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-6,870,535,078,389,188,000 | Kramers-Kronig-Beziehungen
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Die Kramers-Kronig-Beziehungen, auch Kramers-Kronig-Relation, setzen Real- und Imaginärteil bestimmter meromorpher Funktionen in Form einer Integralgleichung miteinander in Beziehung. Sie stellen damit einen Spezialfall der Hilbert-Transformation dar. Die Beziehungen wurden nach ihren Entdeckern Hendrik Anthony Kramers und Ralph Kronig benannt.
Eine wichtige Anwendung der Kramers-Kronig-Beziehungen ist der Zusammenhang zwischen der Absorption und der Dispersion der Ausbreitung von Licht in einem Medium. Weitere Anwendungen gibt es in der Hochenergiephysik.
Mathematische Formulierung[Bearbeiten]
Sei F : \mathbb{C} \rightarrow \mathbb{C} eine meromorphe Funktion, deren Polstellen in der unteren Halbebene liegen. Dieser Forderung an die Lage der Polstellen entspricht physikalisch das Kausalitätspostulat. Ferner seien \operatorname{Re}\, F|_\mathbb{R} bzw. \operatorname{Im}\, F|_\mathbb{R} Real- und Imaginärteil der Funktion F. Es sei vorausgesetzt, dass diese beiden Funktionen gerade bzw. ungerade sind. Das bedeutet, dass F durch Fourierintegration nicht aus einer beliebigen komplexen, sondern aus einer reellen Funktion gebildet werden kann.
In der Physik betrachtet man oft statt F die Funktion F/i, wodurch sich die Voraussetzungen bezüglich gerade und ungerade vertauschen. Schließlich sei \lim_{|z| \rightarrow \infty} |F(z)| = 0. Dann gelten für x \in \mathbb{R} die folgenden als Kramers-Kronig-Beziehungen bezeichnete Gleichungen:
\operatorname{Im}\, F(x) = -\frac{2}{\pi} \cdot \;\mathrm{CH}\, \int_{0}^{+\infty} \frac{x\cdot\operatorname{Re}\,F(t)}{t^2-x^2}\mathrm{d}t
\operatorname{Re}\, F(x) = \frac{2}{\pi} \cdot \;\mathrm{CH}\, \int_{0}^{+\infty} \frac{t\cdot\operatorname{Im}\,F(t)}{t^2-x^2}\mathrm{d}t
\mathrm{CH} bezeichnet den cauchyschen Hauptwert des auftretenden Integrals.
Real- und Imaginärteil der Funktion F bedingen sich also gegenseitig durch Integration. Dies findet Anwendungen in der Optik und in der Systemtheorie wenn F die Suszeptibilität eines Systems angibt, siehe Kausalität. Anwendungen finden sich auch in der Hochenergie-Physik bei den Dispersionsrelationen der S-Matrix.
Motivation (Ein Randwertproblem)[Bearbeiten]
Auf der reellen Achse \mathbb R sei eine stetige reelle Funktion \,f vorgegeben, die analog zu \operatorname{Re}\,F als gerade vorausgesetzt werden soll. Dazu soll eine in der ganzen oberen Halbebene holomorphe komplexe(!) Funktion \,F so konstruiert werden, dass \operatorname{Re}\, F|_\mathbb{R} \stackrel{!}{=} f gilt.
Es soll also ein Randwertproblem gelöst werden, wobei im Innern des betrachteten Gebietes \,G, d. h. oberhalb von \mathbb R, wegen der Holomorphie-Bedingung die Cauchy-Riemannschen Differentialgleichungen erfüllt werden müssen und auf dem Rand, \partial G=\mathbb R\,, eine stetige reelle Funktion, f, vorgegeben ist, die dort angenommen werden soll.
Eine holomorphe Funktion kann nach dem Residuensatz dargestellt werden als:
F(z) = \frac{1}{2 \pi i} \left( \int_{HK_r} \frac{F(t)}{t - z} \mathrm{d}t + \int_{-r}^r \frac{F(t)}{t - z} \mathrm{d}t \right),
wobei HK_r(0) den (positiv orientierten) Halbkreis in der oberen Halbebene mit Zentrum 0 und Radius r > 0 bezeichnet. Fällt nun F im Unendlichen schnell genug ab, so reduziert sich im Grenzübergang r \rightarrow \infty die Darstellung zu einem Integral über der reellen Achse, also:
F(z) = \frac{1}{2 \pi i} \int_{-\infty}^{+\infty} \frac{F(t)}{t-z} \mathrm{d}t
Im Falle \operatorname{Im}\, z = 0, und weil \,f bzw. \operatorname{Re}\,F(t) eine gerade Funktion sein soll, ergibt sich schließlich
\operatorname{Im} \, F(z) = - \frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{\operatorname{Re}\,F(t)}{t-z} \mathrm{d}t
= - \frac{2}{\pi} \int_{0}^{+\infty} \frac{z \cdot \operatorname{Re}\,F(t)}{t^2-z^2} \mathrm{d}t,
wobei das auftretende Integral als Cauchyscher Hauptwert zu interpretieren ist (Singularität für t = z) und mit der Hilbert-Transformation von f übereinstimmt. Der Residuensatz wird hierbei auf den Integrationsweg [-r,z-\varepsilon] \cdot HK_\varepsilon(z) \cdot [z+\varepsilon, r] \cdot HK_r(0) angewendet. Diese Gleichung entspricht der einen Kramers-Kronig-Beziehung.
