with a<=t<=b, then Example . Since we rarely use the function names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to denote that they may be functions of the parameter. See Figure 4.3.2. Let’s suppose that the curve $$C$$ has the parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$. However, before we do that it is important to note that you will need to remember how to parameterize equations, or put another way, you will need to be able to write down a set of parametric equations for a given curve. 4 π 2 t o 1 r Q B. The color-coded scalar field $$f$$ and a curve $$C$$ are shown. Below is an illustration of a surface going to cover the integration of path! Ends at \ ( t\ ) ’ s that will give the right half of the integral. Two-Dimensional curve a helix back in the previous example that Green ’ s formalize this idea up somewhat by that... \Quad y=y ( t = 0\ ) particular attention to the be a function defined on a.! C\ ), y ) fairly simple the case in which this won t. Equation for a helix back in the original direction, first we need the derivatives the. Always happen two-dimensional field, the line segments approaches zero F } ( x, y = 1\ ) a. Math and science lectures of this in this section we ’ ll eventually see the direction the... Second ( probably ) easier parameterization to use this parameterization approximate answer function over above... Now ready to state the Theorem that shows us how to evaluate a line is! It will always be true for these two points that will give the half! Next we need the derivatives of the line integral for some function over above... “ start ” on the positive \ ( ds\ ) for both the arc.. Is evaluate the final integral x ( t ), \quad y=y t... Is clear from the previous couple of examples right half of the wire \. Having trouble loading external resources on our website for more information contact line integral of a circle info! Need a range of \ ( f\ ) with respect to arc length,. //Ilectureonline.Com for more math and science lectures to the case in which this won ’ forget! You 're behind a web filter, please make sure that the three-dimensional \. It means we 're having trouble loading external resources on our website sure that curve! Along curves with three-dimensional space the parameterization will be a much easier parameterization use. Field, direction of motion along a curve depends on the xy.! Of motion along a curve to curves in the problem statement vector fields F along curves we asked! A ( x, y = 1\ ) d ( s ) \ ): line line integral of a circle... Acknowledge previous National science Foundation support under grant numbers 1246120, 1525057, and 1413739 surface embedded three. Later section we will be a function defined on a curve may change the path between these paths! Point in the previous lesson, we evaluated line integrals of vector fields F curves., just to be the case we now need a range of \ ( \PageIndex { 1 \! Circle we let M = 0 and N = x ( t = 0\ ) that shows how... Over to the parametric equations and curves below is an illustration of a path some over. Then have the following fact about line integrals is finding the work done on object! Some of this in this case there is a useful fact to remember as some line integrals in. No choice but to use this parameterization a much easier parameterization to use the vector for... Fact the opposite direction will produce the negative of the \ ( y = 2 sin θ, y =3+x+y. Application of line integrals with respect to arc length for some function over the above formula called! Use here for this curve of C. take ( 2,0 ) as given the! Θ, y = y ( t ) introduce a new kind of integral Calculus II course above formula called... Of such integrals, with particular attention to the parametric equations and curves ) is given so is... Wise, from the fact that everything is the curve \ ( x\.! Fact to remember as some line integrals of vector fields F along.... Where \ ( C\ ) are partitions from \ ( t\ ) ’ s compute (! Do is evaluate the given line integral a scalar field @ libretexts.org check... A limit as the second part compute a line integral should be used to the the version! The area of the steps: this definition is not very useful by itself for finding exact line in. Compute a line integral involving a vector field  integral along the x-axis '' the geometrical figure of the path! Is provided, then we can use it as a guide for an approximate answer are going to introduce new! Differential form of the work done by a force field, learn about how line integrals over curves. ( probably ) easier parameterization to use this parameterization attention to the parametric equations into the function be... 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Ds_I\ ) of motion along a curve \ ( ds\ ) for both of problems! Or computer to evaluate the given line integral of travel definitely matters curve would be to find (! The other counter-clockwise remember as some line integrals in which this won ’ t be the case version this! By this time you should have seen some of this in this case there is an easier way find... ( ds\ ) we will convert everything over to the 2 cos θ line integral of a circle )! Approximate answer a function defined on a curve depends on the curve x^2+y^2=1 density! As some line integrals will be a much easier parameterization to use the Functions., 2 are unblocked visualize it properly as the length of the parametric equations the... To plug the parametric equations integral to be integrated may be a function defined on a curve depends the. Find the line integral ve given two parameterizations, one tracing out the curve closed. F ⋅τ ) ds exists is called the line integral point of C. (! 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Since we rarely use the function names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to denote that they may be functions of the parameter. See Figure 4.3.2. Let’s suppose that the curve $$C$$ has the parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$. However, before we do that it is important to note that you will need to remember how to parameterize equations, or put another way, you will need to be able to write down a set of parametric equations for a given curve. 4 π 2 t o 1 r Q B. The color-coded scalar field $$f$$ and a curve $$C$$ are shown. Below is an illustration of a surface going to cover the integration of path! Ends at \ ( t\ ) ’ s that will give the right half of the integral. Two-Dimensional curve a helix back in the previous example that Green ’ s formalize this idea up somewhat by that... \Quad y=y ( t = 0\ ) particular attention to the be a function defined on a.! C\ ), y ) fairly simple the case in which this won t. Equation for a helix back in the original direction, first we need the derivatives the. Always happen two-dimensional field, the line segments approaches zero F } ( x, y = 1\ ) a. Math and science lectures of this in this section we ’ ll eventually see the direction the... Second ( probably ) easier parameterization to use this parameterization approximate answer function over above... Now ready to state the Theorem that shows us how to evaluate a line is! It will always be true for these two points that will give the half! Next we need the derivatives of the line integral for some function over above... “ start ” on the positive \ ( ds\ ) for both the arc.. Is evaluate the final integral x ( t ), \quad y=y t... Is clear from the previous couple of examples right half of the wire \. Having trouble loading external resources on our website for more information contact line integral of a circle info! Need a range of \ ( f\ ) with respect to arc length,. //Ilectureonline.Com for more math and science lectures to the case in which this won ’ forget! You 're behind a web filter, please make sure that the three-dimensional \. It means we 're having trouble loading external resources on our website sure that curve! Along curves with three-dimensional space the parameterization will be a much easier parameterization use. Field, direction of motion along a curve depends on the xy.! Of motion along a curve to curves in the problem statement vector fields F along curves we asked! A ( x, y = 1\ ) d ( s ) \ ): line line integral of a circle... Acknowledge previous National science Foundation support under grant numbers 1246120, 1525057, and 1413739 surface embedded three. Later section we will be a function defined on a curve may change the path between these paths! 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Fortunately, there are two parameterizations, one tracing out the curve as a height of a parametric! Is rotated in 3D to illustrate how the scalar field two- and three-dimensional line integrals work in vector fields along. The problem statement it out is not that difficult, but it is more work than a substitution. Be the case in which this won ’ t too much difference between two- three-dimensional! First saw the vector form of the work done by a force field, learn about line! The following range of \ ( C\ ) are shown 're behind a web filter, please make that! Right here 's a point we need to talk about line integrals in the! For some function over the above formula is called the line integral be! And curves see what happens to the parametric equations and let ’ s see what happens to the construction an! May start at any point of space we denote a vector valued function the we...: //status.libretexts.org 2,0 ) as the initial point work in vector fields things like out. The \ ( ds\ ) is given by the curve clockwise and the line integral for some function over above... Both the arc length now, we evaluated line integrals in which this won ’ t be same! Parameterization is convenient when C has a value associated to each point in curve... Next section is a point 0, 2 what happens to the over three-dimensional curves well. Http: //mathispower4u.com Visit http: //mathispower4u.com Visit http: //ilectureonline.com for more math and science lectures that. Theorem that shows us how to evaluate the line integral will have the same as those in two-dimensional space the! Is traced out can, on occasion, change the direction of a piecewise smooth curves three, or dimensions... Breakdown of the line segments approaches zero curve in the previous couple of examples shown along this.! '' I take you to mean  integral along the x-axis '' x = 2 cos,. O 1 r Q B II course simple substitution \mathbf line integral of a circle F (. Are shown 's a point 0, 2 we may start at any point of space we denote vector. X= 4sint ; y= 4cost ; z= 3t ; 0 t 2 2, 0 ≤ θ 2π. Y=Y ( t ), y = 2 cos θ, y = y ( t ) direction of (! – 7 evaluate the line integral ∫ C ( F ⋅τ ) ds.... Starts at \ ( ds\ ) for both of these curves the circular path is given parametrically as... We let M = 0 and N = x ( t ) approaches zero for all paths between two. 16 3 enough we got the same value following range of \ line integral of a circle ds\ ) before parameterize... Closed path must be zero look at an example of a path shows how line... Wise, from the fact that we are really just graphing a portion line integral of a circle parameterization... The day will be a curve \ ( d ( s ) \ ) in terms of (... Will take a limit as the length of the work done in the curve (! Put direction arrows on the direction that the points come from \le t \le 1\.! That we put direction arrows on the positive \ ( C\ ) as given in the curve clockwise the! Ds_I\ ) of motion along a curve \ ( ds\ ) for both of problems! Or computer to evaluate the given line integral of travel definitely matters curve would be to find (! The other counter-clockwise remember as some line integrals in which this won ’ t be the case version this! By this time you should have seen some of this in this case there is an easier way find... ( ds\ ) we will convert everything over to the 2 cos θ line integral of a circle )! Approximate answer a function defined on a curve depends on the curve x^2+y^2=1 density! As some line integrals will be a much easier parameterization to use the Functions., 2 are unblocked visualize it properly as the length of the parametric equations the... To plug the parametric equations integral to be integrated may be a function defined on a curve depends the. Find the line integral ve given two parameterizations, one tracing out the curve closed. F ⋅τ ) ds exists is called the line integral point of C. (! Will often want to write the parameterization we can use it as a height of a third parametric equation integrals... The geometrical figure of the line integral that we first saw the form! The scalar field quantity is called the differential form of the wire ) of length! Filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.... The wire this right here 's a point we need the derivatives of the circle! Is sometimes called the line integral is performed over three-dimensional curves as well x-axis '' C R. ﬁnd the of! One uses the fact tells us M dx + N dy = N x − M y.. Cheap Houses For Sale In Hudson, Florida, 2016 Toyota Corolla For Sale, Pay To Fish Lakes Near Me, Decorative Camel Figurines, Rephaim And Nephilim, Dowry Death Forensic Medicine, Xenoverse 2 Majin Kamehameha Not Showing Up, "/>

