ࡱ> npklm%` o$bjbjٕ .(j@41LHt97.8޻޻޻޻$eh=޻޻޻޻޻, 0]$%hz07oooc0= 7t9t9t9db؛$t9t9t9؛><  Lecture Notes on Calculus Part 2 (Lectures 10 ,11,12) by Reinaldo Baretti Machn  HYPERLINK "http://www.geocities.com/serienumerica3" www.geocities.com/serienumerica3  HYPERLINK "http://www.geocities.com/serienumerica2" www.geocities.com/serienumerica2  HYPERLINK "http://www.geocities.com/serienumerica" www.geocities.com/serienumerica  HYPERLINK "mailto:reibaretti2004@yahoo.com" reibaretti2004@yahoo.com  HYPERLINK "http://usa.nedstatbasic.net/cgi-bin/viewstat?name=Boladura"  INCLUDEPICTURE "http://usa.nedstatbasic.net/cgi-bin/nedstat.gif?name=Boladura" \* MERGEFORMATINET  References: 1.  HYPERLINK "http://www.amazon.com/Elements-Calculus-Analytic-Geometry-George/dp/0201076640/ref=sr_1_14?ie=UTF8&s=books&qid=1202309043&sr=8-14" \o "Elements of Calculus and Analytic Geometry" Elements of Calculus and Analytic Geometry by George B. Thomas 2.  HYPERLINK "http://www.amazon.com/Essential-Calculus-Applications-Advanced-Mathematics/dp/0486660974/ref=sr_1_1?ie=UTF8&s=books&qid=1202309215&sr=8-1" \o "Essential Calculus with Applications (Dover Books on Advanced Mathematics)" Essential Calculus with Applications (Dover Books on Advanced Mathematics) by Richard A. Silverman Lecture 10. Applications Example 1. Periodic motion in one dimension . Let a particles position be given by x(t) = A cos(t) (x and A ~meters) ,  ~radians/s , t~s . data: A = 0.15 m  =2 /s . Find the velocity and acceleration of the particle as functions of time. Apply the chain rule with u= t = 2t , du/dt= 2 rad/s . v a" dx/dt = A (d (cos u)/du) du /dt = A (-sin(u) ) (2) v(t) = -.30 sin(t) ~m/s acceleration a" dv/dt = d 2x/dt2 = -.30 (d sin(u)/du)(du/dt) a(t)= -.60 cos(u) ~m/s2 Differentiation and plots using Matlab A=.15 ; w=2; tau=2*pi/w ; t= [0:tau/200:tau]; x=A*sin(w*t); plot(t,x),xlabel('t~s'),ylabel('x ~ m')  position x over one period tau syms A w t; x=A*sin(w*t); v=diff(x,t) accel=diff(x,t,2) v = A*cos(w*t)*w accel = -A*sin(w*t)*w^2 ***plot code for the velocity A=.15 ; w=2; tau=2*pi/w ; t= [0:tau/200:tau]; v =A*cos(w*t)*w ; plot(t,v),xlabel('t~s'),ylabel('v~m/s')  Velocity over one period tau acceleration code A=.15 ; w=2; tau=2*pi/w ; t= [0:tau/200:tau]; accel = -A*sin(w*t)*w^2; plot(t,accel),xlabel('t~s'),ylabel('accel~m/s^2')  acceleration over one period tau b) damped motion let x=A exp(-bt) sin (  t + /4) x and A ~meters) ,  ~radians/s , b ~1/s Example 2. A particle is falling and the velocity is given by v (t) = - A ( 1- exp(-bt) ) , v~ meters/sec ,A= 28m/s t~ seconds , b=.2/sec . Find the acceleration, plot v and a. a= dv/dt =-28 d(1- exp(-bt))/dt= -28 d( -exp(-bt) ) /dt = 28 exp(-bt) d(-bt)/dt = -b(28) exp(-bt) =-5.6 exp(-.2t) ~ m/s2 Using MATLAB syms t b; v=-28*(1-exp(-b*t)); a=diff(v,t) a = -28*b*exp(-b*t) Plot using Matlab A=28 ; b=.2 ;tf=3*(1/b); v=-A*(1-exp(-b*t)); accel = -28*b*exp(-b*t) ; t=(0:tf/100:tf); plot(t,v,t,accel) ,xlabel('t(sec)'),ylabel('v(m/s) and accel(m/s^2)')  Example 3 : Maxima and minima Let the trajectory of a particle thrown up vertically be given by y = -4.9t2 + 10t +2 ( y~ m , t ~s ) Find the maximum value of y .The first derivative is the speed dy/dt= v = -9.8t +10 ~m/s ,this corresponds to the speed . At t=0 , v=+10 m/s. See plots. Matlab code tf=2; t= [0:tf/200:tf]; y=-4.9*t.^2+10*t+2 ; v=-9.8*t+10; a= -9.8; plot(t,y,t,v),xlabel('t~s'),ylabel('y(m),v(m/s)');  A plot of y vs t easily reveals the maximum value and the approximate instant t when it occurs. The speed that starts at +10 m/s,decreases for an instant becomes zero at about tH"0 is zero . The particle starts to fall and v<0. The maximum value of y is about 7 meters. In the language of calculus; to find the maxima or minima of y(t) we take the first derivative , equate it to zero and solve for the independent variable , t in this case. From the above results dy/dt = -9.8t +10 equated to zero , gives 0=-9.8t +10 and t m = 10/9.8 = 1.02 seconds . Evaluate y(t= t m) = -4.9 (1.02)2 -10(1.02) +2 = 7.10m We can see that the second derivative is(negative/positive) at a (maximum/minimum). Consider the next figure. y1 is a maximum and y0 0 and (dy/dx)1 = (y2-y1)/dx <0 . The approximation to the second derivative is d2 y/dx2 ={ (dy/dx)1  (dy/dx)0 } /dx ={y2-2y1+y0}/dx2 or = { (y2-y1) + (y0-y1) }/ dx2 < 0 since both quantities in the inner parenteheses are negative. A similar argument would show that at a minima d2 y/dx2 > 0. Example 4: A balloon is increasing its volume V= (4 /3) R3 at the rate (dV/dt) = 1.00e-3 m3/s . At what rate is the radius r increasing. thakin the derivatives of (4 /3) R3 gives 1.00e-3 m3/s = (4 /3)3 R2 (dR/dt)= 4 R2 (dR/dt) dR/dt= 1.00e-3 /{4 R2 } Example 5:Sliding ladder.  A ladder of length 5 meters is sliding. Find a relation between the horizonatal speed vx and the speed vy of the vertical end. Star with x2 + y2 = 52 =25 m2 Take derivatives with respect to time t- using the implicit derivative method. We have 2 x (dx/dt) + 2y(dy/dt) =0 . Then , vy = (dy/dt) = -(x/y) (dx/dt) = -(x/y) vx =- (x/(25-x2)1/2) vx . Let vx =1.8m/s =constant . A plot of vy (x) is given now. % MATLAB code falling ladder vx= 1.8; x=[0.1:.1:4.9]; vy=-(x./(25-x.^2).^1/2)*vx; plot(x,vy),xlabel('x~m'),ylabel('vy~m/s')   Example 6: The variational principle. The sole electron in a hydrogenic atom with atomic number Z ,has in the ground state, the approximate probabilty function  (r) = ( 3 / a)1/2 exp(-a r). The total energy is a functional of the parameter  a- E = -(1/2) +"  "  4 r2 dr + +"  (-Z/r)  4 r2 dr . " is the Laplacian operator. The first term gives the average kinetic energy while the second ,proportional to -Z ,is the average potential