Man braucht jetzt zur Lösung des Randwertproblems nur die Beziehung \operatorname{Re}\, F|_\mathbb{R} {=} f einzusetzen.
Für ungerade Funktionen f verfährt man analog und erhält die andere Kramers-Kronig-Beziehung. Eine beliebige Funktion kann immer durch die Vorschrift \, f=f_+ + f_-, mit f_\pm (t) = \frac{1}{2}\left (f(t)\pm f(-t)\right ), in einen geraden bzw. ungeraden Anteil zerlegt werden.
Anwendungen[Bearbeiten]
Die Kramers-Kronig-Beziehungen finden dort Anwendung, wo eine reelle gerade Funktion zu einer holomorphen Funktion ergänzt werden soll, was meistens der Vereinfachung der auftretenden Rechnungen dient, insbesondere bei Wellenfunktionen, also hauptsächlich in der Signalverarbeitung und in der Optik, aber auch in der Statistischen Physik im Zusammenhang mit dem Fluktuations-Dissipations-Theorem. Auf diese Weise hängt die Absorption von elektromagnetischen Wellen in einem Medium mit dem Brechungsindex zusammen. Es reicht also, die Abhängigkeit einer der beiden Größen von der Frequenz zu kennen, um die andere berechnen zu können.
Die von der Kreisfrequenz abhängige Absorption lässt sich als Funktion einer von der Kreisfrequenz abhängigen Permittivität \varepsilon(\omega) ausdrücken:[1]
\operatorname{Re}(\varepsilon(\omega))=1+\frac{2}{\pi} \cdot \;\mathrm{CH}\, \int \limits_{0}^{\infty} {{\Omega \cdot \operatorname{Im}(\varepsilon(\Omega))} \over {\Omega^2-\omega^2}} \,\mathrm{d}\Omega
wobei \Omega die Kreisfrequenz als Integrationsvariable und \mathrm{CH} der cauchysche Hauptwert (engl. Cauchy principal value) des Integrals sind.
Eine alternative Betrachtungsweise ergibt sich mit dem Absorptionskoeffizienten \alpha, dem Brechungsindex n und der Lichtgeschwindigkeit c:
n(\omega)=1+\frac{c}{\pi} \cdot \;\mathrm{CH}\, \int \limits_{0}^{\infty} {{\alpha(\Omega)} \over {\Omega^2-\omega^2}} \,\mathrm{d}\Omega
Dadurch lässt sich vor allem in der nichtlinearen Optik aus einer einfachen Absorptionsmessung die komplexe Form des Brechungsindex ableiten.
Literatur[Bearbeiten]
Originalarbeiten:
• R. de L. Kronig: On the theory of dispersion of X-rays. In: Journal of the Optical Society of America. 12, Nr. 6, 1926, S. 547-556, doi:10.1364/JOSA.12.000547.
• H.A. Kramers: La diffusion de la lumiere par les atomes, In: 'Atti Cong. Intern. Fisici, (Transactions of Volta Centenary Congress) Como. Bd. 2, 1927, S. 545–557.
Weitere Literatur:
• Mansoor Sheik-Bahae: Nonlinear Optics Basics. Kramers-Krönig Relations in Nonlinear Optics. In: Robert D. Guenther (Hrsg.): Encyclopedia of Modern Optics. Academic Press, Amsterdam 2005, ISBN 0-12-227600-0, S. 234–240.
