# Vastor

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5

1. ## Vastor

FUS RO DAH!!!

1. Someone's playing Skyrim!

2. ## how cot is valid if 1 / infinity =/= 0

in calculus, I learned that 1/infinity will approaches 0, but not equal to 0 like 1/0 is approaching infinity, but not equal to infinite tan 90 = sin(90)/cos(90) = 1/0 = infinite? 1/tan 90 = 1/infinite = 0? how does cot graph is valid (instead of using the "undefined" value or line up an asymptote there) just because the assumed value of 1/tan(90) = 0, while tan 90 = infinite..
3. ## Noob proof question.

In general, what I mean about the "flow-in-between" is how much for a proof is sufficient to be a proof? let's give a look at this question: 1. Prove the following: (iv) $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$ (v) $x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})$ My Answer: (iv) $x^3 - y ^3 = (x-y)(x^2 + xy + y^2)$ $= x(x^2 + xy + y^2) - y(x^2 + xy + y^2)$ $= x^3 + x^2 + xy^2 - x^2y - xy^2 - y^3)$ $= x^3 - y^3$ (v) This gonna be a bit of mess, considering that I don't really know what I'm doing. $x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})$ $x^5 - y^5 = (x - y)(x^{4} + x^{3}y + x^2y^2 + xy^{3} + y^{4})$ if n = 5 $x^3 - y^3 = (x - y)(x^{2} + xy + y^{2})$ if n =3 As we can see, certain pattern developed here based on the first equation. Thought so, I have a bit of problem managing (+ ... +) in the original equation. So I tried this. when n = 3, now let $K = xy$ $x^3 - y^3 = (x-y)(x^2 + K + y^2)$ $= x(x^2 + K + y^2) - y(x^2 + K + y^2)$ $= x^3 + xK + xy^2 - x^2y - yK - y^3 = x^3 + y^3 + 0$ let $M = xK - yK + xy^2 - yx^2 = 0$ when n = n/unknown, now let $K = x^{n-2}y + ... + xy^{x - 2}$ $x^n - y^n = (x-y)(x^{n-1} + K + y^{n-1})$ $= x(x^{n-1} + K + y^{n-1}) - y(x^{n-1} + K + y^{n-1})$ $= x^n + xK + xy^{n-1} - x^{n-1}y - yK - y^n = x^n + y^n + 0$ let $M = xK - yK + xy^{n-1} - yx^{n-1} = 0$ when n =5 $M = xK - yK + xy^4 - yx^4 = 0$ is supposed to be. $x^5 - y^5 = (x-y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4)$ $thus K = x^3y + x^2y^2 + xy^3$ $x^5 - y^5 = (x-y)(x^4 + K + y^4)$ $= x^5 + xk - yk + xy^4 - x^4y - y^5$ $= x^5 - y^5 + M = x^5 - y^5$ where $M = xK - yK + xy^4 - yx^4 = 0$. My Question: (iv) I found another way to prove this question, by assigning random number to the variable on both equation. The equation would be true if both equation produce same result and false otherwise. Thus, which one is better(more sufficient) as the proof for this question? or (based on your explanation, if I understand correctly), it's doesn't matter either way as long as others can read the "flow-in-between" the proof? If so, how about question (v), is the proof sufficient enough? (I got the hard feeling it's not) Well, at least it's because I try to mimic from this site on how to proof the series part:- http://www.purplemath.com/modules/inductn3.htm But, (unfortunately?) the above solution is the best I can think of, even though I realised & noted several problem that I faced: - based on the site, There is only 1 unknown/variable which make solving the equation much simpler, should I make y = 2x or insert any random number for each x and y? I did, and I not found much or any pattern to manipulate, but end up with a mess of number. - secondly, I don't know how to manage the "series's symbol" = (+ ... +). Actually, I had a bad time with solving +(...)*(x) - (...)*(y), which end up something un-solvable. - based on second point, I'm not even sure if putting (K = blabla + ... + blabla) on the above solution is "legal" in the first place! Yet, I used it because the site used an equation to represent its series, and I'm sure a variable can represent an equation. Thus, variable represent a series! - based on the site also, I tried to apply the induction process by providing a variable for (n) to represent "anywhere" and (n+1) to represent the next place or "everywhere". Again, I'm facing a big problem on managing the (+ ... +) and created another mess of equation. Anyway, thank you for reply beforehand, really appreciated.
4. ## Noob proof question.