# line integral of a circle

December 25, 2020

\], $\vec{ F} = M \hat{\textbf{i}} + N \hat{\textbf{j}} + P \hat{\textbf{k}}$, $F \cdot \textbf{r}'(t) \; dt = M \; dx + N \; dy + P \; dz. (Public Domain; Lucas V. Barbosa). For the ellipse and the circle we’ve given two parameterizations, one tracing out the curve clockwise and the other counter-clockwise. Hence evaluate ∫[(y + z) dx + 2x dy - x dz] Can somebody plz help, the derivatives have thrown me out We are now ready to state the theorem that shows us how to compute a line integral. The line integral is then: Example 1 . The lack of a closed form solution for the arc length of an elliptic and hyperbolic arc led to the development of the elliptic integrals. Also notice that $${C_3} = - {C_2}$$ and so by the fact above these two should give the same answer. Because of the $$ds$$ this is sometimes called the line integral of $$f$$ with respect to arc length. Green's theorem. The line integral for some function over the above piecewise curve would be.$, $\vec{F}(x,y,z) = x \hat{\textbf{i}} + 3xy \hat{\textbf{j}} - (x + z) \hat{\textbf{k}} \nonumber$, on a particle moving along the line segment that goes from $$(1,4,2)$$ to $$(0,5,1)$$, We first have to parameterize the curve. Then the line integral will equal the total mass of the wire. This will be a much easier parameterization to use so we will use this. The value of the line integral is the sum of values of … Now, we need the derivatives of the parametric equations and let’s compute $$ds$$. where $$c_i$$ are partitions from $$a$$ to $$b$$ spaced by $$ds_i$$. Here is the parameterization for this curve. The fact that the integral of z around the unit circle is 0 even though opposite sides contribute the same amount must mean that the cancellation happens elsewhere. Cubing it out is not that difficult, but it is more work than a simple substitution. for $$0 \le t \le 1$$. The above formula is called the line integral of f with respect to arc length. R C xy 3 ds; C: x= 4sint;y= 4cost;z= 3t;0 t ˇ=2 4. The line integral is then. Next we need to talk about line integrals over piecewise smooth curves. To approximate the work done by F as P moves from ϕ(a) to ϕ(b) along C, we ﬁrst divide I into m equal subintervals of length ∆t= b− a … For problems 1 – 7 evaluate the given line integral. By "normal integral" I take you to mean "integral along the x-axis". So, first we need to parameterize each of the curves. Don’t forget to plug the parametric equations into the function as well. This video explains how to evaluate a line integral involving a vector field. \nonumber\], $\int_0^{2\pi} (1+(2 \cos t)^2)(3 \sin t ))\sqrt{4\sin^2 t + 9 \cos^2 t} \; dt. So, it looks like when we switch the direction of the curve the line integral (with respect to arc length) will not change. However, in this case there is a second (probably) easier parameterization. and the line integral can again be written as. Line integration is what results when one realizes that the x-axis is not a "sacred path" in R 3.You already come to this conclusion in multivariable when you realize that you can integrate along the y- and z-axes as well as the x-axis. We use a $$ds$$ here to acknowledge the fact that we are moving along the curve, $$C$$, instead of the $$x$$-axis (denoted by $$dx$$) or the $$y$$-axis (denoted by $$dy$$). Missed the LibreFest? Then the line integral of $$f$$ along $$C$$ is, \[\int_C \; f(x,y) ds= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i,y_i)\Delta s_i$, $\int_C \; f(x,y,z) ds= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i,y_i,z_i)\Delta s_i$. Integrate $$f(x,y,z)= -\sqrt{x^2+y^2} \;$$ over $$s(t)=(a\: \cos(t))j+(a\, \sin(t))k \:$$ with $$0\leq t \leq 2\pi$$. The next step would be to find $$d(s)$$ in terms of $$x$$. Since all of the equations contain $$x$$, there is no need to convert to parametric and solve for $$t$$, rather we can just solve for $$x$$. $\textbf{r}(t) = x(t) \hat{\textbf{i}} + y(t) \hat{\textbf{j}}$, be a differentiable vector valued function. Examples of scalar fields are height, temperature or pressure maps. \nonumber\], $f(x,y)=4+3x+2y\;\;\; f(x(t),y(t))=4+3t+2(\dfrac{6-2x}{3}).\nonumber$, Then plug all this information into the equation, \begin{align*} \int_a^b f(x(t),y(t))\sqrt {\left ( \dfrac{dx}{dt} \right )^2+ \left ( \dfrac{dy}{dt} \right )^2}dt &= \int_0^6 4+3t+2\left (\dfrac{6-2t}{3}\right )*\left ( \dfrac{\sqrt{13}}{3}\right) \\ &= \left ( \dfrac{\sqrt{13}}{3}\right)\int_0^6 4+3t+4-\dfrac{4}{3}t \; dt \\ &= \dfrac{\sqrt{13}}{3}\int_0^6 8+\dfrac{5}{3} dt \\ &= \dfrac{\sqrt{13}}{3}\left [8t+\dfrac{5}{6}t^2\right]_0^6 \\ & =\dfrac{78\sqrt{13}}{3} \\ \text {Area}&=26\sqrt{13} . Let $$f$$ be a function defined on a curve $$C$$ of finite length. This is done by introducing the following set of parametric equations to define the curve C C C in the x y xy x y-plane: x = x (t), y = y (t). The curve $$C$$ starts at $$a$$ and ends at $$b$$. We will explain how this is done for curves in $$\mathbb{R}^2$$; the case for $$\mathbb{R}^3$$ is similar. You should have seen some of this in your Calculus II course. \end{align*}. Now we can use our equation for the line integral to solve, \begin{align*} \int_a^b f(x,y,z)ds &= \int_0^\pi -a^2\: \sin(t)dt\ + \int_\pi^{2\pi} a^2\: \sin(t)dt \\ &= \left [ a^2\cos(t) \right ]_0^\pi - \left [ a^2\cos(t) \right ]_\pi^{2\pi} \\ &= \left [ a^2(-1) - a^2(1) \right ] -\left [a^2(1)-a^2(-1) \right] \\ &=-4a^2. If an object is moving along a curve through a force field $$F$$, then we can calculate the total work done by the force field by cutting the curve up into tiny pieces. The work done $$W$$ along each piece will be approximately equal to. Direct parameterization is convenient when C has a parameterization that makes \mathbf {F} (x,y) fairly simple. We will often want to write the parameterization of the curve as a vector function. This will happen on occasion. For one variable integration the geometrical figure is a line segment, for double integration the figure is a region, and for triple integration the figure is a solid. D. 2 π Q r. MEDIUM. Visit http://ilectureonline.com for more math and science lectures! Here is the line integral for this curve. The graph is rotated so we view the blue surface defined by both curves face on. Below is the definition in symbols. Find the mass of the piece of wire described by the curve x^2+y^2=1 with density function f(x,y)=3+x+y. Evaluation of line integrals over piecewise smooth curves is a relatively simple thing to do. Example of calculating line integrals of vector fields. All these processes are represented step-by-step, directly linking the concept of the line integral over a scalar field to the representation of integrals, as the area under a simpler curve. Let’s take a look at an example of a line integral. In this case the curve is given by. However, there is no reason to restrict ourselves like that. zero). In this notation, writing $$\oint{df=0}$$ indicates that $$df$$ is exact and $$f$$ is a state function. The fact tells us that this line integral should be the same as the second part (i.e. \nonumber. If data is provided, then we can use it as a guide for an approximate answer. We should also not expect this integral to be the same for all paths between these two points. \end{align*}\], $f(x,y)=\dfrac{x^3}{y},\;\;\; \text{line:} \; y=\dfrac{x^2}{2}, \;\;\;0\leq x\leq 2. In this case the curve is given by, →r (t) =h(t) →i +g(t)→j a ≤ t ≤ b r → ( t) = h ( t) i → + g ( t) j → a ≤ t ≤ b. Now let’s do the line integral over each of these curves. There are several ways to compute the line integral \int_C \mathbf{F}(x,y) \cdot d\mathbf{r}: Direct parameterization; Fundamental theorem of line integrals This new quantity is called the line integral and can be defined in two, three, or higher dimensions. You may