase A by 25% and get B, I will not get A if I reduce B by 25%. The bases are different in the two cases.
e.g. A = 80
25% of A is 20 so I increase A by 25% to get B = 100
Now if I decrease B by 25%, I will decrease B by 25 (25% of 100). This will give me 75 which is not the same as A (which is 80). | {
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In the first step, I found 25% of 80 and added that. In the second step, I found 25% of 100 and subtracted that. These two numbers are different.
You will get the correct answer if you do 6250*(4/5)*(4/5)*(4/5)*(4/5)
4/5 is the inverse of 5/4 (which is 25% increase in x).
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 13 Feb 2014 Posts: 3 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 25 Oct 2014, 12:01 You can use the formula A=p(1 + r/100)^n with r = 25% You will get P as 2560. Intern Joined: 23 Aug 2014 Posts: 42 GMAT Date: 11-29-2014 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 30 Dec 2014, 10:59 Is it true that the number of trees in the beginning has to be a multiple of 4 (as additions are always 1/4 th of previous and will not be non-integers)? In that case I'm left with D or E. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7996 Location: Pune, India Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 31 Dec 2014, 02:53 deeuk wrote: Is it true that the number of trees in the beginning has to be a multiple of 4 (as additions are always 1/4 th of previous and will not be non-integers)? In that case I'm left with D or E. Yes, that's correct. But you might still need to work with two options. On the other hand, working with 6250 will definitely yield the answer. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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06 Jan 2015, 10:39
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Trees increase by 1/4 the number of trees in preceding year. Hence, correct answer must be divisible by 4. Based on divisibility rules, if last 2 digits are divisible by 4 then the number is divisible by 4. Thus, we can eliminate A, B, C. The answer has to be D or E.
Again, trees increase by 1/4 the number of trees in preceding year. Hence, the number of trees increase by 5/4 times the number of trees the preceding year.
If x = initial number of trees = Answer D or E = 2560 or 2752
Year 1 = 5/4x
Year 2 = (5/4)(5/4)x
Year 3 = (5/4)(5/4)(5/4)x
Year 4 = (5/4)(5/4)(5/4)(5/4)x
Only for Answer D: (5/4)(5/4)(5/4)(5/4) 2560 = 6250
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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02 Apr 2015, 03:36
@VeritasPrepKarishma
VeritasPrepKarishma wrote:
imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
A. 1250
B. 1563
C. 2250
D. 2560
E. 2752
Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.
Thanks
The number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population. | {
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If x increases by 25%, how we denote it? (5/4)*x
If next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x
and so on...
For more on this, check: http://www.veritasprep.com/blog/2011/02 ... e-changes/
So if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250
As for your next question, the numbers given would be such that the calculation will not be tough.
Say, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.
Something like 2^8 * x = 2560
and if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", note that you still need to work with
(5/4)^4 * x = 6250
since you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by $$5^{11}$$ which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by $$5^{11}$$ but you wouldn't really need to bother about it.
Karisma,
However, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.
Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?
Thank you a lot for your help!
Regards,
Eli.
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02 Apr 2015, 19:43
elisabettaportioli wrote:
However, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.
Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?
Thank you a lot for your help!
Regards,
Eli.
The base for each year changes.
Say, you have 100 trees and you increase them by 20% each year.
At the end of the first year, you will have 100 + (20/100)*100 = 100*(1 + 20/100) = 100*120/100 = 100*(6/5) = 120 trees.
So if you have to increase a number by 20%, you just need to multiply it by 6/5 every time no matter what the number is.
By the same logic, next year, you will have 120 * (6/5) = 144 trees when you increase them by 20%.
You can write 120 as 100 * (6/5) and then multiply it by another 6/5 to increase 120 by 20%.
This is the same as 100 * (6/5) * (6/5) = 144 trees.
So every time you multiply it by 6/5, you increase the base. The next time you multiply it by 6/5, you are increasing the new base by 20%. This is the concept of successive percentage changes and I have discussed it in the link I mentioned in my post above.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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02 Apr 2015, 20:15
elisabettaportioli wrote:
Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?
