Elusoft - Mathematical functions
(degree) €arccos €arcus cosinus (degree) €arctan €arcus tangens (degree) €round €round €floor €rounding to next integer€ €ceil €round up to next integer €sqrt €Square root €sin €sinus (degree) €cos €cosinus (degree) €tan €tangens (degree) €sqr €to the power of two €log €decade logarithm €exp
Algebra Trigonometry
cosine cosinus cotangent cotangente degree of a polynomial degré d'un polynôme NYS Language RBE‐RN at the Steinhardt NYU Metro Center March 2014 6 English
Knudsen’s Cosine Law and Random Billiards
Knudsen’s Cosine Law and Random Billiards R Feres∗ and G Yablonsky† Abstract This article lays the groundwork for understanding the details of gas transport in the so-called Knudsen regime, particularly during the early,
Trig Cheat Sheet - Lamar University
©2005 Paul Dawkins Trig Cheat Sheet Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 p
Le cercle trigonom etrique 1 Le cercle trigonom etrique et la
3 Cosinus et sinus : On appelle cercle trigonom etrique dans le plan muni d’un rep ere orthonorm e O;~i;~j le cercle de centre O et de rayon 1 A tout nombre r eel ton associe le poiny M(t) du cercle tel que tsoit une mesure en radian de l’angle ~i;OM~ On d e nit ainsi le sinus et le cosinus : cos(t) : l’abscisse de M sin(t) : l
Commonly Used Taylor Series
Math 142 Taylor/Maclaurin Polynomials and Series Prof Girardi Fix an interval I in the real line (e g , I might be ( 17;19)) and let x 0 be a point in I, i e , x 0 2I : Next consider a function, whose domain is I,
Exercice 1 f - Les MathémaToqués
♠ Exercice 5 Donner la valeur exacte du cosinus et du sinus des nombres réels suivants Nombre réel x − π 3 5π 53π 3 π 6 − 5π 2 77π 6 cosx sinx ♠ Exercice 6 Sachant que sinx= 3 5 et que π
Technical data sheet EncoderAnalyzer Rev12 us
Cosinus = cosinus output of the encoder the cosinus output is 90 degree phase shift to the sinus output /Cosinus = inverted cosinus output of the encoder the cosinus output is 90 degree phase shift to the sinus output U = commutation signal U at block commutation /U = inverted commutation signal U at block commutation
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Commonly Used Taylor Series
series when is valid/true 11x= 1 +x+x2+x3+x4+:::note this is the geometric series.
just think ofxasr= 1X n=0x nx2(1;1)e x= 1 +x+x22! +x33! +x44! +:::so: e= 1 + 1 +12! +13! +14! e (17x)=P1 n=0(17x)nn!=1X n=017 nxnn!= 1X n=0x nn!x2Rcosx= 1x22! +x44! x66! +x88! :::notey= cosxis an evenfunction (i.e.,cos(x) = +cos(x)) and the taylor seris ofy= cosxhas only evenpowers. 1X n=0(1)nx2n(2n)!x2Rsinx=xx33! +x55! x77! +x99! :::notey= sinxis an oddfunction (i.e.,sin(x) =sin(x)) and the taylor seris ofy= sinxhas only oddpowers. 1X n=1(1)(n1)x2n1(2n1)!or=1X n=0(1)nx2n+1(2n+ 1)!x2Rln(1 +x) =xx22 +x33 x44 +x55 :::question: isy= ln(1 +x)even, odd, or neither?= 1X n=1(1)(n1)xnn or=1X n=1(1)n+1xnn x2(1;1]tan1x=xx33
+x55 x77 +x99 :::question: isy= arctan(x)even, odd, or neither?= 1X n=1(1)(n1)x2n12n1or=1X n=0(1)nx2n+12n+ 1x2[1;1]1Math 142Taylor/Maclaurin Polynomials and SeriesProf. GirardiFix an intervalIin the real line (e.g.,Imight be (17;19)) and letx0be a point inI, i.e.,
x 02I :Next consider a function, whose domain isI,
f:I!R and whose derivativesf(n):I!Rexist on the intervalIforn= 1;2;3;:::;N. Denition 1.TheNth-order Taylor polynomialfory=f(x) atx0is: pN(x) =f(x0) +f0(x0)(xx0) +f00(x0)2!
(xx0)2++f(N)(x0)N!(xx0)N;(open form) which can also be written as (recall that 0! = 1) pN(x) =f(0)(x0)0!