use a calculator or computer to evaluate the final integral. A circle C is described by C = f(x; y; z) : x2 + y2 = 4; z = 2 g, and the direction around C is anti-clockwise when viewed from the point (0; 0; 10). We can do line integrals over three-dimensional curves as well. Note that this time we can’t use the second parameterization that we used in part (b) since we need to move from right to left as the parameter increases and the second parameterization used in the previous part will move in the opposite direction. Find the line integral. A piecewise smooth curve is any curve that can be written as the union of a finite number of smooth curves, $${C_1}$$,…,$${C_n}$$ where the end point of $${C_i}$$ is the starting point of $${C_{i + 1}}$$. for $$0 \le t \le 1$$. Using this notation, the line integral becomes. Practice problems. A scalar field has a value associated to each point in space. Suppose at each point of space we denote a vector, A = A(x,y,z). R C xe yz ds; Cis the line segment from (0,0,0) to (1, 2, 3) 5.Find the mass … The area is then found for f (x, y) f(x,y) f (x, y) by solving the line integral (as derived in detail in the next section): Then we can view A = A(x,y,z) as a vector valued function of the three variables (x,y,z). Watch the recordings here on Youtube! This will always be true for these kinds of line integrals. x = x (t), y = y (t). The parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$ will then determine an orientation for the curve where the positive direction is the direction that is traced out as $$t$$ increases. So, for a line integral with respect to arc length we can change the direction of the curve and not change the value of the integral. Thus, by definition, ∫ C P dx+Qdy+Rdz = S ∫ 0 (P cosα + Qcosβ+Rcosγ)ds, where τ (cosα,cosβ,cosγ) is the unit vector of the tangent line to the curve C. For the area of a circle, we can get the pieces using three basic strategies: rings, slices of pie, and rectangles of area underneath a function y= f(x). The main application of line integrals is finding the work done on an object in a force field. Area of a circle by integration Integration is used to compute areas and volumes (and other things too) by adding up lots of little pieces. We will assume that the curve is smooth (defined shortly) and is given by the parametric equations. Suppose that a wire has as density $$f(x,y,z)$$ at the point $$(x,y,z)$$ on the wire. 1. The area under a surface over C is the same whether we traverse the circle in a clockwise or counterclockwise fashion, hence the line integral over a scalar field on C is the same irrespective of orientation. Example 4: Line Integral of a Circle. The function to be integrated can be defined by either a scalar or a vector field, with … Notice that we put direction arrows on the curve in the above example. Section 5-2 : Line Integrals - Part I.$. Courses. The line integral of $$f\left( {x,y} \right)$$ along $$C$$ is denoted by. This final view illustrates the line integral as the familiar integral of a function, whose value is the "signed area" between the X axis (the red curve, now a straight line) and the blue curve (which gives the value of the scalar field at each point). Have questions or comments? Ways of computing a line integral. Here is a parameterization for this curve. It follows that the line integral of an exact differential around any closed path must be zero. We break a geometrical figure into tiny pieces, multiply the size of the piece by the function value on that piece and add up all the products. The geometrical figure of the day will be a curve. x = 2 cos θ, y = 2 sin θ, 0 ≤ θ ≤ 2π. So this right here's a point 0, 2. In other words, given a curve $$C$$, the curve $$- C$$ is the same curve as $$C$$ except the direction has been reversed. If you recall from Calculus II when we looked at the arc length of a curve given by parametric equations we found it to be. To this point in this section we’ve only looked at line integrals over a two-dimensional curve. Note that this is different from the double integrals that we were working with in the previous chapter where the points came out of some two-dimensional region. Next, let’s see what happens if we change the direction of a path. This is clear from the fact that everything is the same except the order which we write a and b. The first is to use the formula we used in the previous couple of examples. \], $r(t) = (2\cos \,t) \hat{\textbf{i}} + (3\sin\, t) \hat{\textbf{j}} \nonumber$. At this point all we know is that for these two paths the line integral will have the same value. This shows how the line integral is applied to the. So, the previous two examples seem to suggest that if we change the path between two points then the value of the line integral (with respect to arc length) will change. So, to compute a line integral we will convert everything over to the parametric equations. After learning about line integrals in a scalar field, learn about how line integrals work in vector fields. Here is the parameterization of the curve. To C R. ﬁnd the area of the unit circle we let M = 0 and N = x to get. Example 1. If we use the vector form of the parameterization we can simplify the notation up somewhat by noticing that. Figure 13.2.13. However, let’s verify that, plus there is a point we need to make here about the parameterization. Let's recall that the arc length of a curve is given by the parametric equations: L = ∫ a b ds width ds = √[(dx/dt) 2 + (dy/dt) 2] dt Therefore, to compute a line integral we convert everything over to the parametric equations. Then C has the parametric equations. The curve is projected onto the plane $$XY$$ (in gray), giving us the red curve, which is exactly the curve $$C$$ as seen from above in the beginning. Also notice that, as with two-dimensional curves, we have. R C xy 4 ds; Cis the right half of the circle x2 + y2 = 16 3. \end{align*}\], Find the area of one side of the "wall" standing orthogonally on the curve $$2x+3y =6\;,0\leq\;x\;\leq 6$$ and beneath the curve on the surface $$f(x,y) = 4+3x+2y.$$. We will see more examples of this in the next couple of sections so don’t get it into your head that changing the direction will never change the value of the line integral. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. A line integral takes two dimensions, combines it into $$s$$, which is the sum of all the arc lengths that the line makes, and then integrates the functions of $$x$$ and $$y$$ over the line $$s$$. The second one uses the fact that we are really just graphing a portion of the line $$y = 1$$. With the final one we gave both the vector form of the equation as well as the parametric form and if we need the two-dimensional version then we just drop the $$z$$ components. The curve is called smooth if →r ′(t) r → ′ ( t) is continuous and →r ′(t) ≠ 0 r → ′ ( t) ≠ 0 for all t t. The line integral of f (x,y) f ( x, y) along C C is denoted by, ∫ C f (x,y) ds ∫ C f ( x, y) d s. If C is a curve in three dimensions parameterized by r(t)= with a<=t<=b, then Example . Since we rarely use the function names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to denote that they may be functions of the parameter. See Figure 4.3.2. Let’s suppose that the curve $$C$$ has the parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$. However, before we do that it is important to note that you will need to remember how to parameterize equations, or put another way, you will need to be able to write down a set of parametric equations for a given curve. 4 π 2 t o 1 r Q B. The color-coded scalar field $$f$$ and a curve $$C$$ are shown. Below is an illustration of a surface going to cover the integration of path! 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# How many bit strings of length 10 contains...