Thank you a lot for your help!
Regards,
Eli. | {
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Thank you a lot for your help!
Regards,
Eli.
hi,
you are correct that the increase is on the preceeding year and that is exactly what is happening when you are multiplying x by 5/4..
first year say it was x.. 2nd year it becomes 5/4*x.. now for next year when we write the value as 5/4*5/4*x, the increase is on 5/4*x and not x..
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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13 Apr 2015, 08:38
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1/4 = 0.25 = 25%
using the formula of compound interest calculation
p(1+r)^n= A
p = principal amount (to be identified), r = rate (0.25), n = numbers years (4), A= amount at the end of term(6250 given)
p(1+0.25)^4=6250
p(1.25)^4= 25*25*10 = 5^2*5^2*10=5^4*10
p(125/100)^4 = 5^4 x 10
p(5/4)^4 = 5^4 x 10
p(5^4)/4^4 = 5^4 x 10
p= (5^4 x 10 x 4^4)/5^4
p= 10 x 4^4
p= 2560
Ans D
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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19 May 2015, 12:37
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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20 May 2015, 03:32
BrainLab wrote:
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
Dear BrainLab
Perfect logic but for easier calculation, you may want to work with ratio here (1/4 increase per annum) instead of percentages (25% increase per annum). Both convey the same thing but the equation
$$(\frac{5}{4})^4*X = 6250$$
will take lesser time to solve (especially if you know that $$5^4 = 625$$) than $$(1.25)^4*X = 6250$$
Hope this was useful!
Japinder
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# In the rectangular coordinate system above, if the equation of m is y
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In the rectangular coordinate system above, if the equation of m is y [#permalink]
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15 Dec 2017, 02:09
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In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
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In the rectangular coordinate system above, if the equation of m is y [#permalink]
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Bunuel wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
The attachment 2017-12-15_1259.png is no longer available
With the formula (My diagram does not apply)
I did not use it, but to do so is standard. So I quote the formula from Bunuel HERE
Distance between two parallel lines $$y=mx+b$$ and $$y=mx+c$$ can be found by the formula:
$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$ | {
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Equation for line $$m$$, $$y = x$$, to match the exact form of $$y = mx + b$$ can be written:
$$y =(1)x + 0$$ (all lines that pass through the origin have x- and y-intercepts of 0)
$$b = 0$$ = y-intercept, $$m = 1$$ = slope
Find the equation of line $$n$$, $$y = mx + c$$
Parallel lines have identical slopes. Line $$n$$ has slope $$m = 1$$
Use the point on line $$n$$ from the graph: (1, 0)
Plug its coordinates into slope-intercept form to find $$c$$
$$y = mx + c$$
$$(0) = 1(1) + c$$
$$-c = 1$$, so $$c = -1$$
Equation for line $$n: y = x - 1$$
Distance between parallel lines
$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$
$$m = 1$$, $$b = 0$$, $$c = -1$$
$$D=\frac{|0-(-1)|}{\sqrt{1^2+1}}$$
$$D=\frac{1}{\sqrt{2}}$$
Does not match the answers. Rationalize the denominator; multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
Attachment:
linesmandn.png [ 18.05 KiB | Viewed 868 times ]
Without the formula see diagram
The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A
That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB
A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$
The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):
$$x\sqrt{2}= 1$$
$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines
That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
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In the rectangular coordinate system above, if the equation of m is y [#permalink]
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14 Apr 2018, 06:59
generis wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
2017-12-15_1259.png
Attachment:
linesmandn.png
Without the formula see diagram
The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A
That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB
A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$
The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):
$$x\sqrt{2}= 1$$
$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines
That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?
thanks.
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Re: In the rectangular coordinate system above, if the equation of m is y [#permalink]
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14 Apr 2018, 07:59
1
MT1988 wrote:
generis wrote: | {
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### Show Tags
14 Apr 2018, 07:59
1
MT1988 wrote:
generis wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
2017-12-15_1259.png
Attachment:
linesmandn.png
Without the formula see diagram
The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A
That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB
A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$
The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):
$$x\sqrt{2}= 1$$
$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines
That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?
thanks.
