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    AdrianCavinder's Avatar
    AdrianCavinder Posts: 55, Reputation: 1
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    #1

    Jan 9, 2012, 07:48 AM
    Factorials by inductive definition
    I have to show that [9/6] = (9x8x7)/(1x2x3) by inductive definition for [n/r]. I seem to have lost the thread and can't understand how they reach this conclusion. Any help would be appreciated: as simple as possible! Thanks.
    corrigan's Avatar
    corrigan Posts: 115, Reputation: 18
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    #2

    Jan 9, 2012, 03:38 PM
    If you mean the binomial coefficient,


    that's easy enough,







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    AdrianCavinder Posts: 55, Reputation: 1
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    #3

    Jan 9, 2012, 08:32 PM
    Hi Corrigan, Thanks for the reply. I had understood that from the formula the answer would be 9!/(6!x3!), but what I don't understand is why. Why does [9/6] become a factorial answer? I need help in understanding how [9/6] translates to a factorial. Can you help me understand that? Thanks
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    corrigan Posts: 115, Reputation: 18
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    #4

    Jan 10, 2012, 10:11 AM
    I'm sorry, I don't quite understand where you are lost. The answer, isn't a factorial.

    If you don't understand how we went from to , the on the numerator cancelled out the same multiples in the denominator.

    If you don't understand why "9 choose 6" translates into which then translates into , that can be a little more complicated. In math why can be more complicated than how. Now that I think about it, I think this is probably to what you asking. Let me see if I can help you.

    If I have items and I pick one, there are possibilities. If I still have items and I pick two, and order is important, than there are possibilities, since there are possibilities for the first pick and possibilities for the second pick. Continuing in this way, if I have items and I pick , and order is important, than there are possibilities. Since the binomial coefficient is concerning picking when order is not important, than I take the number of possibilities when order is important and divide it by the number of possibilities of arranging items, which is , and that gives us which simplifies to . The formula is just an easy way to remember the formula, by it's nature, there will be factors that are cancelled out. Something important to notice is that "9 choose 6" is equal to "9 choose 3" since the number of ways to choose 6 from 9 is the same number of ways to not choose 3 from 6, that's why ends up canceling out.

    I hope this helps, I know it can be hard to follow. If you have any more questions, feel free to ask. :)
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    AdrianCavinder Posts: 55, Reputation: 1
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    #5

    Jan 10, 2012, 09:31 PM
    Hi Corrigan. That's very helpful indeed! I think I must have asked too many 'why' questions at school.
    So just to make sure I've understood, I'd like to work through an example and you could let me know if I have understood correctly.
    To follow on from your example [9/2], the number of possibilities, if order is important, is 72 or 9x8. That is, not only could we have ab but also ba, not only ac but ca etc. (taking the 9 possibilities to be a.b.c.d.e.f.g.h.I.). However, if order is not important, then obviously there would be half as many options i.e. 36 possibilities. (it wouldn't matter if it were ab or ba).
    Bringing the formula into the equation (so to speak), [9/2] = 9!/2!(9-2)! = (9x8)/2 = 36
    I was puzzled why you had to 'take the number of possibilities when order is important and divide it by the number of possibilities of arranging 2 items, but it dawned on me why, so no more questions, just a big thanks.
    Just to put a 'face' to the message, I'm Adrian, 57 years old, British, married, living in India for the past 30 years. I run a tuition centre and recently took on a student studying for AS Level Mathematics. As it has been 40 years since I last looked at this level of mathematics, I get into difficulty explaining the 'why' because I don't fully understand myself.
    If I may, I would appreciate being able to throw any other 'whys' that come up during this course. Your explanation was very easily understood. Cheers.
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    corrigan Posts: 115, Reputation: 18
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    #6

    Jan 12, 2012, 10:21 AM
    Sounds like you got it. If you have any more questions, I haunt the Math & Science and Mathematics sections.
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    AdrianCavinder Posts: 55, Reputation: 1
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    #7

    Jan 16, 2012, 09:45 PM
    Hi again. I'm stumped on the last binomial question. I've written pages of figures but I have a feeling I'm missing a connection with the terms somewhere. A hint would be appreciated.
    Given that a + b(1 +x)^3 + c(1 + 2x)^3 + d(1 + 3x)^3 = x^3 for all values of x, find the values of the constants a, b, c and d.
    Do I expand the binomials and then group the like b terms, c terms and d terms? I've done that and seem to just get bogged down with figures.
    Do I substitute values for x and then solve by simultaneous equations? Again, I tried that but it seemed ridiculously long and ultimately I didn't get the right answers.
    a = -1/6 b = 1/2 c = -1/2 and d = 1/6.
    Any help would be most appreciated. Cheers.
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    corrigan Posts: 115, Reputation: 18
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    #8

    Jan 17, 2012, 02:29 AM
    You're going to get bogged down with figures. This is one of those problems that you really need to be carful with your calculations. A simple mistake on any step will give you a bad solution. Solving this doesn't require substituting anything for , we're just focusing on the coefficients for each of the terms. The reason that polynomials are used is so the coefficients of terms with different powers don't "mingle". For example, if I have , I know that and , (since ), then I can solve for .