1, Prove the following: (i) $if, ax = a$ for some number $a \neq 0, then, x = 1.$ (ii) $x^2 - y^2 = (x-y)(x+y)$ My Answer: (i) if $ax = a, a \neq 0$ $ax * a^{-1} = a * a^{-1}$ $x * a * a^{-1} = a * a^{-1} = 1$ $x * 1 = x = 1$ (ii) let $y = x +k$ then $x^2 - y^2 = x^2 - (x+k)^2$ $= x^2 - (x^2 + 2xk + k^2)$ $= x^2 - x^2 -2xk - k^2$ $= -2xk -k^ = -k(2x+k)$ then $(x-y)(x+y) = (x - (x+k))(x + x +k)$ $= (x-x-k) (2x +k)$ $= (0-k)(2x+k) = -k(2x +k)$ Thus, $x^2 - y^2 = (x-y)(x+y)$ My Question: - Is my proof right & accurate? - How to tell either the proof that I make right or no? (I can't afford to ask here every time I make one) More Detail on my question:- on first Exercise, I found it to be very obvious but I still have some doubt (which is probably unnecessary) on the proof. On second one, I found it to be very obvious too if I'm going from $(x+y)(x-y) => x^2 - y^2$ but not its reverse! It give me headache on how to proof the existence of "xy - xy = 0" in the calculation. So, I try a different approach and get the above proof but is it appropriate? Does it doesn't matter how people proof it as long as you can put the "flow-in-between", then consider the proof as accurate? By multiple afford searching some clue on google, I end up become more confused as I see proof got it's own group and it's specific method on how to do. Yeah, it's pretty understandable with their given example yet for meI found it very hard to apply it on my very own question. Some note: - Those above are my first two proof problems. - Problem taken from Spivak Calculus, if anyone found better book for me, please recommend!
5. ## Mathematician's lament.

Yep, I agree with you on how the article is not really accurate on what maths really is & how maths interested others, but lets face reality, the education happen to be (terribly) bad and mathematics got bad names because of it. The typical "I can't do Math, there's too much formula to be memorized" complains started to piss me off on the education system. I can't agree on this, there are students who able to do those table without have any a clue what the table is about, this is the problem the writer talk about. Especially, in high school, we use calculator anyway. and I'm sure that almost every average person have enough for memory multiplication table, they 'can't get far' in math because they have no interest in the first place. Until we can spark some awe into others and make them an actively thinking person, we can't get anyone far in mathematics. From my perception, I see mathematics as a (shortcut) Language + Philosophy with good Axiom. So yeah, it's not really an Art because it's what standardized by mathematician themselves. but teaching them through memorizing (plug & chugged the exercise and repeat) is very bad step to begin with. Furthermore, there's too lack of mathematical reasoning discussion in the so-called 'Maths Class'. Yet, even on these kind of chapter, (I got a) Teacher told us, "if it's a Statement, u need to answer with 'if x then y, blabla' to get the MARKS". teachers are moving in wrong direction when they start taking care about marks than teaching math which sometime make me even forgot either I am still in a Math Class. They just sampling the answer and put in our mouth, tadaaa, mathematics! The effect? Well, I got an older sister who got good mark in both Math and Add. Math, A+ or so in the SPM (O-lvl exam), but now she start decide to pick a course that use least math possible & stay away from it. Because there are people who can memorized all of the formula but end up knowing (almost) nothing what they are actually doing. Whether the exam is too standardized or 'Math Class' is just a fraud or both. The only 'benefit' from these education system is less competition for a mathematician, duh...
6. ## Understanding Electricity

given voltage at transmission wire is 24V a.c., given the resistance of the wire is 30"Ohm". so based on V/R = I, the current is 0.8A. the power loss due to heating? P = I^2 * R P = 0.8^2 * 30 = 19.2W, right? above are given question & answer from past year official exam. now my question, if we set voltage at 48V a.c., resistance is still 30"Ohm". so I = 1.6A the power loss due to heating = 1.6^2 * 30 = 76.8W. so, how "Increasing voltage reduce power loss in transmission" statement make sense?
7. ## A bit advice for me?

well let's see... half of the "must learn" subject is useless and most teacher are distracting rather than helping. + typical school day stress... thought so, really need to score good to get on on what I wanted/needed to learn. well, it's not that bad, but seriously you should have answer the most recent at that time which is not what you are answering.(You're answering the one which is around 2months ago.)
8. ## Understanding Electricity