Got it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.
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Re: In the rectangular coordinate system above, if the equation of m is y [#permalink]
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### Show Tags
14 Apr 2018, 16:33
1
Quote:
Bunuel wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
2017-12-15_1259.png
MT1988 wrote:
MT1988 wrote:
generis wrote:
Without the formula see diagram
The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A
That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB
A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$
The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):
$$x\sqrt{2}= 1$$
$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines
That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?
thanks.
Got it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.
BOOM. Excellent.
I came prepared with a diagram
to help, but behold, no need.
Here is a post that discusses
that create 45° angles.
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In the rectangular coordinate system above, if the equation of m is y [#permalink]
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15 Apr 2018, 03:40
Bunuel wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
The attachment 2017-12-15_1259.png is no longer available
My method to solve this:-
Attachments
File comment: for line y = x, the angle between x-axis and the line m will be 45 degrees.
The min distance will the perpendicular distance between the two lines.
In the image, by mistake I've written 1/root(2) = 2/root(2), it should be root(2)/2. Please excuse that mistake
IMG_20180415_160317.jpg [ 1.99 MiB | Viewed 347 times ]
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In the rectangular coordinate system above, if the equation of m is y &nbs [#permalink] 15 Apr 2018, 03:40
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# A company has two types of machines, type R and type S
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Re: A company has two types of machines, type R and type S [#permalink]
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04 Nov 2016, 15:05
1
enigma123 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?
A. 3
B. 4
C. 6
D. 9
E. 12
We have a combined worker problem for which we can use the following formula:
work (1 machine) + work (2 machine) = total work completed | {
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work (1 machine) + work (2 machine) = total work completed
Since we are completing one job, we can say:
work (1 machine) + work (2 machine) = 1
We are given that a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours.
Thus, the rates for the two machines are as follows:
rate of machine R = 1/36
rate of machine S = 1/18
We are also given that the company used the same number of each type of machine to do the job in 2 hours. If we let x = the number of each machine used, we can multiply each rate by x and we have:
rate of x number of R machines = x/36
rate of x number of S machines = x/18
Finally, since we know some number of R and S machines worked for two hours, and since work = rate x time, we can calculate the work done by each type of machine.
work done by x number of R machines = 2x/36 = x/18
work done by x number of S machines = 2x/18 = x/9
Now we can determine x using the combined worker formula:
work (machine R) + work (machine S) = 1
x/18 + x/9 = 1
x/18 + 2x/18 = 1
3x/18 = 1
x/6 = 1
x = 6
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Work/Rate problem- A company has.. [#permalink]
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01 Jan 2017, 06:10
A company has two types of machines, Type R and Type S. Operating at a certain rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company uses the same number of each type of machine to do the job in 2 hours, how many machines of Type R were used?
Options:
A. 3
B. 4
C. 6
D. 9
E. 12
Can anybody give me a quickest and simpler way to solve this? I took almost 3.5 min on this question to solve during a mock test. Still ended up guessing as I wasn't sure of the approach.
Thanks.
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Re: Work/Rate problem- A company has.. [#permalink]
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### Show Tags
01 Jan 2017, 06:47
5
1
baalok88 wrote:
A company has two types of machines, Type R and Type S. Operating at a certain rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company uses the same number of each type of machine to do the job in 2 hours, how many machines of Type R were used?
Options:
A. 3
B. 4
C. 6
D. 9
E. 12
Can anybody give me a quickest and simpler way to solve this? I took almost 3.5 min on this question to solve during a mock test. Still ended up guessing as I wasn't sure of the approach.
Thanks.