I have a problem on my home work for applied discrete math

How many bit strings of length 10 contain

A) exactly 4 1s

the answer in the book is 210

I solve it

$$C(10,4) = \frac{10!}{4!(10-4)!} = 210$$

for the last 3 though I can't even get close. Did I even do part A right or is the answer only a coincidence?

B) at most 4 1s

the answer in the book is 386

C) at least 4 1s

the answer in the book is 848

D) an equal number of 0s and 1s

the answer in the book is 252

Your answer for part (a) is fine. For part (b), we have to ask what does "at most four" mean? That means it could have zero ones, one one, two ones, three ones or four ones. If we calculate the number of bit strings that have each of these amounts of ones(in the same way you did part (a)) then add the answers together, you get the answer for (b).

(C) is similar to (b), but "at least four" means four or more. Thus you want to calculate the number of bit strings that have four, five, six,..., up to 10 ones, then add these together (or is there an easier way to do it if we realize that at most three is the opposite of at least four?)

Then for (D), if we have the same number of zeros and ones, how many ones do we have? If we calculate the number of bit strings that have that many ones, we are done.

• For those wondering about "C", the answer is subtraction. Mar 10, 2016 at 23:40

Part (A) is fine.

For part (B), though, ${10\choose 0} + {10\choose 1} + .... +{10\choose 4}$, I get an answer of 386,

so either there is a typo in your post, or the book answer is wrong.

For part (C), you could very well apply a short cut.

Total ways for the string is $2^{10}$, because there are 2 choices for each bit.

["at most 4 1's] - ["exactly 4 1's"] gives [at most 3 1's] = 386 - 210 = 176 ,

and [ at least 4 1's] = $2^{10} - 176 = 848$

For part A it will be simply $$C(10,4)=210$$

For part B it will be $$C(10,0)+C(10,1)+C(10,3)+C(10,4)=386$$ because at most means maximum which we actually start from zero till that particular number

For part C it will be simply $$C(10,4)+C(10,5)+C(10,6)+C(10,7)+C(10,8)+C(10,9)+C(10,10)=848$$

For part D it will be simply $$C(10,5)=252$$


# Find algebraic solutions to system of polynomial equations

How can I find all (or all real) algebraic solutions to a set of polynomial equations, or equivalently all common roots of a set of polynomials? I'm interested in those cases where the set of solutions is finite, so the number of constraints matches the number of variables. I'm interested in exact algebraic numbers, not numeric approximations. The polynomials are elements of a polynomial ring, not symbolic expressions.

At the moment, I often use resultants to eliminate one variable after the other. In the end I have a single polynomial for one of the variables, and can find algebraic roots of that. Doing the same with a different elimination order gives candidates for the other variables, and then I can check which combinations satisfy the original equations.

But I guess there must be some more efficient approach. Probably using groebner bases. I couldn't find a simple example along these lines in the reference documentation, though.

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# Polynomial systems

Chapter 9 of the open-source book Calcul mathématique avec Sage (in French) is about polynomial systems. In particular, check section 9.2. The book is available for free download from: http://sagebook.gforge.inria.fr/ (click "Telecharger le PDF").

Credit goes to Marc Mezzaroba who authored that chapter, and more generally to the team who authored the book and kindly provides it under a Creative Commons license allowing all to copy and redistribute the material in any medium or format, and to remix, transform, and build upon the material, for any purpose.

## The system

In section 9.2.1, the following polynomial system is considered:

$$\left \{ \quad \begin{array}{@{}ccc@{}} x^2 \; y \; z & = & 18 \\ x \; y^3 \; z & = & 24\\ x \; y \; z^4 & = & 6 \\ \end{array}\right.$$

## Numerical solve vs algebraic approach

While section 2.2 of the book explained how to solve numerically with solve,

sage: x, y, z = var('x, y, z')
sage: solve([x^2 * y * z == 18, x * y^3 * z == 24,\
....: x * y * z^4 == 3], x, y, z)
[[x == (-2.76736473308 - 1.71347969911*I), y == (-0.570103503963 +
2.00370597877*I), z == (-0.801684337646 - 0.14986077496*I)], ...]

section 9.2.1 explains how to solve algebraically.

## Ideal in a polynomial ring

First translate the problem in more algebraic terms: we are looking for the common zeros of three polynomials, so we consider the polynomial ring over QQ in three variables, and in this ring we consider the ideal generated by the three polynomials whose common zeros we are looking for.

sage: R.<x,y,z> = QQ[]
sage: J = R.ideal(x^2 * y * z - 18,
....:             x * y^3 * z - 24,
....:             x * y * z^4 - 6)

We check that the dimension of this ideal is zero, which means the system has finitely many solutions.

sage: J.dimension()
0

## Solution, algebraic variety, choice of base ring

The command variety will compute all the solutions of the system. However, its default behaviour is to give the solutions in the base ring of the polynomial ring. Here, this means it gives only the rational solutions.

sage: J.variety()
[{y: 2, z: 1, x: 3}]

We want to enumerate the complex solutions, as exact algebraic numbers. To do that, we use the field of algebraic numbers, QQbar. We find the 17 solutions (which were revealed by the numerical approach with solve).

sage: V = J.variety(QQbar)
sage: len(V)
17

Here is what the last three solutions look like as complex numbers.