Hey everyone, 4.(a) A student finds that the brightness of three bulbs is different in series circuit and in parallel circuit, although the three bulbs and the new dry cells used are identical. Each bulb is lablelled 3V 0.3A. Figures 3 and 4 show the brightness of the bulbs in series circuit and in parallel circuit respectively. (iii) Compare the brightness of the bulbs in each circuit. Relate the brightness of the of the bulbs to the potential difference and the current flow for the bulb s in series and parallel circuits. The confusion is based on the given answer: -> The total potential difference in a series circuit is the sum of the potential differences across each bulb. -> The current in a complete circuit of a series circuit is the same as the current flow through each bulb. -> Therefore, the brightness of each bulb in the series circuit is the same. (makes sense!) -> The potential difference across each bulb in a parallel circuit is the same as the potential difference across the cell. -> The total current in the parallel circuit is the sum of the currents that pass through each bulb. -> The brightness of the bulbs in the parallel circuit is more than that in the series circuit. (WTHell?!) I'm informed that Brightness = Power, right? so let analyse something here... let's give the voltage of the cell 3V, internal resistance ignored. The information that we have:- Bulb:- voltage: 3V current: 0.3A resistance: 10'Ohm' (R = V/I) power : 0.9W (P=VI) So, it's mean that the bulb will consume 0.9joule of electrical energy every second if it is connected to a 3V supply. Series circuit:- voltage: sum of bulb voltage. current: I of bulb = I of another bulb. Parallel Circuit:- voltage: V of bulb = V of another bulb. current: sum of bulb current. Given that P=VI, now, doesn't it's weird? I mean: For series circuit: The power of each bulb $P = (1/3)V * I$ 1/3 due to the sum of bulb voltage = total voltage. For parallel circuit: the power of each bulb $P = V * (1/3)I$ 1/3 due to the sum of bulb current = total current. Conclusion: The power should be equal, thus the brightness are just the same. I'm not done yet, it's posibble to "make" the bulb on parallel circuit glow brighter. based on Ohm's law, the resistance should be constant. So, voltage is directly proportional to current. Thus increasing the voltage would increase the current. and for P=VI.This create great amount of power difference between series(0.1W) and parallel(0.9W). but let's see the calculation: Series $P = 1V * 0.1A(10R=\frac{1V}{I}) = 0.1W$ Parallel $P = 3V * 0.3A = 0.9W$ The question is, where does the current come from? Induced from the Ohm's law equation? It's not make sense, I mean there given information that told both circuit use identical bulb(equal resistance) and identical cell(equal voltage supply). doesn't that mean total current should be equal too? hope anyone can give some insight. btw, theres one more. (b) An electrical circuit can be connected in series and in parallel. Given the advantage and disadvantage of a parallel circuit. given answer:- Advantage = The circuit can still be used if one of the bulbs is blown. Disadvantage = The brightness of the bulbs is the same as the brightness of a bulb in a single circuit. No problem here, just want to share for the amusing "Disadvantage" answer... ;P

10. ## Understanding Electricity

great, I got that. though for 35, I got another question, does the arrangement of the bulb (in series circuit) affect it brightness? I mean: either this, or the other way (the bulb exchange places). could it affect the brightness? based on your answer to 35, I don't get how my teacher claim there are brightness difference when changing the position due to the voltage amount would be depend on resistance anyway. btw,(for 35 too) in parallel circuit, the voltage is the same. so, any bulb that has higher current would create more power, right? thanks for others who replied too.
11. ## Understanding Electricity

22. The electromotive force and the internal resistance of the dry cell in the circuit below are 2V and 0.1(Ohm) respectively. What is the reading of the voltmeter when the switch is closed if the resistor is replaced with a larger resistance? a. 0V b. 2V c. less than 2V d. more than 2V I answered "d", but the given answer is "c". I thought that more resistance, more 'joule per coulomb' needed to push the charge, right? V = R*I, right? 35. Two bulbs rated 240V 60W and 240V 100W. are connected in series and then in parallel to the mains supply. Which of the following correctly describes the brightness of the 60W bulb compared with the 100 W bulb? Brightness (In series : In parallel) A. 60W > 100W : 60W > 100W B. 60W > 100W : 60W < 100W C. 60W < 100W : 60W < 100W D. 60W < 100W : 60W > 100W the answer is B, which mean 60W bulb is brighter in series circuit but dimmer in parallel circuit. I try using the Ohm's Law relationship (R=V/I) but failed to recognized which one is needed to determine the brightness, thought so, I got a presumption where the brightness depend on energy it receive (per second). But then I got messed up. help? thanks
12. ## Mathematic Coursework

