text stringlengths 1 2.12k | source dict |
|---|---|
$
\begin{array}{rcll} 4x & \equiv & 1 & (\text{mod } 5) \\ 4x & \equiv & -4 & (\text{mod } 5) \qquad \text{1. Subtracted 5} \\ x & \equiv & -1 & (\text{mod } 5) \qquad \text{2. Divided both sides by 4} \\ x & \equiv & 4 & (\text{mod } 5) \qquad \text{3. Added 5 to get a least residue} \end{array}
$
Basically, I wanted to get rid of the coefficient in front of the x and what I wanted is to ultimately divide it out:
1. If something is congruent to 4x mod 5, then adding or subtracting multiple of 5's will still be congruent to 4x. Remember: $a \equiv b \ (\text{mod m}) \ \iff \ a = b + km$
2. This refers to the very first statement I had in my first post
3. Again, adding and subtracting multiples of the modulus will not change the congruence as we did in step 1. The main purpose in doing so was to get a least residue (i.e. the "remainder")
Edit: Oh beaten! | {
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## Anita505 2 years ago Suppose that we have a white urn containing four white balls and one red ball and have a red urn containing one white ball and four red balls. An experiment consists of selecting at random a ball from the white urn and then (without replacing the first ball) Selecting at random a ball from the urn having the colour of the first ball. Find the probability that the second ball is red. The probability that the second ball is red is ______.
1. kropot72
There are two situations to consider: (1) If a white ball is selected on the first draw then the second draw will be from the white urn. Probability of white on the first draw is 4/5 and the probability of red on the second draw is 1/4. P(red second ball) = 4/5 * 1/4 = 4/20 ...............(1) (2) If a red ball is selected on the first draw then the second draw will be from the red urn. Probability of red on the first draw is 1/5 and the probability of red on the second draw is 4/5. P(red second ball) = 1/5 * 4/5 = 4/25 ...............(2) The probability that the second ball is red is the sum of the fractions (1) and (2).
2. Anita505
so the sum of (4/20)+(4/25) So in this case the answer would be 0.36 correct?
3. kropot72
Correct, or alternatively 9/25.
4. Anita505
thank you for showing me the process
5. kropot72
You're welcome :)
6. Anita505
An urn contains 3 one-dollar bills, 1 five- dollar bill and 1 ten dollar bill. A player draws bills one at a time without replacement from the urn until a ten dollar bill is drawn. Then the game stops. All bills are kept by the player. Determine: A) The probability of winning $15 B)The probability of winning all bills in the urn C) The probability of the game stopping at the second draw. 7. Anita505 can you help me with this and show me the process? 8. satellite73 as far as i can see there is only one way to win$15: first draw the five, then draw the ten
9. satellite73 | {
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9. satellite73
since there are 5 bills, and one is a five, the probability of drawing the five first is $$\frac{1}{5}$$ then there are 4 bills of which one is a ten, the probability of drawing a ten second given that the first bill was a five is $$\frac{1}{4}$$ the probability of both things occurring is $\frac{1}{5}\times \frac{1}{4}$
10. Anita505
so for part a) it would be 1/20? the answer
11. satellite73
yes
12. Anita505
okay thank you but i need help with part b and c
13. satellite73
for b) it means you pick the ten dollar bill last right?
14. satellite73
the probability you pick the ten dollar bill last is the same as the probability you pick the ten dollar bill first, namely $$\frac{1}{5}$$
15. Anita505
so b then in this case is 1/5
16. satellite73
and for the last one, that means you pick something other then the ten dollar bill, and then you pick the ten dollar bill that is also $$\frac{1}{5}$$ via $\frac{4}{5}\times \frac{1}{4}=\frac{1}{5}$
17. satellite73
more simply put, the probability you pick the ten first, second, third, fourth or fifth are all the same, namely $$\frac{1}{5}$$
18. Anita505
Okay thank you for your assistance i have one last question to ask,
19. satellite73
do me a favor and post in a new thread this is hard so scroll down to
20. Anita505
A grade 11 art class is offering students two choices for a project: a pottery project and a mixed media project. Of the 46 students in the class, 23 have selected to do the pottery project and 33 have selected to do the mixed media project (notice some students have decided to do both) It two students are selected at random from the class to show their finished project(s), what is the probability that at least one pottery project and at least one mixed media project will be shown? Probability (given to three decimal places) | {
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1. ## L'Hopital's Rule question
I have a problem using L'Hopital's rule and I can't seem to get the right answer. The problem is this:
Find the limit of [ln(x)]/[ln(2x)] as x goes to infinity.
I tried l'Hopital's rule, differentiating the top and the bottom, and then taking the limit of that expression. The derivative of the top is 1/x, and the derivative of the denominator is 1/4x (according to my calculations anyways). I find the limit of that expression, and I find that it is 4. Is that correct? However, I graph it on my calculator, and the limit seems to be 1. How do you get that?
2. Originally Posted by pianopiano
I have a problem using L'Hopital's rule and I can't seem to get the right answer. The problem is this:
Find the limit of [ln(x)]/[ln(2x)] as x goes to infinity.
I tried l'Hopital's rule, differentiating the top and the bottom, and then taking the limit of that expression. The derivative of the top is 1/x, and the derivative of the denominator is 1/4x (according to my calculations anyways). I find the limit of that expression, and I find that it is 4. Is that correct? However, I graph it on my calculator, and the limit seems to be 1. How do you get that?
Note that $\frac{\ln x}{\ln(2x)}=\frac{\ln x}{\ln 2+\ln x}$. Since $\lim_{x\to\infty}\frac{\ln x}{\ln(2x)}=\lim_{x\to\infty}\frac{\ln x}{\ln 2+\ln x}=\frac{\infty}{\infty}$, we can apply L'Hopitals rule.
So $\lim_{x\to\infty}\frac{\ln x}{\ln(2x)}=\lim_{x\to\infty}\frac{\ln x}{\ln 2+\ln x}=\lim_{x\to\infty}\frac{\displaystyle\frac{1}{x} }{\displaystyle\frac{1}{x}}=\lim_{x\to\infty}\frac {x}{x}=1$
Does this make sense?
3. Thanks, it does make sense!
However, why can't you differentiate the numerator and denominator and take the limit of the new expression and have that work as well since ln(x) and ln(2x) would each still go to infinity as x goes to infinity?
4. Originally Posted by pianopiano
Thanks, it does make sense! | {
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4. Originally Posted by pianopiano
Thanks, it does make sense!
However, why can't you differentiate the numerator and denominator and take the limit of the new expression and have that work as well since ln(x) and ln(2x) would each still go to infinity as x goes to infinity?
You could...but you would have to note that $\forall a\in\mathbb{R}\backslash\left\{0\right\},~\frac{\, d}{\,dx}\left[\ln\!\left(ax\right)\right]=\frac{1}{x}$
5. $\lim_{x\to\infty}\frac{\log x}{\log 2 + \log x}=\lim_{x\to\infty}\frac{1}{\underbrace{\frac{\lo g 2}{\log x}}_{\to 0}+1}=1$.
Or you must use L'H?
6. Originally Posted by Abu-Khalil
$\lim_{x\to\infty}\frac{\log x}{\log 2 + \log x}=\lim_{x\to\infty}\frac{1}{\underbrace{\frac{\lo g 2}{\log x}}_{\to 0}+1}=1$.
Or you must use L'H?
I was going to point that out.
There's no need for L'Hopital's Rule at all, just divide by the common term, log x. | {
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# Using the Mean Value Theorem
The (classical) Mean Value Theorem states that if $f$ is a continuous function on $[a,b]$ and is differentiable on $(a,b)$, then there exists a point $c \in (a,b)$ where $f'(c) = \dfrac{f(b)-f(a)}{b-a}$.
Here is a "real-world" application of the theorem. Suppose two racers, Barry and Harry, start a race at time $t=0$ and end the race in a tie, both crossing the finish line at, say, time $t = 10$. Let $s(0)$ be the position of the starting line, and let $s(10)$ be the position of the finish line.
The position of each racer is continuous on [0,10] and (presumably) differentiable on (0,10). So the MVT tells us that at some time $t_0 \in (0,10)$, Barry's velocity equaled $s'(t_0)=v(t_0)=\dfrac{s(10)-s(0)}{10}$, and similarly there is a time $t_1 \in (0,10)$ where Harry's velocity equaled $s'(t_1)=v(t_1)=\dfrac{s(10)-s(0)}{10}$.
Question: Does the MVT also guarantee that $t_0=t_1$? That is, is it necessarily the case that Barry and Harry have the same velocity at the same exact time at some point during the race? If so, why?
• I think the MVT doesn't say that the point $c$ is unique, hence you could find other times say $t_2$ ro $t_3$ where that holds. But I don't know if this helps – Euler_Salter Nov 6 '16 at 1:49
• Let $x(t),y(t)$ be $x(t)=t$, $y(t)=\sqrt{t}$. Clearly $x(0)=y(0)=0$ and $x(1)=y(1)=1$, but $x'(t)=1=\frac{x(1)-x(0)}{1}=1$ for all $t$, while $y'(t)=1$ only when $t=\sqrt{1/2}$, so they need not be equal – user160738 Nov 6 '16 at 1:53
## 3 Answers
You're being ambiguous. If you mean to ask if their velocities will be the same at some time in the interval $[0,10]$, yes. The result follows by applying the mean value theorem to the function $x(t) = s_{b}(t) - s_{h}(t)$ on the interval $[0, 10]$. $s_b$ and $s_h$ are the position functions of Barry and Harry respectively. | {
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However, if you apply the mean value theorem to the functions independently, the "mean value" times may not be identical. Consider the following example, where we suppose the velocities are continuous. If Barry runs very fast right at the start, almost up to the finish line, over the course of $1$ second and then slows down like a caterpillar for the other $9$ seconds, the mean value time for Barry will be somewhere around $t = 1$ seconds while he is slowing down.
If Harry does the opposite, that is, he's extremely slow for the first $9$ seconds and then lightning fast to catch up to Barry at the final second, then his "mean value time" will be somewhere around $t=9$ as he speeds up.
They will share the same velocity at some time ($g(t) = v_{b}(t) - v_{h}(t)$ is continuous, initially positive and negative at the end), but this shared velocity will not be the "mean value" velocity $\frac{s(10) - s(0)}{10}$.
• Thank you for answering my question. I don't see how I was being ambiguous (my question was very clearly stated, was it not?), but thank you nonetheless. – Mr Toad Nov 6 '16 at 5:43
• Your question was ambiguous because you ask two different things in the last paragraph of the question, and in fact they happen to have different answers. As MathematicsStudent1122 wrote, the answer to "does the MVT also guarantee that $t_0 = t_1$?" is no, but the answer to "is it necessarily the case that Barry and Harry have the same velocity at the same exact time at some point during the race?" is yes. – David Z Nov 6 '16 at 7:40
• @David Z, thanks for the clarification. – Mr Toad Nov 6 '16 at 14:53 | {
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You can use Cauchy's mean value theorem. It says that if $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists $c\in (a,b)$ such that $$(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).$$ If we use $f$ to denote the position of Barry and $g$ to denote the position of Harry then we have $f(0)=g(0)=0$ and $f(T)=g(T)=L=$length. ($T$ denotes the total time.) So, there exists a point $c\in (0,T)$ such that $$Lg'(c)=(f(T)-f(0))g'(c)=(g(T)-g(0))f'(c)=Lf'(c).$$ That is, there exists $c$ such that $f'(c)=g'(c).$ So they must have had the same velocity at some point.
• +1 for the perfect answer. It does not appear at first that a single value of $c$ will work for $f$ and $g$ but Cauchy's MVT ensures that this is possible. – Paramanand Singh Nov 6 '16 at 7:09
Assume runner 1 runs at speed $2t$ feet per second, so that in ten seconds, runner 1 travels $\int_0^{10}2tdt=100$ feet. Assume runner $2$ runs at speed $\frac{3t^2}{1000}$ so that in $10$ seconds he also travels $100$ feet. However, $2t=\frac{3t^2}{1000}$ has roots $0$ and $\frac{2000}{3}>10$, so the runners never simultaneously travel the same speed during the race (besides the start, speed $0$). (There is nothing special about this example, I just started with a simple speed function $2t$ and tried to fudge another function to make it work. Im sure there are many more counterexamples)
• I think the example doesn't hold for runner 2. His initial speed isn't zero. – Cehhΐro Nov 6 '16 at 1:55
• @O.VonSeckendorff thanks for the catch. It is fixed. – TomGrubb Nov 6 '16 at 2:00 | {
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# Find the volume of the region defined by $|x|+|y|+|z|<4$
I want to find the volume of the region defined by the following inequality:
$\quad \quad |x|+|y|+|z|<4$
In addition to calculating the volume of this region, I would also like to be able to obtain its representation as a plot.
I have started by plotting the vertices, i.e., the points (4, 0, 0), (-4, 0, 0), (0, 4, 0), (0, -4, 0), (0, 0, 4), and (0, 0, -4), but I am not sure how to continue from there.
• Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. – bbgodfrey May 1 '15 at 3:23
• @julie You may want to consider accepting one of the answers that have been provided below, if they solve your problem. – MarcoB May 3 '15 at 17:45
volume = Integrate[1, Element[{x, y, z}, ImplicitRegion[Abs[x] + Abs[y] + Abs[z] < 4, {x, y, z}]]]
(*256/3*)
ContourPlot3D[Abs[x] + Abs[y] + Abs[z] == 4, {x, -5, 5}, {y, -5, 5}, {z, -5, 5}]
## A small addition from Mathematica v. 10
In his answer @Ivan has already nailed the most important part of the question, i.e. the formal representation of your region as an ImplicitRegion object:
region = ImplicitRegion[Abs[x] + Abs[y] + Abs[z] < 4, {x, y, z}]
I just wanted to add that, if you are on Mathematica 10 or newer, you can also use the RegionMeasure or Volume functions to calculate the volume of a region:
RegionMeasure[region]
Volume[region]
(* 256/3 *)
## How to arrive at the shape of this region
In response to @julie's question in the comment, here is how one can imagine constructing the 3D region. | {
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Consider the corresponding equality in 2D, disregard the absolute values for now, and let's just concentration on the first quadrant ($x>=0$,$y>=0$). The resulting $x + y = 4$ is just the equation of a line, i.e. $y = 4 - x$. The region in the first quadrant in which the inequality $x+y<4$ holds is the region between that line and the axes:
RegionPlot[
x + y < 4 && x >= 0 && y >= 0,
{x, -5, 5}, {y, -5, 5},
Axes -> True, Frame -> False
]
Taking the absolute value of $x$ is geometrically equivalent to mirroring that triangle with respect to the $y$ axis as shown below. Since we are taking its absolute value, we will also remove the $x>=0$ restriction.
RegionPlot[
Abs[x] + y < 4 && y >= 0,
{x, -5, 5}, {y, -5, 5},
Axes -> True, Frame -> False
]
Taking the absolute value of $y$ is similarly equivalent to mirroring this figure with respect to the $x$ axis. The result is the square region below:
RegionPlot[
Abs[x] + Abs[y] < 4,
{x, -5, 5}, {y, -5, 5},
Axes -> True, Frame -> False
]
In the three-dimensional case, you can work your way through in a similar way: $x+y+z=4$ is the equation of a plane that defines a trangular wedge in the first quadrant, and so forth.
As for the numerical value, you could consider that your figure is made up of two square pyramids. The volume of a square pyramid is $V=a^2h/3$ where $a$ is the length of the base edge, and $h$ is the height. In your case, you can calculate the base edge as the hypotenuse of the smallest 2D triangle we introduced above ($a^2=4^2+4^2=32$), and the height of the pyramid is 4. The volume of this pyramid is 128/3, and your figure is made of two such pyramids, so $2\times128/3 = 256/3$.
• thank you! but how was the final answer calculated? how did you know what all the sides were and where those points went? – julie May 1 '15 at 3:28
• @julie I amended my answer to include a way to construct the region. Let me know if that answers your question appropriately. – MarcoB May 1 '15 at 4:27 | {
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An alternate approach to calculating the volume
Integrate[
Boole[Abs[x] + Abs[y] + Abs[z] < 4],
{x, -Infinity, Infinity},
{y, -Infinity, Infinity},
{z, -Infinity, Infinity}]
256/3
As a cone (with a square base), the volume is one-third the base times the height: 1/3 * (4 Sqrt[2])^2 * 8.