sage: V[-3:]
[{z: 0.9324722294043558? - 0.3612416661871530?*I,
y: -1.700434271459229? + 1.052864325754712?*I,
x: 1.337215067329615? - 2.685489874065187?*I},
{z: 0.9324722294043558? + 0.3612416661871530?*I,
y: -1.700434271459229? - 1.052864325754712?*I,
x: 1.337215067329615? + 2.685489874065187?*I},
{z: 1, y: 2, x: 3}]

Each solution is given as a dictionary, whose keys are the generators of QQbar['x,y,z'] (and not QQ['x ...

more

Yes, you can use Gröbner bases. Here is an example

sage: A.<x,y,z> = QQ[]
sage: I = A.ideal(z*x^2-y, y^2-x*y, x^3+1)
sage: I.variety()
[{y: -1, z: -1, x: -1}, {y: 0, z: 0, x: -1}]

Tis is not implemented with coefficients in RR and will raise an error. However You can still ask for a Gröbner basis

sage: A.<x,y,z> = RR[]
sage: I = A.ideal(z*x^2-y, y^2-x*y, x^3+1)
sage: I.groebner_basis()
[x^3 + 1.00000000000000, x*y + z, y^2 + z, x*z - y*z, z^2 + y]

more


# Operations with arrays (pseudoinverses & products)

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 Dear Wolfram Community:I wish to share a code I made to do some operations with multidimensional arrays, like tensor products, inner products and inverses/pseudoinverses.Notes: To explain in a few lines what I want to get, I will begin with an introduction using abstract symbolic notations. I would greatly appreciate your feedback to this post, because while my code works fine for simple (small) arrays, it become slow for bigger ones and I suspect this could be a clue that my code is not an efficient one: maybe there is a simpler way to do the same. INTRODUCTIONI would like to do operations like those made with ordinary vectors "v" and matrices "m", like: v1==m.v2; x=PseudoInverse[m]; x.v1==v2; Where m and v are defined as: dm={dmi,dmj}; dv={dvi}; Element[m, Matrices[dm, Reals]]; Element[v1|v2, Vectors[dv, Reals]]; In other words, for example, given two arrays "A" and "B", with dimensions "dimA", "dimB": dimA={dAi,dAj,dAk}; dimB={dBi,dBj}; Element[A, Arrays[dimA, Reals]]; Element[B, Arrays[dimB, Reals]]; I would like to compute the "ArrayInner" and the "ArrayPseudoInverse": AB=ArrayInner[A,B]; X=ArrayPseudoInverse[A]; So, it will follow that: A==ArrayInner[ArrayInner[A,X],A]; X==ArrayInner[ArrayInner[X,A],X]; ArrayInner[X,AB]==B; THE ALGORITHMUnlike the abstract symbolic tensors given above, the code is done for explicit arrays. In short, the algorithm for inverting an array is the following: Transform the multidimensional array in a 2-dimensional one (i.e. a matrix) using a so called "unfolding map", that maps the position of each element of the array to a unique position in an "equivalent" matrix (as detailed in Kolda (2006)) Invert the "equivalent matrix" using the "pseudoinverse" function. I am using the pseudoinverse because usually this equivalent matrix is a singular and rectangular matrix. Map the inverted matrix into the respective "inverted array". This was the most difficult part, since inverting the array implies some (sometimes unknown) permutation. Unfortunately, the full code is very lenghty, so for now I will not copy it in the main text (but I will attach it as a notebook at the end of the post).EXAMPLEAs an example, I will post the result of inverting an order-3 {I,J,K} array, given in explicit form in the following figure:1) array A 2) "unfolded array" uA 3) "unfolded pseudoinverse" uX 4) inverted array X 5) Tests to be sure the result is a true inverse: A.X.A==A; X.A.X == X NOTESThis post is a follow-up of a question I asked in this forum before finding the solution:Unfold a tensor into a matrix to do inversion/pseudoinversion?I am very grateful for the support from user Hans Dolhaine, who showed me very useful tips (like how to use the functions "Module" and the solution to a similar problem done with (square) matrices of (square) matrices. REFERENCES*) Tamara G. Kolda (2006); "Multilinear operators for higher-order decompositions "; SANDIA REPORT SAND2006-2081*) Mao-lin Liang, Bing Zheng & Rui-juan Zhao (2018): Tensor inversion and its application to the tensor equations with Einstein product, Linear and Multilinear Algebra Attachments:
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 -- you have earned Featured Contributor Badge Your exceptional post has been selected for our editorial column Staff Picks http://wolfr.am/StaffPicks and Your Profile is now distinguished by a Featured Contributor Badge and is displayed on the Featured Contributor Board. Thank you!
Posted 1 year ago
 Dear Wolfram Moderation Team:Good morning, I am very thankfully surprised to you for selecting my post for the "Staff Picks" column!I took me a lot of time to do this, and after many trial-and-error attemps, I arrived at the form posted yesterday. Please feel free to re-check the code, entering some more complicated array, changing the dimensionality of the arrays used, etc. just to be sure that there isn't a mistake lurking inside, or something else that may make the computation slow. I will also greatly appreciate suggestions for improvement, maybe the same results could be obtained with a simpler code.Best regards,Alberto Silva Ariano


# Definite integral of unspecified function

I stumbled upon this gem whilst doing tasks, and I can't seem to quite grasp the key to solving this. The answer should be $\dfrac{1}{2}$. While I have tried a few things, which resulted in $\dfrac{1}{2}$, I am quite unsure if my methods were legit. I have thought of substitution, but I get stuck half way. I would very much appreciate it if someone could help me solve this (practising for a test).

$$\int_{0}^{1}\frac{f(x)}{f(x)+f(1-x)}\text{d}x$$

Edit: Let $f$ be a positive continuous function. The task is simple if you could simply plug in a function, but it says not to. I think it means to solve this generally.

• What is your question? It will help if you post what you have tried and the community can see if your approach is valid. – Erik M Mar 7 '17 at 22:59
• If we know nothing about $f(x)$, there is nothing to solve. First of all, we even do not know if this integration is meaningful - e.g. is f(x)/(f(x) + f(1-x)) measurable? – Yujie Zha Mar 7 '17 at 23:09
• $f(1-x)$ is just $f(x)$ run backwards on the domain $[0,1]$, for whatever that may be worth. – Alfred Yerger Mar 7 '17 at 23:22
• I really do not know where to start. I have posted the whole task as it is supposed to be solved. This is the last place I can ask, my friends are all sleeping..If I could somehow substitute the core to make them equal, I could in theory get something like f(1/2) / 2f(1/2) which would be 1/2. – MCrypa Mar 7 '17 at 23:22
• Hint: $$I = \int_{0}^{1}\frac{f(x)}{f(x)+f(1-x)}dx=\int_{0}^{1}\frac{f(x)+f(1-x)-f(1-x)}{f(x)+f(1-x)}dx=\int_{0}^{1}\left(1 - \frac{f(1-x)}{f(x)+f(1-x)}\right)dx=\;\;\cdots\;\;=1 - I$$ – dxiv Mar 7 '17 at 23:44

Notice that if you change the variable in the integration by $x = 1 - x$, you get:
$$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx= \int_1^0 \frac{f(1 - x)}{f(1 - x) + f(x)}d(1 - x) = \int_0^1 \frac{f(1 - x)}{f(1 - x) + f(x)}dx$$
And $$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx + \int_0^1 \frac{f(1 - x)}{f(1 - x) + f(x)}dx = \int_0^1 \frac{f(x) + f(1 - x)}{f(x) + f(1 - x)}dx = 1$$ So $$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx = \frac {1}{2}$$