    This is how I would solve it, get comfy because this is going to be long.

    Since we'll be raising a lot of binomials to the third power, remember that . Since the first term in every binomial is one, this gets a little easier, so . Well just be replacing with , , and . I'm just telling you this so when it looks like I'm skipping steps, that's what I'm doing.



    Now we're going to group the terms by powers of

    That takes care of the left side, now for the right side just keep in mind that since anything times zero is zero, and anything plus zero doesn't change anything. So this gives us:

    So from this we know that:
    (i)
    (ii)
    (iii)
    (iv)
    If we take (iii) and subtract (ii), we get
    and solve for
    (v)
    Now take (v) and use it to substitute for in (iv), so



    (vi)
    Now we have b and c in terms of d. We can't substitute these into (iv) because that would give us something like which does us no good. We can substitute b and c into (ii) or (iii). Since (ii) has easier coefficients, we'll use (ii). So by (v), (vi) and (ii) we get







    So now we have d and the rest is just cleaning up.







    I tried to make this as easy to follow as I could. If you have anymore questions just ask. :)
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    AdrianCavinder Posts: 55, Reputation: 1
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    #9

    Jan 17, 2012, 04:44 AM
    Wow! Thanks, that was very helpful. Understood completely. I got as far as the expansion of the terms, then got bogged down because I didn't appreciate your one line x^3 = 0 + 0x + 0x^2 + x^3: that was the 'light'.
    After 3 frustrating days of pages of figures. I finally found an alternative way to solve the question, although I knew it was not the correct method as I didn't expand the terms. I took values of x which cancelled a term each time i.e. x = -1, x = -1/2, x = -1/3 and x = 0. Then I did basically the same thing as you did by substituting and subtracting.
    Really appreciate you taking the time to help answer these questions. Thanks.
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    AdrianCavinder Posts: 55, Reputation: 1
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    #10

    Jan 31, 2012, 07:47 AM
    HI again. Sorry, I have another problem I can't seem to work out:
    A funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of liquid in the funnel is 12cm, the liquid is dripping from the funnel at a rate of 0.2cm^3s^-1. At what rate is the depth of the liquid in the funnel decreasing at this instant?
    Now, from the chain rule, I understand that dV/dt = dV/dh x dh/dt, where V = Volume h = depth and t = time.
    From the formula, V = (1/3)πr^2h, dV/dh = (1/3)πr^2, so 0.2 = (1/3)πr^2 x dh/dt. As the depth is twelve, I believe the radius to be 8 because of the proportion rule 30:12 = 20:8
    So dh/dt = 0.6/64π = 0.00298cms^-1, but the answer is 0.004cms^-1.
    Can you show me where I'm zagging when I should be zigging? Cheers.
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    corrigan Posts: 115, Reputation: 18
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    #11

    Jan 31, 2012, 11:33 PM
    Hi AdrianCavinder. I'm sorry I don't have the time to fully work this out right now. I have to be at work tomorrow and it's getting late over here. But just at a glance, before you take the derivative of the volume in terms of height, you need to make the volume a function of height because the radius is changing while the height is changing. Since the diameter is 20 when the height is 30, then the radius is 1/3 the height. So







    So just take the change in volume over time and divide it by the change in volume over height, and you should get the answer you need. I'm sorry I don't have more time now, I'll take a look at it after work tomorrow.
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    AdrianCavinder Posts: 55, Reputation: 1
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    #12

    Feb 1, 2012, 12:01 AM
    Hi Corrigan. Thanks so much. I didn't sink in to relate r to h, so with the information you gave, I worked out the right answer. Really helpful and thanks for keeping it simple! Some explanations are far too complex - rocket science stuff!
    Just as a side point, seems like some of my figures come out looking weird when posted. For example, I type an apostrophe ' and all this gobbledygook appears. Even when I type pi π that looks weird too. Is it because I'm using a mac or something? Sorry, not exactly a mathematics question. Just when I type it, it looks fine, only when it gets posted does it do a Dr, Jekyll on me.
    Cheers.
    corrigan's Avatar
    corrigan Posts: 115, Reputation: 18
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    #13

    Feb 1, 2012, 08:03 PM
    Glad I could help. As far as the gobbledeegook, formulas look best on this site when put in math brackets. At they beginning of a formula you type [*math] (without the asterisk) and at the end you type [/*math] (again without the asterisk)), and the formulas need to be typed in Latex, it's a scientific typeset language. A lot of Latex is pretty intuitive, but you'll need to learn the special commands. It's how I type stuff like this:



    If you are studying in a scientific field, it will help you to learn it. I don't know about other fields, but Latex is the standard typeset used in mathematics. If that seems a little daunting, and I remember learning Latex for me had a very steep learning curve, then you can just type it out. For some reason this site doesn't handle cut and paste very well, so you'll just have to type it. As far as the apostrophe, I don't know, maybe your keyboard is on some setting that isn't conducive to this site. You might want to pose that question in one of the technical topics, I'm really not qualified to answer that. I do know that it isn't that you are on a Mac because I'm typing this on a Mac.
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    AdrianCavinder Posts: 55, Reputation: 1
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    #14

