mm-2489 === Subject: Really Hard Problems Alright so my calculus teacher likes to give us amazingly complex problems without any hints on how to complete them. I really need help so anything you have to offer will be great. The problems are as follows. 1. Find the points P and Q on the parabola y = 1-x^2 so that triangle ABC formed by the x-axis and the tangent lines at P and Q is an equilateral triangle. 2. Find the two points on the curve y = x^4 - 2x^2 - x that have a common tangent line. 3. If g(x) = f(b+mx) + f(b-mx) where f is differentiable at b, find g'(0). THANK YOU!! === Subject: Re: Really Hard Problems >Alright so my calculus teacher likes to give us amazingly complex >problems without any hints on how to complete them. I really need help >so anything you have to offer will be great. The problems are as >follows. >1. Find the points P and Q on the parabola y = 1-x^2 so that triangle >ABC formed by the x-axis and the tangent lines at P and Q is an >equilateral triangle. Make use of symmetry here. The two tangent lines will strike the graph at the same y-values, but the x-values will be symmetric about the y-axis. That is, if one strikes the graph at the point (x, y), then the other strikes the graph at ( - x, y). Furthermore, the tangent lines will intersect on the y-axis and make angles of 30 degrees with the y-axis. You can use this information to get the slopes of the lines, and then get the points on the graph where the lines strike the graph. >2. Find the two points on the curve y = x^4 - 2x^2 - x that have a >common tangent line. With a grapher, I saw two points that were good candidates for a common tangent line. Hopefully, by looking at the graph, you will see these two points as well. My intuition turned out to be correct when I computed the tangent lines for each of these points. >3. If g(x) = f(b+mx) + f(b-mx) where f is differentiable at b, find >g'(0). Differentiate the function g (making use of the chain rule), and then plug in x = 0. Brian >THANK YOU!! === Subject: Re: Really Hard Problems <9jhgl19t9lsc6dvndtctd1hpnpv1omu1me@4ax.com> And I also tried to do number 2 analytically but I got stuck. I used two points (x1, y1) and (x2, y2). I knew that the derivates had to equal so I set up an equation using the derivative... 4(x1)^3 - 4(x1) - 1 = 4(x2)^3 - 4(x2) - 1 I got it to... (x1)^3 - x1 = (x2)^3 - x2 I dont know where to go from there. === Subject: Re: Really Hard Problems <9jhgl19t9lsc6dvndtctd1hpnpv1omu1me@4ax.com> Wow, that seemed too easy for number two. I tried what you said and got (-1, 0) and (1, -2). Is that correct? === Subject: Integration question How would you integrate xe^(2x)? What steps would you follow to come up with the answer? === Subject: Re: Integration question Integrate by parts. Let u = x and let dv = e^(2x)dv. Then du = dx and v = e^(2x)/2 . Substituting into the integrate by parts formula becomes integral = uv - $(vdu) = xe^(2x)/2 - $ [e^(2x)/2]dx integral = xe^(2x)/2 - e^(2x)/(2 times 2) integral = xe^(2x)/2 - e^(2x)/4 integral = [e^(2x)/2][x - 1/2] or, if you prefer, integral =[e^2x)/4][2x - 1] Thass it! (Don't forget the constant of integration.) gwh > How would you integrate xe^(2x)? What steps would you follow to come up with > the answer? === Subject: Re: Integration question > How would you integrate xe^(2x)? What steps would you follow to come up with > the answer? Use integration by parts. === Subject: Re: Integration question >> How would you integrate xe^(2x)? What steps would you follow to come up >> with >> the answer? > Use integration by parts. Ack - is that really the only way to do it? The actual problem in it's entirety is: integrate: xe^(2x) / (2x+1)^2 I set u to xe^(2x), dv to 1/(2x+1)^2, and just used my calculator to get xe^(2x) and I was able to solve the problem. Could it be that we are expected to use integration by parts to get xe^(2x) for the purpose of using integration by parts to do the rest of the problem? That makes sense I guess (bleah). === Subject: Re: Integration question > How would you integrate xe^(2x)? What steps would you follow to come up > with > the answer? >> Use integration by parts. >Ack - is that really the only way to do it? The actual problem in it's >entirety is: >integrate: >xe^(2x) / (2x+1)^2 You should