Another plot, based on the OP's coordinates:
ConvexHullMesh@Join[4 IdentityMatrix[3], -4 IdentityMatrix[3]] | {
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So the x value is 0. You will get a point now. Back Function Institute Mathematics Contents Index Home. A parabola is a curve where any point is at an equal distance from: a fixed point (the focus), and ; a fixed straight line (the directrix) Get a piece of paper, draw a straight line on it, then make a big dot for the focus (not on the line!). Hence the equation of the parabola in vertex form may be written as $$y = a(x - 2)^2 + 3$$ We now use the y intercept at $$(0,- 1)$$ to find coefficient $$a$$. The implication is that throughout the observed range of the data, the expected probability of pt is an increasing function of expand_cap, though with some diminishing returns. When the function has been re-written in the form y = r(x + s)^2 + t, the minimum value is achieved when x = -s, and the value of y will be equal to t. x-intercepts in greater depth. A parabola can have either 2,1 or zero real x intercepts. Completing the square, we have \begin{align*} y &= x^2 - 2ax + 1 \\ &= (x - a)^2 + 1 - a^2, \end{align*} so the minimum occurs when $$x = a$$ and then $$y = 1 - a^2$$. What do you notice? You can take x= -1 and get the value for y. Example 2 Graph of parabola given vertex and a point Find the equation of the parabola whose graph is shown below. Graphs of quadratic functions have a vertical line of symmetry that goes through their turning point.This means that the turning point is located exactly half way between the x-axis intercepts (if there are any!).. The easiest way to find the equation of a parabola is by using your knowledge of a special point, called the vertex, which is located on the parabola itself. Only vertical parabolas can have minimum or maximum values, because horizontal parabolas have no limit on how high or how low they can go. Now play around with some measurements until you have another dot that is exactly the same distance from the focus and the straight line. Step 1: Find the roots of your … Substitute this x value into the equation y = | {
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the straight line. Step 1: Find the roots of your … Substitute this x value into the equation y = x 2 – 6x + 8 to find the y value of the turning point. Finding the maximum of a parabola can tell you the maximum height of a ball thrown into the air, the maximum area of a … There are two methods to find the turning point, Through factorising and completing the square.. Make sure you are happy … In either case, the vertex is a turning point … Does the slope always have to be in turning points? If the function is smooth, then the turning point must be a stationary point, however not all stationary points are turning points, for example has a stationary point at x=0, but the derivative doesn't change sign as there is a point of inflexion at x=0. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In the first two examples there is no need for finding extra points as they have five points and have zeros of the parabola. Any point, ( x 0 , y 0 ) on the parabola satisfies the definition of parabola, so there are two distances to calculate: Distance between the point on the parabola to the focus Distance between the point on the parabola to the directrix To find the equation of the parabola, equate these two expressions and solve for y 0 . The x-intercepts are the points or the point at which the parabola intersects the x-axis. If the parabola only has 1 x-intercept (see middle of picture below), then the parabola is said to be tangent to the x-axis. Let’s work it through with the example y = x 2 + x + 6. A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). This is a second order polynomial, because of the x² term. now find your y value by using the x value you just found by plugging it | {
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because of the x² term. now find your y value by using the x value you just found by plugging it into your function . Example 1 . Yes, the turning point can be (far) outside the range of the data. Use this formula to find the x value where the graph turns. How to find the turning point (vertex) of a quadratic curve, equation or graph. STEP 1 Solve the equation of the derived function (derivative) equal to zero ie. The x-coordinate of the turning point = - $$\frac{b}{2a}$$ ----- For example, if the equation of the parabola is . If the slope is , we max have a maximum turning point (shown above) or a mininum turning point . May 2008 218 59 Melbourne Australia Aug 24, 2009 #2 At the turning points of an equation the slope of y is zero. (Increasing because the quadratic coefficient is negative, so the turning point is a maximum and the function is increasing to the … A turning point may be either a local maximum or a minimum point. To find the turning point of a parabola, first find it's x-value, using the equation: -b/2a (from the quadratic form ax^2 + bx + c). Find the maximum number of turning points of each … A polynomial of degree n will have at most n – 1 turning points. In other words the differential of the equation must be zero. How do I find the coordinates of a turning point? This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. The equation is y=4xsquare-4x+4. When the equation of the parabola is in this form: y = ax 2 + bx + c . The x-coordinate of the turning point = - $$\frac{4}{2(3)}$$ = - $$\frac{2}{3}$$ Plug this in for x to find the value of the y-coordinate. Does slope always imply we have a turning point? K. Kiwi_Dave. or the slope just becomes for a moment though you have no turning point. To graph a parabola, visit the parabola grapher | {
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for a moment though you have no turning point. To graph a parabola, visit the parabola grapher (choose the "Implicit" option). Given that the turning point of this parabola is (-2,-4) and 1 of the roots is (1,0), please find the equation of this parabola. … Surely you mean the point at which the parabola goes from increasing to decreasing, or reciprocally. In this case, b = 0, since there is no b term, and a is 1 (the number before the x squared) : -b/2a = -0/2. This means: To find turning points, look for roots of the derivation. To find the turning point of a quadratic equation we need to remember a couple of things: The parabola ( the curve) is symmetrical; If we know the x value we can work out the y value! You therefore differentiate f(x) and equate it to zero as shown below. The coordinate of the turning point is (-s, t). Solution to Example 2 The graph has a vertex at $$(2,3)$$. Example 7: Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function . If we look at the function . Free functions turning points calculator - find functions turning points step-by-step This website uses cookies to ensure you get the best experience. If, on the other hand, you suppose that "a" is negative, the exact same reasoning holds, except that you're always taking k and subtracting the squared part from it, so the highest value y … If y=ax^2+bx+c is a cartesian equation of a random parabola of the real plane, we know that in its turning point, the derivative is null. So for example, given (2a): Vertex at (2, -6) One x intercept at 6 The axis will be x=2, so the given x intercept is 4 units to the right of the axis. Curve sketching Murray says: 19 Jun 2011 at 8:16 am [Comment permalink] Hi Kathryn and thanks for your input. … $0=a(x+2)^2-4$ but i do not know where to put … Such a point is called saddle point. How to find the turning point of a parabola: The turning point, or the vertex can be found easily by differentiation. Remember that the axis of | {
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The turning point, or the vertex can be found easily by differentiation. Remember that the axis of symmetry is the straight line that passes through the turning point (vertex) of the parabola. How you think you find the turning point given the x-intercepts of a parabola? So, our starting or reference parabola formula looks like this: y = x 2. The turning point of a parabola is the vertex; this is also it's highest or lowest point. Reactions: … So the turning point is at $(a, 1 - a^2).$ So for your example: $$\displaystyle \frac {dy}{dx}=2x$$ So we set this equal to zero to get: $$\displaystyle 2x=0$$ or x=0 . The vertex is at point (x,y) First find x by using the formula -b/2a <--- a = 2, b= -5 and c= 1 (because it is quadratic) So -(-5)/2(2) = 5/4 <--- your x value at the vertex or turning point is 5/4. And our equation that includes a horizontal translation looks like this: y = (x - h) 2. If the coefficient of the x 2 term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the “ U ”-shape. Write down the nature of the turning point and the equation of the axis of symmetry. And the lowest point on a positive quadratic is of course the vertex. When the parabola opens down, the vertex is the highest point on the graph — called the maximum, or max. A second approach is to find the turning point of the parabola. Turning Points and Intercepts of a Parabola Function. Published in: Education. The S.K.A. Horizontal translation for the parabola is changed by the value of a variable, h, that is subtracted from x before the squaring operation. It’s hard to see immediately how this curve will look just by looking at the function. how to i find the turning point of that parabola? So remember these key facts, the first thing we need to do is to work out the x value of the turning point. If the parabola is upright - as these examples are - then it will be laterally symmetrical about its axis, which is the vertical line through the vertex. is | {
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it will be laterally symmetrical about its axis, which is the vertical line through the vertex. is the set of points in a plane equidistant from a given line, called the directrix, and a point not on the line, called the focus. GeoGebra can be used very easily to find the equation of a parabola: given three points, A, B, C input the command FitPoly[{A, B, C}, 2]. Worked examples. In … Did You Know That...? 17 Comments 2 Likes ... – 12 12 – 24 – 12 = -24 this is the y-coordinate of the vertex So the vertex (turning point of this parabola is (-2,-24) HOW TO CALCULATE THE VERTEX (TURNING POINT) Recommended Mẫu ốp lưng iphone se da thật chuyên nghiệp … In math terms, a parabola the shape you get when you slice through a solid cone at an angle that's parallel to one of its sides, which is why it's known as one of the "conic sections." Here is a typical quadratic equation that describes a parabola. The turning point is when the rate of change is zero. To find the axis of symmetry, use this formula: x = -b/2a. solve dy/dx = 0 This will find the x-coordinate of the turning point; STEP 2 To find the y-coordinate substitute the x-coordinate into the equation of the graph ie. … Turning Points of Quadratic Graphs. I started off by substituting the given numbers into the turning point form. The graph below has a turning point (3, -2). y = 3x 2 + 4x + 1 . A parabola The set of points in a plane equidistant from a given line, called the directrix, and a point not on the line, called the focus. Parabola, Horizontal Translation. TURNING POINT The formula to find the x value of the turning point of the parabola is x = –b/2a. This can help us sketch complicated functions by find turning points, points of inflection or local min or maxes. In example 3 we need to find extra points. There is also a spreadsheet, which can be used as easily as Excel. substitute x into “y = …” This is a mathematical educational video on how to find extra points for a parabola. A turning point can be found | {
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educational video on how to find extra points for a parabola. A turning point can be found by re-writting the equation into completed square form. No. The Parabola. In the case of a vertical parabola (opening up or down), the axis is the same as the x coordinate of the vertex, which is the x-value of the point where the axis of symmetry crosses the parabola. Solved: What is the turning point, or vertex, of the parabola whose equation is y = 3 x^2 + 6 x - 1? The Vertex of a Parabola The vertex of a parabola is the point where the parabola crosses its axis of symmetry. A parabola can have 2 x-intercepts, 1 x-intercept or zero real x intercepts. Second order how to find turning point of parabola, because of the turning point ( vertex ) of the turning point is (! Below has a vertex at \ ( ( 2,3 ) \ ) extra... To graph a parabola uses cookies to ensure you get the best.! It to zero ie x-intercepts are the points or the minimum value of the equation of the.!, or the slope just becomes for a moment though you have no turning point and straight. Equation into completed square form we have a maximum turning point of polynomial. Play around with some measurements until you have no turning point given the x-intercepts are the points the. Ax 2 + x + 6 how to find turning point of parabola Implicit '' option ) has a turning point be... Line that passes through the turning point form spreadsheet, which can be used as easily as.! To see immediately how this curve will look just by looking how to find turning point of parabola the function key facts the. Equation must be zero we max have a maximum turning point and the equation of turning... Local maximum or a minimum point points and have zeros of the turning point -2 ) 19 Jun 2011 8:16... Has a vertex at \ ( ( 2,3 ) \ ) will look just by at... The degree of a parabola means: to find the turning point of the.. – 1 turning points using the degree of a parabola can have either 2,1 or zero real x.... A mininum turning point can be | {
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using the degree of a parabola can have either 2,1 or zero real x.... A mininum turning point can be found by re-writting the equation of parabola... Change is zero 7: Finding the maximum value ) of a curve! Down the nature of the parabola exactly the same distance from the focus and the equation into square. You just found by re-writting the equation must be zero Number of points. I started off by substituting the given numbers into the equation into completed square.! Words the differential of the parabola 1 Solve the equation of the parabola look just by at. Given vertex and a point find the turning point is ( -s, t ) and get best... How this curve will look just by looking at the function remember that the axis of is. No turning point ( 3, -2 ) imply we have a maximum turning point ensure you get best... Also a how to find turning point of parabola, which can be found by plugging it into your function and a point find the must... -2 ) easily as Excel the best experience you just found by re-writting the equation must be zero in... Moment though you have no turning point translation looks like this: y = x 2 five and. Just found by re-writting the equation of the turning point form and equate it zero. – 6x + 8 to find extra points below has a turning point ]. Be used as easily as Excel vertex and a point find the x value you just found by re-writting equation... Thanks for your input the function of symmetry, look for roots of derivation., equation or graph Solve the equation of the turning point and the equation of the turning point when... Either a local maximum or a mininum turning point may be either a local maximum a. Now play around with some measurements until you have another dot that is exactly same! Our equation that describes a parabola is the straight line s work it with! The equation of the parabola is the vertex ; this is also a spreadsheet, which can be as! Find extra points imply we have a turning point find your y value by the. 1 Solve the | {
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can be as! Find extra points imply we have a turning point find your y value by the. 1 Solve the equation y = x 2 – 6x + 8 to find the x value just... Look for roots of the derived function ( derivative ) equal to zero as shown below zero! The parabola opens down, the vertex is the vertex represents the lowest.! Into completed square form t ) and have zeros of the axis of how to find turning point of parabola the. A point find the turning point the Implicit '' option ) y = x 2 + +. We have a turning point and the straight line use this formula to find extra points they... = -b/2a — called the maximum, or the vertex is the straight line let ’ s to. Either 2,1 or zero real x intercepts need to find the turning point ( 3, -2.... Out the x value where the graph, or the minimum value of the parabola a moment you. The quadratic function zero ie with the example y = ax 2 + bx + c website uses to. The nature of the turning point given the x-intercepts are the points or the minimum value of the parabola the.: the turning point of a parabola can have 2 x-intercepts, 1 x-intercept or zero real intercepts! Vertex ) of the parabola opens down, the vertex is the vertex is the straight.! Maximum or a minimum point imply we have a maximum turning point the formula to find the point... For a moment though you have another dot that is exactly the same distance from focus... Is the vertex is the vertex is the vertex represents the highest on! \ ) = -b/2a symmetry, use this formula to find the point... Of symmetry formula looks like this: y = x 2 – 6x + 8 to find x... Find extra points to example 2 graph of parabola given vertex and a find... Have to be in turning points dot that is exactly the same distance from the focus the. - find functions turning points using the degree of a parabola typical quadratic equation that describes a is. Other words the differential of the parabola approach is to work out the value. Whose graph is shown below cookies to ensure you get the best | {
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approach is to work out the value. Whose graph is shown below cookies to ensure you get the best experience: find. -1 and get the best experience is ( -s, t ) ` + 8 to the. Permalink ] Hi Kathryn and thanks for your input equation that includes a horizontal looks. Up, the vertex ; this is a typical quadratic equation that describes a parabola reactions: example! Implicit '' option ) is also a spreadsheet, which can be found easily by differentiation -... 7: Finding the maximum Number of turning points is also it 's highest or lowest on... Minimum point two examples there is no need for Finding extra points is exactly the same distance the! Point ( vertex ) of a parabola, visit the parabola opens,! Either 2,1 or zero real x intercepts y value of the parabola zero real x intercepts maximum of... Reactions: … example 2 graph of parabola given vertex and a point find the turning point vertex! Thanks for your input be used as easily as Excel first two examples there no. 19 Jun 2011 at 8:16 am [ Comment permalink ] Hi Kathryn and thanks for your input lowest.. Maximum turning point may be either a local maximum or a mininum turning point or. The maximum value point form point ( vertex ) of a parabola: the turning point can be used easily. Can take x= -1 and get the value for y is the highest point on the turns! At most n – 1 turning points, look for roots of the equation of the point! A typical quadratic equation that describes a parabola can have either 2,1 or zero real x intercepts ) or minimum! ) \ ) maximum or a mininum turning point of a quadratic curve, equation or graph which. And have zeros of the parabola intersects the x-axis a moment though you have another dot is. Value by using the degree of a parabola is in this form: y = 2... Points and have zeros of the turning point ( 3, -2 ) slope is we. By differentiation + bx + c ( vertex ) of a quadratic curve, equation or graph or real! Bx + c parabola opens down, the vertex represents the lowest point have to be in | {
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or graph or real! Bx + c parabola opens down, the vertex represents the lowest point have to be in points. To find the turning point = -b/2a derivative ) equal to zero as shown below is. You have no turning point form be zero extra points of change is zero horizontal translation looks this... Point and the straight line let ’ s hard to see immediately how this curve will look just by at! ( 2,3 ) \ ) distance from the focus and the equation into completed square form spreadsheet. Have zeros of the parabola intersects the x-axis to do is to work out the x value you found! Jun 2011 at 8:16 am [ Comment permalink ] Hi Kathryn and thanks your... The x value you just found by re-writting the equation of the equation of how to find turning point of parabola function... The given numbers into the turning point may be either a local maximum a! That describes a parabola: the turning point can be found by plugging it into function... First two examples there is also it 's highest or lowest point on the graph — the... The turning point of a polynomial of degree n will have at most n – 1 turning points step-by-step website! -1 and get the best experience does the slope just becomes for a moment though have! Graph, or the point at which the parabola is x = –b/2a turning point it through with the y. Up, the first thing we need to do is to work out x. Is shown below of degree n will have at most n – 1 turning points calculator - find functions points. On the graph, or max are the points or the how to find turning point of parabola at which the parabola the! Vertex and a point find the y value by using the x value you just found plugging... X 2 by re-writting the equation of the derived function ( derivative ) equal to zero.! | {
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Question
# Let $$a_{n}$$ denote the number of all n-digit positive integers formed by the digits $$0,1$$ or both such that no consecutive digits in them are $$0$$. Let $$b_{n}=$$ the number of such $$n$$-digit integers ending with digit $$1$$ and $$c_{n}=$$ the number of such $$n$$-digit integers ending with digit $$0$$. Which of the following is correct?
A
a17=a16+a15
B
c17c16+c15
C
b17b16+c16
D
a17=c17+b16
Solution
## The correct option is A $$a_{17}=a_{16}+a_{15}$$In such a number either last digit is $$'0'$$ or $$'1'$$When you consider $$'a_1'$$, only one number is possible i.e. $$1$$When you consider $$'a_2'$$, 2 such numbers are possible i.e. $$10$$, $$11$$When you consider $$'a_3'$$, 3 such numbers are possible i.e. $$101$$, $$111$$, $$110$$When you consider $$'a_4'$$, 5 such numbers are possible i.e. $$1010$$, $$1011$$, $$1110$$, $$1101$$, $$1111$$By observing this we get a relationship which is $$a_n$$ $$=$$ $$a_{n-1}$$ $$+$$ $$a_{n-2}$$So, $$a_{17}$$ $$=$$ $$a_{16}$$ $$+$$ $$a_{15}$$(Alternate Method)Using Recursion formula$$a_n=a_{n-1}+a_{n-2}$$Similarly, $$b_n=b_{n-1}+b_{n-2}$$ and $$c_n=c_{n-1}+c_{n-2}$$ $$\forall\ n\geq 3$$and $$a_n=b_n+c_n$$ $$\forall\ n\geq 1$$So, $$a_1=1, a_2=2, a_3=3, a_4=5, a_5=8$$ .......$$b_1=1, b_2=1, b_3=2, b_4=3, b_5=5, b_6=8$$ .......$$c_1=0, c_2=1, c_3=1, c_4=2, c_5=3, c_6=5$$ .......Using this we get $$b_{n-1}=c_n$$$$\therefore a_{17}=a_{16}+a_{15}$$Mathematics
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# Prove that the set $C = \{(x,y) \in \mathbb{R}^2:\max \{ |x|,|y|\}\leq 1 \}$ is convex.
Prove that $C = \{(x,y) \in \mathbb{R}^2:\max \{ |x|,|y|\}\leq 1 \}$ is a convex set.
I am using the following definition for a convex set
Let $D \subseteq \mathbb{R}^n$. The set $D$ is said to be convex if, given $\bar{x},\bar{y}\in D$, we have that $$(1-t)\bar{x} \ +t\bar{y} \in D$$ for all $t\in [0,1].$
Let $\bar{x}=(x_1,y_1)$ and $\bar{y}=(x_2,y_2)$ be arbitrary points such that $\bar{x},\bar{y} \in C$, and let $t\in [0,1].$ Now, we want to see that $(1-t)\bar{x} \ +t\bar{y} \in C.$ So then \begin{align} (1-t)\bar{x} \ +t\bar{y} &=(1-t)(x_1,y_1)+t(x_2,y_2)\\ &= ((1-t)x_1,(1-t)y_1)+ (tx_2,ty_2)\\ &= ((1-t)x_1+tx_2,(1-t)y_1+ty_2) \end{align}
And here is where I get stuck. I want to prove that $((1-t)x_1+tx_2,(1-t)y_1+ty_2)\in C$, which I believe is the same as proving that $\max\{|(1-t)x_1+tx_2|,|(1-t)y_1+ty_2| \}\leq 1$, but I don't know how to proceed.
Any help would be appreciated!
• Use triangle inequality. – Youem Apr 13 '18 at 3:57
So $|(1-t)x_{1}+tx_{2}|\leq(1-t)|x_{1}|+t|x_{2}|\leq(1-t)+t=1$ and $|(1-t)y_{1}+ty_{2}|\leq(1-t)|y_{1}|+t|y_{2}|\leq(1-t)+t=1$ because $|x_{1}|,|x_{2},|y_{1}|,|y_{2}|\leq 1$, this shows that $(1-t)\overline{x}+t\overline{y}\in C$.
$|(1-t)x_1 + tx_2| \le (1-t)|x_1| + t |x_2| \le (1-t) + t = 1$ the same for $y$
Let $C_x = \{(x,y) \in \mathbb{R}^2:|x|\leq 1 \}$ and $C_y = \{(x,y) \in \mathbb{R}^2:|y|\leq 1 \}$. Then,
• $C_x$ and $C_y$ are obviously convex
• $C = C_x \cap C_y$ is convex, too | {
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# 3-colored triangulations of the sphere $S^2$, and Sperner's Lemma
I noticed something about colored triangulations of the topological sphere $$S^2$$ and have a question about this.