We are given rates of each machine and that the number of machines of each type being used is the same = x.
$$\frac{x}{36} + \frac{x}{18} = \frac{1}{2}$$
$$\frac{3x}{36} = \frac{1}{2}$$
$$6x = 36$$
$$x = 6$$
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Re: A company has two types of machines, type R and type S [#permalink]
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07 Jan 2017, 21:07
2
Rate(R) = $$\frac{1}{36}$$
Rate(S) = $$\frac{1}{18}$$
Combined rate of both machines ,
Rate(RS)=Rate(R) + Rate(S) = $$\frac{1}{12}$$
we have given time , T = 2hrs , Workdone = 1 (for single job)
Plug in the values :
[Rate] * [Time] * [No. of machines] = Workdone
$$\frac{1}{12}$$ * 2 * [No. of machines] = 1
[No. of machines] = 6
Ans : C
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Re: A company has two types of machines, type R and type S [#permalink]
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18 Apr 2017, 05:05
Bunuel wrote:
enigma123 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?
A. 3
B. 4
C. 6
D. 9
E. 12 | {
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A. 3
B. 4
C. 6
D. 9
E. 12
Rate of A - $$\frac{1}{36}$$ job/hour, rate of x machines of A - $$\frac{1}{36}x$$ job/hour;
Rate of B - $$\frac{1}{18}$$ job/hour, rate of x machines of B - $$\frac{1}{18}x$$ job/hour, (same number of each type);
Remember that we can sum the rates, hence combined rate of A and B is $$\frac{1}{36}x+\frac{1}{18}x=\frac{3}{36}x=\frac{1}{12}x$$ job/hour.
We are told that together equal number (x in our case) of machines A and B can do the job (1 job) in 2 hours --> $$Time*Rate=2*\frac{1}{12}x=1=Job$$ --> $$x=6$$.
When I am double checking, why am I not getting the right answer?
If 1 R machine is taking 36 hours, then 6 R machines will take 6 hours.
If 1 S machine is taking 18 hours, then 6 S machines will take 3 hours.
Together, they will take 6-3=3 hours.
Please tell me where am I wrong :/
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Re: A company has two types of machines, type R and type S [#permalink]
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16 Aug 2017, 18:41
narendran1990 wrote:
Took 36 as the unit of work to be completed by R & S. [LCM of 36 & 18].
So, R does 1 unit per hour & S does 2 units per hour. Since an equal number of R & S machines have to be deployed, consider 'x' as the number of equipment. Additionally, the work is to be completed in 2 hours, so 18 units have to be manufactured.
Therefore, x+2x = 18, 3x = 18, x=6. [Correct Answer]
I was trying to solve using similar kind of method. But i could not understand how you arrived on 18? It will be great if you could elaborate.
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A company has two types of machines, type R and type S [#permalink]
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09 Nov 2017, 21:26
There is no need to compute the combined rate in this case, explicitly that is.
Since the number of machines is the same, we can use a simple equation:
$$x*\frac{1}{36}+ x* \frac{1}{18} = \frac{1}{2}$$ (here x machines work at the respective rates for 2 hours)
Solving for x, we get 6.
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Solving for x, we get 6.
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Re: A company has two types of machines, type R and type S [#permalink]
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13 Nov 2018, 19:46
hannahkagalwala wrote:
Bunuel wrote:
enigma123 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?
A. 3
B. 4
C. 6
D. 9
E. 12
Rate of A - $$\frac{1}{36}$$ job/hour, rate of x machines of A - $$\frac{1}{36}x$$ job/hour;
Rate of B - $$\frac{1}{18}$$ job/hour, rate of x machines of B - $$\frac{1}{18}x$$ job/hour, (same number of each type);
Remember that we can sum the rates, hence combined rate of A and B is $$\frac{1}{36}x+\frac{1}{18}x=\frac{3}{36}x=\frac{1}{12}x$$ job/hour.
We are told that together equal number (x in our case) of machines A and B can do the job (1 job) in 2 hours --> $$Time*Rate=2*\frac{1}{12}x=1=Job$$ --> $$x=6$$.
When I am double checking, why am I not getting the right answer?
If 1 R machine is taking 36 hours, then 6 R machines will take 6 hours.
If 1 S machine is taking 18 hours, then 6 S machines will take 3 hours.
Together, they will take 6-3=3 hours.
Please tell me where am I wrong :/
It is not correct to subtracting time taken by one machine from that of other to calculate final time,
Time taken are for complete job, and if other machines are also performing the same job, then there will be less amount of work of each machine type and hence leaser time will be taken.