    Feb 7, 2012, 11:17 PM
    Hi. Back again I'm afraid. I'm stumped with this question: I have to determine the min and max values of y and the least possible values of x at which these occur. Question is: y = 8 - 3cos^2x. I found the min and max to be 5 and 8, but finding the x value is proving difficult. I would have thought it would have been: cos^2x = 1 or -1, which would give cosx = + or - 1, which should be 180 and then 360. But the answer says 90 and 180. In a previous question, y = 29 - 20sin(3x - 45), I found the min and max to be 9 and 49 and the x value to be: 3x - 45 = 90 or x = 45 and 3x - 45 = -90 or x = -15 which becomes 195 (180 - (-15)). However the answer sheet says 45 and 105. Am I making a fundamental mistake? Your help would be appreciated.
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    corrigan Posts: 115, Reputation: 18
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    #15

    Feb 8, 2012, 09:11 PM
    Hi adriancavinder, you have a lot of the fundamentals down, it refreshing to see someone asking for help on their homework and they've actually attempted the problem.

    For the first question, the part that you are missing is that while the value of oscillates between +1 and -1, the value of oscillates between +1 and 0. The reason for that is that when a real number is squared, it can't be negative, so when is negative, is positive. If you were to graph the graph would look like a miniature version of the cosine graph between 0 and 1 with a period of 180 degrees. So the minimum value of is when [math] \cos x =0 [\math] which is 90 degrees. As far as the maximum value of , that happens when which is at 180 degrees. (I guess your teacher doesn't want zero degrees). If this is a little confusing, try graphing and you should see the period.

    For the second question, the part is spot on, don't change it. Where you messed up on is when you you took (180-(-15)). The period of is 1080 degrees, not 360 degrees like it is for , and the graph is shifted 45 degrees to the left. So it gets very ugly very quickly. I noticed that you were setting the value of for at -90. Perhaps an easier way might be to set the value of for at 270 degrees. Then we don't have to deal with the whole mess of finding where is shifted to. Then we get
    .

    What math are you taking, and have they covered radians? When I studied this stuff it was all in radians which is just it's a much more intuitive way of describing angles, especially when doing stuff like this. Anyway, I hope this was helpful, if any of it seems confusing, just ask and I'll be happy to help.
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    AdrianCavinder Posts: 55, Reputation: 1
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    #16

    Feb 8, 2012, 11:07 PM
    Hi Corrigan. Thanks once again. It had dawned on me that the graph couldn't oscillate below 0, but I hadn't realised the effect that had on the period.
    Better just clear up a confusion: although we should always consider ourselves students throughout life, I haven't been an actual 'attending school' student for 40 years. I have been teaching the British IGCSE math curriculum for the past 3 years. I oversee a tuition centre here in India. Recently I started teaching AS level mathematics for the first time (sorry, not sure where in the planet you're situated, but basically I was comfortable teaching up to grade 10 mathematics but am now finding some of the 11th grade stuff challenging). This is perhaps why my questions are slightly different: I want to be able to understand the 'why' so I can teach the subject properly. By the way, I am about halfway through the course and radians only appear right at the very end.
    I hope you can now better understand why your help is so appreciated. I promise to bug you only when I'm really stuck. Cheers.
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    corrigan Posts: 115, Reputation: 18
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    #17

    Feb 9, 2012, 09:33 PM
    It's no trouble. I'm in the United States, Texas specifically. If you need anything else, just ask.
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    AdrianCavinder Posts: 55, Reputation: 1
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    #18

    Feb 27, 2012, 04:18 AM
    Hi Corrigan. Well I have a problem which I can't seem to fathom. I am asked to differentiate [(sq rt x) + 1]^2 / x with respect to x. I tackled the problem as follows:
    First I expanded the top term to become [x + 2(sq rt x) + 1]. Then I multiplied by x^-1 to get 1 + 2(sq rt x)^-1 + x^-1.
    When I differentiated, I got -x^-3/2 - x^-2, but the answer sheet says 1 - x^-3/2 - 2x^-3. Can't see where I'm going wrong. Your help would be appreciated.
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    AdrianCavinder Posts: 55, Reputation: 1
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    #19

    Feb 27, 2012, 04:27 AM
    Sorry, I have one more question that's bugging me.
    I had to find the inverse of f(x) = 4 - x^2, where x is greater than or equal to 0, which I found to be sq rt (4 - x).
    However, it then goes on to ask me to show how a solution to f(x) = f^-1(x) can be obtained from x^2 + x - 4 = 0. Finding the solution is no problem (1.56), and when I draw the graph I have no problem understanding the connection, but 'showing' the link is causing me to waste reams of paper! As I understand it, I would have to show a link between x^2 + x - 4 and 4 - x^2 = sq rt (4 - x), but I come out with x^4 - 8x^2 + x +12 = 0 and I just can't catch the link. Again, any help would be most welcome. Thanks.

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