Hey everyone, I'm doing my coursework for mathematics, the marks deduced from wrong answer is considerably huge. I am prone to mistake so I will take safe step by checking my answer here. Hope anyone can give a hand! note: I'm student of SPM, O-lvl equivalence. so, basically there no "hardcore math" here. though so it's recommend to use for higher mark. (silly marking) Section A, Part II. Cute Bakery produce 600 packets of muffin per day with each packet containing 5 pieces. Each muffin has a mass of 20g. Shop operates 5 days a week, eight hours a day, has three employees who each of them receives a salary of RM 150 per day. As a financial planner for this bakery, you are asked to make some suggestion about the selling price of the muffins produced, and the number of additional staff when needed. The table below shows the ratio of the price list and the ingredients used in making muffins. (note: the mass and total price based on my calculation, the others are given) On each day of operation, a total of 25kg of flour used. Minimum profit target for each day of operation as 20% (a) calculate the total cost of the ingredients in order to produce a cupcake $\frac{RM2.00 * 25 + RM8.00 * 20 + RM1.80 * 10 + 4.80 * 15}{5 * 600}$ = total cost of ingredients to produce 1 cupcake some note:- - the mass that I calculated based on the given ratio. - the 5 cupcake * 600 packets = 3,000 cupcake. (b) Utilities costs and arcillary materials amounting to RM 500 per day. i. Calculate the total operating costs per day for these shops. $RM300 + RM 500 + RM 150 * 3$ = total operating costs per day some note:- - RM150 * 3 workers = total salary per day. - RM300 = total cost of ingredients per day. ii. Calculate the total mass of cakes produce every day. $25kg + 20kg + 10kg + 15kg = 70kg$ = total mass of ingredients. X $3,000 * 20g = 60kg$ = total mass of muffin. / some note:- - note that "Each muffin has a mass of 20g" - 3,000 muffin * mass = total mass of muffin. P.S. so, total cupcake ingredient mass that are baked in this shop were reduced like 10kg everyday?! who smell the scorched cupcake? =/ iii. Suggest a reasonable selling price for each packet of muffin offered. $\frac{RM1250}{600} * \frac{120}{100} = RM2.50$ = price for each packet. some note:- - "RM1250" = total cost of operation per day. - "600" = total packet per day. - "120/100" = 20% profit. Note: "Minimum profit target for each day of operation was 20%." stop for now, there are more though. Thank you.
13. ## Understanding Vector

Thanks, I took some time to figure out about this because of my presumption told me that the collinearity of the point must be ABC or CBA instead of CAB or others. Almost get this, but an example of application in any kind of calculation would help me understand more, I mean $\vec{OC}$ used to determine the magnitude of something. So, what does the 'unit vector' used for? Anyway, based on my understanding, correct me if I'm wrong, the unit vector for of cartesian coordinate (i, j) is actually the "unit vector" that we are talking about. so basically it used as a scale?
14. ## Understanding Vector

Hey eveyone, until now I can't figured out what is "unit vector" in term of this formula: $\hat{r} = \frac{\vec{AB}}{||\vec{AB}||}$ especially when referring to the graph, unlike one of the components that it used $||\vec{AB}||$, which referring to the "magnitude/length of a vector". another question, 10. Given that $\vec{OA} = \lambda i$ and $\vec{OB} = 2i + 3j$ and $\vec{OC} = 3i - 4j$. Find the value of $\lambda$ if A, B and C are collinear. (note that i and j are the cartesian coordinates unit vector) per said that "A, B and C are collinear", I plot it like this, assuming the O is the origin. I just not understand what is the *lambda* are really referring too, did I'm missing something here? (btw, the answer is 17/7)
15. ## Confusion on the given tutorial.

Hey guys, now, I'm having problem get some intuitive on Chain Rule. Chain Rule Proof where it said: $\frac{g(x + h) - g(x)}{h} - g'(x) = v \rightarrow 0$, as $h \rightarrow 0$ but then, when I tried to: $\frac{g(x + 0) - g(x)}{0} - g'(x) = \frac{g(x) - g(x)}{0} - 1 = \frac{0}{0} - 1 = 0?$ doesn't $\frac{0}{0}$ = undefined?
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