You are right till, 6 R machines will take 6 hours.
And 6 S machines will take 3 hours.
Now total time taken will be 1/(1/6 +1/3) = 1/(1/2) = 2 hours. | {
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Now total time taken will be 1/(1/6 +1/3) = 1/(1/2) = 2 hours.
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Re: A company has two types of machines, type R and type S &nbs [#permalink] 13 Nov 2018, 19:46
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# Let $p$ and $q$ be two distinct primes. Pick the correct statements from
Let $p$ and $q$ be two distinct primes. Pick the correct statements from the following:
a. $Q(\sqrt p)$ is isomorphic to $Q(\sqrt q)$ as fields.
b. $Q(\sqrt p)$ is isomorphic to $Q(\sqrt{−q})$ as vector spaces over Q.
c. $[Q(\sqrt p,\sqrt q) : Q] = 4$.
d. $Q(\sqrt p,\sqrt q) = Q(\sqrt p + \sqrt q)$.
(c) & (d) are correct. (a) is not correct for fields but correct for vector spaces. (b) not sure. i think it is correct by the same arguement of (a).am i right? | {
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-
Is this a homework question, or just one for personal development, as this may effect the type of response that people give? – David Ward Sep 17 '12 at 8:16
@DavidWard I don't know why it should affect the type of response. If one should not give a full answer to a homework question, many good questions might have no full answer in this site. I think this is against the policy of this site. – Makoto Kato Sep 17 '12 at 18:13
@MakotoKato My reasoning is that if it is a homework question, then the answer itself is of importance to the person asking the question. However, if the question is not homework, then for a question as above, I would feel happy to give the answer, but then lead someone through the reasoning on how to get there. – David Ward Sep 18 '12 at 8:26
@DavidWard I'm sorry that I don't understand your reasoning. Could you explain why you don't want to give the answer to a homework question? – Makoto Kato Sep 18 '12 at 8:43
@MakotoKato My reasoning is as follows: Firstly suppose that the question asked is from an assessed piece of work. Then as the question is a multiple-choice question, the actual answer will attract some if not all of the marks (depending on the desire of the assessor). Thus knowledge of the answer is of critical importance. On the other hand, if the question asked is purely for personal development, whilst leading the member towards the solution, it can be beneficial to actually know where you are being led. It just gives some motivation for where the reasoning is heading. – David Ward Sep 18 '12 at 15:26
a. Suppose $\mathbb{Q}(\sqrt p)$ is isomorphic to $\mathbb{Q}(\sqrt q)$ as fields. Then $\mathbb{Q}(\sqrt q)$ has an element $\alpha$ such that $\alpha^2 = p$. Let $\alpha = a + b\sqrt q$, where $a, b \in \mathbb{Q}$. We denote the conjugate of $\alpha$ by $\alpha'$. Since $\alpha + \alpha' = 0$, $a = 0$. Since $\alpha^2 = p$, $p = b^2 q$. Hence $p = q$. This is a contradiction. Hence $a$. is not true. | {
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b. Since the both fields have dimension 2 as vector spaces over $\mathbb{Q}$, $b.$ is true.
c. As we see in the above, $\sqrt p$ is not contained in $\mathbb{Q}(\sqrt q)$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}(\sqrt q)] = 2$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}] = 4$. Thus $c.$ is true.
d. Let $K = \mathbb{Q}(\sqrt p,\sqrt q)$. Clearly $K/\mathbb{Q}$ is Galois. Let $\sigma \in Gal(K/\mathbb{Q})$. Then $\sigma(\sqrt p) = \sqrt p$ or $-\sqrt p$, $\sigma(\sqrt q) = \sqrt q$ or $-\sqrt q$. Since $|Gal(K/\mathbb{Q})| = 4$ by $c.$, $Gal(K/\mathbb{Q}) = \{1, \sigma_1, \sigma_2. \sigma_3\}$, where
$$\sigma_1(\sqrt p) = \sqrt p,\ \ \ \sigma_1(\sqrt q) = -\sqrt q$$
$$\sigma_2(\sqrt p) = -\sqrt p,\ \sigma_2(\sqrt q) = \sqrt q$$
$$\sigma_3(\sqrt p) = -\sqrt p,\ \ \ \sigma_3(\sqrt q) = -\sqrt q$$
Let $\alpha = \sqrt p + \sqrt q$.