energy. The result is (in atomic units (au) ; 1 au = 27.2 eV) E (a) = a2 / 2 - Z a . The minimum of E is sought. dE/da = a  Z = 0 , thus a0 = Z. So the minimum energy is E (a=Z) = Z2/2  Z2 = - Z2/ 2 (au) and ground state (r) = ( 3 / Z)1/2 exp(-Z r). END OF LECTURE Lecture 11. The Taylor series A function y(x) can be expanded about a point x0 in a power series of the difference (x-x0). y(x) = "n=0 cn (x-x0)n = c0 + c1(x-x0)1+c2 (x-x0)2+c3 (x-x0)3+& (1) The coefficients are ,cn = (1/n!) (dn y/dxn )x=x0 . It can be verified by taking succesive derivatives of (1) and evaluating at x=x0. Start with the first coefficient , let x=x0 , then y(x0) = c0 . Now take the first derivative of (1) dy/dx = 0 + c1 + 2 c2 (x-x0) + 3 c3 (x-x0)2 + , (2) evaluating at x=x0 gives c1 = (dy/dx)x0 . (3) Taking the derivative of (2), d2y/dx2 = 2 c2 + 2(3) c3 (x-x0) +.. and evaluating at x0 yields c2 = (1/2) (d2y/dx2)x0 . (4) Continuing this process shows that cn = (1/n!) (dn y/dxn )x0 . (5) Thus the Taylor series expansion about the point x0 is y(x) = y(x0) + "n=1 (x-x0)n (1/n!) (dn y/dxn )x0 (6) The derivatives may also be expressed in terms of a Taylor series. (dy/dx)x = (dy/dx)x0 + "n=1 (x-x0)n (1/n!) (dn+1y/dxn+1 )x0 (7) (d2 y/dx2 )x = (d2 y/dx2 )x0 + "n=1 (x-x0)n (1/n!) (dn+2y/dxn+2 )x0 (8) Notice that the summation terms in (7) and (8) are proportional to ( dn+1y/dxn+1 )x0 and (dn+2y/dxn+2 )x0 . Example 1: Let y=f(x)=sin(x) expand about x0= 0 . The derivatives are df /dx =f1= cos(x) , f2=-sin(x) , f3=-cos(x), f4= sin (x), f5= cos (x), f6= sin (x). At the origin (x0=0) their values are y(0)=0, f1= 1, , f2= 0 , f3= - 1, f4= 0, f5= 1 , f6= 0... Substitution in (6) gives the Taylor expansion of sin(x) about the origin sin(x) = x x3/3! + x5/5! - x7/7! +& (-1)n xn /(2n+1)! (8) Also , sin(t) =(t)  (t)3/3! + (t)5/5!  (t )7/7! +& (-1)n (t )2n+1 /(2n+1)! The following FORTRAN code calculates sin( /4) using n=10 . c sin(x) by Taylor series pi=2.*asin(1.) x=pi/4. fact=1. sum=x sign=1. n=10 do 10 i=3,n,2 sign=-1.*sign fact=fact*float(i)*float(i-1) sum=sum+sign*x**i/fact 10 continue print*,'x,sum,sin(x)=',x,sum,sin(x) stop end RUN x,sum,sin(x)= 0.785398185 0.707106769 0.707106769 The series result coincides up to nine digits with the exact value 1/ 21/2 . Example 2. Let y(x)=f(x)=cos(x) and x0=0. The derivatives are df /dx =f1= -sin(x) , f2=-cos(x) , f3=sin(x), f4= cos (x), f5= -sin (x), f6= -cos (x). At the origin (x0=0) their values are y(0)a"f0 =1, f1=0, , f2= -1 , f3= 0 , f4= 1, f5= 0 , f6= -1. The series is cos(x) = f0 + x*f1 +(x2/2) f2 + (x3/6) f3 + (x4/24) f4 + (x5/120)f5 + (x6/720) f6 +. cos(x)=1 - (x2/2) +(x4/24) - (x6/720) + (-1)n x2n/(2n)! Example 3. Let y(x)=f(x) = (1 + x ) 1/2 , x0 =0 . y(0)=1. and successive derivatives are ( see MATLAB CODE below) MATLAB CODE syms x ; y=(1+x)^(1/2) ; d1= simplify(diff(y,x)) d2=simplify(diff(y,x,2)) d3=simplify(diff(y,x,3)) x=0; d1=eval(d1) d2=eval(d2) d3=eval(d3) d1 = 1/2/(1+x)^(1/2) or dy/dx=(1/2) (1+x)-1/2 d2 =-1/4/(1+x)^(3/2) or d2 y/dx2 =(-1/4)(1+x)-3/2 d3 =3/8/(1+x)^(5/2) or d3 y/dx3 =(3/8)(1+x)-5/2 Evaluation at x0=0 gives (dy/dx)0= (1/2) , (d2 y/dx2)0= -1/4 , (d3 y/dx3)0= 3/8 ..The Taylor series is (1 + x ) 1/2 = 1 + (1/2)x (1/2)(1/4)x2 +(1/6)(3/8) x3 +. or in general (1+x)n = 1 + n x + (1/2!)n(n-1)x2 +(1/3!)(n(n-1)(n-2)x3 ... The same fromula applies if u=xp where p is a power of x. Then (1+u)n = 1 + n u + (1/2!)n(n-1)u2 +(1/3!)(n(n-1)(n-2)u3... Example : Find the square root of 52. Express 52 as a sum, (52)1/2 = ( 49 +3 )1/2 = 491/2(1+3/49)1/2 =7(1+x)1/2 where x=3/49. So (52)1/2 = 7(1+x)1/2 =7{ 1 + (1/2)(3/49) (1/2)(1/4)(3/49)2 +(1/6)(3/8) (3/49)3 } 7{ } =7 (1.03015792) = 7.21110535 From the calculator ,whose algorithm we dont know ,one obtains (52)1/2 = 7.211 102 551 Example 4: y(x)=f(x) = ln (1+x) The derivatives of the natural log are dy/dx = (1+x)-1 , d2 y/dx2 =(-1)(1+x)-2 , d3y/dx3 =(-1)(-2)(1+x)-3 . Evaluated at x0=0 , ln(1+0) =0 , dy/dx=1 , d2 y/dx2 =-1 , d3y/dx3 =(-1)(-2) The Taylor series is ln (1+x) = 0 + x +(1/2)(-1) x2 +(1/3!)(-1)(-2) x3 +(1/4!)(-1)(-2)(-3)x4 +... ln (1+x) = x (1/2) x2 +(1/3)x3 (1/4) x4 +...... -9 Notice that ln (1-x) = -x (1/2) x2 -(1/3)x3 (1/4) x4 -10 We can use the Taylor series expansion (9) ln (1+x) = x (1/2) x2 +(1/3)x3 (1/4) x4 +x5/5+ (where x <1), to generate both the natural log and log(base 10) logarithms. Assume we dont have calculators. For exp(x) use the Taylor power series. First we calculate some natural log between one and 10. Short log Table v0= 1 = = exp (0) ; ln (v0)=0 v1= 1.648721 = exp(.5) ; ln(v1)=0.5 v2= 2.718281 = exp(1) ; ln(v2)=1 v3= 4.481689 = exp(1.5) ; ln(v3)=1.5 v4=7.389056 = exp(2) ; ln(v4)= 2.0 Given an arbitrary number 1 d" v d" 10 the algorithm locates it between the vi or it could be between 10 and v4 . Let v3 < v < v4 . Then v= v3 (1+x) , where x<1. The natural log is ln(v) = ln(v3) + ln(1+x) = 1.5 + (x  (1/2) x2 +(1/3)x3 (1/4) x4 +x5/5+ ) From the natural log we can obtain the common log. The common log ( base 10) is given by log10 (v) = ln(v) /ln(10) . 10 = v4 ( 1+x) = 7.389056 (1+ .3533528 ). ln(10)= ln v4 + (x (1/2) x2 +(1/3)x3 (1/4) x4 +x5/5+ ) , where x= .3533528 . ln(10)= 2.302585 Table v,calcnatlog,log(v)= 0.100000E+01 0.000000E+00 0.000000E+00 callogbase10,log10(v)= 0.000000E+00 0.000000E+00 v,calcnatlog,log(v)= 0.200000E+01 0.693147E+00 0.693147E+00 callogbase10,log10(v)= 0.301030E+00 0.301030E+00 v,calcnatlog,log(v)= 0.300000E+01 0.109861E+01 0.109861E+01 callogbase10,log10(v)= 0.477121E+00 0.477121E+00 v,calcnatlog,log(v)= 0.400000E+01 0.138620E+01 0.138629E+01 callogbase10,log10(v)= 0.602021E+00 0.602060E+00 v,calcnatlog,log(v)= 0.500000E+01 0.160944E+01 0.160944E+01 callogbase10,log10(v)= 