know better than to give part of the problem when you are trying to solve an integral. Your problem is a little tricky and does eventually require integration by parts. Start with t = 2x + 1 to get: (1/4) int ((1 / t - 1 / t^2) * e^(t - 1) ) dt = (1/(4e)) int ((1 / t - 1 / t^2) * e^t ) dt Then do the int ( (1/t) * e^t ) dt piece by itself using integration by parts with u = 1/t and dv = e^t dt. That will give you a partial answer and a new integral, but when you plug it in above, presto, the other integral will drop out and you are home free. --Lynn === Subject: Re: Integration question > You should know better than to give part of the problem when you are > trying to solve an integral. I only gave part of the problem because that is the only part I was having trouble with. I don't need help solving the entire problem. Part of the problem requires that I find the integral of xe^(2x), and that is the only part I need help with :) === Subject: Re: Integration question >> You should know better than to give part of the problem when you are >> trying to solve an integral. > I only gave part of the problem because that is the only part I was having > trouble with. I don't need help solving the entire problem. Part of the > problem requires that I find the integral of xe^(2x), and that is the only > part I need help with :) PS please don't think me ungratefull - I actually examined and appreciate the answers I was given by both of you :) === Subject: Re: Integration question >> How would you integrate xe^(2x)? What steps would you follow to come >> up with the answer? > Use integration by parts. > Ack - is that really the only way to do it? Ack?? It's quick, simple and easy. > The actual problem in it's entirety is: integrate: > xe^(2x) / (2x+1)^2 Oh? That's not the problem, that's an entirely different problem. > I set u to xe^(2x), dv to 1/(2x+1)^2, and just used my calculator to get > xe^(2x) and I was able to solve the problem. Could it be that we are > expected to use integration by parts to get xe^(2x) for the purpose of using > integration by parts to do the rest of the problem? That makes sense I guess > (bleah). I don't know, calculators are so stupid they haven't learned to show their work. Not only that, you're not stating math, you're just stating some digit pushing you did with a calculator that's means nothing unless I know, ie you state, the equation or theorem you're working with. As I've told students before, turn that &)#(*$&^ machine off until you understand the math. I don't talk computer, I talk math. You want to do math, then show me some math, you know, complete equations, integral xe^2x / (2x + 1)^2 * dx = ... So your calculator says that integral xe^2x / (2x + 1)^2 * dx = xe^2x ? That's an absurdity, have you tried checking it's work? Does d(xe^2x)/dx = xe^2x / (2x + 1)^2 * dx Calculators are dumb. What calculator for your brand new problem would suggest you start with the substitution u = 2x + 1 ? However you're not going to like the results, it's an ulgack solution. === Subject: rates calculator,Gully Road,Porirua. IRD donations rebates/sorted xx etc. myfile.> DON05...TaxIRD.ratescalc5.GullyRd05 -date: NZDT. Thur. 20.10.05 01:40. 21:12. SENT. I rang IRD automated INFO-express. For the purpose of test , I asked them to calculate my donations rebate-.. Not a wise decision! I then listened to 20 options and finally was invited to key in my total donations, $100.00. Then, 3 minutes later, the telephone said my donations rebate is (... mumble.. $33.33. ) And I missed it! Having to make so many telephone selections, there was a ninety per cent chance that someone will make a mistake and stuff up. I query how many hundred pages of programming code and hours of computer processing time should be needed to calculate 100.00 / 3. In other words, all those web 'rates calculators' and so on are a waste of time. (Dominionpost, Gully Road .. rates, 19 October, 2005, A3. ) The retirement commissioner recommended the 0800 get sorted calculators. Individual savers can work out the benefits(?) of their beginning a life insurance policy today. (e.g. blurred food.) I entered my 6-monthly premium that i have now paid over 30 years. What per cent per annum has my policy appreciated relative to discounted