Clearly $\alpha, \sigma_1(\alpha) = \sqrt p - \sqrt q, \sigma_2(\alpha) = -\sqrt p + \sqrt q, \sigma_3(\alpha) = -\sqrt p - \sqrt q$ are distinct. Hence $K = \mathbb{Q}(\alpha)$. Thus $d.$ is true.
-
please explain why α+α′=0? In the proof of a ? – GA316 Aug 9 '15 at 4:01 | {
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# How do I describe formally complexity of 2-sum problem algorithm?
I have algorithm that finds if there are two elements in sorted array that have sum zero.
1.ZeroSumPair(A[1..n]) // A[1..n] <-- sorted
2. l <- 1, r <- n
3. while(l < r)
4. while(l < r or A[l] + A[r] > 0)
5. r--
6. if(A[l] + A[r] = 0)
7. return true
8. l++
9. return false
Intuitively I know that this algorithm is $$O(n)$$, but how do I deduct it using proof with summations like in CLRS book?
I've also saw How to develop an $O(N)$ algorithm solve the 2-sum problem?, but I didn't see any formal proof.
• I would try putting in a counter t. Then increment t by 1 every time you do an operation. Then try to prove a loop invariant for the outer-while loop that claims $t = O(n)$. With the correct loop invariant on $t$, this can be proven by induction relatively easily. Perhaps something like $t \leq c(l + (n - r))$ for some constant $c$ might work. Then at termination (in the worst case) you know $l = r$, thus $t \leq cn \implies t = O(n)$. – ryan Mar 27 at 18:28
Consider this slightly modified algorithm where I've added an "operation counter" t. This will be incremented every time we do a comparison or assignment.
1. ZeroSumPair(A[1..n]) // A[1..n] <-- sorted
2. l <- 1, r <- n
3. t <- 3 // 2 for first two assignments and 1 for initial while check
4. while(l < r)
5. t++ // 1 for initial while check
6. while(l < r or A[l] + A[r] > 0)
7. t++ // 1 for decrementing r
8. r--
9. t++ // 1 for following while check
10. t++ // 1 for if comparison
11. if(A[l] + A[r] = 0)
12. return true
13. t++ // 1 for incrementing l
14. l++
15. t++ // 1 for following while check
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This is a bit verbose, but it will work. Now we must simply prove that, at termination we have $$t = O(n)$$. We can do this inductively with a loop invariant.
Let's use the following loop invariant for the loop on line 4.
$$t = 3 + 4(l-1) + 2(n-r)$$
### Base Case
Initially $$t = 3$$, $$l = 1$$, and $$r = n$$. Thus we have:
$$3 = 3 + 4(1 - 1) + 2(n - n) = 3$$
### Inductive Case
Let $$t'$$, $$l'$$, and $$r'$$ be the values of $$t$$, $$l$$, and $$r$$ at the end of our previous iteration. At the end of our current iteration we have $$l = l' + 1$$, $$r = r' - k$$ for some $$k$$, and $$t = t' + 4 + 2k$$. Thus we have:
\begin{align*} t & = t' + 4 + 2k\\ & = 3 + 4(l' - 1) + 2(n - r') + 4 + 2k\\ & = 3 + 4(l - 2) + 2(n - (r + k)) + 4 + 2k\\ & = 3 + 4(l - 1) - 4 + 2(n - r) - 2k + 4 + 2k\\ & = 3 + 4(l - 1) + 2(n - r) & \square \end{align*}
Thus, we can conclude the loop invariant holds. At the end of the loop (in the worst case) we have $$l = r \leq n$$. We then have: \begin{align*} t & = 3 + 4(l - 1) + 2(n - r)\\ & = 3 + 4l - 4 + 2n - 2l\\ & = 2(n + l) - 1\\ & \leq 2(n + n) - 1\\ & = 4n - 1\\ & = O(n) & \square \end{align*}