0.698970E+00 0.698970E+00 v,calcnatlog,log(v)= 0.600000E+01 0.179175E+01 0.179176E+01 callogbase10,log10(v)= 0.778149E+00 0.778151E+00 v,calcnatlog,log(v)= 0.700000E+01 0.194550E+01 0.194591E+01 callogbase10,log10(v)= 0.844919E+00 0.845098E+00 v,calcnatlog,log(v)= 0.800000E+01 0.207944E+01 0.207944E+01 callogbase10,log10(v)= 0.903090E+00 0.903090E+00 v,calcnatlog,log(v)= 0.900000E+01 0.219722E+01 0.219722E+01 callogbase10,log10(v)= 0.954243E+00 0.954243E+00 v,calcnatlog,log(v)= 0.100000E+02 0.230258E+01 0.230259E+01 callogbase10,log10(v)= 0.999997E+00 0.100000E+01 FORTRAN code c natural log table algorithm //7 significant digits dimension f(0:10) ,alogn(0:10) data f/1.,1.648721,2.718281,4.481689 ,7.389056,6*0.0/ data alogn /0.,0.5,1.,1.5,2. ,6*0.0/ data vi,vf,nstep/1.,10.,10/ dv=(vf-vi)/float(nstep) aln10=2.302585 do 10 i=0,nstep v=vi+dv*float(i) if(v.ge.f(0).and.v.le.f(1))then jn=0 q=v/f(0) endif if(v.ge.f(1).and.v.le.f(2))then jn=1 q=v/f(1) endif if(v.ge.f(2).and.v.le.f(3))then q=v/f(2) jn=2 endif if(v.ge.f(3).and.v.le.f(4))then jn=3 q=v/f(3) endif if(v.ge.f(4))then jn=4 q=v/f(4) endif x=q-1. c print*,'v,f(jn),q,x=',v,f(jn),q,x apxlog=x-x**2/2.+x**3/3.-x**4/4.+x**5/5.-x**6/6. $+x**7/7. -x**8/8. alogv=alogn(jn)+apxlog print 100,v,alogv,log(v) print 110, alogv/aln10 ,log10(v) print*, ' ' 10 continue 100 format(1x,'v,calcnatlog,log(v)=',3(2x,e13.6)) 110 format(1x,'callogbase10,log10(v)=',2(2x,e13.6)) stop end As for questions of convergence of the series see for example ,  HYPERLINK "http://en.wikipedia.org/wiki/Taylor_series" http://en.wikipedia.org/wiki/Taylor_series The Taylor series expansion assumes that the function y(x) is not infinite nor its derivatives in a region around x0. Intuitively if x0 = 0 , x has to be small. Or if y=(a+x)n one has to assume that x compared with a. A function like (1+x)1/2 is not defined in the real numbers realm if x<-1 This defines a radius of convergence if /x/ >1 the series will not converge. The question is fully dealt within the Theory of Functions using complex variables. Take ln(1+x) obviously x cannot be -1 because ln(0)=". A few important series converge for all values of x. Those are the cases of exp(x) series , sine(x) and cos(x) series. END OF LECTURE 11 Lecture 12. Differential equations A simple harmonic oscillator obeys the idealized equation of motion m (d2x/dt2 ) = - k x . (1) It reads; mass (kg) times acceleration (m/s2) equals minus the force constant (k~ newtons/meter ) times the displacement (x~ meters) . Our main purpose is to solve (1) numerically , subject to the specification of the initial postion x0 and initial velocity v0 . However we will take a glimpse at the solution using a standard analytical procedure Cast (1) in the form ( d2 x/dt2 )  (0)2 x = 0 , (0)2 = (k/m) ~radians/second (2) and introduce a generic solution x(t) = C