Einzelnachweise[Bearbeiten]
1. Safa Kasap, Peter Capper: Springer Handbook of Electronic and Photonic Materials. Springer, 2006, ISBN 9780387260594, S. 49. | {
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-8,167,911,554,242,513,000 | 1 Review of complex numbers
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1 1 Review of complex numbers 1.1 Complex numbers: algebra The set C of complex numbers is formed by adding a square root i of 1 to the set of real numbers: i = 1. Every complex number can be written uniquely as a + bi, where a and b are real numbers. We usually use a single letter such as z to denote the complex number a + bi. In this case a is the real part of z, written a = Re z, and b is the imaginary part of z, written b = Im z. The complex number z is real if z = Re z, or equivalently Im z = 0, and it is pure imaginary if z = (Im zi, or equivalently Re z = 0. In general a complex number is the sum of its real part and its imaginary part times i, and two complex numbers are equal if and only if they have the same real and imaginary parts. We add and multiply complex numbers in the obvious way: (a 1 + b 1 i + (a + b i = (a 1 + a + (b 1 + b i; (a 1 + b 1 i (a + b i = (a 1 a b 1 b + (a 1 b + a b 1 i. For example, ( + 3i( 5 + 4i = 7i. In this way, addition and multiplication are associative and commutative, multiplication distributes over addition, there is an additive identity 0 and additive inverses ( (a + bi = ( a + ( bi, and there is a multiplicative identity 1. Note also that Re(z 1 + z = Re z 1 + Re z, and similarly for the imaginary parts, but a corresponding statement does not hold for multiplication. Before we discuss multiplicative inverses, let us recall complex conjugation: the complex conjugate z of a complex number z = a + bi is by definition z = a bi. It is easy to see that: Re z = 1 (z + z; Im z = 1 (z z. i Thus z is real if and only if z = z and pure imaginary if and only if z = z. More importantly, we have the following formulas which can be checked by 1
2 direct calculation: z 1 + z = z 1 + z ; z 1 z = z 1 z ; z n = ( z n ; z = z; z z = a + b, where in the last line z = a + bi. Thus, z z 0, and z z = 0 if and only if z = 0. We set z = z z = a + b, the absolute value, length or modulus of z.for example, + i = 5. Note that, for all z 1, z C, z 1 z = z 1 z 1 z z = z 1 z z 1 z = (z 1 z (z 1 z = z 1 z, and hence z 1 z = z 1 z. The link between the absolute value and addition is somewhat weaker; there is only the triangle inequality z 1 + z z 1 + z. If z 0, then z has a multiplicative inverse: z 1 = z z. In terms of real and imaginary parts, this is the familiar procedure of dividing one complex number into another by rationalizing the denominator: if at least one of c, d is nonzero, then a + bi c + di = ( a + bi c + di ( c di c di = (a + bi(c di c + d. Thus it is possible to divide by any nonzero complex number. For example, to express ( + i/(3 i in the form a + bi, we write ( ( + i + i 3 + i 3 i = ( + i(3 + i = 3 i 3 + i 3 + = 4 + 7i = i. If z 0, then z n is defined for every integer n, including the case n < 0, and the formula z n = ( z n still holds.
3 1. Complex numbers: geometry Instead of thinking of a complex number z as a + bi, we can identify it with the point (a, b R. From this point of view, there is no difference between a complex number and a -vector, and we sometimes refer to C as the complex plane. The absolute value z is then the same as (a, b, the distance from the point (a, b to the origin. Addition of complex numbers then corresponds to vector addition. However, multiplication of complex numbers is more complicated. One way to understand it is to use polar coordinates: if z = a + bi, where (a, b corresponds to the polar coordinates (r, θ, then r = z and a = r cos θ, b = r sin θ. Thus we may write z = r cos θ + (r sin θi = r(cos θ + i sin θ. This is sometimes called the polar form of z; r = z is, as we have seen, called the modulus of z and θ is called the argument, sometimes written θ = arg z. Note that the argument is only well-defined up to an integer multiple of π. If z = r(cos θ + i sin θ, then clearly r is real and nonnegative and cos θ +i sin θ is a complex number of absolute value one; thus every complex number z is the product of a nonnegative real number times a complex number of absolute value 1. If z 0, then this product expression is unique. (What happens if z = 0? For example, the polar form of 1 + i is (cos(π/4 + i sin(π/4. Given two complex numbers z 1 and z, with z 1 = r 1 (cos θ 1 +i sin θ 1 and z = r (cos θ + i sin θ, we can ask for the polar form of z 1 z : z 1 z = r 1 (cos θ 1 + i sin θ 1 r (cos θ + i sin θ = r 1 r ((cos θ 1 cos θ sin θ 1 sin θ + i(cos θ 1 sin θ + cos θ sin θ 1 = r 1 r (cos(θ 1 + θ + i sin(θ 1 + θ, where we have used the standard addition formulas for sine and cosine. (We will see in a minute where these addition formulas come from. Thus the modulus of the product is the product of the moduli (this is just the formula z 1 z = z 1 z which we have already seen, but the really interesting formula is that the arguments add: arg(z 1 z = arg z 1 + arg z. Of course, this has to be understood as up to possibly adding an integer multiple of π. In this way, we can interpret geometrically the effect of multiplying by a complex number z. If z is real, multiplying by z is just ordinary scalar multiplication and has the usual geometric interpretation. If z = cos θ + i sin θ has absolute value one, then multiplying a complex number x + iy by 3