Thus it is linear. There might be an easier way to do this, but this way makes sense to me pretty clearly. | {
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• I see that this looks formal but I feel like it is “hacking”, amending the algorithm in order to prove its complexity, is this real thing in computer science? Not talking about practice here – kuskmen Mar 27 at 21:53
• @kuskmen you don't need $t$ explicitly in your code. I put it there for clarity. You can have an implicit "operation count" which is realistically what you're trying to determine. Then prove (through loop invariants) that the "operation count" is bounded by $n$. That's all I did here with $t$ being the operation counter. – ryan Mar 27 at 22:49
• I see, I have to admit I am bad in mathematical proofs or any sort of proofs as you can tell. Can you tell why $t=t'+ 4 + 2k$ in the inductive step? – kuskmen Mar 28 at 8:15
• Assume $t$ starts at $t'$ just before the loop executes. Now hop into the loop. Increment by 1 for the initial inner while loop check (line 5). Now assume the inner while loop executes $k$ times. There are 2 increments in the inner while loop (line 7 and line 9), so $t$ is then incremented $2k$ times. Next, exit the inner while loop, increment by 1 for the if comparison (line 10). Assume the if condition is false because we do worst case analysis, then increment by 1 for incrementing l (line 13). Increment 1 last time for the following while check (line 15). Add them all up and you get that. – ryan Mar 28 at 16:11
• @kuskmen, this may also help you: cs.stackexchange.com/a/23595/68251 – ryan Mar 28 at 17:45 | {
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# Show that the equilibrium point $(0,0)$ is asymptotically stable and an estimate of its basin of attraction
Consider the system \begin{aligned} \dot{x} &=-y-x^3+x^3y^2\\ \dot{y}&=x-y^3+x^2y^3\end{aligned} Show that the equilibrium point $$(0,0)$$ is asymptotically stable and an estimate of its attractiveness basin.
Clearly the point $$(0,0)$$ is a point of equilibrium and also $$D_f(x,y)=\begin{pmatrix}-3x^2+3y^4 & -1+2yx^3\\ 1+2xy^3 &-3y^2+3y^2x^2 \end{pmatrix}$$, so $$D_f(0,0)=\begin{pmatrix}0 & -1\\1 &0 \end{pmatrix}$$ and has eigenvalues $$\pm i$$ So, I can not conclude anything of stability for this point, I need a Liapunov function and I do not know what it is or how to find it, could someone help me please? Thank you very much.
• Always try the simplest one first. For example, quadratic. More precisely, $V(x, y) = x^2+y^2$. Oct 30 '17 at 5:44
In addition to what has been said by @Evgeny and @MrYouMath: the set $$M=\left\{ (x,y)\in\mathbb R^2 :\; x^2+y^2<2 \right\}$$ is a positively invariant set of the considered system since $\forall (x,y)\in M$ $$\dot V=-x^4-y^4+x^2y^2(x^2+y^2)<-x^4-y^4+2x^2y^2=-(x^2-y^2)^2\leq 0;$$ it is also a subset (guaranteed estimation) of the domain of attraction.
• Nice one! This estimate is even better than $\| (x, y) \|_{\infty} < 1$ . Oct 30 '17 at 18:06
• +1: Very nice observation. This might be useful in the future. Oct 31 '17 at 14:03
• How Do I show that $M$ is positively invariant with this fact?