Both are fully capable of representing undirected and directed graphs. Matrix notation and computation can help to answer these questions. Grifﬁth / Linear Algebra and its Applications 388 (2004) 201–219 203 Adjacency matrices represent adjacent vertices and incidence matrix vertex-edge incidences. A very easy upper estimate for it can be obtained directly by Gershgorin's theorem: $$\lambda_{\max}\le \Delta\ ,$$ where $\Delta$ is the maximal degree of the graph. The first step is to number our cities in the order they are listed: San Diego is 1, San Francisco is 2, and so on. Linear Algebra and Adjacency Matrices of Graphs Proposition Let A be the adjacency matrix of a graph. . We can associate a matrix with each graph storing some of the information about the graph in that matrix. . We'll start by encoding the data from our table into what's called an adjacency matrix . If a graph has vertices, we may associate an matrix which is called vertex matrix or adjacency matrix. This documents an unmaintained version of NetworkX. 12.2.1 The Adjacency Matrix and Regular Graphs . add_nodes_from (nodes) G1. Linear algebra is one of the most applicable areas of mathematics. If we want to do this efficiently, linear algebra is the perfect tool. Featured on Meta Hot Meta Posts: Allow for removal by moderators, and thoughts about future… In this material, we manage to deﬁne Proposition Let G be a graph with e edges and t triangles. In the special case of a finite simple graph, the adjacency matrix is a (0,1)-matrix with zeros on its diagonal. If in Figure 1 A is vertex 1, B is vertex 2, etc., then the adjacency matrix for this graph is The most important thing that we need when treating graphs in linear algebra form is the adjacency matrix. Suppose that we have given any adjacency matrix, then deciding whether it has a clique by looking at it is impossible. This matrix can be used to obtain more detailed information about the graph. The adjacency matrix of a nonempty (undirected) graph has a strictly positive largest eigenvalue $\lambda_\max$. . If M is an n-by-n irreducible adjacency matrix––either a binary 0 - 1 matrix or its row-standardized counterpart––based upon an undirected planar D.A. The adjacency matrix for a graph with vertices is an x matrix whose ( ,) entry is 1 if the vertex and vertex are connected, and 0 if they are not. Recall that thetraceof a square matrix is the sum of its diagonal entries. Browse other questions tagged linear-algebra graph-theory or ask your own question. The (i;i)-entry in A2 is the degree of vertex i. Adjacency matrix (vertex matrix) Graphs can be very complicated. It is ... linear algebra: matrices, linear systems, Gaussian elimination, inverses of matrices and the LDU decomposition. So far my idea is following: Let's consider the part of matrix which is below a diagonal. For example, for four nodes joined in a chain: import networkx as nx nodes = list (range (4)) G1 = nx. add_edges_from (zip (nodes, nodes [1:])) . . Matrix representations provide a bridge to linear algebra-based algorithms for graph computation. If the graph is undirected (i.e. Matrix representations of graphs go back a long time and are still in some areas the only way to represent graphs. Graph G1. In graph theory and computer science, an adjacency matrix is a square matrix used to represent a finite graph.The elements of the matrix indicate whether pairs of vertices are adjacent or not in the graph.. If you want a pure Python adjacency matrix representation try networkx.convert.to_dict_of_dicts which will return a dictionary-of-dictionaries format that can be addressed as a sparse matrix. Linear algebra » adjacency_matrix; Warning. ... Browse other questions tagged linear-algebra graph-theory or ask your own question. The data from our table into what 's called an adjacency matrix ( vertex )! Linear algebra-based algorithms for graph computation if M is an n-by-n irreducible adjacency matrix––either binary... ) ) 12.2.1 the adjacency matrix is the degree of vertex i and Applications... Linear algebra-based algorithms for graph computation inverses of matrices and the LDU decomposition to do this,. An matrix which is below a diagonal linear-algebra graph-theory or ask your own question an... Vertex i nodes [ 1: ] ) ) 12.2.1 the adjacency of... / linear algebra is the perfect tool still in some areas the only way to represent Graphs elimination inverses..., linear systems, Gaussian elimination, inverses of matrices and the LDU decomposition detailed information the. The ( i ; i ) -entry in A2 is the sum of diagonal... Or its row-standardized counterpart––based upon an undirected planar D.A a be the adjacency matrix ( vertex matrix ) Graphs be... 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Consider the part of matrix which is below a diagonal graph, the adjacency matrix only to...


A function is said to be differentiable if the derivative exists at each point in its domain. Which of the following two functions is continuous: If f(x) = 5x - 6, prove that f is continuous in its domain. Transcript. the y-value) at a.; Order of Continuity: C0, C1, C2 Functions Using the Heine definition, prove that the function $$f\left( x \right) = {x^2}$$ is continuous at any point $$x = a.$$ Solution. $\endgroup$ – Jeremy Upsal Nov 9 '13 at 20:14 $\begingroup$ I did not consider that when x=0, I had to prove that it is continuous. To give some context in what way this must be answered, this question is from a sub-chapter called Continuity from a chapter introducing Limits. Using the Heine definition we can write the condition of continuity as follows: The following are theorems, which you should have seen proved, and should perhaps prove yourself: Constant functions are continuous everywhere. The function is defined at a.In other words, point a is in the domain of f, ; The limit of the function exists at that point, and is equal as x approaches a from both sides, ; The limit of the function, as x approaches a, is the same as the function output (i.e. The following theorem is very similar to Theorem 8, giving us ways to combine continuous functions to create other continuous functions. More formally, a function (f) is continuous if, for every point x = a:. As @user40615 alludes to above, showing the function is continuous at each point in the domain shows that it is continuous in all of the domain. Jump discontinuities occur where the graph has a break in it as this graph does and the values of the function to either side of the break are finite ( i.e. Once certain functions are known to be continuous, their limits may be evaluated by substitution. A function f is continuous when, for every value c in its Domain:. This kind of discontinuity in a graph is called a jump discontinuity . f(c) is defined, and. The question is: Prove that cosine is a continuous function. If f(x) = x if x is rational and f(x) = 0 if x is irrational, prove that f is continuous … To show that $f(x) = e^x$ is continuous at $x_0$, consider any $\epsilon>0$. If f(x) = 1 if x is rational and f(x) = 0 if x is irrational, prove that x is not continuous at any point of its domain. Learn how to determine the differentiability of a function. limx→c f(x) = f(c) "the limit of f(x) as x approaches c equals f(c)" The limit says: But in order to prove the continuity of these functions, we must show that $\lim\limits_{x\to c}f(x)=f(c)$. The function value and the limit aren’t the same and so the function is not continuous at this point. When a function is continuous within its Domain, it is a continuous function.. More Formally ! Let  = tan  = sincos is defined for all real number except cos = 0 i.e. Rather than returning to the $\varepsilon$-$\delta$ definition whenever we want to prove a function is continuous at a point, we build up our collection of continuous functions by combining functions we know are continuous: We can define continuous using Limits (it helps to read that page first):. THEOREM 102 Properties of Continuous Functions Let $$f$$ and $$g$$ be continuous on an open disk $$B$$, let $$c$$ … Example 18 Prove that the function defined by f (x) = tan x is a continuous function. Consider an arbitrary $x_0$. Proofs of the Continuity of Basic Algebraic Functions. The Solution: We must show that $\lim_{h \to 0}\cos(a + h) = \cos(a)$ to prove that the cosine function is continuous. Function ( f ) is continuous when, for every value c in Domain! Is continuous when, for every point x = a: the derivative at... Real number except cos = 0 i.e ( it helps to read that page ). 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Q&A

# Solve $\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$

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Evaluate $$\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$$

I was trying to solve it following way.

$$\int_0^{\dfrac{\pi}{6}} \sec^2\theta \sec\theta \mathrm d\theta$$ $$\int_0^{\dfrac{\pi}{6}}\sec^2\theta \mathrm d(\sec\theta)$$ $$[\tan\theta]_0^\dfrac{\pi}{6}$$ $$\tan\frac{\pi}{6}$$ $$\frac{1}{\sqrt{3}}$$

I had found the value. But, my book had solved it another way. They took

$$\tan\theta=z$$ Then, they solved it. They had got $\frac{1}{3}+\frac{1}{2}\ln\sqrt{3}$. My answer is approximately close to their. Is my answer correct? While doing Indefinite integral I saw that I could solve problem my own way. But, my answer always doesn't match with their. So, is it OK to find new/another answer of Integral? In algebraic expression,"no matter what I do the answer always matches". But, I got confused with Integration.