4 z is the same as rotating the point (x, y by the angle θ. For a general z, multiplying the complex number x + iy by z is a combination of these two operations: rotation by the angle θ followed by scalar multiplication by the nonnegative real number z. Using the formula for multiplication, it is easy to see that if z has polar form r(cos θ + i sin θ, then z n = r n (cos nθ + i sin nθ; z 1 = r 1 (cos( θ + i sin( θ = r 1 (cos θ sin θ. Here the first formula, which is easily proved by mathematical induction, holds for all z and positive integers n, and the second holds for z 0. From this it is easy to check that, for z 0, the first formula holds for all integers n. This formula is called De Moivre s Theorem. We can use De Moivre s Theorem to find powers and roots of complex numbers. For example, we have seen that 1 + i = (cos(π/4 + i sin(π/4. Thus (1 + i 0 = ( 0 (cos(0π/4 + i sin(0π/4 = 10 (cos(5π + i sin(5π = 10 ( 1 = 104. De Moivre s Theorem can be used to generate identities for sin nθ and cos nθ via the binomial theorem. For example, cos 3θ + i sin 3θ = (cos θ + i sin θ 3 = cos 3 θ + 3i cos θ sin θ 3 cos θ sin θ i sin 3 θ = cos 3 θ 3 cos θ sin θ + i(3 cos θ sin θ sin 3 θ. Equating real and imaginary parts, we see that cos 3θ = cos 3 θ 3 cos θ sin θ, and likewise sin 3θ = 3 cos θ sin θ sin 3 θ. It is more interesting to find roots. Let z = r(cos θ+i sin θ be a complex number, which we assume to be nonzero (since the only n th root of 0 is zero why?, and let w = s(cos ϕ + i sin ϕ. Then w n = z if and only if s n = r and nϕ = θ + kπ for some integer k. Thus s = r 1/n, and ϕ = θ/n + kπ/n for some integer k. But sometimes these numbers will be the same for different values of k: if ϕ 1 = θ/n + k 1 π/n and ϕ = θ/n + k π/n, then r 1/n (cos ϕ 1 + i sin ϕ 1 = r 1/n (cos ϕ + i sin ϕ if and only if ϕ 1 and ϕ differ by an integer multiple of π, if and only if k 1 π/n and k π/n differ by an integer multiple of π, if and only if n 4
5 divides k 1 k. Moreover, we can find a complete set of choices by taking the arguments θ/n, θ/n + π/n,..., θ/n + (n 1π/n. Thus we see: If n is a positive integer, then a nonzero complex number has exactly n distinct n th roots given by the formula above. Examples: the two square roots of i = cos(π/ + i sin(π/ are ( π ( π cos + i sin = i; ( π ( π cos 4 + π + i sin 4 + π = i. For another example, to find all of the fifth roots of 3 + i, first write ( i = + 1 ( i = cos π 6 + i sin π. 6 Thus the fifth roots are given by ( π (cos 1/ kπ 5 + i sin ( π 30 + kπ, k = 0, 1,, 3, 4. 5 We can apply the above to the complex number 1 = cos 0 + i sin 0. Thus there are exactly n complex numbers z such that z n = 1, called the n th roots of unity: namely ( kπ cos n + i sin ( kπ n, k = 0, 1,..., n 1. It is easy to see that, once we have found one n th root w of a nonzero complex number z, then all of the n th roots of z are of the form ( ( ( kπ kπ cos + i sin w, n n for k = 0, 1,..., n 1, i.e. any two n th roots of a given nonzero complex number differ by multiplying by an n th root of unity. Warning: the usual rules for fractional exponents that hold for positive real numbers do not usually hold for complex roots; this is connected with the fact that there is not in general one preferred n th root of a complex number. For example, 1 = i = 1 1 ( 1( 1 = 1 = 1. 5
6 1.3 Complex numbers: the complex exponential function Given a power series a nx n, we can try to substitute in complex values for x and see what we get. Here we shall just consider the usual power series for the exponential function e x x n =. We begin by substituting a purely n! imaginary complex number it, where t is real. This gives e it = i n t n n! = i n t n (n! + i n+1 t n+1 (n + 1!, where we have simply broken the sum up into summing over even and odd positive integers. Using i n = ( 1 n, and hence i n+1 = ( 1 n i, we see that the sum is equal to e it = ( 1 n t n (n! + i ( 1 n t n+1 (n + 1! = cos t + i sin t. This beautiful fact is known as Euler s formula. For example, e iπ = 1. We can thus write the polar form r(cos θ + i sin θ for a complex number as re iθ. Assuming the usual