– P.G
Nov 20 '20 at 13:44
• @P.G Just as it is done in the proof of the Lyapunov stability theorem
– AVK
Nov 21 '20 at 7:02
As @Evgeny suggested you could use the Lyapunov function candidate
$$V(x,y)=\dfrac{1}{2}\left[x^2+y^2\right].$$
$V(x,y)$ is clearly positive definite at the origin and radially unbounded (which we would need for assessing global asymptotic stability). | {
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The time derivative of $V(x,y)$ is given by $$\dot{V}=x\dot{x}+y\dot{y}=x(-y-x^3+x^3y^2)+y(x-y^3+x^2y^3)$$ $$\dot{V}=-x^4-y^4+x^2y^2(x^2+y^2)=-(1-y^2)x^4-(1-x^2)y^4.$$
As the lower order terms $-x^4-y^4$ are negative definite we can conclude that the equilibrium point is asymptotically stable in a region around the origin. Using the comment given by @Evgeny we can see that this expression is negative semidefinite if $(x,y)$ lie inside the unit circle $D=\{(x,y)\in \mathbb{R}^2|x^2+y^2<1\}$. This is the basin of attraction (correction due to @Artem).
We cannot say if the origin is globally asymptotically stable because $\dot{V}$ is not negative definite for all regions around the origin. This does not mean that the origin cannot be globally asymptotically stable. It just means we can only show (local) asymptotic stability of the origin with $V(x)$ as our Lyapunov function candidate.
• Actually, you can do better if regroup terms correctly: $\dot{V} = -x^4(1-y^2)-y^4(1-x^2)$. This also gives a rough estimate of basin of attraction since this expression is negative when $\vert x \vert < 1$ and $\vert y \vert < 1$. Oct 30 '17 at 13:18
• @Evgeny: You are right. I felt that I should be able to factor this expression. Oct 30 '17 at 13:19
• Also, speaking about radial unboundedness: it's not a necessary condition really, see recent discussion in comments. Oct 30 '17 at 18:09
• It is not necessary for asymptotic stability but it is necessary for global asymptotic stability (see the comment in the brackets in my answer). I think that is what the example showed. Or do you mean something else? Oct 30 '17 at 18:14
• Even for the global asymptotic stability it is not necessary. You can prove the global asymptotic stability with a Lyapunov function which is not radially unbounded. Moreover, you can transform the image of any Lyapunov function such that it will become bounded but still will prove that something is globally asymptotic stable. Oct 30 '17 at 20:29 | {
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This problem can be handled with an optimization procedure, having in mind that generally is a non convex problem. The result depends on the test Lyapunov function used so we will generalize to a quadratic Lyapunov function
$$V(p) = p^{\dagger}\cdot M\cdot p = a x^2+b x y + c y^2,\ \ \ p = (x,y)^{\dagger}$$
and
$$f(p) = \{-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3\}$$ with $$a>0,c>0, a b-b^2 > 0$$ to assure positivity on $$M$$. We will assure a set involving the origin $$Q_{\dot V}$$ such that $$\dot V(Q_{\dot V}) < 0$$. The optimization process will be used to guarantee a maximal $$Q_{\dot V}$$.
After determination of $$\dot V = 2 p^{\dagger}\cdot M\cdot f(p)$$ we follow with a change of variables
$$\cases{ x = r\cos\theta\\ y = r\sin\theta }$$
so $$\dot V = \dot V(a,b,c,r,\theta)$$. The next step is to make a sweep on $$\theta$$ calculating
$$S(a,b,c, r)=\{\dot V(a,b,c,r,k\Delta\theta\},\ \ k = 0,\cdots, \frac{2\pi}{\Delta\theta}$$
and then the optimization formulation follows as
$$\max_{a,b,c,r}r\ \ \ \ \text{s. t.}\ \ \ \ a > 0, c> 0, a c -b^2 > 0, \max S(a,b,c,r) \le -\gamma$$
with $$\gamma > 0$$ a margin control number.
Follows a MATHEMATICA script which implements this procedure in the present case.
f = {-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3};
V = a x^2 + 2 b x y + c y^2;
dV = Grad[V, {x, y}].f /. {x -> r Cos[t], y -> r Sin[t]};
rest = Max[Table[dV, {t, -Pi, Pi, Pi/30}]] < -0.1;
rests = Join[{rest}, {r > 0, a > 0, c > 0, a c - b^2 > 0}];
sols = NMinimize[Join[{-r}, rests], {a, b, c, r}, Method -> "DifferentialEvolution"]
rest /. sols[[2]] | {
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