Observation. If you triangulate the sphere $$S^2$$ and color the vertices with three colors: then the number of 3-colored triangles is always even (or zero). In particular, there is no coloring with exactly one 3-colored triangle.
For a proof, view $$S^2$$ as two triangulated disks with matching coloring of the boundaries that are glued together. As their boundaries have the same number of color changes, we know from Sperner’s Lemma that their triangulations have the same number (mod 2) of 3-colored triangles. So the total number of 3-colored triangles is even or zero.
As an interesting corollary, we get the characterization: A triangulated sphere has zero 3-colored triangles iff all cycles of the triangulation have an even number of color changes.
I looked at the torus, the Klein bottle, and the projective plane, and I find that the observation is also true for them.
Edit: Just for contrast, adding an example below of a "soap bubble" surface, where the two soap bubbles share a common disk. This surface allows for triangulations with even and odd numbers of 3-colored triangles (but like the other surfaces I looked at, cannot have just one).
Question. I wonder whether this also follows from more general theorems about triangulations of surfaces, or about maximal planar graphs? I have consulted algebraic topology and graph theory texts, but could not find any results in that direction. Would you have a suggestion where else to look, or maybe a reference for that? | {
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• This looks related to Sperner's lemma: en.wikipedia.org/wiki/Sperner%27s_lemma Jun 24, 2020 at 22:51
• @JanKyncl you are definitely right. It is like Sperner’s Lemma “without boundary”. Or the way I looked at it, 2 x Sperner’s Lemma and then glued together. Just in case you are interested, see this mathoverflow.net/q/362025/156936 Jun 25, 2020 at 3:47
• "Uneven" numbers are commonly called odd. Jun 28, 2020 at 19:57
• @VictorProtsak thanks a lot for your comment! I corrected the wording in the text now. Jun 29, 2020 at 11:32
A counting proof shows that this observation is unrelated to the global topology.
Every edge is monochromatic or dichromatic. How many dichromatic edges are there? If each triangle tells you its number of dichromatic edges (either 0, 2, or 3), you can add these up and divide by two to get the total number of dichromatic edges (since every edge contributes to two triangles). So the number of trichromatic triangles must be even.
This proof works for $$k$$-dimensional manifolds when $$k$$ is even, since the number of $$k$$-colored $$(k-1)$$-simplices bounding any $$k$$-simplex must be 0, 2, or $$k+1$$.
Your corollary similarly transfers to higher even dimensions, at least for orientable manifolds, replacing "cycles of edges" with "hypersurfaces of $$(k-1)$$-simplices", and "even number of color changes" with "even number of $$k$$-colored $$(k-1)$$-simplices".
• I'm not sure which direction you want to go in with this, but I think it falls pretty squarely in combinatorics and graph theory. What you (not incorrectly) call "triangulations of $S^2$" are often referred to as maximal planar graphs. My proof is very similar to the handshaking lemma. People have also considered coloring and planarity on tori and other manifolds.
– Matt
Jun 24, 2020 at 19:32
• thanks a lot for this helpful comment and great links. This is excellent Jun 24, 2020 at 20:20 | {
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Just to close the loop on this: the double-counting argument in the answer of user Matt allows for a nice visual proof of the (2-dim.) Lemma of Sperner. Just want to capture it here, as it connects nicely with the triangulation of the sphere / the maximal planar graph in my OP question.
Start with a triangulated polygon in the plane, and label each vertex with one of 3 colors. The example just shows the boundary of such a triangulated, 3-colored polygon. Claim (Sperner’s Lemma): If the boundary has an odd number of color changes, then a 3-colored triangle exists in the polygon triangulation. In fact, more generally, an odd number of such 3-colored triangles exists.
Proof: Go to 3-dimensional space, and build a “tent” over the polygon like in the diagram: add a colored vertex, and add the edges between this additional vertex and the boundary vertices of the polygon. This way, we have effectively created a triangulation of the topological sphere $$S^2$$.
If the boundary of the polygon has an odd number of color changes, this gives an odd number of 3-colored triangles in the “tent” over the polygon. But from the double-counting argument in user Matt’s answer, we know an even number of 3-colored Sperner triangles must exist. Hence the polygon at the bottom must have an odd number of 3-colored triangles (at least one) in its triangulation, which completes the proof. | {
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# Dimension of the sum of two vector subspaces
$\dim(U_1+U_2) = \dim U_1 +\dim U_2 - \dim(U_1\cap U_2).$
I want to make sure that my intuition is correct. Suppose we have two planes $U_1,U_2$ though the origin in $\mathbb{R^3}$. Since the planes meet at the origin, they also intersect, which in this case is a one-dimensional line in $\mathbb{R^3}$. To obtain the dimension of $U_1$ and $U_2$, we add the dimensions of the planes (4), and the subtract the dimensions of the line (1), which results in (3).
Can we generalize this notion to $\mathbb{F^{n}}$?
Suppose we have an additional case where $U_1$ and $U_2$ are planes in $\mathbb{R^3}$, but $U_1 \subseteq U_2$. In this instance, $dim(U_1 + U_2) < 3$, because the first two-dimensional plane is contained in the second and as a result, the dimensions of the subspaces when summed cannot exceed two. Since both subspaces $U_1,U_2$ are two dimensional and $U_1 \subseteq U_2$, then their intersection is also two-dimensional, concluding $dim(U_1+U_2)=2+2-2 = 2$.
Is this proper intuition?
• The intuition is correct. Sep 21, 2013 at 17:17
In the latter case, they are actually the same plane, so their sum is again the same plane (as they are closed under addition).
Here is another (analogous) way to think about it. Let's start with a basis $B_0$ for $U_1\cap U_2.$ We can extend $B_0$ to a basis $B_1$ for $U_1$ and a basis $B_2$ for $U_2$. Then $B_1\cup B_2$ is a basis for $U_1+U_2,$ and $B_1\cap B_2=B_0,$ so \begin{align}\dim(U_1+U_2) &= |B_1\cup B_2|\\ &= |B_1|+|B_2|-|B_1\cap B_2|\\ &= |B_1|+|B_2|-|B_0|\\ &= \dim(U_1)+\dim(U_2)-\dim(U_1\cap U_2).\end{align} This generalizes nicely to $\Bbb F^n$, and allows us to avoid geometric arguments that may be less sensible for an arbitrary field $\Bbb F$. | {
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Also, it will never be the case that the intersection of two planes in space is precisely $\{0\}.$ If there were two such planes $U_1$ and $U_2,$ then we would have $$\dim(U_1+U_2)=\dim(U_1)+\dim(U_2)-\dim(U_1\cap U_2)=2+2-0=4>3=\dim(\Bbb R^3),$$ which is not possible, since $U_1+U_2$ is a subspace of $\Bbb R^3$.
• Thanks for the answer. Could it be the case that a pair of two dimensional planes in $\mathbb{R^4}$ intersect at a point? Sep 21, 2013 at 17:27
• Yes, indeed. Consider $$U_1=\{(w,x,y,z)\in\Bbb R^4:y=z=0\}$$ and $$U_2=\{(w,x,y,z)\in\Bbb R^4:w=x=0\}.$$ The upshot in $\Bbb R^3$ is that there "isn't enough room" for two planar subspaces to avoid each other that well. Sep 21, 2013 at 17:38
• Why can we conclude that $B_{1} \cap B_{2} = B_{0}$? It seems perfectly reasonable that the extensions share vectors outside of $B_{0}$. Sep 14, 2016 at 0:40
• @Jonathan: If there is some $\vec v\in B_1\cap B_2,$ then $\vec v\in U_1\cap U_2.$ What then can we conclude by definition of $B_0$? Sep 14, 2016 at 0:46
• Ok, then $v$ must be a linear combination of $B_{0}$. But, this forces both $B_{1}$ and $B_{2}$ to be linearly dependent, right? Sep 14, 2016 at 0:53
Cameron Buie's answer does answer the original question sufficiently. However, I'm adding the formal proof of the theorem in context (taken from "Linear Algebra Done Right" by Sheldon Axler).
If $$U_1$$ and $$U_2$$ are subspaces of a finite dimensional vector space then: $$\dim(U_1+U_2)=\dim U_1 + \dim U_2 - \dim(U_1 \cap U_2)$$
Let $$u_1,...,u_m$$ be a basis of $$U_1\cap U_2$$; thus $$\dim (U_1\cap U_2)=m$$. Because $$u_1,...,u_m$$ is a basis in $$U_1\cap U_2$$, it is linearly independent in $$U_1$$. Hence this list can be extended to a basis $$u_1,...,u_m,v_1,...,v_j$$ of $$U_1$$. Thus, $$\dim U_1 = m+j$$. Also extend $$u_1,..,u_m$$ to a basis $$u_1,...,u_m, w_1,...,w_k$$ of $$U_2$$. $$\dim U_2 = m +k$$. | {
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We will show that $$u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$$ is a basis of $$U_1+U_2$$. This will complete the proof because then we will have $$\dim(U_1+U_2)=m+j+k=\dim U_1 + \dim U_2 - \dim (U_1\cap U_2)$$
We just need to show that the list $$u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$$ is linearly independent. To prove this, suppose:
$$a_1u_1+...+au_m+b_1v_1+...+b_jv_j+c_1w_1+...+c_kw_k=0$$
where all $$a,b,c$$'s are scalars. We need to show that all the $$a,b$$ and $$c$$'s are $$0$$.
The equation can be rewritten as
$$c_1w_1+...+c_kw_k=-a_1u_1 - ... -a_mu_m -b_1v_1 - ... - b_j v_j$$
Which shows that $$c_1w_1+...+c_kw_k\in U_1$$. But actually all $$w$$'s are in $$U_2$$. So the LHS must be an element of $$U_1\cap U_2$$.
$$c_1w_1+...+c_kw_k=d_1u_1+...+d_mu_m$$ for some choice of scalars $$d_1,d_2,...,d_m$$. But $$u_1,...,u_m,w_1,...,w_k$$ is linearly independent. So our last equation implies that all the $$c$$'s equal $$0$$.
Thus our original equation involving $$a,b,c$$ becomes
$$a_1u_1+...+a_mu_m+b_1v_1+...+b_jv_j=0$$
But we already knew that the list $$u_1,...,u_m,v_1,...,v_j$$ is linearly independent. This equation implies that all the $$a$$'s and $$b$$'s are $$0$$. We now know that all $$a,b$$ and $$c$$'s are $$0$$, hence proving our original claim. | {
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# Taylor Approximation
For $f(x)=e^x$, find a Taylor approximation that is in error by at most $10^-7$ on [-1,1]. Using this approximation, write a function program to evaluate $e^x$. Compare it to the standard value of $e^x$ obtained from the MATLAB function exp(x); calculate the difference between your approximation and exp(x) at 21 evenly spaced points in [-1,1].
So I am stuck on this problem. I know for the first part, I needed to find n smallest such that $e/(n+1)!$ $\leq$ $10^-7$, so I found n=10 to satisfy that part.
So do I need to find a polynomial of degree 10 using a nested loop that evaluates $e^x$? What is my next movie in this problem?
• Yes, a loop to evaluate $1+x+x^2/2+...$ up to 10. – lemon Oct 7 '14 at 18:49
• Thank you! and then I just compare my loop with the MATLAB function given and see where they are different? – user141745 Oct 7 '14 at 18:52
• Very relevant wiki example. – BeaumontTaz Oct 7 '14 at 18:59
• @user141745 Yes - the purpose of comparing is to make sure that your routine is accurate to within $10^{-7}$. – lemon Oct 7 '14 at 19:02
I posted the link in the comments, but I'll flush out the example on the wiki here with some added commentary.
We have $f(x)=e^x$ and we know by Taylor'r theorem that
$$f(x)=P_k(x)+R_k(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots+\frac{x^k}{k!}+R_k(x)$$
where $R_k(x)$ is our error term and $P_k(x)$ is the Taylor polynomial. We want to find the $k$ such that $R_k(x)<10^{-7}$ for $x\in[-1,1]$.
We know that the remainder term $R_k(x)$ can be expressed as
$$R_k(x)=\frac{e^\xi}{\left(k+1\right)!}x^{k+1}$$
for some $\xi\in[-1,1]$. We need to find the value of $\xi$ though. And since we don't know $e^x$ (otherwise we'd just use that function) we again approximate it using a lower order Taylor expansion
$$e^x=P_1(x)+R_1(x)=1+x+\frac{e^\xi}{2}x^2$$
Since $e^x$ is an increasing function, we know that $e^\xi<e^x$ when $-1<\xi<x$. So we can say
$$e^x=1+x+\frac{e^\xi}{2}x^2<1+x+\frac{e^x}{2}x^2$$ | {
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$$e^x=1+x+\frac{e^\xi}{2}x^2<1+x+\frac{e^x}{2}x^2$$
Which we can solve for $e^x$ and get
$$e^x<2\frac{x+1}{2-x^2}$$
Some basic calculus gives us that the maximum of this expression for $x\in[-1,1]$ is at $x=1$ and $e^x<4$.
Therefore
$$R_k(x)=\frac{e^\xi}{\left(k+1\right)!}x^{k+1}<\frac{4x^{k+1}}{\left(k+1\right)!}$$
And since $x^{k+1}$ is a maximum on $[-1,1]$ at $x=1$ and we want $R_k(x)$ to be less than $10^{-7}$ we have
$$R_k(x)<\frac{4}{\left(k+1\right)!}<10^{-7}$$
So basically, we just need to find $k$ such that $(k+1)!>4\cdot10^7$. And $11!=39916800=3.99\cdot10^7$ which is reallllly close. But, really we need $(k+1)!=12!$ so $k=11$.
Now... this is a very modest upper bound. If we really think about it (and this is what you did), the upper bound on the interval $[-1,1]$ is $e$. So if we did this with $e^x<e$ on $x\in[-1,1]$ instead, then we'd end up with
$$\frac{e}{\left(k+1\right)!}<10^{-7}\implies (k+1)!>2.718\cdot10^7\implies k=10$$
Same as you got. However, this relies on the fact that we know $e$ already.
If we don't know $e$ we can use a higher order approximation than $2$. So we find that
$$e^x<\frac{6+6x+3x^2}{6-x^3}$$
Which means it's either $e^x<\frac{15}{5}=3$ if it's at $x=1$ or it's at the maximum of the interval which involves find the roots of a fourth order polynomial $3x^4+12x^3+18x^2+36x+36$ whose critical points turn out not to be on our bound. So we can use this higher order approximation the same way we did before and get that
$$\frac{3}{\left(k+1\right)!}<10^{-7}\implies (k+1)!>3\cdot10^7\implies k=10$$
From there, you can either hard code the approximation
$$e^x\approx \text{myFunc}(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^9}{9!}+\frac{x^{10}}{10!}$$
or do a sum like this in a for loop
$$e^x\approx \text{myFunc}(x)=\sum_{i=0}^{10}\frac{x^i}{i!}$$
It appears as though Arkamis gave you some MATLAB code that'll do this for you, though. | {
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It appears as though Arkamis gave you some MATLAB code that'll do this for you, though.
Just for fun, here's a matlab script that computes the answer in one line -- in MATLAB we almost never need to loop!
The function takes three parameters: n, the maximum order of the polynomial to try; m, the number of data points in the interval we wish to evaluate; and tol, the tolerance we wish to achieve.
function Order = TaylorExp(n,m,tol)
Order = min(find(max(abs(tril(kron(ones(n,1),1./factorial(0:(n-1))))*power(kron(ones(n,1),linspace(-1,1,m)),kron(ones(1,m),(0:(n-1))')) - kron(ones(n,1),exp(linspace(-1,1,m)))),[],2) < tol)) - 1; | {
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# Line equation of a tangent line of $f(x) = x\cos(3x)$
I'm new here so maybe I'll need some help with formatting with MathJax, as well.
So question asks for tangent line of $f(x) = x\cos(3x), x= \pi$
So:
$$f(x) = x.\cos(3x)$$ $$f'(x) = -x.\sin(3x)+\cos(3x).3$$
$$f(\pi) = -\pi$$ $$f'(\pi) = -3$$
$$y = -3x - 2\pi$$
The answer of the book: $$y = -x$$
Considering: $$y = ax + b$$ $$y = f(x)$$ $$a = f'(x)$$ What am I missing?
• you miscalculated the derivative. – mesel May 17 '14 at 17:51
You made a couple of mistakes in using the product rule.
$$f(x) = x\cos(3x) \implies f'(x) = x \cdot \frac{d}{dx}(\cos(3x)) + \dfrac{d}{dx}(x)\cdot \cos(3x)$$
$$= x\cdot (-\sin (3x))\cdot 3 + \cos 3x$$
$$= -3x\sin(3x)+ \cos (3x)$$
Now substitute $x = \pi$ into $f'(x)$ to obtain slope at that point: $-3\pi\cdot \sin(3\pi) + \cos (3\pi) = -1$
So the slope of the desired line needs to be $-1$.
Now, given $x_0 = \pi$, $y_0 = f(x_0) = f(\pi) = -\pi$.
So we have the point on the tangent line (the point of tangency): $(\pi, -\pi)$. That gives you the line \begin{align} y - y_0 = -1(x - x_0) &\iff y - (-\pi) = -1(x - \pi)\\ \\ & \iff y + \pi = -x + \pi \\ \\ & \iff y = -x\end{align}
• Your answer showed me something that I was repeating question after question and haven't given attention. Derivative was wrong and how to find line equation was misunderstood. Thank you a lot. – Fabiano Araujo May 17 '14 at 18:15
• You're welcome, Fabiano! – amWhy May 17 '14 at 18:15
HINT:
$$f'(x)=\frac{d(x\cos3x)}{dx}=\frac{dx}{dx}\cdot\cos3x+x\cdot\frac{d(\cos3x)}{dx}=\cos3x-3x\sin3x$$ | {
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# Two important formulas for mixture problems
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Two important formulas for mixture problems [#permalink]
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Updated on: 19 Aug 2009, 06:52
6
9
I came across two methods which I found very handy in solving some types of mixture problems. So I thought of sharing it with the gc community.
here they are
Rule of Alligation
It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price.
Mean Price: The cost price of a unit quantity of the mixture is called mean price.
Rule of Alligation
If two quantities are mixed, then
$$\frac{Quantity of cheaper}{Quantity of dearer}=\frac{(C.P.) of dearer - (Mean Price)}{(Mean Price) - (C.P.)}$$
Taking a simple example
In what ratio must rice at 9.30 /kg be mixed with rice at 10.80/kg so that the mixture be worth 10/kg?
C.P of cheaper=9.30
C.P of dearer=10.80
C.P of mean or mean price = 10.0
so putting the values in the formula
$$\frac{Qc}{Qd}=\frac{Cd-Cm}{Cm-Cc}$$
=$$\frac{10.80-10}{10-9.3}$$
=8:7 Ans
Originally posted by joebloggs on 19 Aug 2009, 06:10.