rules for exponents, we can see in another way that the arguments add under multiplication: r 1 e iθ 1 r e iθ = r 1 r e iθ 1 e iθ = r 1 r e iθ 1+iθ = r 1 r e i(θ 1+θ. In particular, e iθ 1 e iθ = e i(θ 1+θ. Equating real and imaginary parts, we see that this fact is equivalent to the usual addition formulas for the sine and cosine functions, and indeed is perhaps the best way of explaining why these somewhat mysterious looking addition formulas are true. Euler s formula is a fundamental link between the basic constants of mathematics, e and π, and between the exponential and trigonometric functions. For example, since e it = cos( t + i sin( t = cos t i sin t, we see that cos t = Re e it = 1 (eit + e it ; sin t = Im e it = 1 i (eit e it. Now suppose that we can substitute an arbitrary complex number z = x + iy in the expression for e x, and that the usual rules for exponentiation apply. Then e x+iy = e x e iy = e x (cos y + i sin y. 6
7 In particular, if r is a positive real number, then e ln r+iθ = r(cos θ + i sin θ. Thus every nonzero complex number z = r(cos θ + i sin θ has a logarithm. In fact the possible solutions to e w = z are w = ln r + (θ + nπi, n an integer. We call any of these values a logarithm of z and write w = log z. Of course, log is not a well-defined function. Note that, for real x, the exponential function e x is one-to-one and its values are the positive real numbers; hence ln x is defined for positive x. For complex z, the exponential function is no longer one-to-one: e z 1 = e z exactly when z = z + nπi for some integer n, and the values of the complex exponential are all nonzero complex numbers. Thus log z is defined for all z 0, but only up to adding an arbitrary integer multiple of πi. For example, log( 1 = log(e iπ = iπ + nπi = (n + 1πi, for any integer n. It is then natural to try to define an expression of the form z α, where z is a nonzero complex number and α is any complex number, by the formula z α = e α log z. Since log z is only well-defined up to adding an integer multiple of πi, this says that, given any particular choice of value, say w, for z α, then we nαπi is also a value for z α, for every integer n. If α = 1/k for some integer k, or more generally if α is a rational number, then there are only finitely many possible values for the expression z α ; for example, if α = 1/k, we just see the k different k th roots of z described earlier. But if α is not rational, then the expression z α will define infinitely many different complex numbers! For example, = log e = (ln +nπi e = e ln +n πi = e n πi = (cos(n π + i sin(n π, where, in the second line, the expression means the usual real valued expression e ln, and n is an arbitrary integer. For another example, i = e i log = e i(ln +nπi = e i ln nπ = e nπ (cos(ln + i sin(ln. 7
8 1.4 Homework 1. Write in the form a + bi: (a ( + i (3 + i; (b (1 + 4i( + 4i; (c ( 3i( + 3i.. Write in the form a + bi: 3. Write in polar form: (a + i 1 + 4i 3i ; (b ; (c 3 + i + 8i 3 + i. (a 1 3i; (b 5 + 5i; (c πi. 4. Write the complex number 3 i in polar form (using inverse trigonometric functions if necessary. 5. (i Write in the form a + bi: (a e πi/4 ; (b e 1+πi ; (c e 3+i. 6. Evaluate (in the form a + bi: ( 3 i 7 ; (1 + i Find all complex numbers z such that z 5 = i. (You can leave z in polar form. How many different ones are there? 8. Find all solutions in complex numbers z of the equation (z = z 5. (Note: you should find exactly four different solutions. 9. What are all possible values (in the form a + bi of the following expressions? (a log(1 + i (b i i (c i e (d (1 + i π. How many are real? Pure imaginary? What are all possible values of e i, interpreting e i as (a the value of the complex exponential function on i? (b as the complex number e raised to the power i? 10. Beginning with the formula cos t = 1 (eit + e it, find a formula for cos 1 x in terms of log and square roots. (Hint: let x = cos t and z = e it, so that e it = 1/z. Multiply both sides of the above formula by z and apply the quadratic formula to solve first for z and then for t. 8
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Results 1 to 2 of 2
Math Help - Translation of axes
1. #1
Member
Joined
Jan 2009
Posts
197
Translation of axes
Find the new equation of the circle of the equation x^2+y^2-4x+6y+9=0 after the translation that moves the origin to the point (2,-3)
Attempt
x=x-2
y=y+3
(x-2)^2 +(y+3)^2 -4(x-2)+6(y+3)+9=0
x^2+y^2-8x+12y+48=0
Am I correct?
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2. #2
Super Member
Joined
May 2006
From
Lexington, MA (USA)
Posts
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Thanks
842
Hello, mj.alawami!
Find the new equation of the circle of the equation x^2+y^2-4x+6y+9\:=\:0
after the translation that moves the origin to the point (2,-3).