exp ( i t) where i=(-1)1/2 is the imaginary root ,  ~ 1/time and C are constants to be determined. dx/dt = C exp ( i t) d(i t)/dt = i  C exp ( i t) = i  x(t) d2 x/dt2 = i  ( dx/dt) =( i ){ i  x(t)} = i2 2 x(t) = - 2 x(t) (3) Comparing the last result in (3) with the right hand side in (2) we can identify  = 0 = (k/m)1/2 . We could also choose x(t) = C exp (- i t) , i.e. with a negative sign in the exponential argument. After two derivatives the result is still eq.(2) Hence the general analytic solution of (2) is x(t) = C1 exp( i 0 t ) + C2 exp(- i 0 t ) . (4) We are still some steps away from applying the solution given by (4) because the two coefficients C1 , C2 are functions of the two initial conditions . Vice versa one could assign arbitrary values to C1 , C2 and they would determine the initial conditions x0 , v0 . A general result may be stated. A differential equation of the n-th order has n initial conditions or equivalently, its general solution contains n arbitary constants. Using the Taylor series of the exponential, and the cos(x) and sin(x) series one can show that ( i = (-1)1/2 ) exp(ix) = "2n (ix)2n /(2n)! + "(2n+1) (ix)2n+1 /(2n+1)! (4) = cos(x) + i sin(x) . (5) Using (5) in (4) one can also write another form of the general solution x(t) = A sin(0 t ) + B cos (0 t ) . (6) How are A and B related to the initial conditions ? x(0)=x 0 = A(0) + Bcos(0)= B ~ length~ e.g. meters (7) v(0) = v0 = 0 A cos(0) - 0 sin(0) A = v0 / 0 ~length ~ e.g. meters . (8) There are still other choices for the general solution. One choice is x(t) = D cos(0 t  ) .  is the initial phase angle. Expanding the expression x(t) = D sin ( ) sin(0t ) + D cos() cos( 0t) (9) and comparing with ( 6) gives the correspondence D sin ( ) = A = v0 / 0 ; D cos()= B= x 0 . From which follows that tan() = (v0 / ( x 0 0) ) or  = arctan { v0 / ( x 0 0 ) } ,(10-a) and D = ( x 02 + ( v0 / 0 )2 ) 1/2 . (10-b) After the digression we deal with the solution of the differential equation (d2x/dt2 ) = - (k/m) x = - (0)2 x ( 11) by Taylor series. It is best achieved by two series, one for the position and another for the velocity. In the generic Taylor series y(x) = y(x0) + "n=1 (x-x0)n (1/n!) (dn y/dxn )x0 ( 12 ) Let y(x)! x(t) , i.e. the independent variable is time ~t and the dependent variable is x in meters. 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The oscillation period is = (2)/0. In the Taylor series "t has to be chosen small relative to  ; "t << . In. eq ( 13 ) the terms III , IV and up to whatever order is desired are deduced from the original differential equation. In eq ( 14 ) the term II and onwards are deduced from the original differential equation. From ( 11 ) d3x/dt3 = - (0)2 (dx/dt) = - (0)2 v (15) and d4x/dt4 = - (0)2 (dv/dt)= - (0)2 (d2x/dt2 )= - (0)2 [- (0)2 x] = ( 0)4 x (16) Numerical example: Let