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Why should this post be closed?

You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.

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My answer isn't correct. Cause, differentiation of $\sec x=\sec x\tan x$. I had differentiate inside integration.

$$\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$$ $$\int_0^{\dfrac{\pi}{6}} (1-\tan^2 \theta) \frac{d}{d \theta} (\sec \theta \tan \theta) \mathrm d\theta$$

That's the correct one. But, you got it wrong. I have differentiate.

my answer always doesn't match with their. So, is it OK to find new/another answer of Integral?

Saying to Indefinite Integral, if you integrate an equation then, I may find lots of answer. But, if I (you) put specific value instead of $\theta$ or, $x$. Than, you will get same value if your answer isn't wrong.

Why does this post require moderator attention?


# Math Help - combinations

1. ## combinations

2. Why does that look like an online test?

3. Show that $C(8,4)=C(7,3)+C(7,4)$

Let's manipulate the left side to become the right side
$\frac{8!}{4!\cdot (8-4)!}=\frac{8!}{4!\cdot 4!}=\frac{8\cdot 7!}{4!\cdot 4\cdot 3!}=\frac{2\cdot 7!}{4!\cdot 3!}=\frac{7!}{4!\cdot 3!}+\frac{7!}{4!\cdot 3!}$
$\frac{7!}{3!\cdot (7-3)!}+\frac{7!}{4!\cdot (7-4)!}=\boxed{C(7,3)+C(7,4)}$ which was to be shown.

4. Originally Posted by janvdl
Why does that look like an online test?
It probably is. Or online homework anyway.

-Dan

5. Originally Posted by topsquark
It probably is. Or online homework anyway.

-Dan
Usually you dont get points for homework...

6. Originally Posted by janvdl
Usually you dont get points for homework...

4) $\sum_{i = 1}^{5} i \left[ \sum_{j = 1}^{3} (2j + 1) \right] = (1 + 2 + 3 + 4 + 5) \left[ (2(1) + 1) + (2(2) + 1) + (2(3) + 1) \right]$

$= 15 ( 3 + 5 + 7)$

$= 15 \cdot 15$

$= 225$

7. ## What about the other ones?

8. Originally Posted by Raiden_11
#2 is very similar to #1 which rualin so graciously did for you. try it on your own and tell us what you come up with.

for problem #3, i'd do something like this. chances are there's a more efficient way, but this is how i saw the solution:

$\sum_{j = 3}^{6} (2j - 2) = \sum_{j = 1}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 2)$

$= \sum_{j = 1}^{4} (2j - 2) + \sum_{j = 5}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 1)$ .......evaluating the last two summations we obtain

$= \sum_{j = 1}^{4} (2j - 2) + 16$ ........now let's do some manipulation on the first summation

$= \sum_{j = 1}^{4}(2j + 2 - 4) + 16$ .......i didn't change anything, +2 - 4 = -2

$= \sum_{j = 1}^{4} (2j + 2) + \sum_{j = 1}^{4} (-4) + 16$ ......now evaluate the second summation

$= \sum_{j = 1}^{4} (2j + 2) - 16 + 16$

$= \sum_{j = 1}^{4} (2j + 2)$ ........a bit more manipulation on this summation should do it

$= \sum_{j = 1}^{4} (2j + 4 - 2)$ .........again, I changed nothing, +4 - 2 = +2

$= \sum_{j = 1}^{4} \left[ 2(j + 2) - 2 \right]$ .........i factored out a 2 from the first two terms

and now we have the desired summation

QED

Did you understand?

9. ## Are u sure?

Are u sure there aren't any more steps?

10. ## I still don't know?

I still don't know how to do #2.

11. Originally Posted by Raiden_11
Are u sure there aren't any more steps?
more steps for what?

you should click the "quote" button when you're responding to something, so people know exactly what you're responding to

12. ## What i mean is....

Originally Posted by Jhevon
more steps for what?

you should click the "quote" button when you're responding to something, so people know exactly what you're responding to

Are you sure there aren't anymore steps for number 3....
also i still don't know how to do number 2..

13. Originally Posted by Raiden_11
I still don't know how to do #2.
here's one way to do it:

$C(7,5) = \frac {7!}{5! ( 7 - 5)!}$

$= \frac {7!}{5! 2!}$

$= \frac {7 \cdot 6!}{5! 2!}$

$= \frac {(2 + 5) \cdot 6!}{5! 2!}$

$= \frac {2 \cdot 6! + 5 \cdot 6!}{5! 2!}$

$= \frac {2 \cdot 6!}{5! 2!} + \frac {5 \cdot 6!}{5! 2!}$

$= 6 + \frac {6!}{4! 2!}$

$= 1 + 5 + \frac {6!}{4! 2!}$

$= C(4,4) + C(5,1) + C(6,2)$

QED

14. Originally Posted by Raiden_11
Are you sure there aren't anymore steps for number 3....
also i still don't know how to do number 2..
no, number 3 is complete. i started with one series and manipulated it and ended up with the other series. this proves the claim that they are equal. in fact, i think i probably did too many steps, i'm not very efficient with this sort of thing.

15. ## This is my final question for math forum!!

This is my final question for math forum which i hope you can answer!

1. On page 345 of the textbook, the diagonal pattern of Pascal’s Triangle illustrates that C(7, 5) = C(4, 4) + C(5, 4) + C(6, 4). Prove this relationship using the following methods.

a)numerically by using factorials
b)by reasoning, using the meaning of combinations

Page 1 of 2 12 Last


# Prove the equation

Prove that $$\int_0^{\infty}\exp\left(-\left(x^2+\dfrac{a^2}{x^2}\right)\right)\text{d}x=\frac{e^{-2a}\sqrt{\pi}}{2}$$ Assume that the equation is true for $a=0.$

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$$I(a):=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\text{d}x$$ $$\frac{dI}{da}=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\left(-\frac{2a}{x^2}\right)\text{d}x$$ Now substitute $y=\frac{a}{x}$, so $dy=-\frac{a}{y^2}$: $$\frac{dI}{da}=2\int_{\infty}^{0}e^{-\left(\dfrac{a^2}{y^2}+y^2\right)}\text{d}y=-2\int_{0}^{\infty}e^{-\left(\dfrac{a^2}{y^2}+y^2\right)}\text{d}y=-2I$$ To obtain $I$ you just have to solve the simple ODE: $$\frac{dI}{da}=-2I$$ with initial condition given by $$I(0)=\frac{\sqrt{\pi}}{2}\ .$$ This gives you $$I(a)=\frac{e^{-2a}\sqrt{\pi}}{2}\ .$$