cash flow? The program involves major assumptions. The life insurance company reports from time to time what is the current value of your policy. during the (17.9.05) election campaign every party invited us to go online and calculate how much personal income tax you would possibly pay if their party became the government. we will reduce the tax rate on the average wage band to 19%. Perhaps// it was 19.5% already, according to Statistics NZ Yearbook, 1998. NZ First party will demand the national superanuation will be at 66% of average wage. Is it 65% already?. So, what do these (so-called calculators) prove? Garbage in-turn the handle-garbage out! GIGO. Do you trust the one-line answer? Have readers heard of Floating-Point divide error? FDIV. the earlier Pentium computer processor chips used a strange inconsistent erroneous table to perform division by decimal numbers. The results were not reliable. how many metres in 1 mile? 1760yds *3 ft * 12 in * .0254 = 1609.344 metres exactly. a certain windows pc calculator, called Excalibur, uses conversion factor and may be several figures out. in the dominion 5.6.91 and nz science monthly magazine many times showed the data or formulas,- sometimes the complete calculator steps- that i use. There's an argument, proof. I also have over 40 sequences in on-line encyclopedia of here are my exact odds of lower tier prizes in lotto big wednesday. 45-choose-6. (similar to advertising.) Hit -- Kk.eno or Ll.otto: (Standard NZ Lotteries Commission shortcuts,) Balls in barrel = ?45 big wednesday. no. of lucky balls drawn 6 Balls = 45 lucky / unlucky = 6 / 39 Enter combo or how many group nos. you chose, 6 enter pattern. enter selection, Enter minimum correct = ?0 combo (group) = 6 minimum correct = 0 t =(Balls, combo) = 8145060 n_C_r = n! / (r!.(n-r)!) = 45! / ( 6!.39! ). select FIXED FONT. (I hope the table displays ok.) c (lucky, c)x (unlucky,combo-c); /correct-- wrong nr; s; odds = 1:t/nr | cumul 1:t/s == == (Exact.) ( at least.) ( binom. coeffs.) 6 (6,6)x(39,0) 1 1 8145060 8145060 5 (6,5)x(39,1) 234 235 34807.9487 34659.8298 4 (6,4)x(39,2) 11115 11350 ** 732.79892 ** 717.626432 0 1 2 3 4 5.. 3 (6,3)x(39,3) 182780 194130 44.5620965 41.95673 2 (6,2)x(39,4) 1233765 1427895 6.60179207 5.70424296 1 (6,1)x(39,5) 3454542 4882437 2.35778288 1.66823658 0 (6,0)x(39,6) 3262623 8145060 2.49647599 1 ? you need at least 4 correct to win a prize ?apart from dud ticket, replace. ?therefore, overall odds 1 in: 717 5/8th. approx. ? total 11,350 / 8145,060. I wouldnot be surprised lotto will lie about it. Test me! ARM BBC BASIC V version 1.05 (C) Acorn 1989 ..to.keno.20/80cmt *spool end yours sincerely, / Don S. McDonald (PG. 322** Wellington, New Zealand) === Subject: speed of the deed 'I'm gonna put it in hard.. I'm gonna wreck myself inside of you..' The speed of the deed * the mass of the ass = the torque of the pork The torque of the pork ^ the speed of the deed = the heat of the meat You're parents lied to you.. we all lied to you.. there's no mister nice guy.. -sEungkIm === Subject: sin is inver.. sin is inverse T where T = |8| where -|x|<0 -sEungkIm === Subject: Re: sin is inver.. > sin is inverse T where T = |8| where -|x|<0 You are sinning. === Subject: Why is sin^(-1)A considered the angle whose sin is A? Why is sin^(-1)A considered the angle whose sin is A? That doesn't make any sense to me. If sin^2 A is the same as (sin A)^2, then sin^-1(A) should be the same as (sin A) ^-1 or csc A? I hope this makes sense, because I can't type it the way, I would write it. I'd like to get your input on this. Kees === Subject: Re: Why is sin^(-1)A considered the angle whose sin is A? Writing the arcsine of an angle as sin^(-1) is simply a conventional shorthand form, that's all. It is not intended to represent the reciprocal of the sine of the angle. One would write that as 1/(sin A) , for example. gwh > Why is sin^(-1)A considered the angle whose sin is A? That doesn't make any > sense to me. If sin^2 A is the same as (sin A)^2, then sin^-1(A) should be > the same as (sin A) ^-1 or csc A? > I hope this makes sense, because I can't type it the way, I would write it. > I'd like to get your input on this. > Kees === Subject: Re: Why is sin^(-1)A considered the angle whose sin is A? > Why is