+f(1)(x0)1! (xx0)+f(2)(x0)2! (xx0)2++f(N)(x0)N!(xx0)N -a nite sum, i.e. the sum stops:Formula (open form) is in open form. It can also be written in closed form, by using sigma notation, as
pN(x) =NX
n=0f (n)(x0)n!(xx0)n:(closed form) Soy=pN(x) is a polynomial of degree at mostNand it has the form pN(x) =NX
n=0c n(xx0)nwhere the constantsc n=f(n)(x0)n! are specially chosen so that derivatives match up atx0, i.e. the constantscn's are chosen so that: pN(x0) =f(x0)
p (1)N(x0) =f(1)(x0)
p (2)N(x0) =f(2)(x0)
p (N)N(x0) =f(N)(x0):
The constantcnis thenthTaylor coecientofy=f(x) aboutx0. TheNth-order Maclaurin polynomialfory=f(x) is just theNth-order Taylor polynomial fory=f(x) atx0= 0 and so it is pN(x) =NX
n=0f (n)(0)n!xn:Denition 2.
1TheTaylor seriesfory=f(x) atx0is the power series:
P1(x) =f(x0) +f0(x0)(xx0) +f00(x0)2!
(xx0)2++f(n)(x0)n!(xx0)n+:::(open form) which can also be written as P1(x) =f(0)(x0)0!
+f(1)(x0)1! (xx0)+f(2)(x0)2! (xx0)2++f(n)(x0)n!(xx0)n+::: -the sum keeps on going and going: The Taylor series can also be written in closed form, by using sigma notation, as P1(x) =1X
n=0f (n)(x0)n!(xx0)n:(closed form) TheMaclaurin seriesfory=f(x) is just the Taylor series fory=f(x) atx0= 0.1Here we are assuming that the derivativesy=f(n)(x) exist for eachxin the intervalIand for eachn2N f1;2;3;4;5;:::g.
2 Big Questions 3.For what values ofxdoes the power (a.k.a. Taylor) series P1(x) =1X
n=0f (n)(x0)n!(xx0)n(1)converge (usually the Root or Ratio test helps us out with this question). If the power/Taylor series in formula (1)
does indeed converge at a pointx, does the series converge to what we would want it to converge to, i.e., does
f(x) =P1(x) ? (2)Question (2) is going to take some thought.
Denition 4.TheNth-order Remainder termfory=f(x) atx0is: RN(x)def=f(x)PN(x)
wherey=PN(x) is theNth-order Taylor polynomial fory=f(x) atx0. So f(x) =PN(x) +RN(x) (3) that is f(x)PN(x) within an error ofRN(x):We often think of all this as:
f(x)NX n=0f (n)(x0)n!(xx0)n -a nite sum, the sum stops atN :We would LIKE TO HAVE THAT
f(x)??=1X n=0f (n)(x0)n!(xx0)n -the sum keeps on going and going:In other notation:
f(x)PN(x) and the question isf(x)??=P1(x) wherey=P1(x) is the Taylor series ofy=f(x) atx0. Well, let's think about what needs to be forf(x)??=P1(x), i.e., forfto equal to its Taylor series. Notice 5.Taking the limN!1of both sides in equation (3), we see that f(x) =1X n=0f (n)(x0)n!(xx0)n -the sum keeps on going and going: if and only if limN!1RN(x) = 0: Recall 6.limN!1RN(x) = 0 if and only if limN!1jRN(x)j= 0 .So 7.If
limN!1jRN(x)j= 0 (4) then f(x) =1X n=0f (n)(x0)n!(xx0)n:So we basically want to show that (4) holds true.How to do this? Well, this is where Mr. Taylor comes to the rescue!2
2According to Mr. Taylor, his Remainder Theorem (see next page) was motivated by coeehouse conversations about works of Newton
on planetary motion and works of Halley (ofHalley's comet) on roots of polynomials. 3Taylor's Remainder Theorem
Version 1: for a xed pointx2Iand a xedN2N.3
There existscbetweenxandx0so that
R N(x)def=f(x)PN(x)theorem=f(N+1)(c)(N+ 1)!(xx0)(N+1):(5) So eitherxcx0orx0cx. So we do not know exactly whatcis but atleast we know thatcis betweenxandx0 and soc2I.Remark: This is aBig Theoremby Taylor. See the book for the proof. The proof uses the Mean Value Theorem.