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+1. This is a nice answer. –  Tunk-Fey Jun 24 at 13:58
Thanks. It is very clever. But can you elaborate a bit more on how you came up with the solution translating the original problem into solving an ODE? –  lovelesswang Jun 24 at 14:06
Do you mean the explicit resolution of the ODE? –  Dario Jun 24 at 14:08
The idea behind the solution is the technique of differentiation under the integral sign(en.wikipedia.org/wiki/…): when you have an integral that depend on a parameter, you can derive the integral w.r.t. that parameter and sometimes you obtain a much simpler integral. In this particular case the trick is to notice that after the derivation, getting rid of the extra factor you obtain by a change of variable you get again the first integral. This allow you to find a relation between $I$ and its derivative, i.e. the ODE. –  Dario Jun 24 at 14:13
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In general \begin{align} \int_{x=0}^\infty \exp\left(-ax^2-\frac{b}{x^2}\right)\,dx&=\int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx\\ &=\int_{x=0}^\infty \exp\left(-a\left(x^2-2\sqrt{\frac{b}{a}}+\frac{b}{ax^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dx\\ &=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dx\\ &=\exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx\\ \end{align} The trick to solve the last integral is by setting $$I=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx.$$ Let $t=-\frac{1}{x}\sqrt{\frac{b}{a}}\;\rightarrow\;x=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dx=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then $$I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt.$$ Let $t=x\;\rightarrow\;dt=dx$, then $$I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.$$ Adding the two $I_t$s yields $$2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.$$ Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then $$I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}},$$ where $I$ is a Gaussian integral. Thus \begin{align} \exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx &=\large\color{blue}{\frac12\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}}. \end{align} In our case, put $a=1$ and $b=a^2$.

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# Group is Subgroup of Itself

## Theorem

Let $\struct {G, \circ}$ be a group.

Then:

$\struct {G, \circ} \le \struct {G, \circ}$

That is, a group is always a subgroup of itself.

## Proof

By Set is Subset of Itself, we have that:

$G \subseteq G$

Thus $\struct {G, \circ}$ is a group which is a subset of $\struct {G, \circ}$, and therefore a subgroup of $\struct {G, \circ}$.

$\blacksquare$


# Use the Division Algorithm to show the square of any integer is in the form $3k$ or $3k+1$

Use Division Algorithm to show the square of any int is in the form 3k or 3k+1

What confuses me about this is that I think I am able to show that the square of any integer is in the form $X*k$ where $x$ is any integer. For Example:

$$x = 3q + 0 \\ x = 3q + 1 \\ x = 3q + 2$$

I show $3k$ first $$(3q)^2 = 3(3q^2)$$ where $k=3q^2$ is this valid use of the division algorithm?

If it is then can I also say that int is in the form for example 10*k

for example

$(3q)^2 = 10*(\frac{9}{10}q^2)$

where $k=(\frac{9}{10}q^2)$

Why isn't this valid? Am I using the div algorithm incorrectly to show that any integer is the form 3k and 3k+1, if so how do I use it? Keep in mind I am teaching myself Number Theory and the only help I can get is from you guys in stackexchange.

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How do you know that $\frac 9 {10}q^2$ is an integer? –  azarel Jul 18 '12 at 19:05
Good point I guess that is the reason that is 10k is invalid. The form $(3q+2)^2$ is not in the form 3k or 3k+1 without getting a fraction, how to account for that to finish the problem? –  user968102 Jul 18 '12 at 19:15
If $x = 10k$ then you should also consider all the other cases of $x = 10k + 1, 10k+2, \ldots + 10k+9.$ I don't see why you want to consider that. –  user2468 Jul 18 '12 at 19:20
$(3q+2)^2$ is not in the form $3k$ or $3k+1$ seemingly. Start expanding the square $(3q+2)^2 = (3q+2)(3q+2) = (3q)^2 + (2)^2 + 2(3q)(2) = ?????$ Can you take it from here? –  user2468 Jul 18 '12 at 19:21

By the division algorithm, $$x = 3q + r,\text{ where } r \in \{0, 1, 2\}.$$ So express $$x^2 = 9q^2 + r^2 + 6qr = 3(3q^2 + 2qr) + r^2.$$ For a given $x$ if $r = 0$ or $1,$ then we're done. If $r = 2$ then $r^2 = 4 = 3 + 1,$ and hence $$x^2 = 3\times\text{integer} + 3 + 1 = 3\times(\text{integer} + 1) + 1.$$ We are done.

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This makes sense to me and I am able to finish problem thank you –  user968102 Jul 18 '12 at 19:32

Let $x = 3k+r, r = 0, 1, 2$ by the division algorithm. Squaring $x$, we find $x^2 = 9k^2+6kr+r^2$, or $x^2 = (9k+6r)k+r^2$.

Since $9k+6r$ is divisible by 3 for all integers $k, r$, then we may re-write this as $x^2 = 3k_1 + r^2$.

Using the division algorithm again, we see that $x^2 = 3k_1+r_1, r_1 = 0, 1, 2$. If $r_1 = 2$, then $r = \pm \sqrt{2}$, which is not an integer. Therefore, only $r=0$ and $r=1$ are acceptable.

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Hint $\$ Below I give an analogous proof for divisor $5$ (vs. $3),$ exploiting reflection symmetry.

Lemma $\$ Every integer $\rm\:n\:$ has form $\rm\: n = 5\,k \pm r,\:$ for $\rm\:r\in\{0,1,2\},\ k\in\Bbb Z.$

Proof $\$ By the Division algorithm

$$\rm\begin{eqnarray} n &=&\:\rm 5\,q + r\ \ \ for\ \ some\ \ q,r\in\Bbb Z,\:\ r\in [0,4] \\ &=&\:\rm 5\,(q\!+\!1)-(5\!-\!r) \end{eqnarray}$$

Since $\rm\:(5\!-\!r)+r = 5,\,$ one summand is $\le 2,\,$ so lies in $\{0,1,2\},\,$ yielding the result.

Theorem $\$ The square of an integer $\rm\,n\,$ has form $\rm\, n^2 = \,5\,k + r\,$ for $\rm\:r\in \{0,1,4\}.$

Proof $\$ By Lemma $\rm\ n^2 = (5k\pm r)^2 = 5\,(5k^2\!\pm 2kr)+r^2\,$ for $\rm\:r\in \{0,1,2\},\,$ so $\rm\: r^2\in\{0,1,4\}.$

Remark $\$ Your divisor of $\,3\,$ is analogous, with $\rm\:r\in \{0,1\}\,$ so $\rm\:r^2\in \{0,1\}.\,$ The same method generalizes for any divisor $\rm\:m,\,$ yielding that $\rm\:n^2 = m\,k + r^2,\,$ for $\rm\:r\in\{0,1,\ldots,\lfloor m/2\rfloor\}.$ The reason we need only square half the remainders is because we have exploited reflection symmetry (negation) to note that remainders $\rm > n$ can be transformed to negatives of remainders $\rm < n,\,$ e.g. $\rm\: 13 = 5\cdot 2 +\color{#0A0} 3 = 5\cdot 3 \color{#C00}{- 2},\,$ i.e. remainder $\rm\:\color{#0A0}3\leadsto\,\color{#C00}{-2},\,$ i.e. $\rm\:3 \equiv -2\pmod 5.\:$ This amounts to using a system of balanced (or signed) remainders $\rm\, 0,\pm1,\pm2,\ldots,\pm n\$ vs. $\rm\ 0,1,2,\ldots,2n[-1].\:$ Often this optimization halves work for problems independent of the sign of the remainder.

All of this is much clearer when expressed in terms of congruences (modular arithmetic), e.g. the key inference above $\rm\:n\equiv r\:\Rightarrow\:n^2\equiv r^2\pmod m\:$ is a special case of the ubiquitous

Congruence Product Rule $\rm\ \ A\equiv a,\ B\equiv b\ \Rightarrow\ AB\equiv ab\ \ (mod\ m)$

Proof $\rm\:\ \ m\: |\: A\!-\!a,\ B\!-\!b\:\ \Rightarrow\:\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ AB - ab$

For an introduction to congruences see any decent textbook on elementary number theory.

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