sin^(-1)A considered the angle whose sin is A? That doesn't make > any sense to me. If sin^2 A is the same as (sin A)^2, then sin^-1(A) > should be the same as (sin A) ^-1 or csc A? > I hope this makes sense, because I can't type it the way, I would write > it. > I'd like to get your input on this. > Kees sin^n(x) is a power for all exponents EXCEPT -1. That notation is reserved for the function inverse idea. === Subject: Re: Why is sin^(-1)A considered the angle whose sin is A? > Why is sin^(-1)A considered the angle whose sin is A? That doesn't make any > sense to me. If sin^2 A is the same as (sin A)^2, then sin^-1(A) should be > the same as (sin A) ^-1 or csc A? > I hope this makes sense, because I can't type it the way, I would write it. > I'd like to get your input on this. > Kees As everyone else pointed out, your understanding is correct ... it's just ambiguous notation. When you're talking about numbers, raising to the -1 is 1/something. When you're talking about functions, the -1 notation has nothing to do with exponentiation -- it's simply a notation that designates the inverse function. For example if f(x) = 4x, then f^(-1) refers to the inverse function, given by g(x) = 1/4. === Subject: Re: Why is sin^(-1)A considered the angle whose sin is A? > Why is sin^(-1)A considered the angle whose sin is A? That doesn't make any > sense to me. If sin^2 A is the same as (sin A)^2, then sin^-1(A) should be > the same as (sin A) ^-1 or csc A? > I hope this makes sense, because I can't type it the way, I would write it. > I'd like to get your input on this. > Kees As the other reply states it is just (bad) notation. If you really need to write the reciprocal of sin X you would need to write (sin X)^-1 which does equal sec X. === Subject: Re: Why is sin^(-1)A considered the angle whose sin is A? > Why is sin^(-1)A considered the angle whose sin is A? That doesn't make > any > sense to me. If sin^2 A is the same as (sin A)^2, then sin^-1(A) should be > the same as (sin A) ^-1 or csc A? > I hope this makes sense, because I can't type it the way, I would write > it. > I'd like to get your input on this. > [...] > As the other reply states it is just (bad) notation. If you really need to > write the reciprocal of sin X you would need to write (sin X)^-1 which does > equal sec X. You mean csc X. (sec X = 1 / cos X.) --- Christopher Heckman === Subject: Re: Why is sin^(-1)A considered the angle whose sin is A? > Why is sin^(-1)A considered the angle whose sin is A? That doesn't make any > sense to me. If sin^2 A is the same as (sin A)^2, then sin^-1(A) should be > the same as (sin A) ^-1 or csc A? The notation is bad for the reason you state, but it is also widely used. Better is arc sin A for the angle (in a certain range) whose sin is A. -- The world will little note, nor long remember, what we say here. === Subject: Re: Simple problem I can't figure out. You're right, the straightforward way to get the answer (76) is to take pencil and paper in hand and write down all of the possible combinations. The process is called partitioning and has been studied at length for many years. I do not have a closed-form solution to offer you. I might mention that your child's teacher either screwed up by assigning such a problem to a 6-year old or else actually placed some simplifying restrictions on the problem which would render it solvable by that age group. The partitioning problem was investigated by the world-famous mathematician Ramanujan some yeara ago and his approach is still an amazing one to a plodding old engineer like myself. Good luck and keep us advised-- I'm curious about this assignment....... Grover Hughes Boxington Headmaker My six year old has been given a problem for his 1/2 term homework. I'd > like to help him out, but I can't get my head round how work it out. > He's been asked to find all the possible arrangements of 12 blocks. For > example. 12 blocks in one column, 2 columns of 6, 12 columns of 1 block. > I've been scratching my head as to how to work it out (I've an idea that I > should work out how many ways to make 1, 2, 3 etc, then add them all > together) > Can anybody point me in the right direction, or show me a formula that I can > apply. > Box === Subject: Re: Simple problem I can't figure out. | My six year old has been given a problem for his 1/2 term homework. I'd | like to help him out, but I can't get my head round how work it out. | | He's been asked to find all the possible arrangements of 12 blocks. For | example. 