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I want to make sure that this:
Reads: "The limit of a as it approaches infinity approximates (with increasing accuracy) the closed form of the bracketed series". Is this correct? Is there a better way of saying this?
164.11.204.51 02:35, 14 February 2007 (UTC)
The "right-hand side" means part of the equation after the equals sign. You don't have an equals sign (except for the range of the summation, which doesn't count), so you don't have an equation; you have an expression. The translation of an expression into English is not a sentence, but a noun phrase. So the verb "approximates" doesn't belong; you shouldn't have a verb at all. (Well, you can have them in subordinate clauses, but not as a predicate.) -- Trovatore 07:59, 14 February 2007 (UTC)
Hi! I did the intergral of Cosine squared X or Cos2X and well... a peer of mine got me double guessing myself.. I worked it out (if correctly!) to come out to be (1/4)(sin [2x] + x) + C i was wondering if that is correct or is it just [(Sin[2x])/4] + X + C. I know I know it sounds like a silly question--
Agester 03:34, 14 February 2007 (UTC)s with just the order of operations but help on this would greatly help me progress without thinking i'm a total idiot...
P.S. the difference is just the if the whole segment is over 4 or just the sine part
Thanks for input! --
Agester 03:00, 14 February 2007 (UTC)
You can always verify by differentiation. Dlong 03:12, 14 February 2007 (UTC)
Dlong 03:12, 14 February 2007 (UTC)
hmm... i see.. that does make sense! thank you very much! That saves me from a massive headache! -- Agester 03:34, 14 February 2007 (UTC)
I'm trying to figure out whether me or the book is right on this problem:
Let and
Find
I got does not exist (since one of the factors is which evaluates to negative one over zero), but the book has . Anyone know what's going on?
Curtmack of the Asylum 19:27, 14 February 2007 (UTC)
So I'm in highschool calculus and my head's about to explode. I'm trying to learn something from the examples in a book so I can do the hw, but I just don't get what happened. There's this equation in which I'm trying to separate out W in terms of t (time). F and k are constants. The original equation is:
so the book says to integrate both sides. They put the integral sign on both sides of the equation. Then they create a 1/k to the left of the integral on the left side, and counterbalance it by multiplying the numerator inside the integral by k (so it's just multiplying by a fancy form of 1, since the constant isn't affected by the integral sign). They do this because, "the differential of the denominator, d(F-kW), equals -kdW." so then it becomes integral of (-kdW)/(F-kW). Which they then say, because it's supposedly the "Integral of the reciprocal function," is ln|f-kW| But how the heck is it a reciprocal? it's essentially a derivative of a function over the function, not one over the function. How is that supposed to work? Thanks for any help, Sasha
Oh. Ok. So how would one perform similar mathematical magic to a new equation: integral of dM/(100-kM)? Do you multiply it by k over k again? Isn't it sort of deriv over function again? I'm having trouble really getting what the book is saying to do in these cases. And why.
And thanks for the help on that first one.
So I've hit a new one, if you guys don't mind helping again. The new equation is E = RI + L(dI/dt). E, L, and R are constants. I'm trying to solve for I in terms of t. Apparently, says the answers at the back of the book, this can be rearranged and integrated into
I can't figure this one out at all. How does one isolate the I? I tried moving the RI to the left, then multiplying by dt and fiddling around with other things, but I'm just digging myself into a whole. Is the integral of RI + L(dI/dt) equal to the integral of RI plus the integral of L(dI/dt)? Do you think this'll get me anywhere?
Awesome. That helped a lot. I just never know in what direction to start with this stuff.
Mathematics desk | ||
---|---|---|
< February 13 | << Jan | February | Mar >> | February 15 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
I want to make sure that this:
Reads: "The limit of a as it approaches infinity approximates (with increasing accuracy) the closed form of the bracketed series". Is this correct? Is there a better way of saying this?
164.11.204.51 02:35, 14 February 2007 (UTC)
The "right-hand side" means part of the equation after the equals sign. You don't have an equals sign (except for the range of the summation, which doesn't count), so you don't have an equation; you have an expression. The translation of an expression into English is not a sentence, but a noun phrase. So the verb "approximates" doesn't belong; you shouldn't have a verb at all. (Well, you can have them in subordinate clauses, but not as a predicate.) -- Trovatore 07:59, 14 February 2007 (UTC)
Hi! I did the intergral of Cosine squared X or Cos2X and well... a peer of mine got me double guessing myself.. I worked it out (if correctly!) to come out to be (1/4)(sin [2x] + x) + C i was wondering if that is correct or is it just [(Sin[2x])/4] + X + C. I know I know it sounds like a silly question--
Agester 03:34, 14 February 2007 (UTC)s with just the order of operations but help on this would greatly help me progress without thinking i'm a total idiot...
P.S. the difference is just the if the whole segment is over 4 or just the sine part
Thanks for input! --
Agester 03:00, 14 February 2007 (UTC)
You can always verify by differentiation. Dlong 03:12, 14 February 2007 (UTC)
Dlong 03:12, 14 February 2007 (UTC)
hmm... i see.. that does make sense! thank you very much! That saves me from a massive headache! -- Agester 03:34, 14 February 2007 (UTC)
I'm trying to figure out whether me or the book is right on this problem:
Let and
Find
I got does not exist (since one of the factors is which evaluates to negative one over zero), but the book has . Anyone know what's going on?
Curtmack of the Asylum 19:27, 14 February 2007 (UTC)
So I'm in highschool calculus and my head's about to explode. I'm trying to learn something from the examples in a book so I can do the hw, but I just don't get what happened. There's this equation in which I'm trying to separate out W in terms of t (time). F and k are constants. The original equation is:
so the book says to integrate both sides. They put the integral sign on both sides of the equation. Then they create a 1/k to the left of the integral on the left side, and counterbalance it by multiplying the numerator inside the integral by k (so it's just multiplying by a fancy form of 1, since the constant isn't affected by the integral sign). They do this because, "the differential of the denominator, d(F-kW), equals -kdW." so then it becomes integral of (-kdW)/(F-kW). Which they then say, because it's supposedly the "Integral of the reciprocal function," is ln|f-kW| But how the heck is it a reciprocal? it's essentially a derivative of a function over the function, not one over the function. How is that supposed to work? Thanks for any help, Sasha
Oh. Ok. So how would one perform similar mathematical magic to a new equation: integral of dM/(100-kM)? Do you multiply it by k over k again? Isn't it sort of deriv over function again? I'm having trouble really getting what the book is saying to do in these cases. And why.
And thanks for the help on that first one.
So I've hit a new one, if you guys don't mind helping again. The new equation is E = RI + L(dI/dt). E, L, and R are constants. I'm trying to solve for I in terms of t. Apparently, says the answers at the back of the book, this can be rearranged and integrated into
I can't figure this one out at all. How does one isolate the I? I tried moving the RI to the left, then multiplying by dt and fiddling around with other things, but I'm just digging myself into a whole. Is the integral of RI + L(dI/dt) equal to the integral of RI plus the integral of L(dI/dt)? Do you think this'll get me anywhere?
Awesome. That helped a lot. I just never know in what direction to start with this stuff.