k = 40N/m , mass= 0.500 kg , x0=.07 m , v0 = 0. We have 0 = ( k/m)1/2 = 8.94 rad/s , f = 0/(2)= 1.42 hertz The period is  = 1/f =0.704 seconds From 10-a and 10-b we get =0 rad , D = x 0=0.07 m The analytical answer is x(t)= .07 cos(0) , (17-a) v(t) = -.07 0 sin(0 t) . (17-b) The procedure of calculation by series finds x1 = x(0+"t) from the values of x0 , v0 , t =t 0 =0 . Then x2 =x( 0+2"t) is found from the series computed at x1 , v1 , t1 =0+ "t. The same procedure applies for v(t). The two series are recalculated at every point. A Fortran code is provided for the above example. "t a" (1./1000.) *  . The output is printed below where the analytic answer is compared to the numerical answer. FORTRAN CODE c harmonic oscillator /Taylor series for x(t) and for v(t) c SI units k~ newton/meter , m~ kg , x~ meters , v~ meters/s real k, m data k, m,x0,v0/40.,0.5,.07,0.0 / xa(t)=.07*cos(omega*t) va(t)=-.07*omega*sin(omega*t) d2x(x,v)=-(k/m)*x d3x(x,v)= - omega**2*v d4x(x,v)= omega**4*x pi=2.*asin(1.) period=2.*pi*sqrt(m/k) omega=2.*Pi/period dt=(1./1000.)*period nstep=int(1.5*period/dt) kp=int(float(nstep)/60.) kount=kp print*,'period ,dt,nstep=' ,period ,dt,nstep print*,' ' print 100,0.,x0,xa(0.),v0,va(0.) do 10 i=1, nstep t=dt*float(i) x1=x0+dt*v0+.5*dt**2*d2x(x0,v0)+(1./6.)*dt**3*d3x(x0,v0) v1= v0+dt*d2x(x0,v0)+.5*dt**2*d3x(x0,v0)+(1./6.)*dt**3*d4x(x0,v0) if(i.eq.kount)then print 100,t,x1,xa(t),v1,va(t) kount=kount+kp end if x0=x1 v0=v1 10 continue 100 format('t,x1,x(t),v1,v(t)=',5(1x,e11.4) ) stop end RUN period ,dt,nstep= 0.702481508 0.000702481542 1499 t,x1,x(t),v1,v(t)= 0.0000E+00 0.7000E-01 0.7000E-01 0.0000E+00 0.0000E+00 t,x1,x(t),v1,v(t)= 0.1686E-01 0.6921E-01 0.6921E-01 -0.9406E-01 -0.9406E-01 t,x1,x(t),v1,v(t)= 0.3372E-01 0.6684E-01 0.6684E-01 -0.1860E+00 -0.1860E+00 t,x1,x(t),v1,v(t)= 0.5058E-01 0.6296E-01 0.6296E-01 -0.2737E+00 -0.2737E+00 t,x1,x(t),v1,v(t)= 0.6744E-01 0.5765E-01 0.5765E-01 -0.3552E+00 -0.3552E+00 t,x1,x(t),v1,v(t)= 0.8430E-01 0.5103E-01 0.5103E-01 -0.4286E+00 -0.4286E+00 t,x1,x(t),v1,v(t)= 0.1012E+00 0.4325E-01 0.4325E-01 -0.4923E+00 -0.4923E+00 t,x1,x(t),v1,v(t)= 0.1180E+00 0.3449E-01 0.3449E-01 -0.5448E+00 -0.5448E+00 t,x1,x(t),v1,v(t)= 0.1349E+00 0.2495E-01 0.2495E-01 -0.5850E+00 -0.5850E+00 t,x1,x(t),v1,v(t)= 0.1517E+00 0.1484E-01 0.1484E-01 -0.6119E+00 -0.6119E+00 t,x1,x(t),v1,v(t)= 0.1686E+00 0.4395E-02 0.4395E-02 -0.6249E+00 -0.6249E+00 t,x1,x(t),v1,v(t)= 0.1855E+00 -0.6150E-02 -0.6150E-02 -0.6237E+00 -0.6237E+00 t,x1,x(t),v1,v(t)= 0.2023E+00 -0.1655E-01 -0.1655E-01 -0.6083E+00 -0.6083E+00 t,x1,x(t),v1,v(t)= 0.2192E+00 -0.2658E-01 -0.2658E-01 -0.5792E+00 -0.5792E+00 t,x1,x(t),v1,v(t)= 0.2360E+00 -0.3601E-01 -0.3601E-01 -0.5369E+00 -0.5369E+00 t,x1,x(t),v1,v(t)= 0.2529E+00 -0.4462E-01 -0.4462E-01 -0.4824E+00 -0.4824E+00 t,x1,x(t),v1,v(t)= 0.2698E+00 -0.5222E-01 -0.5222E-01 -0.4170E+00 -0.4170E+00 t,x1,x(t),v1,v(t)= 0.2866E+00 -0.5863E-01 -0.5863E-01 -0.3421E+00 -0.3421E+00 t,x1,x(t),v1,v(t)= 0.3035E+00 -0.6371E-01 -0.6371E-01 -0.2594E+00 -0.2594E+00 t,x1,x(t),v1,v(t)= 0.3203E+00 -0.6734E-01 -0.6734E-01 -0.1709E+00 -0.1709E+00 t,x1,x(t),v1,v(t)= 0.3372E+00 -0.6945E-01 -0.6945E-01 -0.7847E-01 -0.7847E-01 t,x1,x(t),v1,v(t)= 0.3541E+00 -0.6998E-01 -0.6998E-01 0.1573E-01 0.1573E-01 t,x1,x(t),v1,v(t)= 0.3709E+00 -0.6892E-01 -0.6892E-01 0.1096E+00 0.1096E+00 t,x1,x(t),v1,v(t)= 0.3878E+00 -0.6630E-01 -0.6630E-01 0.2009E+00 0.2009E+00 t,x1,x(t),v1,v(t)= 0.4046E+00 -0.6217E-01 -0.6217E-01 0.2877E+00 0.2877E+00 t,x1,x(t),v1,v(t)= 0.4215E+00 -0.5663E-01 -0.5663E-01 0.3680E+00 0.3680E+00 t,x1,x(t),v1,v(t)= 0.4383E+00 -0.4981E-01 -0.4981E-01 0.4399E+00 0.4399E+00 t,x1,x(t),v1,v(t)= 0.4552E+00 -0.4185E-01 -0.4185E-01 0.5019E+00 0.5019E+00 t,x1,x(t),v1,v(t)= 0.4721E+00 -0.3295E-01 -0.3295E-01 0.5524E+00 0.5524E+00 t,x1,x(t),v1,v(t)= 0.4889E+00 -0.2330E-01 -0.2330E-01 0.5904E+00 0.5904E+00 t,x1,x(t),v1,v(t)= 0.5058E+00 -0.1312E-01 -0.1312E-01 0.6150E+00 0.6150E+00 t,x1,x(t),v1,v(t)= 0.5226E+00 -0.2638E-02 -0.2638E-02 0.6257E+00 0.6257E+00 t,x1,x(t),v1,v(t)= 0.5395E+00 0.7900E-02 0.7900E-02 0.6221E+00 0.6221E+00 t,x1,x(t),v1,v(t)= 0.5564E+00 0.1826E-01 0.1826E-01 0.6044E+00 0.6044E+00 t,x1,x(t),v1,v(t)= 0.5732E+00 0.2820E-01 0.2820E-01 0.5730E+00 0.5730E+00 t,x1,x(t),v1,v(t)= 0.5901E+00 0.3751E-01 0.3751E-01 0.5286E+00 0.5286E+00 t,x1,x(t),v1,v(t)= 0.6069E+00 0.4596E-01 0.4596E-01 0.4722E+00 0.4722E+00 t,x1,x(t),v1,v(t)= 0.6238E+00 0.5337E-01 0.5337E-01 0.4051E+00 0.4051E+00 t,x1,x(t),v1,v(t)= 0.6407E+00 0.5957E-01 0.5957E-01 0.3288E+00 0.3288E+00 t,x1,x(t),v1,v(t)= 0.6575E+00 0.6442E-01 0.6442E-01 0.2450E+00 0.2450E+00 t,x1,x(t),v1,v(t)= 0.6744E+00 0.6780E-01 0.6780E-01 0.1557E+00 0.1557E+00 t,x1,x(t),v1,v(t)= 0.6912E+00 0.6965E-01 0.6965E-01 0.6284E-01 0.6284E-01 t,x1,x(t),v1,v(t)= 0.7081E+00 0.6991E-01 0.6991E-01 -0.3146E-01 -0.3146E-01 t,x1,x(t),v1,v(t)= 0.7250E+00 0.6859E-01 0.6859E-01 -0.1250E+00 -0.1250E+00 t,x1,x(t),v1,v(t)= 0.7418E+00 0.6571E-01 0.6571E-01 -0.2158E+00 -0.2158E+00 t,x1,x(t),v1,v(t)= 0.7587E+00 0.6134E-01 0.6134E-01 -0.3016E+00 -0.3016E+00 t,x1,x(t),v1,v(t)= 0.7755E+00 0.5558E-01 0.5558E-01 -0.3806E+00 -0.3806E+00 t,x1,x(t),v1,v(t)= 0.7924E+00 0.4856E-01 0.4856E-01 -0.4510E+00 -0.4510E+00 t,x1,x(t),v1,v(t)= 0.8093E+00 0.4043E-01 0.4043E-01 -0.5111E+00 -0.5111E+00 t,x1,x(t),v1,v(t)= 0.8261E+00 0.3139E-01 0.3139E-01 -0.5596E+00 -0.5596E+00 t,x1,x(t),v1,v(t)= 0.8430E+00 0.2163E-01 0.2163E-01 -0.5955E+00 -0.5955E+00 FH`bzRTZ\^gd@gd) "$&(*DFHJLRT\^lnprtx~02Tfh|~޹°ʨ }hpCJH*aJhpCJH*aJhpCJaJh{CJH*aJh5~CJaJh{CJaJh)CJH*aJh)CJH*aJh)CJaJhnCJaJh)h)CJH*aJhQh)CJaJh)h)CJH*aJh)h)CJaJ1RT"$&ԹԹԀtiZh(hT4@CJaJmH PsH 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