12 blocks in one column, 2 columns of 6, 12 columns of 1 block. | I've been scratching my head as to how to work it out (I've an idea that I | should work out how many ways to make 1, 2, 3 etc, then add them all | together) | | Can anybody point me in the right direction, or show me a formula that I can | apply. | | I believe the problem solution is called partitions (partitioning?). The algorithm for finding the number of partitions for any number is (written in REXX): __________________________________________________________________________ /**/ arg n . /*get argument. */ if n=='' then n=12 /*if not present, then assume 12.*/ @.=1 /*define all elements of array =1*/ do j=0 to n /*process all partitions up to N.*/ if j<2 then _=1 /*handle two special cases. */ else do /*handle the general case. */ _=0 /*start with zero partitions. */ do k=0 to j-1 /*process partitions 0 --> j-1. */ _=_+$sigma(j-k)*@.k /*add them up (so far). */ end _=_/j /*now, divide the sum by J. */ @.j=_ /*also, save the result. */ end end say 'n='n 'partitions='_ exit /*procedure to find the SIGMA for any X (sum of all the divisors). */ $sigma: procedure; parse arg x; z=0 i=$isq(x); do j=1 while j<=i; if x//j==0 then do; y=x%j if x==1 | y==i then y=0; z=z+j+y end; end; return z /*procedure to find the integer square root of any X [floor(sqrt(x)]. */ $isq: procedure;arg x; x=trunc(x); r=0; q=1 do while q<=x; q=q*4; end do while q>1; q=q%4; _=x-r-q; r=r%2 if _>=0 then do; x=_; r=r+q; end end; return r _________________________________________________________________________ I didn't add comments for the $sigma and $isq subroutines as those routines are ancillary to the partitions function. But for those who need to know: // is divide, but return the remainder, % is integer divide, and DO loops test for the range immediately, and the WHILE clause check for conditions at the TOP of the loop. _______________________________________________________________Gerard S. === Subject: Re: Simple problem I can't figure out. This type of problem is known as partitioning and there have been lots of manhours spent on searching for closed-form solutions to the answer. I have written a program which gives me the answer of 76 for your particular case, i.e., 12. I don't happen to have a closed-form formula, but someone else in this newsgroup may be able to supply one. I think you might find it rewarding to work it out yourself, the direct way. I'll get you started. In the table which follows, I've not bothered to stick in plus signs or other dividers; I think you'll see what's going on right away. Have fun! 12 = 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 2 2 2 1 1 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 2 1 1 2 2 2 2 2 2 This exhausts the 2's; now do the 3's: 3 1 1 1 1 1 1 1 1 1 3 2 1 1 1 1 1 1 1 3 2 2 1 1 1 1 1 3 2 2 2 1 1 1 3 2 2 2 2 1 3 3 1 1 1 1 1 1 3 3 2 1 1 1 1 and so on; really, it won't take all that long--just be sure to check that each rwo adds to 12 (this is to catch errors of omission!) Grover Hughes Boxington Headmaker My six year old has been given a problem for his 1/2 term homework. I'd > like to help him out, but I can't get my head round how work it out. > He's been asked to find all the possible arrangements of 12 blocks. For > example. 12 blocks in one column, 2 columns of 6, 12 columns of 1 block. > I've been scratching my head as to how to work it out (I've an idea that I > should work out how many ways to make 1, 2, 3 etc, then add them all > together) > Can anybody point me in the right direction, or show me a formula that I can > apply. > Box === Subject: Re: Please help (Sin, Cos, Tan) > I really have no idea how to start this: > Total internal reflection between two materials is seen to occur if a > critical angle of 60 degrees is exceeded. If the index of reflection of the > material that the light is entering is 1.98, find the index of refraction of > the material in which the light is originating. > If I could just get a formula for it I think I could solve. Use