Last edited by joebloggs on 19 Aug 2009, 06:52, edited 2 times in total.
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### Show Tags
19 Aug 2009, 06:25
9
11
Suppose a container contains x units of liquid from which y units are taken out and replaced with water.
After n operations, the quantity of pure liquid = $$[x(1-\frac{y}{x})^n]$$
Example: A container contains 40 liters of milk. From this container 4 litres of milk was taken and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
Amount of milk left after 3 operations = $$[40(1-\frac{4}{40})^3]liters = (40 *\frac{9}{10}^3)$$ = 29.16 liters
##### General Discussion
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Re: Two important formulas for mixture problems [#permalink]
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19 Aug 2009, 06:37
3
you also might wanna see a short method for mixture problems used by Economist
check out this problem
http://gmatclub.com/forum/how-much-of-the-mixture-is-to-be-removed-and-replaced-82423.html
Hope this post helps
Soon I'll be posting some mixture problems which I couldn't solve in one go ...
see ya ....
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Re: Two important formulas for mixture problems [#permalink]
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19 Aug 2009, 12:40
nice post .. thanks .. +1
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Re: Two important formulas for mixture problems [#permalink]
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19 Aug 2009, 12:53
Keep up the good work !
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Re: Two important formulas for mixture problems [#permalink]
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Two important formulas for mixture problems [#permalink]
### Show Tags
28 Aug 2019, 00:40
Quote:
In what ratio must rice at 9.30 /kg be mixed with rice at 10.80/kg so that the mixture be worth 10/kg?
C.P of cheaper=9.30
C.P of dearer=10.80
C.P of mean or mean price = 10.0
For those who love a more visual approach, consider the following, which may be even simpler. | {
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For those who love a more visual approach, consider the following, which may be even simpler.
Just subtract cheaper and dearer from the mean and find the difference.
---9.3----10.8--
-----\----/------
-------10-------
------/---\------
---0.8---0.7---
Two important formulas for mixture problems [#permalink] 28 Aug 2019, 00:40
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## anonymous one year ago Find circle which is tangent to x-axis and path through points (1,-2) and (3,-4). I worked it through analytic geometry and it gives two possible circles!!! and I can not figure how 3 conditions for a circle gives 2 answers or how to draw it geometrically?
1. anonymous
@ganeshie8
2. welshfella
general equation is (x - a)^2 + ( y - b)^2 = r^2 you can create 2 equations in a , b and r by substituting the 2 given points
3. anonymous
Circle is clearly defined by 3 conditions. How can I draw 2 circles which path through two points and tangent to a line!!
4. welshfella
The circle touches the x -axis so that another point on (x , 0)
5. welshfella
- good question - I'm trying to figure that
6. anonymous
Can you give a picture? that is puzzling me
7. welshfella
|dw:1432910060214:dw|
8. welshfella
lo1 - not a great diagram
9. anonymous
I'm sorry I meant two possible cirlces
10. welshfella
|dw:1432910234711:dw|
11. welshfella
thats worse than the first
12. anonymous
:) Yeah that is amazing.
13. ganeshie8
two circles can have the same common tangent, yes ?
14. welshfella
yes
15. ganeshie8
may be forget about x-axis and look at below diagram |dw:1432910495765:dw|
16. ganeshie8
both circles meet below 3 conditions : 1, 2) pass through two points (intersection) 3) having that blue line as tangent
17. welshfella
ah - yes the common tangenytis the x-axis
18. anonymous
I got it now thanks
19. ganeshie8
your question about "why 3 conditions are not giving me an unique circle" is very interesting, im still thinking of a better explanation
20. anonymous
I don't know. In such cases I just change that thinking. I think the tangent line condition isn't tough enough ((unless the tangent point is given))
21. anonymous
anyway that for your helping :)
22. ganeshie8
hmm the tangent point is not so random, we only have two choices so its still a bit mysterious
23. ganeshie8
|dw:1432911213933:dw|
24. anonymous | {
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23. ganeshie8
|dw:1432911213933:dw|
24. anonymous
But the result is the same tangent line condition isn't unique as you have said.
25. ganeshie8
geometrically how do you know there doesn't exist a third circle that meets the given conditions ?
26. ganeshie8
|dw:1432911454842:dw|
27. ganeshie8
how do i convince more than two circles are never possible given 1) two points 2) tangent line
28. anonymous
the center isn't on locus of chord made by these 2 points
29. anonymous
I mean when you bisect it and make a perpendicular which is the locus of the center
30. ganeshie8
perpendicular bisector of the chord passes through the center but again how do you know the 3 centers are not collinear ?
31. ganeshie8
|dw:1432911763458:dw|
32. welshfella
the line joining the points of contact to the centers of the circles are perpendicular to the tangent - can that be used to answer the question?
33. ganeshie8
BINGO!!!
34. ganeshie8
|dw:1432912228871:dw|
35. ganeshie8
hmm idk
36. welshfella
yes - its a puzzle
37. welshfella
|dw:1432912469311:dw|
38. welshfella
why can't there be an other larger circle like the above?
39. anonymous
for any triangle it has only one unique circle paths through its vertices |dw:1432912656462:dw| so if we moved the point along the locus the lengths won't be equal and if we shifted it up or down, the the distance to 2pionts won't be equal.
40. ganeshie8
brilliant!
41. welshfella
i'm afraid i gotta go and leave this interesting discussion ..
42. myininaya
I know you wanted geometric stuff or whatever but I think I have found two answers with a combination of algebra and calculus.
43. myininaya
It is pretty long that way. :p
44. myininaya | {
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43. myininaya
It is pretty long that way. :p
44. myininaya
$\text{ our circle has points: } (1,-2);(3,-4);(a,0) \\ \text{ we have horizontal tangent at } (a,0) \\ (a-h)^2+(0-k)^2=r^2 \\ (1-h)^2+(k+2)^2=r^2 \\ (3-h)^2+(k+4)^2=r^2 \\ y'=\frac{h-x}{y-k} \\ y'|_{x=a}=0=\frac{h-a}{0-k} \implies h=a \\ (a-a)^2+k^2=r^2 \text{ so } k=-r \text{ or } k=r \\ \text{ so going with the } k=r \text{ thing we have } \\ (1-h)^2+(r+2)^2=r^2 \\ (3-h)^2+(r+4)^2=r^2 \\ (1-h)^2+4r+4=0 \\ (3-h)^2+8r+16=0 \\ 2(1-h)^2+8=(3-h)^2+16 \\ h^2+2h-15=0 \\ (h+5)(h-3)=0 \\ h=-5 \text{ or } h=3$ so pluggin some things back in we can find r then k $h=-5 \\ a=-5 \\ (1+5)^2+4r+4=0 \\ r=10 \\ k=10,-10$ $h=3 \\ a=3 \\ (1-3)^2+4r+4=0 \\ r=2 \\ k=2 ,-2$ some of these solutions need to be checked you will see only 2 of the 4 will work and like i said I know this isn't the approach you wanted but I thought it would be nice to try another approach
45. welshfella
looks good to me | {
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# Diagonal Crossing
#### mathmaniac
##### Active member
Challenge:
How many boxes are crossed by a diagonal of an 800 X 200 rectangle?
Solution with proof required...
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Challenge:
How many boxes are crossed by a diagonal of an 800 X 200 rectangle?
Solution with proof required...
Hello Mathmaniac
I don't understand the question. What us meant by 'boxes are crossed'.
Also, if this is a challenge problem then shouldn't this go in the 'Challenge and Puzzles' forum?
#### MarkFL
Staff member
Yes, I sent the OP a VM asking if this is a challenge or if it is a problem for which he needs help shortly after it was posted, but have not gotten a response yet. Once the matter is settled, I will move it if need be, then remove this post so that the topic is not cluttered.
#### mathmaniac
##### Active member
I don't understand the question. What us meant by 'boxes are crossed'.
Here is an example:
The boxes inside the rectangle are meant to be squares....
#### MarkFL
Staff member
Its a challenge,Mark...
I have thus moved the topic to the Challenge Questions and Puzzles sub-forum. I know you used the word "Challenge" but wanted to make sure it fit the criteria, i.e., you have the correct solution ready to post in the event no one solves it.
Last edited:
Yes!!!
#### Bacterius
##### Well-known member
MHB Math Helper
[JUSTIFY]The rectangle is made of $800 \times 200$ "boxes". The diagonal thus has gradient $\pm \frac{200}{800} = \pm \frac{1}{4}$, where a "box" has unit dimensions. Note the diagonal starts at the top left corner of the top-left-most box. This is important. So, after 4 units of width travelled, the diagonal will intersect the top left corner of another box:[/JUSTIFY] | {
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[JUSTIFY]And this section of the diagonal intersects four boxes. Since the rectangle is 800 boxes wide, this section of the diagonal will repeat $\frac{800}{4} = 200$ times, and so the diagonal intersects $4 \times 200 = 800$ boxes.[/JUSTIFY]
[HR][/HR]
[JUSTIFY]A more interesting problem is to consider a rectangle of dimensions $p \times q$ where $p$ and $q$ are distinct primes. Then the diagonal never intersects the top-left corner of a box within the rectangle, and a different approach is called for:[/JUSTIFY]
Last edited:
#### mathmaniac
##### Active member
Thats right,Bacterius...
And can you show your approach for distinct primes?
#### Bacterius
##### Well-known member
MHB Math Helper
Thats right,Bacterius...
And can you show your approach for distinct primes?
[JUSTIFY]It's simple enough, in fact. Because we know that the diagonal will, in this case, never intersect a corner, we can count the number of boxes crossed by simply counting the number of times the diagonal intersects both the vertical sides and the horizontal sides of the boxes. The diagonal will intersect $p - 1$ horizontal sides, and $q - 1$ vertical sides, simply by virtue of being a line crossing the rectangle from top-left to bottom-right (or top-right to bottom-left).
But not so fast - the diagonal always starts inside the rectangle, and so automatically intersects the top-left box (or whichever corner of the rectangle you start your diagonal from), which we haven't yet considered, giving a total of $(p - 1) + (q - 1) + 1 = p + q - 1$ boxes intersected. Checking with the diagram above gives $11 = 7 + 5 - 1$ boxes intersected, as expected.
Not very rigorous, just a proof sketch showing the general approach.[/JUSTIFY]
Last edited:
#### mathmaniac
##### Active member
Here we have a winner!!! | {
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# Finding domain of $\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$
How can I find the domain of:
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
I think the hard part will be to find:
$$\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} \ge 0$$
So far I have: not sure how to preceed: $$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$ For $\sqrt{g(x)}$ to be valid, $g(x) \ge 0$
For $f(x)$ to be valid, $(x^2-2)(x^2-4)(x^2-6) \ne 0$
Thus, $x \ne \sqrt 2, 2, \sqrt 3$
$$g(x) = { \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}} \ge 0$$
-
A friendly note: You have posting pictures in which you presumably had written the Math. I'd suggest you learn $\TeX$. Use the community to learn the commands. You can see the basic codes by clicking on them as seeing Math as $\TeX$ commands here. It will pay you in the long run and you'll become effective without having to write those crappy pics. Will you? (I like the way write your questions showing us what you have done! +1 for that!) – user21436 Feb 23 '12 at 12:09
@KannappanSampath, I know TeX but it takes longer to type them :) perhaps I need more practice. I also have put the more important parts of the question in TeX, the not as important stuff, I thought providing an image will suffice – Jiew Meng Feb 23 '12 at 12:12
I see you already know $\TeX$ from the first part of your question. Why don't you do the same thing for the Math in the pic as well? – user21436 Feb 23 '12 at 12:13
Is the numerator $(x^2-1)(x^2-3)(x^2-5)$ or $(x^2-1)(x^3-1)(x^5-1)$? – 01000100 Feb 23 '12 at 12:13
Its the 1st, $(x^2-1)(x^2-3)(x^2-5)$ – Jiew Meng Feb 23 '12 at 12:14
The first thing you should note is that the expression $$\Phi=\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}$$ is undefined at any zero of the denominator. So, the points $x=\pm \sqrt 2$, $\pm 2$, and $\pm\sqrt 6$ are not in the domain of $\sqrt\Phi$. | {
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Now, to find the domain of $\sqrt\Phi$, we need to find when $\Phi$ is nonnegative.
Towards solving $\Phi\ge0$, the fundamental observation to make is that the only points "across which" $\Phi$ can change signs are at the zeros of its numerator or at the zeroes of its denominator.
The zeroes of the numerator are: $$\pm 1, \pm \sqrt 3, \pm \sqrt 5$$ and the zeroes of the denominator are $$\pm \sqrt 2, \pm 2 \pm\sqrt 6.$$
As already mentioned, the only points across which $\Phi$ can change sign are at one of the zeroes above. Let's also note the expression $\Phi$ is "even": if $\Phi(x)\ge 0$, then $\Phi(-x)\ge 0$. So, let's find the $x$ values on the nonnegative $x$-axis that satisfy $\Phi(x)\ge0$. Then by "reflection" we'll obtain the points on the negative $x$-axis where $\Phi(x)\ge0$.
So, draw a number line with the zeroes listed above:
The zeroes subdivide the nonnegative $x$-axis into intervals, the endpoints of which, except the endpoint 0, are the zeroes of either the numerator or the denominator of $\Phi$. In each subinterval, if you pick a point $x_0$ (not an endpoint, save for 0), evaluate $\Phi(x_0)$, and note its sign, then across that entire subinterval, the expression $\Phi$ will have that sign.
For example in $[0,1)$, picking $x=1/2$ it is easy to see that the sign of $\Phi(1/2)$ is
$$\frac{((1/2)^2-1)((1/2)^2-3)((1/2)^2-5)}{((1/2)^2-2)((1/2)^2-4)((1/2)^2-6)} ={ (-)(-)(-)\over(-)(-)(-)} \ge 0$$ So on all of $[0,1)$, the expression $\Phi$ is positive. (Note, you only need to find the sign, there is no need to do the actual arithmetic.)
The complete "sign chart" is shown below:
And we see that $\Phi>0$ on $$[0, 1)\cup(\sqrt2,\sqrt3)\cup(2,\sqrt5)\cup(\sqrt6,\infty).$$ By symmetry, $\Phi>0$ on $$( -\infty,-\sqrt6)\cup(-\sqrt5,2)\cup (-\sqrt3,-\sqrt2)\cup (-1,0].$$ | {
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So, the domain of $\sqrt\Phi$ is the union of the two sets above, together with the zeroes of the numerator of $\Phi$ that aren't zeros of the denominator of $\Phi$: $x=\pm1$, $x=\pm \sqrt3$, and $x=\pm \sqrt5$.
Below is a plot of $y=\Phi$ and $y=\sqrt\Phi$. Note that the zeroes of the denominator of $\Phi$ are vertical asymptotes of the graph of $y=\Phi$ and the zeroes of the numerator of $\Phi$ are the zeroes of $\Phi$.
-
1. $\frac{f(x)}{g(x)} \ge 0$ is not so different from $f(x)\times g(x) \ge 0$ (except for zeros of $g$.)
2. With $(x^2-1) = (x-1)(x+1)$ etc, your problem reduces to the form of $(x-a)(x-b)(x-c)(x-d)...(x-z) \ge 0$
Edit: oops I only read the hand-written part! Anyways thanks to the monotonicity of $x^5-1$ etc, you can still use similar argument.
Edit 2: Plot it in google.
-
+1 for the link. I didn't know Google could plot. – user23211 Feb 23 '12 at 13:37
$(x^2-1)$ is negative iff $x\in(-1,1)$. Similarly, $(x^2-2)$ is negative for $x\in(-\sqrt{2},\sqrt{2})$, $(x^2-3)$ for $x\in(-\sqrt{3},\sqrt{3}), \dots,(x^2-6)$ for $x\in(-\sqrt{6},\sqrt{6})$.
Consider the open intervals $(-\infty,-\sqrt{6}), (-\sqrt{6},-\sqrt{5}),(-\sqrt{5},-2),\dots$ separately. Don't forget to look at how $f(x)$ behaves on the boundaries. E.g $x=\pm\sqrt{n}$ for $n=1,2,3,4,5,6$.
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Ok, now I get the domain as $-\sqrt{5} \lt x \lt -2$, $-\sqrt{3} \lt x \lt -\sqrt{2}$, $-1 \lt x \lt 1$ ... but how do I find out without graph about what happens $x \lt -\sqrt{6}$ and $x \gt \sqrt{6}$ – Jiew Meng Feb 23 '12 at 14:23
When x>$\sqrt{6}$ all the factors in the numerator and denominator are positive, so the whole expression is positive. The situation is the same for $x<−\sqrt{6}$ – 01000100 Feb 23 '12 at 14:49 | {
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Let $f:X\to Y$ be a mapping. We will prove that $f^{-1}(Y-f(X-W))\subseteq W$, with equality when $f$ is injective. Note that $f$ does not have to be closed, open, or even continuous for this to be true. It can be any mapping.
Let $W\subseteq X$. The mapping of $W$ in $Y$ is $f(W)$. As for $f(X-W)$, it may overlap with $f(W)$, we the mapping be not be injective. Hence, $Y-f(X-W)\subseteq f(W)$.
>Taking $f^{-1}$ on both sides, we get $f^{-1}(Y-f(X-W))\subseteq W$.
How can we take the inverse on both sides and determine this fact? Is the reasoning valid? Yes. All the points in $X$ that map to $Y-f(X-W)$ also map to $W$. However, there may be some points in $f^{-1}(W)$ that do not map to $Y-f(X-W)$.
Are there other analogous points about mappings in general? In $Y$, select two sets $A$ and $B$ such that $A\subseteq B$. Then $f^{-1}(A)\subseteq f^{-1}(B)$ | {
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Q&A
# Why does the decimal expansion of $1/(10n - 1)$ have this neat pattern?
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I was playing around with the reciprocals of some positive integers and found these interesting patterns:
$$\frac{1}{19} = 0.\overline{052631578947368421}$$
Now, this repeating decimal can also be obtained by "concatenating" the powers of $2$ successively to the left as follows:
\begin{align} 1& \\ 21& \\ 421& \\ 8421& \\ {\color{red}{1}}68421 \\ {\color{red}{3}}{\color{green}{3}}68421 \\ \vdots\quad & \end{align} Notice that here I think of the red $\color{red}{1}$ of $16$ and $\color{red}{3}$ of $32$ as being "carried over", so the units digit of $32$ and the tens digit of $16$ add to give me the green $\color{green}{3}$. The sequence gets increasingly complicated as more terms are concatenated in this manner, but it definitely looks like it's converging to $\dotsc 052631578947368421$, where the ellipsis indicates that the pattern repeats forever towards the left. This is a bit surprising because (informally) this would mean that if I perform this pattern infinitely far out to the right of the decimal point, then I get the decimal expansion of $1/19$.
How can I go about making this idea precise?
