Taller serie de taylor
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This document contains the work of a group - Lizeth Paola Barrero Riaño, Viviana Marcela Bayona Cardenas, Marcela Lagos Colmenares, Maria Fernanda Vergara Mendoza, Astrid Xiomara Rodríguez Castelblanco, and Victor Hugo Rondon Cordero - on several exercises involving Taylor series approximations. The exercises include determining the Taylor polynomial for a given function, approximating trigonometric functions using McLaurin polynomials, comparing Taylor series approximations to functions, and demonstrating finite difference expressions from Taylor series.
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(4) Parabola theory Module.pdf
A parabola is the locus of a point which moves in such a way that its distance from a fixed point is equal to its perpendicular distance from a fixed straight line.
1.1 Focus : The fixed point is called the focus of the Parabola.
1.2 Directrix : The fixed line is called the directrix of the Parabola.
(focus)
2.3 Vertex : The point of intersection of a parabola and its axis is called the vertex of the Parabola.
NOTE: The vertex is the middle point of the focus and the point of intersection of axis and directrix
2.4 Focal Length (Focal distance) : The distance of any point P (x, y) on the parabola from the focus is called the focal length. i.e.
The focal distance of P = the perpendicular distance of the point P from the directrix.
2.5 Double ordinate : The chord which is perpendicular to the axis of Parabola or parallel to Directrix is called double ordinate of the Parabola.
2.6 Focal chord : Any chord of the parabola passing through the focus is called Focal chord.
2.7 Latus Rectum : If a double ordinate passes through the focus of parabola then it is called as latus rectum.
2.7.1 Length of latus rectum :
The length of the latus rectum = 2 x perpendicular distance of focus from the directrix.
2.1 Eccentricity : If P be a point on the parabola and PM and PS are the distances from the directrix and focus S respectively then the ratio PS/PM is called the eccentricity of the Parabola which is denoted by e.
Note: By the definition for the parabola e = 1.
If e > 1 Hyperbola, e = 0 circle, e < 1
ellipse
2.2 Axis : A straight line passes through the focus and perpendicular to the directrix is called the axis of parabola.
If we take vertex as the origin, axis as x- axis and distance between vertex and focus as 'a' then equation of the parabola in the simplest form will be-
y2 = 4ax
3.1 Parameters of the Parabola y2 = 4ax
(i) Vertex A (0, 0)
(ii) Focus S (a, 0)
(iii) Directrix x + a = 0
(iv) Axis y = 0 or x– axis
(v) Equation of Latus Rectum x = a
(vi) Length of L.R. 4a
(vii) Ends of L.R. (a, 2a), (a, – 2a)
(viii) The focal distance sum of abscissa of the point and distance between vertex and L.R.
(ix) If length of any double ordinate of parabola
y2 = 4ax is 2 𝑙 then coordinates of end points of this Double ordinate are
𝑙2 𝑙2
, 𝑙
and
, 𝑙 .
4a
4a
3.2 Other standard Parabola :
Equation of Parabola Vertex Axis Focus Directrix Equation of Latus rectum Length of Latus rectum
y2 = 4ax (0, 0) y = 0 (a, 0) x = –a x = a 4a
y2 = – 4ax (0, 0) y = 0 (–a, 0) x = a x = –a 4a
x2 = 4ay (0, 0) x = 0 (0, a) y = a y = a 4a
x2 = – 4ay (0, 0) x = 0 (0, –a) y = a y = –a 4a
Standard form of an equation of Parabola
Ex.1 If focus of a parabola is (3,–4) and directrix is x + y – 2 = 0, then its vertex is (A) (4/15, – 4/13)
(B) (–13/4, –15/4)
(C) (15/2, – 13/2)
(D) (15/4, – 13/4)
Sol. First we find the equation of axis of parabola
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1) Write the characteristic equation: det(A - λI) = 0
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A parabola is the locus of a point which moves in such a way that its distance from a fixed point is equal to its perpendicular distance from a fixed straight line.
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(focus)
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The focal distance of P = the perpendicular distance of the point P from the directrix.
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2.6 Focal chord : Any chord of the parabola passing through the focus is called Focal chord.
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The length of the latus rectum = 2 x perpendicular distance of focus from the directrix.
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ellipse
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y2 = 4ax
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(ii) Focus S (a, 0)
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(v) Equation of Latus Rectum x = a
(vi) Length of L.R. 4a
(vii) Ends of L.R. (a, 2a), (a, – 2a)
(viii) The focal distance sum of abscissa of the point and distance between vertex and L.R.
(ix) If length of any double ordinate of parabola
y2 = 4ax is 2 𝑙 then coordinates of end points of this Double ordinate are
𝑙2 𝑙2
, 𝑙
and
, 𝑙 .