Attempt: . \begin{array}{c}x\:=\:x-2 \\ y\:=\:y+3 \end{array}
(x-2)^2 +(y+3)^2 -4(x-2)+6(y+3)+9\:=\:0
x^2+y^2-8x+12y+48\:=\:0
Am I correct?
Yes! . . . Good work!
\begin{array}{cccccc}\text{The original circle is:} & (x-2)^2 + (y+2)^2 \:=\:4 &\Rightarrow& \text{Center: }(2,\,\text{-}3),\;r = 2 \\ \\[-3mm]<br />
\text{The new circle is:} & (x-4)^2 + (y+6)^2 \:=\: 4 &\Rightarrow& \text{Center: }(4,\,\text{-}6),\;r = 2 \end{array}
And this checks out . . .
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162,838,533,270,285,660 | Answers
Solutions by everydaycalculation.com
Answers.everydaycalculation.com » Compare fractions
Compare 10/5 and 21/9
1st number: 2 0/5, 2nd number: 2 3/9
10/5 is smaller than 21/9
Steps for comparing fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 5 and 9 is 45
2. For the 1st fraction, since 5 × 9 = 45,
10/5 = 10 × 9/5 × 9 = 90/45
3. Likewise, for the 2nd fraction, since 9 × 5 = 45,
21/9 = 21 × 5/9 × 5 = 105/45
4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction
5. 90/45 < 105/45 or 10/5 < 21/9
MathStep (Works offline)
Download our mobile app and learn to work with fractions in your own time:
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Related:
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2,812,631,707,506,987,000 | Vectors
Quadrants
Like a Star Trek quadrant?
First off: Nerd alert!
Secondly, yes, a quadrant is a circle cut into four parts.
What does this have to do with electricity?
Voltage and currents are constantly changing magnitude and direction. When changing direction, they actually rotate in a counterclockwise direction. They are tethered to a point of origin.
Figure 29. Quadrant point of origin
Each quadrant contains certain directions.
• Quadrant 1 has 0 to 90 degrees.
• Quadrant 2 has 90 to 180 degrees.
• Quadrant 3 has 180 to 270 degrees.
• Quadrant 4 has 270 to 360 degrees.
This is very important as it helps us to determine which vectors belong in which quadrant.
Polarity
It is also important to understand polarity when dealing with quadrants. A quadrant system is basically an X-Y graph. We use the point of origin as a reference point. On the X axis, anything to the right of the point of origin is positive and anything to the left is negative. On the Y axis, anything above the point of origin is positive and anything underneath it is negative. This means each quadrant has its own polarity, as shown in Figure 30.
Figure 30. Quadrant polarity
This too is extremely important when it comes to adding vectors.
Are you getting excited yet? This is all going to come together in one magical dance.
Video!
This video goes into greater detail about the specifics of all four quadrants.
Attributions
Quadrants and vectors. video by The Electric Academy is under a Creative Commons Attribution Licence.
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JEE Main Mathematics Important Chapters
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Last updated date: 05th Dec 2023
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JEE Main Mathematics: Important Chapters for Scoring High Marks
JEE Main Mathematics is a crucial section that requires focused preparation for success in the examination. Within the vast syllabus, there are certain chapters that carry significant weightage and are considered particularly important. These JEE Main Mathematics important chapters play a pivotal role in determining a candidate's overall score and rank. Mastering these chapters is essential for students aiming to secure a competitive rank in the JEE Main examination. In this article, we will explore the key JEE Main Mathematics important chapters that deserve special attention and discuss the benefits of prioritising them in the preparation journey.
Most Important Chapters for JEE Main Maths
The JEE Main Mathematics syllabus is made up of around 25 chapters. That is the most crucial aspect, yet it requires a lot of effort. The following are the most essential JEE Main 2023 Mathematics chapters:
• Complex Numbers And Quadratic Equations
• Circle, Conic Sections
• Integral Calculus
• Three Dimensional Geometry
• Vector Algebra
• Probability
• Trigonometry
• Permutations And Combinations
• Sequence And Series
Syllabus for JEE Main Maths Important Chapters 2024
The JEE Main Mathematics syllabus is vast, but certain chapters are considered crucial for scoring high marks. Here are the important chapters from the JEE Main Mathematics syllabus:
Most Important Chapters of Maths JEE Main 2024
Complex Numbers and Quadratic Equations
Complex numbers as ordered pairs of reals, Representation of complex numbers in the form a + ib and their representation in a plane, Argand diagram, algebra of complex number, modulus and argument (or amplitude) of a complex number, square root of a complex number, triangle inequality, Quadratic equations in real and complex number system and their solutions Relations between roots and co-efficient, nature of roots, the formation of quadratic equations with given roots.