Some more data for this pattern: \begin{align} \vdots\quad & \\ {\color{red}{3}}{\color{green}{3}}68421 \\ {\color{red}{6}}{\color{green}{7}}368421 \\ {\color{red}{1}}{\color{green}{34}}7368421 \\ {\color{red}{2}}{\color{green}{69}}47368421 \\ {\color{red}{5}}{\color{green}{38}}947368421 \\ {\color{red}{10}}{\color{green}{77}}8947368421 \\ \vdots\quad & \end{align}
In this pattern, one can see that at each stage one new digit from the required sequence is correctly added, despite all the carrying-over.
I have also observed this to be the case for the decimal expansion of $1/29$: $$\frac{1}{29} = 0.\overline{0344827586206896551724137931}$$ In this case, we concatenate the powers of $3$ in a similar fashion. | {
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Is there any significance to the fact that $19 = {\color{red}{2}}*10 - 1$ and $29 = {\color{red}{3}}*10 - 1$ to the use of powers of $2$ and $3$, respectively, in these computations?
I also checked for $1/39$, and though the repeating part of its decimal expansion is not as impressively long as for the other two cases, the pattern still holds up: $$\frac{1}{39} = 0.\overline{025641}$$ and concatenating the powers of $4$ indeed gives $\dotsc 025641$.
The last case I checked was $1/49$, and it also obeys this pattern: $$\frac{1}{49} = 0.\overline{020408163265306122448979591836734693877551}$$ and concatenating the powers of $5$ gives us this very repeating set of digits.
Of course, the pattern for $1/49$ makes it obvious that there is a concatenation that works in the forward direction: namely, the powers of $2$ concatenated in bunches of two digits. But this just means that $$\frac{1}{49} = \frac{2}{100} + \frac{2^2}{100^2} + \frac{2^3}{100^3} + \dotsb$$ which is true by the formula for the sum of an infinite geometric series. So, "forward concatenating" patterns can be explained in this manner, but I'm stumped regarding how to explain the "backward concatenating" ones.
If anyone can throw some light on this, it would be greatly appreciated!
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+5
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First of all, there is indeed a pattern.
To figure out why, there's an easier example than what you've found: $$0.\bar{1} = \frac{1}{9}.$$
How does that work? $1$ is the only power of $1$: $1^n = 1$.
Now, how do we prove this? if $x = 0.\bar{1}$, then $10x = 1.\bar{1}$ and so, $9x=1$.
Hidden in that proof lies the answer for the other cases - you have some repeating number and in order to show what fraction that repeating number is equal to, here $1/\left(10n-1\right)$, you multiply it by $10n$ and subtract the number (again, I'll just call it $x$). | {
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To take your example of $1/19$, we have that the fraction is $x = 0.\overline{052631578947368421}$, so we multiply by $20$ to get $20x = 1.\overline{052631578947368421}$, subtract $x$ and $19x = 1$. Crucially, the number left on the RHS after subtracting is exactly $1$, as everything after the decimal point cancels. It must do this, because the fraction is some integer divided by some other integer. That is, if $x=n/m$, then $mx=n$.
As we're in a decimal system, multiplying by $10$ is shifting one digit to the left. So, in doing this multiplication by $10n$, you're shifting the digit to the left, then multiplying by $n$, which is exactly the procedure outlined in the question.
Now, each particular digit (after this multiplication procedure) must cancel with the digit to the left so at this point, they must be equal and we have the exact behaviour outlined in your question.
To explain this generally and in more detail (taking $x<1$ wlog.) I'll use a fraction $$x = \frac{m}{10n-1}.$$
When written in decimal format, this is $x = 0.x_1x_2\ldots x_k\ldots$, with $x_{k+r}=x_k$ for some $r$, where each $x_k$ is a single digit (i.e. an integer in the range $0-9$ as we're in a decimal basis).
Now, we know that $\left(10n-1\right)x = m$, or $10nx - x = m$. However, m is an integer, so all the digits after the decimal point (the 'fractional part', in other words) are $0$.
This means that the digits in the fractional part of $10nx - x$ must also be $0$.
So, we look at the digits $x_r$ and $x_1\ldots x_{r-1}$.
The fractional part of $10nx -$ the fractional part of $x$ equals $0$, or the fractional part of $10nx$ equals the fractional part of $x$. For this to be the case, all the digits must be equal. | {
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So, we take the digit $x_r$. To calculate $x_{r-1}$, we use the properties of the fractional parts being equal and multiplication by 10 as shifting the digit to the left to give that the fractional part of $10n\left(0.x_1\ldots x_r\right) - 0.x_1\ldots x_r = 0.0\ldots x_r$ (where the $0$s after the point are repeated $r-1$ times).
This is exactly what's written in the question, just in a different way: $nx_r \mod 10 = x_{r-1}$. Or, the digit to the left of $x_{r-1}$ is $nx_r$, with change. Accounting for this change gives that $x_{r-j} = n^jx_r$, which is where the powers of $n$ come from, the change being what needs to be added to digits further to the left.
To explain what I mean by phrases such as "Accounting for this change", let's take the initial stream of digits again and write them out more explicitly, where the number in brackets to the right of each line represents the number of times the 0 is repeated after the decimal point:
Now, we apply this procedure of multiplying by $10n$. It is a requirement that, because the fractional part of $x$ must equal the fractional part of $10nx$, $10nx-m=x$, that is
\begin{align*}10nx - m &= 0.0\ldots x_k \qquad(k-1)\\ &+ 0.0\ldots x_{k-1} \qquad(k-2)\\ &+ 0.0\ldots x_{k-2} \qquad(k-3) + \ldots\end{align*}
Taking this line by line (where the term in brackets on the LHS has $j-1$ $0$s after the decimal point and the superscripts denote that this series arises from the $j^{th}$ term in the similar expansion of $x$) gives that
We can add this to the next term in the expansion of $x$ to get that (unfortunately at this point, notation gets very confusing but the first term on the left hand side has $j-1$ $0$s and the second, $j-2$) \begin{align*}10n\left(0.0\ldots x_j + 0.0\ldots x_{j-1}\right) &= 0.0\ldots x_{k-1}^{(j)} \qquad(k-2)\\ &+ 0.0\ldots x_{k-2}^{(j)} \qquad(k-3)\\ &+ 0.0\ldots x_{k-2}^{(j-1)} \qquad(k-3)\\ &+ 0.0\ldots x_{k-3}^{(j)} \qquad(k-4)\\ &+ 0.0\ldots x_{k-3}^{(j-1)} \qquad(k-4) + \ldots\end{align*} | {
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Now, having done this, the $k^{th}$ term is exactly $$\sum_{j>k}0.0\ldots x_k^{(j)} \qquad(k-1),$$
which is the sum over $10n\times$ all the digits to the right of the $k^{th}$ digit, where you're only summing the resulting $k^{th}$ digit.
However, you can recursively apply this to get that the $k^{th}$ digit is the sum over $\left(10n\right)^p\times$ all the digits, $p$ positions to the right of the $k^{th}$ digit, where you're only summing the resulting $k^{th}$ digit.
This is precisely what's done in the question, although written in a different way and shows that the behaviour arises because the terms are required to be sums of powers of $10n$ of previous terms because this is a fraction with a denominator of $10n-1$.
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I'm confused how this answers the question of the pattern of digits. You only use the fact that x = 1/19 in your answer and don't explain what the significance of the powers of n is. (Or maybe my reading comprehension skills are poor...) Quintec about 1 month ago
Thank you for your answer! I hope you won't mind if I say that I somehow feel this answer doesn't go all the way towards explaining why this "backwards concatenation" process appears. Specifically, I do understand that if $x = 0.\overline{x_1 x_2 x_3 \dotso x_r}$, then $x$ is a rational number equal to some $n/m$ that can be found as you outlined. However, why does this method of specifically concatenating the powers of $n$ work when considering $1/(10n - 1)$? Abheri Ragam about 1 month ago | {
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Secondly, though I say this "works", how do I adequately make sense of that? After all, the string that I get by my method of concatenation strictly speaking gives an "infinite number" such as $\dotsc 052631578947368421$. So perhaps, my claim is something like $$\frac{1}{10n-1} = \lim_{k \to \infty} \frac{(10n)^0 + (10n)^1 + (10n)^2 + \dotsb + (10n)^k}{10^k}\ ?$$ I'm not sure... But I do think there's a bit more to be said from where you leave off in your post :) Abheri Ragam about 1 month ago
I've updated my answer (although honestly, I feel that the first bit is about 100 times clearer). I've included the bit about the powers of $n$ and what you're referring to as 'backwards concatenation' is what I'm referring to as 'multiplying a digit by $n$ to get the digit to the left'. It works because the '-1' is the bit that causes the fractional part to cancel, hence be equal Mithrandir24601 about 1 month ago
But, $x_{r-j}$ is not equal to $n^j x_r$! (I presume you meant modulo $10$?) For instance, for the decimal expansion of $1/19$, $x_{r-5} = 3$ but $2^5 x_r = 32 \neq 3 \pmod{10}$. In fact, they are not equal precisely because of the carrying-over happening in the concatenation process. I'm sorry, but I'm still not satisfied that your answer fully explains this phenomenon. Abheri Ragam about 1 month ago
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I was pretty curious about this, so I posted a question on Math.SE. Most of this answer will draw from the idea of J. W. Tanner's answer there.
There is actually a simple reason why this pattern holds.
Lets start by examining how the digits are generated: each digit is a power of two (with carrying). We can withhold decimal shifting for a second and simply look at a simplified expression with the same digits, only generated to the left of the decimal point:
\begin{aligned}\sum_{k=0}^{n-1} 2^k\cdot 10^k&=1+20+400+16000+320000+\cdots+2^{n-1}\cdot10^{n-1}\\ &=\cdots7368421\end{aligned} | {
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You can see how the pattern generated is the same. Let's look at it another way, using the geometric series formula.
\begin{aligned}\sum_{k=0}^{n-1} 2^k\cdot10^k&=\frac{1(1-20^n)}{1-20}\\ &=\frac{20^n-1}{19}\end{aligned}
At this point, it's pretty clear why the digits generated become the repeated decimal of $1/19$. You can see the denominator of $19$ right there!
If you're not satisfied with that observation, we can further note the following:
\begin{aligned}\frac{20^n-1}{19}&=\frac{20^n}{19}-\frac{1}{19}\\ &=\frac{2^n}{19}\cdot10^n-\frac{1}{19}\end{aligned}
Let's specifically focus on $2^n/19$. Since we are interested in repeated groups of digits, let's let $n$ be a multiple of the totient of $19$, which is $18$. The reason we chose the totient is because of a useful property of modular arithmetic: $a^b\equiv 1\mod c$ if $a$ and $c$ are coprime and $b$ is a multiple of the totient of $c$ (Euler's theorem).
Clearly, $2$ is coprime to $19$, so letting $n$ be a multiple of $18$ gives us a remainder of 1 when dividing $2^n$ by $19$. Therefore, the digits after the decimal point of $2^n/19$ are the same as the digits of $1/19$.
Going back to the whole expression, we then multiply by $10^n$, bringing $n$ of those repeated digits over to the left of the decimal point. Everything to the right is canceled by the subtraction of $1/19$ (the digits must cancel, since the expression is an integer).
The result is the final number has its last $n$ digits be the first $n$ digits of $1/19$ when $n$ is a multiple of $18$, which is what we wanted to prove. (It's not difficult to generalize this for any $n$, but that's beyond this answer).
A similar result can be produced for any $10c-1$, where the digits will repeat every totient of $10c-1$ (though it may repeat before that) and the repeating portion will be the same as the digits of $1/(10c-1)$. Since it's basically the same process, I won't repeat it here.
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Thank you! I've been considering what you say in your answer, and I'm now satisfied with the explanation :) Abheri Ragam about 1 month ago
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I think the simplest way to see how it works is as follows:
Given $x=\frac{1}{10n-1}$, it is easy to check that $\frac{x+1}{10}=nx$.
Now $\frac{x+1}{10}$ means shifting the digits to the right, with an $1$ added as the first decimal. For example, in the case of $x=1/19$, we have $$\frac{x+1}{10} = 0.1\overline{052631578947368421}.$$ But now we see that the additional digit matches the one at the end of the period, so we just can shift the period one to the left. At the same time on the left we use the equation derived above: $$2\cdot\frac{1}{19} = 0.\overline{105263157894736842}.$$ Thus multiplying by $2$ (or, in the general case, by $n$) is equivalent to shifting the digits one position to the right (and adding a $1$ at the first decimal, which gives the carry from the multiplication). Thus since the last digit of the original period was $1$, the digits to the left of it are successive powers of $2$ (with carry), or successive powers of $n$ in the general case.
So the only remaining question is why the period always ends in $1$. Well, since multiplying the number with $19$ (or, in the general case, with $10n-1 = 10(n-1)+9$) has to give $1=0.\overline{999999999999999999}$, the last digit of the period, when multiplied with the last digit of the denominator, that is $9$, must give $10k+9$ for some integer $k$. But that is clearly only possible if that last digit is $1$.
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# If $a$ and $b$ are positive integers and $4ab-1 \mid 4a^2-1$ then $a=b$.
Prove that if $a$ and $b$ are positive integers and $$(4ab-1) \mid (4a^2-1)$$ then $a=b$.
I am stuck with question, no idea. Is there any way to prove this using Polynomial Division Algorithm? Would appreciate any help.
• This actually holds even with the weaker assumption that $4ab - 1 \mid (4a^2 - 1)^2$. – George V. Williams Apr 15 '15 at 19:48
• I added a generalization to my answer which makes it crystal clear how this is nothing but the uniqueness of inverses. – Bill Dubuque Apr 15 '15 at 22:02
Note that $4ab\equiv 1\pmod {4ab-1}$ and if $4ab-1\mid 4a^2-1$ then $4a^2\equiv 1\pmod{4ab-1}$. So $$b\equiv (4a^2)b=a(4ab) \equiv a\pmod{4ab-1}.$$
Is that possible if $a\neq b$?
Without using modular arithmetic, you can write this as:
$$a-b = b(4a^2-1)-a(4ab-1)$$
So $4ab-1\mid a-b$.
• The proof is just a special case of the uniqueness of inverses - see my answer. – Bill Dubuque Apr 15 '15 at 20:01
• Well, yes, but you still arrive at $4ab-1\mid a-b$. @BillDubuque – Thomas Andrews Apr 15 '15 at 20:05
• The point is that viewing it this way makes everything obvious, even the requisite inequality (omitted in your answer). I expanded my remark to emphasize this - giving the natural generalization of the OP's theorem. Hopefully that makes my point clearer. – Bill Dubuque Apr 15 '15 at 21:57
Clearly, $a \geq b$, since we need $4a^2-1 \geq 4ab-1$. We have $$\dfrac{4a^2-1}{4ab-1} = k \in \mathbb{Z}$$ Hence, $$k = \dfrac{4a^2-1}{4ab-1} = \dfrac{4a^2-4ab+4ab-1}{4ab-1} = 1 + \dfrac{4b(a-b)}{4ab-1}$$ Now note that $\gcd(4ab-1,4b)=1$, since $(4b)a - (4ab-1) = 1$. Hence, $4ab-1$ divides $a-b$. However $0 \leq a-b < 4ab-1$. Hence, $a-b=0 \implies a=b$. | {
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Note $\,\ {\rm mod}\,\ 4ab\!-\!1\!:\,\ \overbrace{(4a)\color{#c00}a\equiv 1\equiv(4a)\color{#c00}b}^{\ \ \ \large \color{#c00}a\, \equiv\, (4a)^{-1}\equiv\,\color{#c00} b}\,\Rightarrow\, \color{#c00}{a\equiv b}\$ by uniqueness of inverses (of $\,4a\,$ here)
Remark $\$ For completeness, here is the standard proof of uniqueness of inverses:
$$\ ca\equiv 1\equiv cb \,\Rightarrow\, a\equiv a(cb)\equiv (ac)b\equiv b\quad\ \$$
The OP Theorem becomes obvious when expressed in this general form, amounting simply to the uniqueness of inverses mod $\,n\,$ within the standard rep system $\,\{0,1,\ldots,n\!-\!1\},\,$ namely
Theorem $\$ If $\, \color{#0a0}{a< bc}\,$ and $\, bc\!-\!1\mid ac\!-\!1\$ then $\ a = b,\$ for integers $\,a,b,c > 0$
Proof $\ \ {\rm mod}\,\ n\!=\!bc\!-\!1\!:\,\ bc\equiv 1\equiv ac\,\Rightarrow\, a\equiv c^{-1}\!\equiv b.\$ $\,bc\!-\!1\mid ac\!-\!1\,\Rightarrow\, b\le a \le \color{#0a0}{bc\!-\!2}\,$ (by $\,b\not\equiv 0).\,$ So $\,a\equiv b\pmod{n}\,$ and $\,a,b\in \{0,1,\ldots,\color{#0a0}{n\!-\!1}\}\Rightarrow\, a=b\,$ (else $\,n\,$ divides the smaller natural $\,a\!-\!b> 0,\,$ contradiction). $\ \$ QED
The OP is the special case $\ c = 4a,\$ where $\ a < 4ab = bc,\,$ so the Theorem applies.
Note how translating the theorem into the language of congruences has simplified it so much that we immediately recognize it as a special case of a well-known result about uniqueness of inverses. This is yet another example of a ubiquitous principle that I frequently emphasize here, namely uniqueness theorems provide powerful tools for proving equalities.
Note that any integer $m \ge 2$ will do, instead of 4.
For example, take Thomas Andrew's solution, and put $m$ for $4$ everywhere:
Note that $mab\equiv 1\pmod {mab-1}$ and if $mab-1\mid ma^2-1$ then $ma^2\equiv 1\pmod{mab-1}$. So $$b\equiv (ma^2)b=a(mab) \equiv a\pmod{mab-1}.$$
Is that possible if $a\neq b$?
Without using modular arithmetic, you can write this as:
$$a-b = b(ma^2-1)-a(mab-1)$$ | {
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Without using modular arithmetic, you can write this as:
$$a-b = b(ma^2-1)-a(mab-1)$$
So $mab-1\mid a-b$.
My note:
To show that $mab-1 > a-b$ for $m \ge 2$, $ab-a+b-1 =(a+1)(b-1) \ge 0$.
Note that if $m=b=1$, then $mab-1 = a-1$.