4a
4a
3.2 Other standard Parabola :
Equation of Parabola Vertex Axis Focus Directrix Equation of Latus rectum Length of Latus rectum
y2 = 4ax (0, 0) y = 0 (a, 0) x = –a x = a 4a
y2 = – 4ax (0, 0) y = 0 (–a, 0) x = a x = –a 4a
x2 = 4ay (0, 0) x = 0 (0, a) y = a y = a 4a
x2 = – 4ay (0, 0) x = 0 (0, –a) y = a y = –a 4a
Standard form of an equation of Parabola
Ex.1 If focus of a parabola is (3,–4) and directrix is x + y – 2 = 0, then its vertex is (A) (4/15, – 4/13)
(B) (–13/4, –15/4)
(C) (15/2, – 13/2)
(D) (15/4, – 13/4)
Sol. First we find the equation of axis of parabola
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On Application of Power Series Solution of Bessel Problems to the Problems of...
On Application of Power Series Solution of Bessel Problems to the Problems of...BRNSS Publication Hub
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Nbhm m. a. and m.sc. scholarship test september 20, 2014 with answer key
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SERIES SOLUTION OF ORDINARY DIFFERENTIALL EQUATION
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Taller serie de taylor
- 1. MEMBERSGROUP 4<br />Lizeth Paola Barrero Riaño<br />Viviana Marcela Bayona Cardenas<br />Marcela Lagos Colmenares<br />Maria Fernanda Vergara Mendoza<br />Astrid Xiomara Rodríguez Castelblanco.<br />Victor Hugo Rondon Cordero<br />WORKSHOP<br />1. Determine the nth Taylor polynomial centered on c:<br />Taylor Series<br /> n = 4 , c = 1 = xi<br />Solutión<br />We found each of the derivatives of the given function and evaluated at xi = 1:<br />We replace these values in the Taylor series and resolve to find the polynomial:<br />, n = 4 , c = 1 = xi<br />Solutión<br />We found each of the derivatives of the given function and evaluated at xi = 1<br />Replacing in the Taylor series f (x) and its derivatives we have the following polynomial<br />2. For f (x) = arcsin (x)<br />a) Write the polynomial MclaurinP3 (x) for f (x)<br />Solutión. <br />We found each of the derivatives of the given function and evaluated at xi = 0<br />Series replace the McLaurin<br />Reducing the above terms, McLaurin polynomial for f (x) = arcsin (x) is<br />c) Complete the following table to P3 (x) and f (x) (Use radians).<br />The error calculations are made for each of the values of x from Table 1:<br />X-0,75-0,5-0,2500,250,50,75f(x)-0,8481-0,5236-0,252700,25270,52360,8481P3(x)-0,8203-0,5208-0,252600,25260,52080,8203%E3,2780,53480,0395700,039570,53483,278<br /> TABLA 1<br />c. Draw your graphs on the same coordinate axes.<br />3. Confirm the following inequality with the help of the calculator and fill in the table to confirm numerically<br />Sn being the series which approximates the f (x) given<br />Solutión<br />Approximate the function by McLaurin series, where xi = 0<br />• For S2 the polynomial is:<br />• For S3 the polynomial is:<br />In (x+1)<br />S2<br />S3<br />These calculations are made for each of the values of x in Table 2:<br />x0,00,20,40,60,81S200,18000,32000,42000,48000,5000In (x+1)00,18230,33640,47000,58770,6931S300,18260,34130,49200,65060,8333<br />Plot and analyze results<br />According to the results it is clear that the function In (x +1) is greater than S3 but less than S2, so the inequality is not correct.<br />4. From the Taylor series to demonstrate the finite difference expressions down and centered finite difference.<br />Taylor Series:<br />a. Backward finite difference:<br />With the truncated Taylor series in the second term, obtaining:<br />Solving the first derivative<br />(1)<br />a. Centered finite difference:To obtain the equation is required to have the series of finite difference and finite difference down progressively.Taylor series for progressive finite and differences:<br />With the truncated Taylor series in the second term, obtaining:<br />Finite difference equation for progressive, clearing the first derivative<br /> QUOTE (2)<br />Subtracting equation (1) (2), we obtain:<br />Solving the first derivative<br />5. Using the terms of the Taylor series, approximate the function f(X)=cos(x) en x0=π/3 based on the value of the function f and its derivatives at the point x1=π/4. Start with only the term n = 0 by adding successively a term until the percentage error is less than the tolerance, taking four significant figures.<br />Solutión:<br />Tolerance<br />According to the statement:<br />• We find the derivatives of the function:<br />• Replacing have<br />Now start adding term by term<br />• Next term:<br />• Next term:<br />2927985231775<br />Continue adding terms until as shown in the table:<br />TermsResultεa(%)10.707120.522035.4630.49784.8640,49990.4250,50000.0260.50000<br />The approximation using the sixth term of the series meets the tolerance required.<br />