Circle, Conic Sections
A standard form of equations of a circle, the general form of the equation of a circle, its radius and central, equation of a circle when the endpoints of a diameter are given, points of intersection of a line and a circle with the centre at the origin and condition for a line to be tangent to a circle, equation of the tangent, sections of conics, equations of conic sections (parabola, ellipse, and hyperbola) in standard forms, condition for Y = mx +c to be a tangent and point (s) of tangency.
Integral Calculus
Integral as an anti-derivative, Fundamental Integrals involving algebraic, trigonometric, exponential, and logarithms functions. Integrations by substitution, by parts, and by partial functions. Integration using trigonometric identities.
Evaluation of simple integrals of the type
$\int{\dfrac{dx}{x^2+a^2}}, \int{\dfrac{dx}{\sqrt{x^2 \pm a^2}}}, \int{\dfrac{dx}{a^2-x^2}}, \int{\dfrac{dx}{\sqrt{a^2-x^2}}}, \int{\dfrac{dx}{ax^2+bx+c}}, \int{\dfrac{dx}{\sqrt{ax^2+bx+c}}}, \int{\dfrac{(px+q)dx}{ax^2+bx+c}}, \int{\dfrac{(px+q)dx}{\sqrt{ax^2+bx+c}}}, \int{\sqrt{a^2 \pm x^2}dx}, \int{\sqrt{x^2-a^2}dx}$
Integral as limit of a sum. The fundamental theorem of calculus, properties of definite integrals. Evaluation of definite integrals, determining areas of the regions bounded by simple curves in standard form.
Three Dimensional Geometry
Coordinates of a point in space, the distance between two points, section formula, directions ratios, direction cosines, the angle between two intersecting lines. Skew lines, the shortest distance between them, and its equation. Equations of a line and a plane in different forms, the intersection of a line and a plane, and coplanar lines.
Vector Algebra
Vectors and scalars, the addition of vectors, components of a vector in two dimensions and three-dimensional space, scalar and vector products, scalar and vector triple product.
Probability
Probability of an event, addition and multiplication theorems of probability, Baye's theorem, probability distribution of a random variate, Bernoulli trials, and binomial distribution.
Trigonometry
Trigonometrical identities and equations, trigonometrical functions, inverse trigonometrical functions, and their properties, heights, and distance.
Permutations and Combinations
The fundamental principle of counting, permutation as an arrangement and combination as section, Meaning of P (n,r) and C (n,r), simple applications.
Sequence and Series
Arithmetic and Geometric progressions, insertion of arithmetic, geometric means between two given numbers, Relation between A.M and G.M sum up to n terms of special series; Sn, Sn2, Sn3. Arithmetico-Geometric progression.
Most Important Chapters of Class 11 Maths for JEE Mains: Marks Weightage
For JEE Main, the Class 11 Mathematics syllabus provides a strong foundation for the more advanced topics in Class 12. While the entire syllabus is essential, certain chapters carry more weightage in terms of marks and importance. Here are the most important chapters of Class 11 Maths for JEE Mains and their approximate marks weightage:
Chapters
No. of Questions
Weightage (%)
Limits, Continuity and Differentiability
3
12
Integral Calculus
3
12
Coordinate Geometry
3
12
Statistics and Probability
2
8
Matrices and Determinants
2
8
Vector Algebra
2
8
Three Dimensional Geometry
2
8
Complex numbers and Quadratic Equation
2
8
Sets, Relation and Function
1
4
Sequence and Series
1
4
Binomial Theorem and Its Application
1
4
Differential Equation
1
4
Differential Calculus
1
4
Permutation and Combinations
1
4
Trigonometry
1
4
Mathematical Reasoning
1
4
Statics and Dynamics
1
4
JEE Main Maths Important Chapter 2023-24: Benefits of Solving Papers With Vedantu
There are many advantages to solving Previous Years’ JEE Main papers.To begin, examining past years' question papers is the most effective technique to determine the test format. Second, it aids in determining which chapters are crucial. It also allows you to construct an effective JEE Main study schedule.
As a result, past years' questions are quite useful for applicants to practise with.
As a result, prior years' questions are extremely beneficial for applicants to practise, and there are additional grounds to think that.
1. It Reveals Your Strengths and Weaknesses.
You should thoroughly examine the past years' JEE Main question papers. Just mark the questions that you are familiar with. You'll learn which questions you know and which you don't by doing so. Candidates may then readily specify where they should focus their efforts and begin their hard work. Candidates can categorise the test as challenging, moderate, or easy by reviewing past years' JEE Main question papers. Following that, they might begin improved preparing procedures to get better results.
2. Enhances Your Time Management
Before taking any exam, time is an important consideration, and JEE Main is no different. As they begin tackling prior years' question papers, candidates must manage their time properly. One of the most effective techniques for students to enhance their time management abilities is to study past years' JEE Main question papers.