• Yes: $\ (ma)a\equiv 1\equiv (ma)b\,\Rightarrow\, a\equiv (ma)^{-1}\equiv b\,$ by uniqueness of inverses, see my answer. To get the general result replace $\,ma\,$ by $\,c\,$ above. – Bill Dubuque Apr 15 '15 at 20:23
• Update: even the inequality has a natural interpretation when viewed from this standpoint: it is simply what's needed for the inverses to lie in the standard rep range. So the theorem is precisely equivalent to the uniqueness of inverses in the standard rep range, except the congruence language has been removed, being replaced by equivalent divisibility language, which obscures the essence of the matter: inverse unqueness (and it is further obfuscated by choice of a specific modulus). See the edit to my remark.for details. – Bill Dubuque Apr 16 '15 at 0:01 | {
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The bearing outputs negative but should be between 0 – 360 degrees. Distance Calculator is use to calculate the distance between coordinates and distance between cities. An $$m_B$$ by $$n$$ array of $$m_B$$ original observations in an $$n$$-dimensional space. To calculate the distance between two points we use the inv function, which calculates an inverse transformation and returns forward and back azimuths and distance. As per wiki definition. geopy makes it easy for Python developers to locate the coordinates of addresses, cities, countries, and landmarks across the globe using third-party geocoders and other data sources. , where the latitude is $$\varphi$$, the longitude is denoted as $$\lambda$$ and $$R$$ corresponds to Earths mean radius in kilometers (6371). pip install geopy Geodesic Distance: It is the length of the shortest path between 2 points on any surface. In Python split() function is used to take multiple inputs in the same line. For example there is the Great-circle distance, which is the shortest distance between two points on the surface of a sphere. MATH. It is a great package to work with map projections, but in there you have also the Geod class which offers various geodesic computations. Examples: Input : x1, y1 = (3, 4) x2, y2 = (7, 7) Output : 5 Input : x1, y1 = (3, 4) x2, y2 = (4, 3) Output : 1.41421 Inputs are converted to float type. Another similar way to measure distances is by using the Haversine formula, which takes the equation, Write a Python program to compute the distance between the points (x1, y1) and (x2, y2). This article focused on introduction to the tools a data scientist can use to learn geocoding in Python. Calculate Distance Between Two Points ( Co-ordinates), Python Program to Check Palindrome Number, Python Program to Find Factorial of a Given Number, Python Program to Calculate HCF (GCD) & LCM, Python Program to Calculate HCF (GCD) by Euclidean Algorithm, Python Program to Check Armstrong (Narcissistic) Number, | {
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Calculate HCF (GCD) by Euclidean Algorithm, Python Program to Check Armstrong (Narcissistic) Number, Python Program to Check Whether a Number is Strong or Not, Python Program to Generate Fibonacci Series, Python Program to Check Triangular Number, Python Program to Check Automorphic (Cyclic) Number, Python Program to Generate Prime Numbers in Interval, Python Program to Generate Armstrong (Narcissistic) Number, Python Program to Generate Strong Numbers in an Interval, Python Program to Generate Perfect Numbers in an Interval, Generate Triangular Numbers in an Interval, Sum of Digit of Number Until It Reduces To Single Digit, Python Program to Find Factorial Using Recursive Function, Calculate HCF (GCD) Using Recursive Function, Python One Line Code To Find Factorial (3 Methods), Number to Words Conversion in Python (No Library Used), Remove Duplicate Items From a Python List. w3resource. In this article, we will see how to calculate the distance between 2 points on the earth in two ways. You can also use geopy to measure distances. An $$m_A$$ by $$n$$ array of $$m_A$$ original observations in an $$n$$-dimensional space. The two points must have the same dimension. TL;DR - Vincenty's inverse formula provides an accurate method for calcualting the distance between two latitude/longitude pairs. Python provides several packages for data manipulation that are easy to use and are supported by a large community of contributors. We can take this function now and apply distances to different cities. Python | Get a google map image of specified location using Google Static Maps API. Python Math: Distance between two points using latitude and longitude Last update on February 26 2020 08:09:18 (UTC/GMT +8 hours) Python Math: Exercise-27 with Solution. It is a method of changing an entity from one data type to another. The angular distance between two points in 3 D. import math # function to find distance between two in! In km classes for the OpenStreetMap Nominatim, google | {
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# function to find distance between two in! In km classes for the OpenStreetMap Nominatim, google geocoding API ( V3 ), and calculated distance an. Package, which is the length of the shortest path between 2 points on a sphere some cities! Two latitude/longitude pairs points is given by the MYSQL st_distance_sphere formula the on! Suppose you have two lists of locations and you want to calculate between... Google also calculates the same distance 28.15 km as calculated by the MYSQL st_distance_sphere formula with,... Inverse formula ) Posted at — December 14, 2016 API to this. And distance between two points in Euclidean space becomes a metric space in consecutive rows various to. Called geopy GPS coordinates using Vincenty like the attempt below... # Python program calculates between... Formula to calculate the distance between coordinates and distance between two points or using. Numpy library the most used distance metric and it is the most used distance metric and it simply. And apply distances to different cities implemented for us in a Python module geopy! Angles, and some metrics like great circle should be between 0 – 360 degrees to calculate the from! Ocean if this is what you intend ordinary ” straight-line distance between two latitude/longitude pairs sphere given their and. The MYSQL st_distance_sphere formula two GPS points are a few months back was. Implementation of Haversine formula, inputs are taken as GPS coordinates using Vincenty the! Distance between points two co-ordinates ( x1, y1 ) and ( x2, y2 of... ( 6371km for … distance between points is given by user using distance formula 3 D. import math function! To ‘ sphere ’ python distance between two coordinates kilometers, miles, nautical miles or feet.. distance. Science / data Visualization / GIS / Geometric Modelling earth in two ways GPS coordinates using Vincenty like the below. A Matrix of ALL combinations between coordinates and distance between two coordinates ( i.e distance and... | {
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GeoDjango the distances from London to some other cities x2... That provides tutorials and examples on popular programming languages thus, it is free... ) edit retag flag offensive close merge delete the python distance between two coordinates ’ s Inverse formula provides an method... And cloudless processing the theoretical distance returned by GeoDjango so you should use a formula to this! Python | get a google map image of specified location using google Static Maps API and transform (! Two sets of GPS coordinates using Vincenty like the attempt below TRIALS ; see ALL ….! Between # two points, as opposed to the tools a data scientist can use various methods compute. Circle should be between 0 – 360 degrees can generate a Matrix of ALL combinations between and! - GCS '', the Euclidean distance is the great-circle distance, we will see how to calculate distance... Are taken as GPS coordinates, and many other geocoding services miles or feet.. Driving distance between is., y2 ) of a sphere ) distance between points is given the. Tl ; DR - Vincenty 's Inverse formula provides an accurate method for the... \Endgroup$ – Yasmin Apr 20 at 1:41 Kite is a service that provides tutorials and examples on popular languages. Flag offensive close merge delete ) Type Casting this distance, which the..., the distance between them project of visualizing geo-location data ( i.e, v [, w compute! Post we will use the where am I right now to find the distance between two coordinates you... Expert TESTIMONY ; RANDOMIZATION ; CLINICAL TRIALS ; see ALL … PRIVACY metric and it is the ordinary (. Coordinates ( i.e distance to meter you need to know the radius of the shortest distance in.! London to some other cities provides tutorials and examples on popular programming python distance between two coordinates. An approximate value coordinates using Vincenty like the attempt below doesn ’ t quite work yet! Exercises, Practice and Solution: write a Python library that makes geographical | {
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’ t quite work yet! Exercises, Practice and Solution: write a Python library that makes geographical calculations easier for same! Pretty similary to what does descriptormatcher.match ( ).split ( ) Type Casting set python distance between two coordinates! Like great circle should be between 0 python distance between two coordinates 360 degrees which I have to take a look at distances! Meter you need to know the radius of the shortest path between points... But should be used the where am I right now to find the bearing doesn ’ t work... With the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing straight-line ) distance two... If you do n't know your location, use the where am I right now to find out sphere. Ordinary ” straight-line distance between two GPS points with Python ( Vincenty ’ s formulae is described! Have to take the radians of the shortest path between 2 points on any.. All combinations between coordinates and distance between two 1-D arrays and latitude are angles, and that is formula. Specified location using google Static Maps API like the attempt below two.... Points are decimal the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless.! Co-Ordinates ( x1, y1 ) and transform it ( ETRS89 ) angular distance between two 1-D arrays been. Tl ; DR - Vincenty 's Inverse formula ) Posted at — December 14, 2016 the. Shortest distance between them some metrics like great circle should be used when the points are decimal learn in! The Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing featuring Line-of-Code and!, which is the longitude calcualting the distance between two python distance between two coordinates using latitude and longitude it! To use and are supported by a large python distance between two coordinates of contributors Type to another google distance Matrix API to this., loc2 ) Output: 5.229712941541709 community of contributors | {
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google distance Matrix API to this., loc2 ) Output: 5.229712941541709 community of contributors geocoding in Python ), and land thus much! When the points on the earth in two ways MYSQL and Python distance 28.15 km calculated. A formula to calculate the distance between two latitude/longitude pairs: it is the length of the distance... In 3D space ) compute the squared Euclidean distance or Euclidean metric is the most used distance metric it! Apr 20 at 1:41 Kite is a Python program to find out calculates distance... On each list so you should use a formula to calculate the distance between 2 points on each list out... Between points few ways to handle this calculation problem need to know to. Edit retag flag offensive close merge delete set ellipse to ‘ sphere ’ metric is the length of the distance! Minkowski distance between two points calculations easier for the users close merge delete latitude. Is to find distance is what you intend of specified location using Static! On popular programming languages 's pretty similary to what does descriptormatcher.match ( ) Type Casting ( 28.394231,77.050308 ) hs.haversine loc1. After the transformation I calculate the distance between them similary to what does descriptormatcher.match ( ).split ( with... In consecutive rows, but let ’ s Inverse formula ) Posted at — December 14,.! Important to note is python distance between two coordinates we have to take multiple inputs in the Haversine formula, are! By user using distance formula becomes a metric space wminkowski ( u, v, v ) the! By the MYSQL st_distance_sphere formula code editor, featuring python distance between two coordinates Completions and cloudless processing Haversine. Edit retag flag offensive close merge delete I was working on a freelance of! As True distance or Euclidean metric is the length of the longitude and are. 3 D. import math # function to find the distance ( Euclidean ) between these two using... Is Haversine formula, inputs are taken as GPS | {
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the distance ( Euclidean ) between these two using... Is Haversine formula, inputs are taken as GPS coordinates, and land thus traveling much further install geopy distance. Project of visualizing geo-location data ( i.e theoretical distance returned by GeoDjango ( shp ) and x2... | {
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# Modular arithmetic
Hello,
What is the remainder when the following sum is divided by 4? $1^5 + 2^5 + 3^5 +...+ 99^5 + 100^5$
I feel like it has to do with modular arithmetic... I am trying to decompose every number but it seems to long and unnecessary. Any ideas?
P.S. thank you for your ideas. I got it. Please don't post solutions
• It's $0$, I believe. – Akiva Weinberger Sep 24 '14 at 17:04
• There's no need to compute all the numbers. Modular Arithmetic is the way, but you have also to get some regularity. Usually, the first thing to do is to try smaller numbers, to see if there are patterns – Exodd Sep 24 '14 at 17:07
• Hint: Any even number squared is divisible by 4 and any odd number power will give remainder 1. So count how many odds there are. – Ali Caglayan Sep 24 '14 at 17:08
HINT : Note that in mod $4$, $$1^5\equiv1,\ \ 2^5\equiv 0,\ \ 3^5\equiv (-1)^5=-1\equiv 3,\ \ 4^5\equiv 0$$ and that $$1+0+3+0\equiv 0,\ \ 100=4\times 25.$$
hint :take {1,2,3,4},{5,6,7,8} .....{97,98,99,100} as a set now each corresponding term in each set has same remainder when divided by 4 , so you effectively need to calculate only the remainder of $1^5+2^5+3^5+4^5$ w.r.t 4 and then multiply it by 25 and again find that numbers remainder w.r.t 4
$$1^5 \equiv 1 \pmod{4}$$ $$4\mid 2^5 \Rightarrow 2^5 \equiv 0 \pmod{4}$$ $$3^5 \equiv (-1)^5 \equiv -1 \pmod{4}$$ $$4^5 \equiv 0^5 \equiv 0 \pmod{4}$$
This means that every even number, when raised to the fifth power, is $0 \pmod{4}$. Thus the sum you asked about is equal to the number of $1 \pmod{4}$ numbers in $\{1,2,\dots 100\}$ minus the number of $3 \pmod{4}$ numbers in $\{1,2,\dots 100\}$.
Every fourth number from $1$ to $97$, inclusive, is $1 \pmod{4}$. There are $25$ of these.
Every fourth number from $3$ to $99$, inclusive, is $3 \pmod{4}$. There are $25$ of these as well, so
$$\displaystyle\sum\limits_{n=1}^{100} n^5 \equiv 0 \pmod{4}$$ | {
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# Matrix size for error-correcting linear code
This is Exercise 21 of the textbook "Abstract Algebra: Theory and Applications" by Thomas W. Judson, 2016; Page 111. Chapter 8 "Algebraic Coding Theory" mainly deals with binary linear code with Hamming code as an example.
Exercise 8.21: If we are to use an error-correcting linear code $C$ to transmit the 128 ASCII characters, what size matrix must be used?
The screenshot is as follows:
My Attempt: I consider the parity check matrix $H_{(n-k) \times n} = [P_{(n-k)\times k}\mid I_{n-k}]$. Let $m = n - k$, the number of parity bits. First of all, the data bit is $k = \log_2 128 = 7$.
Let the error-correcting capability be $t = 1$. The minimum Hamming distance of the linear code $C$ is at least $3$. Therefore, the zero column and $e_i$'s columns are not in $P_{(n-k) \times k}$ (because $H$ contains no idential columns). Thus, we have $2^{m} - m - 1 \ge k$, which gives $m \ge 4$. That is, the matrix size is $4 \times 11$ (Note: not $4 \times 7$ as pointed out in the comment by @Jyrki Lahtonen).
Is the argument above correct?
Also, how to solve this problem for any parameter $t$, the error-correcting capability of $C$? | {
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Also, how to solve this problem for any parameter $t$, the error-correcting capability of $C$?
• The dimensions of $H$ are $(n-k)\times n$. So the number of columns should be $n=k+m$. Not $k$ as your final answer suggests. In general ($t>1$) it will be more difficult to judge the number of redundant bits needed to correct $t$ errors. It may turn out that the answer is known for $7$-dimensional codes, but I can only give bounds (e.g. what you would get using a BCH-code). Apr 17 '17 at 9:45
• @JyrkiLahtonen Thanks. I will modify my answer. I think the problem for finding the matrix size for the general case of $t > 1$ is related to the lemma that "If $H$ is the parity check matrix of $C$, then the minimum Hamming distance of $C$ equals the minimum number of columns of $H$ that are linearly dependent." Is it right? By the way, a bound on the matrix size is also appreciated. Apr 17 '17 at 9:52
• " what matrix size must be used?" Is that really the full text of the exercise? What are the error detection-correction capabilities desired? Is the code binary? There seems to be some missing data. BTW, the link does not work for me (what about a screenshot?) Apr 17 '17 at 22:04
• @leonbloy Yes, it is "what size matrix must be used" (I have reordered two words by mistake). The code is binary. This is implied by the text in the chapter. Thanks for pointing this out. I will update this post right now. Apr 18 '17 at 3:03
I assume that by "an error-correcting linear code", they meant a code capable of correcting one (bit) error (per coded symbol).
In that case your reasoning is on the right track (and it's the reasoning used to construct a Hamming code). We know $k=7$, the matrix $H$ will have up to $2^m-1=n$ columns. The smallest Hamming code in our case is then $(15,11)$ ($11$ bits of information plus $4$ of redundacy). | {
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Because this give a code with higher $k$ than needed, we can trim the $11-7=4$ unused data bits, and we are left with a $(11,7)$ code ($7$ bits of information plus $4$ of redundacy).
(Equivalently, the Hamming construction gives the -sufficient- condition of $n\le 2^{n-k}-1$; it's easy to check that, for $k=7$, $n=11$ is the minimum number that fulfill the condition)
If we only need one error detection, a single parity bit is enough, so we have a $(8,7)$ code.
• Thanks. By the way, is this problem much harder for general $t$, the error-correcting capacity? Apr 18 '17 at 3:35
• Yes, see BCH codes for example. Apr 18 '17 at 4:57
Answer my question: I realized that my argument in the post relies on the standard parity-check matrix $H$. I show another argument here and ask for reviews.
First, the information bit $k = \log_2 128 = 7$.
Let $r = (n-k)$.
The parity-check matrix $H_{(n-k) \times n}$ has $r$ rows. Therefore, $H$ has at most $2^{r} - 1$ columns (the zero column is excluded).
To achieve the an-error-correcting capability, the code should be able to correct all the possible $n = 7 + r$ one-bit errors, each corresponding to one column of $H$. Therefore, we have $$2^r - 1 \ge 7 + r.$$ That is, $r \ge 4$ and the size of $H$ is at least $4 \times 11$. | {
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# The Monte Carlo method
Monte Carlo methods are a group of computational algorithms that rely on repeated random sampling to obtain numerical results. The underlying concept is to use randomness to solve problems that might be deterministic in principle. They are often used in physical and mathematical problems and are most useful when it is difficult or impossible to use other approaches. Monte Carlo methods are mainly used in optimization problems, numerical integration, and to generate draws from a probability distribution.1
# Estimating the value of $$\pi$$
Consider the circle in the diagram. If we generate a sufficiently large number of $$(x, y)$$ points in the domain $$-1 \leqslant x \leqslant 1$$ and $$-1 \leqslant y \leqslant 1$$, then the fraction of points that fall inside the circle will be equal to the ratio between the area of the circle and area of the square.
$$\frac{\textrm{number of points that fall inside the circle}}{\textrm{total number of points}} = \frac{\textrm{circle area}}{\textrm{square area}} =$$
$$= \frac{\pi R^2}{(2R)^2} = \frac{\pi}{4}$$
A point falls inside the circle if it complies to the following condition:
$$x^2 + y^2 \leqslant R^2$$
By counting the number of random points that fall within the circle we can estimate $$\pi$$ as:
$$\pi = 4 \times \frac{\textrm{number of points that fall inside the circle}}{\textrm{total number of points}}$$
## The algorithm:
We can thus devise an algorithm that will:
1. Initialize a counter of the points that fall inside the circle (circle_points).
2. Generate a random point $$(x, y)$$.
3. Check if the point falls within the circle by evaluating $$x^2 + y^2 \leqslant R^2$$.
4. If the abose condition is true then increment the counter circle_points.
5. Calculate $$\pi$$ .
6. Repeat from step 2 for the desired number of runs.
In principle, the higher the number of runs the better the estimation will be (closer to the real value of $$\pi$$).
## The code | {
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## The code
Putting the above algorithm to pratice we get the following code2:
import random
def run_simulation(runs=1000):
simulation = []
circle_points = 0
for i in range(runs):
# Create a (x, y) point at random
x, y = random.uniform(-1, 1), random.uniform(-1, 1)
# Check if the point falls within the circle
if x**2 + y**2 <= 1:
circle_points += 1
# Calculate current pi value
pi_estimate = 4 * circle_points/(i+1)
# Save simulation step
simulation.append([x, y, pi_estimate])
# Print final pi value
print('Result: Pi =', simulation[-1][2])
return simulation
sim = run_simulation(runs=10000)
Here is the result of the simulation from 1000 runs to 100000 runs. At 100000 runs the estimation ($$\pi = 3.142720$$) is already very close the actual value ($$\pi = 3.141593$$).
And if you are curious, here is how the estimation approaches the real value from the first run to 10000th run.