Snell's Law: n_1*sin(theta_1) = n_2*sin(theta_2) If total internal reflection occurs going from 1 to 2, then n_1 > n_2 and theta_2=90 degrees for theta_1=critical angle. Bob === Subject: Integration question 10-21-2005 I figured I'd date stamp my questions :-P OK, next problem (and I'll post the entire problem). I think I did this one right, but I had to graph the answer before I was sure because the answer I got was not what the book shows. Problem - evaluate the integral of: cos(x)^3 six(x) dx I choose to let u = sin(x), and du would be cos(x) dx. That gives us: cos(x)^2 sin(x) cos(x) dx Now we need to get rid of the cos(x)^2, using the pythagorean identity sin(x)^2 + cos(x)^2 = 1, therefore cos(x)^2 = 1 - sin(x)^2. Rewrite as: (1-sin(x)^2) sin(x) cos(x) dx which = ( sin(x) - sin(x)^3 ) cos(x) dx Now we can stick in the u's and du: ( u - u^3 ) du Now we can integrate: u^2/2 - u^4/4 And finally stick the sin(x) back in place of u: sin(x)^2/2 - sin(x)^4/4 + C The book did it the other way around by making u = cos(x), and du -sin(x)dx. Since the power of both sin and cos are odd, I guess it doesn't matter which way we go, except you get two very different looking answers. I finally graphed both of them and realized that they were the same graphs that only differed by the constant, which tells me that I did it right after all. What I'm looking for is some kind soul to review the above and say, yes, I actually got this one right :-) === Subject: Re: Integration question 10-21-2005 It is true that if the graphs are the same except for the constant term, then each of the integrals are correct. Keep in mind that there is always a constant to be considered, and it can be anything you wish if not specified by the problem statement itself or by whatever may be known about the physical situation you are modeling. gwh > I figured I'd date stamp my questions :-P > OK, next problem (and I'll post the entire problem). I think I did this one > right, but I had to graph the answer before I was sure because the answer I > got was not what the book shows. Problem - evaluate the integral of: > cos(x)^3 six(x) dx > I choose to let u = sin(x), and du would be cos(x) dx. That gives us: > cos(x)^2 sin(x) cos(x) dx > Now we need to get rid of the cos(x)^2, using the pythagorean identity > sin(x)^2 + cos(x)^2 = 1, therefore cos(x)^2 = 1 - sin(x)^2. Rewrite as: > (1-sin(x)^2) sin(x) cos(x) dx > which = > ( sin(x) - sin(x)^3 ) cos(x) dx > Now we can stick in the u's and du: > ( u - u^3 ) du > Now we can integrate: > u^2/2 - u^4/4 > And finally stick the sin(x) back in place of u: > sin(x)^2/2 - sin(x)^4/4 + C > The book did it the other way around by making u = cos(x), and du -sin(x)dx. > Since the power of both sin and cos are odd, I guess it doesn't matter which > way we go, except you get two very different looking answers. I finally > graphed both of them and realized that they were the same graphs that only > differed by the constant, which tells me that I did it right after all. What > I'm looking for is some kind soul to review the above and say, yes, I > actually got this one right :-) === Subject: Re: Integration question 10-21-2005 > I figured I'd date stamp my questions :-P > OK, next problem (and I'll post the entire problem). I think I did this one > right, but I had to graph the answer before I was sure because the answer I > got was not what the book shows. Problem - evaluate the integral of: > cos(x)^3 six(x) dx Substitute u = cos x. > I choose to let u = sin(x), and du would be cos(x) dx. That gives us: So what? Tell us what your doing, substitution of variable, integration by parts or what? > cos(x)^2 sin(x) cos(x) dx > Now we need to get rid of the cos(x)^2, using the pythagorean identity > sin(x)^2 + cos(x)^2 = 1, therefore cos(x)^2 = 1 - sin(x)^2. Rewrite as: > (1-sin(x)^2) sin(x) cos(x) dx > which = > ( sin(x) - sin(x)^3 ) cos(x) dx > Now we can stick in the u's and du: > ( u - u^3 ) du > Now we can integrate: > u^2/2 - u^4/4 > And finally stick the sin(x) back in place of u: > sin(x)^2/2 - sin(x)^4/4 + C Oh sigh, no equations, mostly stream of thought. > The book did it the other way around by making u = cos(x), and du -sin(x)dx. > Since the power of both sin and cos are odd, I guess it doesn't matter which > way