It is also regarded as an effective approach to practise and improve. It helps to increase both speed and accuracy over time. Good time management may yield excellent rewards and allow candidates to finish their assignments on time.
3.Improves Speed and Accuracy
Also, practising past years' examinations enhances pupils' speed and accuracy while dealing with various issues. Candidates might create shortcut tactics while attempting to solve problems from previous years' JEE Main examinations. These strategies can help students solve Math and Physics issues, particularly numerical problems.
4. Practice Makes One Perfect
Practice, as they say, makes perfect. Whatever that is practised on a regular basis will improve. Candidates that spend a significant amount of time practising previous year's questions are automatically likely to earn a high percentile. Also, practising prior years' JEE Main will show you how well prepared you are. As a consequence, applicants may now quickly assess their degree of preparedness. Furthermore, based on their assessment, candidates might focus on closing the preparation gap.
5. Increase Your Self- Motivation
Examine past years' question papers to increase your confidence. Because most competitive tests have a set question format, being acquainted with it can increase your confidence and prepare you for the actual exam. The JEE Main test necessitates a tremendous amount of preparation.
Many questions, for example, may be puzzling to applicants who have never taken similar ones before. You may find yourself in a difficult scenario if you go into the JEE Main test with only theoretical knowledge and no understanding of the question pattern. You may learn a lot about the test format by solving previous year's exams, which can boost your confidence in your real performance.
Conclusion
Focusing on the JEE Main Mathematics important chapters is a strategic approach to excel in the examination. These chapters carry significant weightage in terms of marks and are frequently asked in the paper. By dedicating ample time and effort to understand the concepts and solve practice problems from these chapters, students can enhance their overall performance and increase their chances of obtaining a competitive rank. However, it is important to remember that a well-rounded preparation covering the entire Mathematics syllabus is essential for success. Balancing thorough study of the important chapters with equal attention to other topics will ensure a comprehensive and effective preparation strategy for the JEE Main Mathematics section.
FAQs on JEE Main Mathematics Important Chapters
1: What are JEE Main Mathematics important chapters 2024?
JEE Main Mathematics important chapters are those topics that carry significant weightage in terms of marks and are frequently asked in the examination. These chapters are crucial for scoring high marks and securing a competitive rank in the JEE Main Mathematics section.
2: How many important chapters are there in JEE Main Mathematics?
There are several important chapters in JEE Main Mathematics, typically ranging from 10 to 15, depending on the year and the paper pattern.
3: Can I skip studying other chapters if I focus only on the important chapters for JEE Main 2024?
While focusing on important chapters is essential, it is not advisable to skip studying other chapters entirely. A balanced preparation covering the entire syllabus is necessary for a good overall score in JEE Main Mathematics.
4: How can I identify the important chapters for JEE Main Mathematics 2024?
The important chapters can be identified based on their historical weightage of marks in previous years' papers and the frequency of questions asked from these topics.
5: Should I start with the important chapters or cover the entire syllabus first for JEE Main 2024?
It is recommended to start with the important chapters first and then gradually move on to cover the entire syllabus. This approach ensures that you allocate sufficient time to master the high-weightage topics.
6: Are the important chapters the same every year for JEE Main Exam?
The important chapters may vary slightly from year to year, depending on the JEE Main examination pattern and trends. | {
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-4,149,961,035,303,303,000 | Functions ordered pairs
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Functions Introduction
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Functions ordered pairs Presentation Transcript
• 1. Aim: What is a function?Do Now: Place your contracts in the basket. Takeout a pen or pencil and a sheet of paper.You have 50 minutes to complete the diagnostic.This counts for Quiz 1. You will not be graded on correctness.Your grade will depend on how many questions you answerand how much work you show.
• 2. Functions as a set of ordered pairs.Domain: The set of all inputs in a function, or the set of x values.Range: The set of all outputs of a function, or the set of y values.DEFINITIONS:
• 3. A function is a relation in which eachelement of the domain is paired withexactly one element in the range.Example: {(1, 1), (2, 4), (3, 9), (4, 16)}Domain RangeMapping Diagram
• 4. Ordered pairs on the grid....Write the ordered pairs for the relationshown in the graph.What is the Domain? (x values)What is the Range? (y values)Is this relation a function?Domain Range
• 5. Is the relation {(1, 3), (4, 3), (-2, -4),(0, -5)}a function or not? Why or why not?Domain Range . .. .
• 6. Function or not a function?{(0, 1), (2, 2), (0, 3), (4, 5)}Domain RangeNot a function because the same value in the domain ispaired with two different values in the range.
• 7. Visually, a relation is a function if it passesthe vertical line test.When a vertical line is drawn if it only hits the graph ofa relation once, then the relation is a function.✔ ✖
• 8. Are these graphs of functions? | {
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