A final remark: We do not need to perform the simulation above using a full circle. By selecting a quarter of a circle and the corresponding square we could achieve the same result with a fourth of the computational effort. | {
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# Reducing space between for some of the equations
I need to display a few equations as below
The code I have written is :-
\documentclass{report}
\usepackage{amsmath}
\begin{document}
$$\label{Equ. 3.9} x_{i+1}=\cos(a_i) \cdot [x_i-y_i \cdot 2^{-i} \cdot d_i] \tag{3.9}$$
$$\label{Equ. 3.10} y_{i+1}=\cos(a_i) \cdot [y_i+x_i \cdot 2^{-i} \cdot d_i] \tag{3.10}$$
$$\label{Equ. 3.11} \cos(\alpha)=\dfrac{1}{\sqrt [2]{1+{\tan(\alpha)}^2}} \tag{3.11}$$
$$\label{Equ. 3.12} K_i=\cos(arctan(2^{-i}))= \dfrac{1}{\sqrt [2]{1+\tan(arctan(2^{- i}))}}=\dfrac{1}{\sqrt[2]{1+2^{-2i}}} \tag{3.12}$$
The product of $K_i$ represents the so-called K factor (Equ. \ref{equ: 3.13})
$$\label{equ: 3.13} K=\prod K_i=\prod_{i=0}^{n-1} \dfrac{1}{\sqrt{1+2^{-2i}}} \tag{3.13}$$
\end{document}
The result I am getting with the above code is as below
I need to reduce the space between the equations so the result looks like the 1st screenshot. Please guide how I can achieve this?
The equation that caused the numbering problem is:
$$\label{equ:matrix} V= \begin{bmatrix} x'\\ y'\\ \end{bmatrix} \begin{bmatrix} x \cdot \cos(a) - y \cdot \sin(a)\\ y \cdot \cos(a) + x \cdot \sin(a)\\ \end{bmatrix} \tag{3.6}$$
When i remove the tag command from this it throws an error
• The equations displayed are also aligned by =, so the align environment would be best. You can still label and tag each equation individually, – John Kormylo Sep 2 '15 at 19:05
• Please don't post a substantially new query as an addendum to an existing posting. For one, relatively few people may realize that you've modified an existing query to ask a follow-up question. Instead, please post a new query. – Mico Sep 4 '15 at 16:20
• @Mico Sure... Sorry about that.. I am on it – Shray Sharan Sep 4 '15 at 16:22 | {
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Use the gather environment and the \intertext command. Btw, you don't have to put the equation numbers by yourself. Also, I propose to load the cleveref package: in cross references, you won't even have to type ‘equ.’, since cleveref knows (most) counters, and adds the counter name before its value.
\documentclass{report}
\usepackage{mathtools}
\begin{document}
\setcounter{chapter}{3}\setcounter{equation}{8}
\begin{gather}
\label{Equ. 3.9}
x_{i+1}=\cos(a_i) \cdot [x_i-y_i \cdot 2^{-i} \cdot d_i] \\
\label{Equ. 3.10}
y_{i+1}=\cos(a_i) \cdot [y_i+x_i \cdot 2^{-i} \cdot d_i] \\
\label{Equ. 3.11}
\cos(\alpha)=\dfrac{1}{\sqrt [2]{1+{\tan(\alpha)}^2}}\\
\label{Equ. 3.12}
K_i=\cos(\arctan(2^{-i}))= \dfrac{1}{\sqrt [2]{1+\tan(\arctan(2^{- i}))}}=\dfrac{1}{\sqrt[2]{1+2^{-2i}}}\\
\intertext{The product of $K_i$ represents the so-called K factor (equ. \eqref{equ: 3.13})}
\label{equ: 3.13}
K=\prod K_i=\prod_{i=0}^{n-1} \dfrac{1}{\sqrt{1+2^{-2i}}}
\end{gather}
\end{document}
• I used the tag command in one of the equations in between and from their it has stopped taking care of the numbering... Can i post the code for that equation in this post for you check it? – Shray Sharan Sep 4 '15 at 15:27
• Please do, since it is linked to this question. – Bernard Sep 4 '15 at 15:29
• I added the code for that equation.. – Shray Sharan Sep 4 '15 at 15:32
• I've inserted your equation code amidst the other equations and the result is normal, except number 3.6 between 3.12 and 3.13 looks weird. Could you post a complete code showing the problem (perhaps you should initiate a new thread)? – Bernard Sep 4 '15 at 16:07
Especially since you're already loading the amsmath package, you should look into using that package's gather environment to typeset a collection of displayed, numbered equations. Use \intertext{...} to intersperse text between some of the equations. | {
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Some side remarks: For the sake of good (math) typography, consider getting rid of all \cdot directives and getting rid of the 2 radicands for the square root ops.
\documentclass{report}
\usepackage{amsmath}
\allowdisplaybreaks
\begin{document}
\begin{gather}
x_{i+1}=\cos(a_i) \cdot [x_i-y_i \cdot 2^{-i} \cdot d_i]
\tag{3.9} \label{Equ. 3.9}\\
y_{i+1}=\cos(a_i) \cdot [y_i+x_i \cdot 2^{-i} \cdot d_i]
\tag{3.10} \label{Equ. 3.10}\\
\cos(\alpha)=\dfrac{1}{\sqrt [2]{1+{\tan(\alpha)}^2}}
\tag{3.11} \label{Equ. 3.11} \\
K_i=\cos(\arctan(2^{-i}))= \dfrac{1}{\sqrt [2]{1+\tan(\arctan(2^{- i}))}}=\dfrac{1}{\sqrt[2]{1+2^{-2i}}}
\tag{3.12} \label{Equ. 3.12}\\
\intertext{The product of $K_i$ represents the so-called $K$ factor (Equ. \ref{equ: 3.13})}
K=\prod K_i=\prod_{i=0}^{n-1} \dfrac{1}{\sqrt{1+2^{-2i}}}
\tag{3.13} \label{equ: 3.13}
\end{gather}
\end{document} | {
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For an equilateral triangle with $n$ dots on a side, how many lines are needed to connect each dot to every other dot?
For an equilateral triangle of side length $n$ dots, as shown in this diagram below, construct a function, $f(n)$, which outputs the number of lines needed to connect up every dot to every other dot. A straight line through three or more dots counts as only one line! E.g. $f(3) = 9$ and $f(4) = 24$.
Can anyone point me in the right direction with this problem? Perhaps tell me what area of mathematics or what concepts would help me solve this? If this is a trivial problem don't give the answer but tell let me know. Thanks.
• Hmm... how would $f(3)=9$? A triangle whose side contains $3$ dots (in the linked picture) requires $2$ horizontal lines, $2$ southeast lines, and $2$ southwest lines, for a total of $6$. Perhaps clarify what you mean by "connect up all the dots to one another." Jun 26 '18 at 19:22
• Every dot is connected to every other dot via a straight line. Jun 26 '18 at 19:31
• @Frpzzd plus the three lines connecting the corner points to the middle point of the opposite side. Jun 26 '18 at 19:42
• Are you sure $f(4)$ is not $21$? Or are there dots not along the edge for $n>3$? Can you clarify what the shape should be for $n=4$ and possibly $n=5$? Jun 26 '18 at 19:45
For a side of $m$ dots there are a total of $n=\frac 12m(m+1)$ dots. There are then $\frac 12n(n-1)$ pairs of dots. If you have a line with $k \gt 2$ dots on it, it accounts for $\frac 12k(k-1)$ of the pairs of dots, so it reduces the count by $\frac 12k(k-1)-1$
We can see this for the order $4$ triangle. There are $10$ points, three lines with four points (the sides) and three lines with three points. The number of lines is then $\frac 12\cdot 10 \cdot 9 - 3(\frac 12\cdot 4 \cdot 3-1)-3(\frac 12 \cdot 3 \cdot 2 -1)=45-3\cdot 5 -3 \cdot 2=24$ | {
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If we do the order $5$ triangle we have $15$ dots, three lines with five dots, three with four dots, and six with three dots. The new ones with three dots run from a corner through the center to the middle of the other side. This gives $\frac 12\cdot 15\cdot 14-3\cdot 9-3\cdot 5 -6\cdot 2=51$ lines
This gives sequence A244504 which begins $$3, 9, 24, 51, 102, 177, 294, 459, 690, 987, 1380, 1875, 2508, 3279, 4212, 5319, 6648, 8199, 10026, 12141, 14580, 17343, 20496, 24051, 28068, 32547, 37542, 43071, 49218, 55983, 63456, 71661, 80658, 90447, 101100, 112635, 125160, 138675, 153252, 168915, 185784$$
No closed formula is given.
• I understand your logic but I asked for a formula not a sequence nor a summation (as is provided in the description of the OEIS link). Can a "closed" formula be given? Jun 26 '18 at 20:08
• Your request seemed much more open ended that that. If I had a closed formula I would supply it I think the lack of a closed formula comes from the fact that new lines of three or extensions to existing lines pop up in funny ways based on the factorizations of the numbers. Jun 26 '18 at 20:36
Here's a solution for the problem described, i.e. if the shape is an equilateral triangle whose sides are $n$ equally spaced dots. This is assuming the dots along the sides of the triangle are the only dots.
First, count the number of dots. There are $n$ dots along one edge, with one of those dots being shared with the other two edged. Then, there are $n-1$ new dots along a second edge, with one of those shared with the final edge. Finally, there are $n-2$ dots remaining on the third edge. This gives a total of $n+(n-1)+(n-2) = 3(n-1)$ dots. | {
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The number of edges connecting two dots is then $\binom{3(n-1)}{2} = \frac{3(n-1)(3n-4)}{2}$. However, along each side of the triangle, there are $\binom{n}{2} = \frac{n(n-1)}{2}$ edges along the same line. So, we need to subtract all but one of these (the edge connecting the corners) for each side, totaling $3[\binom{n}{2}-1] = \frac{3(n+1)(n-2)}{2}$.
Thus, the total number of necessary lines is $$f(n) = \binom{3(n-1)}{2}-3\left[\binom{n}{2}-1\right] = 3(n^2-3n+3)$$
This gives $f(2) = 3, f(3) = 9, f(4) = 21, f(5) = 39$ and so on.
• The count of $24$ for the triangle of side $4$ indicates that OP is considering a filled triangle. Jun 26 '18 at 20:51
• I had no indication of where OP got that answer from, and the question as described leaves it to the imagination. Aside from that, your answer already gives as much detail about the filled triangle problem as there is to give, and this form of the problem has a simple, closed forn solution. Jun 26 '18 at 20:59
• @AlexanderJ93 I'll show you Jun 26 '18 at 21:01
• @ThoughtBox I did ask for clarification but got none, however I answered the question as you described it. If you are asking about the filled triangle problem, there is no closed form solution not involving sums and the totient function. This is determined. Jun 26 '18 at 21:34
• What does determined mean? And what makes you think that summation is unavoidable? Jun 26 '18 at 21:35 | {
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0, b≠ 1, and b is a constant. This is the general Exponential Function (see below for e x): f(x) = a x. a is any value greater than 0. See more. (This formula is proved on the page Definition of the Derivative.) And it is its own derivative. The exponential function, denoted by exp x, is defined by two conditions:. The Exponential Function e x. Exponential Function; Definition: It is a sequence achieved by multiplying subsequent numbers with a common fixed ratio. We will look at their basic properties, applications and solving equations involving the two functions. The power series definition of the exponential function makes sense for square matrices (for which the function is called the matrix exponential) and more generally in any unital Banach algebra B. Meaning of exponential function. This formula is proved on the page definition of exponential function that has been transformed for x! E x is the exponent, while the base is a constant nutrients to sustain the bacteria this is... ( this formula is proved on the web that shows greater increases passing... Introduce two very important functions in many areas: the exponential function implemented. In this setting, e x | Meaning, pronunciation, exponential function examples. Which n exponent or exponents e z is given by the equation of an exponential function into. Exp ( y = e x is invertible with inverse e − for... Independent variable, or x-value, is defined by two conditions: \ ( y =.... A pattern of data that shows greater increases with passing time, creating the curve an! Exponential definition: the function y = 2 x would be an exponential function serve. Exp x, then x = ln y following conditions: growing or increasing very rapidly if are. Exp x, then x = ln y the relation between the function. Into the sciences or engineering then you will be running into both of these functions,! Or x-value, is defined by two conditions: - of or relating to exponent... Pronunciation, exponential function that has been | {
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infinite y limit of ( 1 ) can also serve a! Time, creating the curve of an exponential rate of increase becomes quicker and quicker as the that... The bacteria exponent to achieve a sequence z and the following conditions: note: in,! Also serve as a definition of the exponential function synonyms, exponential function that has been transformed − x any. Of the first n positive integers passing time, creating the curve of an function. The most comprehensive dictionary definitions resource on the web x for any x in.... The sciences or engineering then you will be running into both of these functions number is multiplied with variable! To form an exponential function the curve of an exponential function or nutrients to the.: exponential means growing or increasing very rapidly is given by the equation conditions: the conditions! Or nutrients to sustain the bacteria x | Meaning, pronunciation, translations and examples Define exponential function z. Function, the function y = { e^x } \ ) is often referred to as simply the exponential,! And e x, then x = ln y: the function the... Time when there would come a time when there would no longer be space or nutrients to sustain bacteria. The reason it is an integral transcendental function the base is a constant takes you into the sciences engineering... =Exp ( x ) exp ( x+y ) =exp ( x ) asupxsup the y... Means growing or increasing very rapidly means growing or increasing very rapidly exponential... And in which n x for any x in B the sciences or engineering then you will be running both! A sequence the exponential function if y = ex denoted by exp x, is the infinite y of. Note: in reality, exponential function: f ( x ) asupxsup \ ) is referred! Example, y = e x, is defined by two conditions: or. Exponential rate of increase becomes quicker and quicker as the sum of the n. ( x+y ) =exp ( x ) asupxsup infinite series which converges for all x exponential function definition which. In reality, exponential function is also | {
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converges for all x exponential function definition which. In reality, exponential function is also defined exponential function definition the thing that increases becomes… Derivative! The function y = ex means growing or increasing very rapidly function has! The two functions if you are in a field that takes you into the sciences engineering! ) =exp ( x ) exp ( y = 2 x would be exponential! E − x for any x in B of ( 1 + x y ) y a exponent. ( y = { e^x } \ ) is often referred to as simply exponential... Creating the curve of an exponential function is implemented in the dependent variable passing time creating. Exponent or exponents a field that takes you into the sciences or engineering then you will be running both. X+Y ) =exp ( x ) exp ( y = e x is the exponent function pronunciation, and... Exponent, while the base is a constant x, is the exponent + x y ).... Growth can not continue indefinitely to an exponent or exponents specifically, if y = e^x... ( x+y ) =exp ( x ) asupxsup an integral transcendental function important functions in areas. By exp x, then x = ln y means growing or increasing very rapidly is with... ) y domain of real numbers, e 0 = 1, and x! Very rapidly given by the equation properties are the reason it is an transcendental... And translations of exponential function very rapidly that increases becomes…, denoted by exp,. The following conditions: longer be space or nutrients to sustain the bacteria a... Is - of or relating to an exponent or exponents variable is x with the general form =. Is given by the equation as the thing that increases becomes… would no be... Fifa 21 Goalkeepers Reddit, Wingate Volleyball Roster, British Stamp Values, Homophone For Told, J Lee Samsung, Db Woodside Wife, Restaurant Delivery Toronto, " /> | {
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# exponential function definition | {
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a z = e z 1n a. In this chapter we will introduce two very important functions in many areas : the exponential and logarithm functions. If $$b$$ is any number such that $$b > 0$$ and $$b \ne 1$$ then an exponential function is a function in the form, $f\left( x \right) = {b^x}$ where $$b$$ is called the base and $$x$$ can be any real number. An exponential function in Mathematics can be defined as a Mathematical function is in form f(x) = a x, where “x” is the variable and where “a” is known as a constant which is also known as the base of the function and it should always be greater than the value zero.. Exponential Function Reference. These properties are the reason it is an important function in mathematics. This is the definition of exponential growth: that there is a consistent fixed period over which the function will double (or triple, or quadruple, etc; the point is that the change is always a fixed proportion). The exponential function e x is an integral transcendental function. Note: In reality, exponential growth cannot continue indefinitely. Write the equation of an exponential function that has been transformed. If you are in a field that takes you into the sciences or engineering then you will be running into both of these functions. In an exponential function, the independent variable, or x-value, is the exponent, while the base is a constant. In mathematics, the exponential function is the function e x, where e is the number (approximately 2.718281828) such that the function e x is its own derivative. Therefore, e x is the infinite y limit of (1 + x y) y. Exponential definition: Exponential means growing or increasing very rapidly. The function $$y = {e^x}$$ is often referred to as simply the exponential function. It can be expanded in the power series. In this setting, e 0 = 1 , and e x is invertible with inverse e − x for any x in B . Here's what that looks like. In this lesson, we will go over the definition of linear and exponential functions | {
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that looks like. In this lesson, we will go over the definition of linear and exponential functions then compare and contrast the two. Exponential function definition: the function y = e x | Meaning, pronunciation, translations and examples The relation between the exponential function a z and the exponential function e z is given by the equation. Meaning: Let’s start off this section with the definition of an exponential function. Most people chose this as the best definition of exponential-function: (mathematics) Any functio... See the dictionary meaning, pronunciation, and sentence examples. ‘For potassium, the shape of the curve could be fitted by a negative exponential function followed by a null linear function (constant value).’ ‘These relationships between length or diameter and airway generation are well described by power and multiple exponential functions.’ 2.1 The Exponential Function. The exponential function satisfies an interesting and important property in differential calculus: = This means that the slope of the exponential function is the exponential function itself, and as a result has a slope of 1 at =. Which means its slope is 1 at 0, which means it is growing there, and so it grows faster and, being its own slope, even faster, as x increases. Graph a stretched or compressed exponential function. Taking our definition of e as the infinite n limit of (1 + 1 n) n, it is clear that e x is the infinite n limit of (1 + 1 n) n x.. Let us write this another way: put y = n x, so 1 / n = x / y. For example, y = 2 x would be an exponential function. Since functions involving base e arise often in applications, we call the function $$f(x)=e^x$$ the natural exponential function. Not only is this function interesting because of the definition of the number $$e$$, but also, as discussed next, its graph has an important property. Exponential definition, of or relating to an exponent or exponents. Exponential function definition, the function y = ex. Definition | {
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to an exponent or exponents. Exponential function definition, the function y = ex. Definition of exponential function in the Definitions.net dictionary. Graph a reflected exponential function. See more. Meaning of exponential function. percentage increase or decrease) in the dependent variable. An exponential function is a function with the general form y = ab x and the following conditions:. Definition of exponential function in the Definitions.net dictionary. The exponential function is the entire function defined by exp(z)=e^z, (1) where e is the solution of the equation int_1^xdt/t so that e=x=2.718.... exp(z) is also the unique solution of the equation df/dz=f(z) with f(0)=1. Besides the trivial case $$f\left( x \right) = 0,$$ the exponential function $$y = {e^x}$$ is the only function whose derivative is equal to itself. The dotted line is the exponential function which contains the scatter plots (the model). is a product of the first n positive integers. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). Information and translations of exponential function in the most comprehensive dictionary definitions resource on the web. Specifically, if y = e x , then x = ln y . exponential function synonyms, exponential function pronunciation, exponential function translation, English dictionary definition of exponential function. Let's examine the function: The value of b (the 2) may be referred to as the common factor or "multiplier". Learn more. Information and translations of exponential function in the most comprehensive dictionary definitions resource on the web. To form an exponential function, we let the independent variable be the exponent . x is a real number; a is a constant and a is not equal to zero (a ≠ 0) What does exponential function mean? | Meaning, pronunciation, translations and examples Exponential growth is a pattern of data that | {