we go, except you get two very different looking answers. I finally > graphed both of them and realized that they were the same graphs that only > differed by the constant, which tells me that I did it right after all. What Looking at graphs don't ever tell you different expression are the same. It makes at lot of difference, your method is as about a complex as the tax code, their method is a complicated as we wish the tax code was. ;-) > I'm looking for is some kind soul to review the above and say, yes, I > actually got this one right :-) In addition you didn't tell us what the book got. Do think that just because you got a copy of the book, that everybody else does? What did the book get? integral cos^3 sin dx = integral cos^3 d(-cos x) dx = integral -u^3 du (substituting u = -cos x = -u^4 / 4 + c = -(1/4) cos^4 x + c Is that the book's answer? You want to know if that equals your answer sin(x)^2/2 - sin(x)^4/4 + C ? Have you tried evaluating them at several points? If they're the same, and I mean identical and not just the same out to a few decimal points, those tests have failed to show they're different. BTW, you have the extra bother of accounting for c and C understanding that you cannot assume c = C. If you suspect they're the same, then juggle them around with some trigonometric identities. For example you could finish your work by simplifying your expression. Or you could ask the question is f(x) = (sin^2 x)/2 - (sin x^4)/4 + (cos x^4)/4 constant? Why that f? Anyway does f'(x) = 0 for all x? Why is that a testing question? === Subject: Re: Integration question 10-21-2005 >I figured I'd date stamp my questions :-P >OK, next problem (and I'll post the entire problem). I think I did this one >right, but I had to graph the answer before I was sure because the answer I >got was not what the book shows. Problem - evaluate the integral of: >cos(x)^3 six(x) dx >I choose to let u = sin(x), and du would be cos(x) dx. That gives us: >cos(x)^2 sin(x) cos(x) dx >Now we need to get rid of the cos(x)^2, using the pythagorean identity >sin(x)^2 + cos(x)^2 = 1, therefore cos(x)^2 = 1 - sin(x)^2. Rewrite as: >(1-sin(x)^2) sin(x) cos(x) dx >which = >( sin(x) - sin(x)^3 ) cos(x) dx >Now we can stick in the u's and du: >( u - u^3 ) du >Now we can integrate: >u^2/2 - u^4/4 >And finally stick the sin(x) back in place of u: >sin(x)^2/2 - sin(x)^4/4 + C >The book did it the other way around by making u = cos(x), and du -sin(x)dx. >Since the power of both sin and cos are odd, I guess it doesn't matter which >way we go, except you get two very different looking answers. I finally >graphed both of them and realized that they were the same graphs that only >differed by the constant, which tells me that I did it right after all. What >I'm looking for is some kind soul to review the above and say, yes, I >actually got this one right :-) Glancing at it, it looks OK. But the real lesson in this for you is to see why when faced with int( cos^3(x) sin(x) )dx, that you should immediately see the u^3 du form to be had by letting u = sin(x). When you try u = cos(x) it appears to me you are just guessing what to try for u, and you guessed poorly even though it works. The form of the problem itself often suggests what substitution to use, if indeed a substitution is indicated. --Lynn === Subject: Re: Integration question 10-21-2005 >Glancing at it, it looks OK. But the real lesson in this for you is to >see why when faced with int( cos^3(x) sin(x) )dx, that you should >immediately see the u^3 du form to be had by letting u = sin(x). **Typo alert, of course I meant u = cos(x)** --Lynn === Subject: power series expansion of complex function Is there a smarter (easier) way to expand the function of a complex variable f(z) = 1/(1 + z + z^2) in a power series than the brute force method of differentiating n times to find each of the series coeffs: an = f^(n)(z0)/n! Colin === Subject: Re: power series expansion of complex function > Is there a smarter (easier) way to expand the function of a complex variable > f(z) = 1/(1 + z + z^2) in a power series > than the brute force method of differentiating n times to find > each of the series coeffs: an = f^(n)(z0)/n! > Colin you could try a laurent series, still involves some integration though