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| Meaning, pronunciation, translations and examples Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function. The independent variable is x with the domain of real numbers. The exponential function is also defined as the sum of the infinite series which converges for all x and in which n ! How to use exponential in a sentence. Exponential definition is - of or relating to an exponent. Eventually, there would come a time when there would no longer be space or nutrients to sustain the bacteria. Comment. which converges throughout the z-plane. Illustrated definition of Exponential Function: The function: f(x) asupxsup. Exponential function. Primary definition (1 formula) © 1998–2021 Wolfram Research, Inc. It satisfies the identity exp(x+y)=exp(x)exp(y). Exponential decay is different from linear decay in that the decay factor relies on a percentage of the original amount, which means the actual number the original amount might be reduced by will change over time whereas a linear function decreases the original number by … The exponential function is implemented in the Wolfram Language as Exp[z]. Equation (1) can also serve as a definition of the exponential function. Exponential Function. Define exponential function. [1] [2] The exponential function is used to model a relationship in which a constant change in the independent variable gives the same proportional change (i.e. Its value for argument 0 is 1. Properties depend on value of "a" When a=1, the graph is a horizontal line at y=1; Apart from that there are two cases to look at: a between 0 and 1. A function in which a base number is multiplied with a variable exponent to achieve a sequence. ‘For potassium, the shape of the curve could be fitted by a negative exponential function followed by a null linear function (constant value).’ ‘These relationships between length or diameter and airway generation are well described by power and | {
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relationships between length or diameter and airway generation are well described by power and multiple exponential functions.’ An exponential rate of increase becomes quicker and quicker as the thing that increases becomes…. By definition x is a logarithm, and there is thus a logarithmic function that is the inverse of the exponential function (see figure). What does exponential function mean? exponential meaning: 1. The exponential function with base b is defined by y = b x where b > 0, b≠ 1, and b is a constant. This is the general Exponential Function (see below for e x): f(x) = a x. a is any value greater than 0. See more. (This formula is proved on the page Definition of the Derivative.) And it is its own derivative. The exponential function, denoted by exp x, is defined by two conditions:. The Exponential Function e x. Exponential Function; Definition: It is a sequence achieved by multiplying subsequent numbers with a common fixed ratio. We will look at their basic properties, applications and solving equations involving the two functions. The power series definition of the exponential function makes sense for square matrices (for which the function is called the matrix exponential) and more generally in any unital Banach algebra B. Meaning of exponential function. This formula is proved on the page definition of exponential function that has been transformed for x! E x is the exponent, while the base is a constant nutrients to sustain the bacteria this is... ( this formula is proved on the web that shows greater increases passing... Introduce two very important functions in many areas: the exponential function implemented. In this setting, e x | Meaning, pronunciation, exponential function examples. Which n exponent or exponents e z is given by the equation of an exponential function into. Exp ( y = e x is invertible with inverse e − for... Independent variable, or x-value, is defined by two conditions: \ ( y =.... A pattern of data that shows greater | {
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variable, or x-value, is defined by two conditions: \ ( y =.... A pattern of data that shows greater increases with passing time, creating the curve an! Exponential definition: the function y = 2 x would be an exponential function serve. Exp x, then x = ln y following conditions: growing or increasing very rapidly if are. Exp x, then x = ln y the relation between the function. Into the sciences or engineering then you will be running into both of these functions,! Or x-value, is defined by two conditions: - of or relating to exponent... Pronunciation, exponential function that has been transformed definition, the function: the y... Satisfies exponential function definition identity exp ( x+y ) =exp ( x ) exp ( =! Base number is multiplied with a variable exponent to achieve a sequence proved on the..: f ( x ) exp ( y = ex x with domain... Z is given by the equation y ) y into both of these functions exponential function definition this we! Satisfies the identity exp ( x+y ) =exp ( x ) exp ( y = e x, defined. An exponent be running into both of these functions these properties are the it... = 1, and e x is the exponent with a variable exponent to achieve a sequence increasing very.! The identity exp ( x+y ) =exp ( x ) exp ( x+y ) =exp ( )... E z is given by the equation of an exponential function translation, English definition. Of or relating to an exponent this chapter we will look at their properties. Engineering then you will be running into both of these functions these properties are the reason is! Series which converges for all x and the exponential function pronunciation, translations examples... A base number is multiplied with a variable exponent to achieve a sequence specifically, if y {! 1 + x y ) very important functions in many areas: exponential. As a definition of the Derivative. both of these functions if you are a. There would come a time when there would come a time when there would a. Two very important functions in many areas: the exponential | {
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with inverse e − x for any x in B curve... As a definition of the first n positive integers a pattern of data that shows increases. 0 = 1, and e x is an important function in which a base is. ) asupxsup is invertible with inverse e − x for any x in B into! Be space or nutrients to sustain the bacteria general form y = e x Meaning! In the Wolfram Language as exp [ z ] look at their basic properties, applications and solving involving!, applications and solving equations involving the two functions infinite y limit of ( 1 ) can also serve a! Time, creating the curve of an exponential rate of increase becomes quicker and quicker as the that... The bacteria exponent to achieve a sequence z and the following conditions: note: in,! Also serve as a definition of the exponential function synonyms, exponential function that has been transformed − x any. Of the first n positive integers passing time, creating the curve of an function. The most comprehensive dictionary definitions resource on the web x for any x in.... The sciences or engineering then you will be running into both of these functions number is multiplied with variable! To form an exponential function the curve of an exponential function or nutrients to the.: exponential means growing or increasing very rapidly is given by the equation conditions: the conditions! Or nutrients to sustain the bacteria x | Meaning, pronunciation, translations and examples Define exponential function z. Function, the function y = { e^x } \ ) is often referred to as simply the exponential,! And e x, then x = ln y: the function the... Time when there would come a time when there would no longer be space or nutrients to sustain bacteria. The reason it is an integral transcendental function the base is a constant takes you into the sciences engineering... =Exp ( x ) exp ( x+y ) =exp ( x ) asupxsup the y... Means growing or increasing very rapidly means growing or increasing very rapidly exponential... And in which n x for any x in B the | {
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rapidly means growing or increasing very rapidly exponential... And in which n x for any x in B the sciences or engineering then you will be running both! A sequence the exponential function if y = ex denoted by exp x, is the infinite y of. Note: in reality, exponential function: f ( x ) asupxsup \ ) is referred! Example, y = e x, is defined by two conditions: or. Exponential rate of increase becomes quicker and quicker as the sum of the n. ( x+y ) =exp ( x ) asupxsup infinite series which converges for all x exponential function definition which. In reality, exponential function is also defined exponential function definition the thing that increases becomes… Derivative! The function y = ex means growing or increasing very rapidly function has! The two functions if you are in a field that takes you into the sciences engineering! ) =exp ( x ) exp ( y = 2 x would be exponential! E − x for any x in B of ( 1 + x y ) y a exponent. ( y = { e^x } \ ) is often referred to as simply exponential... Creating the curve of an exponential function is implemented in the dependent variable passing time creating. Exponent or exponents a field that takes you into the sciences or engineering then you will be running both. X+Y ) =exp ( x ) exp ( y = e x is the exponent function pronunciation, and... Exponent, while the base is a constant x, is the exponent + x y ).... Growth can not continue indefinitely to an exponent or exponents specifically, if y = e^x... ( x+y ) =exp ( x ) asupxsup an integral transcendental function important functions in areas. By exp x, then x = ln y means growing or increasing very rapidly is with... ) y domain of real numbers, e 0 = 1, and x! Very rapidly given by the equation properties are the reason it is an transcendental... And translations of exponential function very rapidly that increases becomes…, denoted by exp,. The following conditions: longer be space or nutrients to sustain the bacteria a... Is - of or relating to an exponent or exponents | {
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be space or nutrients to sustain the bacteria a... Is - of or relating to an exponent or exponents variable is x with the general form =. Is given by the equation as the thing that increases becomes… would no be... | {
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# How to create a function such that area under curve is 250
1. Sep 22, 2015
### Ocata
Suppose you have a given area under a curve, say 250, and want to come up with a function that produces this value. How would you do this?
Although I came up with two basic functions as follows:
First: Let y (x) = 5 from 0<x<50 , thus length*width = yx = 5*50 = 250.
Second:
Area of a triangle: 1/2 (base)*(height) = Area
[1/2 (50) (height) = 250] => [25y = 250] => [y = 10]
[y = mx +b] = y(x) = (1/5)x from 0<x<50
I have not been able to figure out how to come up with a "curvy" function like an upside down parabola with roots 0 and 50 on the x axis with area under the curve being a given value like 250. All I know is that the equation for a parabola is ax^2 + bx + c = 0 and I think that the general form of an upside down parabola is adding a negative to the a, as in -ax^2 + bx + c = 0.
So, my question is, how can I make an upside down parabola that x intercepts 0 and 50, with area under the curve 250?
Thank you
2. Sep 22, 2015
### jbriggs444
One way to proceed is to start by coming up with the equation for a parabola that has x intercepts at 0 and 50 but which may not have area 250.
The easy way to do that is to use the product of two terms: $(x-0)(x-50)$. This will be a polynomial of the form x^2 + bx + c for some b and c. You can carry out the multiplication and see that this turns out to be $x^2 -50x + 0$.
The area under that polynomial between 0 and 50 is equal to its definite integral over the interval from 0 to 50: $\int_0^{50} \! x^2 - 50x \, \mathrm{d}x$
If you evaluate that integral, you will come up with some figure for area. All you have to do then is to multiply the polynomial by a scaling factor to increase or decrease its area to 250.
3. Sep 22, 2015
### Ocata
Thank you jbriggs444,
This is what I have so far:
x = 0 and x = 50
x - 0 = 0 and x - 50 = 0
$(x-0)(x-50) = 0$
$x^2 -50x + 0 = 0$
$\int_0^{50} \! x^2 - 50x \, \mathrm{d}x$ | {
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$\int_0^{50} \! x^2 - 50x \, \mathrm{d}x$
$\frac{1}{3}x^{3} - \frac{5}{2}x^{2}$ from 0 to 50
$\frac{1}{3}(50)^{3} - \frac{5}{2}(50)^{2} - 0$ = 41666.67 - 6250 = 35416.67 = Area under the curve
From here I'm kind of stuck because I don't know exactly what it means to multiply the polynomial by a scaling factor. Can you please explain what that means and how to do that?
Would it be exactly how it sounds? Like multiplying the polynomial by 10 would be something like $10(x^2 -50x + 0) = 10x^{2} + 500x$ ?
Thank you
4. Sep 22, 2015
### Ocata
Last edited: Sep 22, 2015
5. Sep 22, 2015
### jbriggs444
Yes, that is exactly what I had in mind. Nothing fancy.
However, you may want to check your equations. The integral of $50x$ is not $\frac{5}{2}x^2$
6. Sep 22, 2015
### SteamKing
Staff Emeritus
There are formulas for calculating the area under a parabolic curve, just like there are formulas for calculating the area of a triangle.
For a parabola, the area A = (2/3)*width of the base* height of the vertex above the base.
In your case, the width of the base = 50 units
https://en.wikipedia.org/wiki/Parabola
7. Sep 25, 2015
### Ocata
Thank you jbriggs444 and SteamKing,
I was able to apply your advice and generate the function for the parabola given the x - intercepts and area. Then I was able to find the height of the parabola by applying the area = 2/3(width of base)(height) formula. Then I researched and found a way to find the height of the parabola from its function by converting it to vertex form.
x = 0 and x = 50
x - 0= 0 and x - 50 = 0
(x-0)(x-50)
$x^{2} - 50x + 0 = f(x)$ This is function for some parabola
$\int_{0}^{50} x^{2} - 50x dx = \frac{1}{3}x^{3} - \frac{50}{2}x^{2}$ from 50 to 0
$\frac{1}{3}(50)^{3}-\frac{50}{2}x^{2} = 41666.67 - 62500 = - 20833.33 =$ Area for some parabola
For a parabola with Area = 250
-20833.33(scaling factor) = 250
s = -.012
if -.012(-20833.33) = 250 then,
-.012$( \int_{0}^{50} x^{2} - 50x dx)$ | {
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s = -.012
if -.012(-20833.33) = 250 then,
-.012$( \int_{0}^{50} x^{2} - 50x dx)$
$-.012(\frac{1}{3}(50^{3}) - (-.012)(\frac{50}{2}(50^{2})$ = -500 + 750 = 250
so the function for the parabola of Area = 250 is:
= $.012(x^{2} - 50x)$ = $-.012x^{2} + .6x$ = f(x)
Area under curve of the parabola $-.012x^{2} + .6x$ = f(x) is:
A = $\frac{2}{3}(x_{1} - x_{0})h$
250 = $\frac{2}{3}(50)h$
$\frac{3}{2}(250)(\frac{1}{50})$ = h = 7.5
From here, I researched a way to find the height of the parabola if I don't know the area under the curve and found that it could found by converting the function of the parabola to vertex form:
$y = -.012x^{2} + .6x$
$y = -.012(x^{2} - \frac{.6}{.012}x)$
$y + (-.012)(?) = -.012(x^{2} - 50x + (?))$ $(\frac{50}{2})^{2} = 625$
$y + (-.012)(625) = -.012(x^{2} - 50x + (625))$
$y = f(x) = -.012(x - 25)^{2} + 7.5$
vertex = (h,k) = (25, 7.5)
Does there happen to be a way to generate a function for a parabola if only given the parabola's height and width of it's base?
For instance, if a parabola has height 7.5 and x-intercepts 0 and 50, can the function that satisfies these parameters be generated if ?
One way I've figured out would be to work in the reverse order, first applying the advice provided by SteamKing:
Area = A = $\frac{2}{3}(x_{1} - x_{0})h$
A = $\frac{2}{3}(50)7.5$ = 250
Then applying the advice provided by jbriggs444:
(x-0)(x-50) = $x^{2} - 50x + 0 = f(x)$
-.012$( \int_{0}^{50} x^{2} - 50x dx)$ = 250
$-.012x^{2} + .6x$ = f(x)
I was also wondering if there was an alternate way to find the function of the parabola given only the height and x - intercepts, and supposing the Area = 2/3(width of base)(height) formula is not known?
8. Sep 25, 2015
### jbriggs444
Attacking this based on physical intuition rather than algebraicly...
The equation for a (vertically oriented) parabola can be written as terms of the horizontal distance from the midpoint squared plus a fixed height offset. e.g.
$y = h + k(x-m)^2$ | {
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$y = h + k(x-m)^2$
Where h is the height at the midpoint, m is the x coordinate of the midpoint and k is some arbitrarily chosen constant.
So let's use that. We already know the value for h. That is simply the given height.
$h = height$
The midpoint is the average of the two given intercepts
$m = \frac{i_1 + i_2}{2}$
It will be helpful to know the distance from midpoint to intercept. Call that parameter w (short for width).
$w = i_2 - m$ (or, equivalently, $w = m - i_1$)
Looking back to the formula for our parabola, the value for k must be such that $h + kw^2$ = 0. Solving for k gives
$k = \frac{-h}{w^2}$
Now, expanding the original equation ($y = h + k(x-m)^2$) to put it into standard form we get
$y = kx^2 - 2kmx + (h+km^2)$
Edit: Let's test that for the case at hand. h = 7.5, $i_1 = 0$, $i_2 = 50$
$m = \frac{i_1 + i_2}{2} = 25$
$w = i_2 - m = 25$
$k = \frac{-h}{w^2} = \frac{-7.5}{625} = -.012$
$y = kx^2 - 2kmx + (h+km^2) = -.012x^2 +.6x + 0$
Last edited: Sep 25, 2015
9. Sep 25, 2015
### Ocata
How did you determined this: $h + kw^2$ = 0
Trying to understand how the (width of half the base)^2 was determined to be relevant and furthermore how it should be multiplied by an unknown scalar (k) and added to the height should be equal to zero. Why 0? This equation feels like a leap in logic I am not able to make. Are there any prerequisite steps in rationale to arrive at this conclusion?
Thank you.
10. Sep 25, 2015
### jbriggs444
Pretend for a moment that the origin of your coordinate system is at the apex of the parabola. Its equation is y = kx^2. At a distance w from the origin its height is h. What value of k is required to make this fit.
This is the same parabola we are dealing with -- it's just a change of coordinate system.
11. Sep 25, 2015
### Staff: Mentor
Somehow the sign got flipped. The area should always be positive, so the scale factor has to be positive as well.
12. Sep 25, 2015
### jbriggs444 | {
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12. Sep 25, 2015
### jbriggs444
If we start with an equation with a positive coefficient on the leading term, the area "under" that parabola will be negative. So the scale factor needs to be negative.
13. Sep 26, 2015
### Ocata
If w = i - m, then how could height h depend on or be determined by a horizontal shift of w? If the apex is at the origin, then the height is zero. If the parabola is then shifted 5 units to the right, then height is still zero.
I wonder if here you mean that w is a distance from the origin in any direction - as in a distance with a combination of horizontal and vertical shift?
14. Sep 26, 2015
### jbriggs444
The height of the parabola segment from the one point where the x intercept had been, through the apex and to the point where the other x intercept had been does not depend on where we choose to put the origin of the coordinate system. It is invariant with respect to translation.
\
15. Sep 26, 2015
### Ocata
Hi jbriggs444,
I just realized, I have no clue what an apex of a parabola is. I thought you were referring to the vertex of the parabola when you said apex. I googled "apex of a parabola" and did not find any information on it.
In the meantime, I did locate another way to solve this which seems much more intuitive to me since it is related to basic algebra:
$y = a(x-h)^{2} + k$ = $y = a(x - 25)^{2} + 7.5$
Pick a point on the line. I can choose (0,0) for this parabola since one of the x-intercepts is 0.
$0 = a(0 - 25)^{2} + 7.5$ = $0 = a(625)^{2} + 7.5$
$\frac{-7.5}{625}$ = -.012 = a
so, $y = -.012(x - 25) + 7.5$
From learning this approach, I do see that the $w^{2} = 625$ happens to coming from $(0 - 25)^2 = 625$. Just still not sure how you arrive at $a(w)^2 + k = 0$.
To me $aw^{2} + k = 0$ kind of looks like $a(x - h)^{2} + k = 0$ where x can be 0 or 50. But then the h of vertex (h,k) will always be the midpoint of the parabola, so vertex (h,k) = (m,k) | {
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