Hood Posted October 27, 2016 Report Share Posted October 27, 2016 On 26/09/2016 at 7:25 AM, Silverblade5 said: I recently encountered a circuit where the resistors are in a triangular mesh that's neither in series nor in parallel. I can't figure out how to network it. Anyone that can help? You can use Wye-Delta transformation. I assume that the resistors are arranged like 'delta' (or like triangle). If it has been solved (which I am assuming) can you show us the question and the answer. 0 Quote Link to comment Share on other sites More sharing options...

Jedal Posted October 28, 2016 Report Share Posted October 28, 2016 Hey boyz wanna give this memer some help on his Geometry? I gotta quiz tomorrow and I aint studyin for that... how do i kno stufffs about isoseles triangles an that kinda crap? 1 Quote Link to comment Share on other sites More sharing options...

Spoolofwhool Posted October 28, 2016 Report Share Posted October 28, 2016 2 hours ago, Orderbound said: Hey boyz wanna give this memer some help on his Geometry? I gotta quiz tomorrow and I aint studyin for that... how do i kno stufffs about isoseles triangles an that kinda crap? Right angle triangles have a right angle. 1 Quote Link to comment Share on other sites More sharing options...

Silverblade5 Posted January 27, 2017 Author Report Share Posted January 27, 2017 @Chaos Calc question: How would I integrate a rational function that can't be expanded using partial fractions do to having an unfactorable denominator? Example: S|1/(x^2+1)|dx 0 Quote Link to comment Share on other sites More sharing options...

Chaos Posted January 27, 2017 Report Share Posted January 27, 2017 Just now, Silverblade5 said: @Chaos Calc question: How would I integrate a rational function that can't be expanded using partial fractions do to having an unfactorable denominator? Example: S|1/(x^2+1)|dx The integral you have is arctan(x). So generally you'll have an arctan term. 0 Quote Link to comment Share on other sites More sharing options...

Silverblade5 Posted January 27, 2017 Author Report Share Posted January 27, 2017 Just now, Chaos said: The integral you have is arctan(x). So generally you'll have an arctan term. Just to clarify: if the denominator is a quadratic that, when solved for x, yields an imaginary number for solutions, it can generally be integrated in terms of arctan? 0 Quote Link to comment Share on other sites More sharing options...

Chaos Posted January 27, 2017 Report Share Posted January 27, 2017 Any irreducible quadratic can be written in the form (x-a)^2 + b^2 (the plus here makes it irreducible). And one over that denominator is a thing that you can manipulate into a 1/(u^2+1) integral, which leads to arctangent. 2 Quote Link to comment Share on other sites More sharing options...

Silverblade5 Posted January 27, 2017 Author Report Share Posted January 27, 2017 @Chaos So, I popped this function into the integral calculator, and it couldn't give me an answer. Think you might be able to help? S|1/(x^3-1)^.5|dx 0 Quote Link to comment Share on other sites More sharing options...

Chaos Posted January 27, 2017 Report Share Posted January 27, 2017 That, to my recollection, does not have a closed-form antiderivative. That x^3 makes it totally different from the x^2 version. So there is not a nice easy Calc 2 way to do that. If it was x^2... 0 Quote Link to comment Share on other sites More sharing options...

Kaymyth Posted January 27, 2017 Report Share Posted January 27, 2017 Harvard scientists managed to smoosh hydrogen down into what they believe is a solid, metallic form. They're going to leave it smooshed between diamond tips, run some lasers through it, and see if it breaks. If it doesn't, they'll release the pressure and test to see whether it will remain in metallic form even after the pressure is released. If it does, this is going to mean major advances in superconducting technology. 3 Quote Link to comment Share on other sites More sharing options...

Glamdring804 Posted January 27, 2017 Report Share Posted January 27, 2017 17 minutes ago, Silverblade5 said: @Chaos So, I popped this function into the integral calculator, and it couldn't give me an answer. Think you might be able to help? S|1/(x^3-1)^.5|dx This is just painful to look at. Don't tell me someone actually gave you this for a homework? 6 minutes ago, Kaymyth said: Harvard scientists managed to smoosh hydrogen down into what they believe is a solid, metallic form. They're going to leave it smooshed between diamond tips, run some lasers through it, and see if it breaks. If it doesn't, they'll release the pressure and test to see whether it will remain in metallic form even after the pressure is released. If it does, this is going to mean major advances in superconducting technology. Wat. Metallic hydrogen? Oh man, this is awesome. We just made ourselves a little piece of Jupiter. 1 Quote Link to comment Share on other sites More sharing options...

Silverblade5 Posted January 27, 2017 Author Report Share Posted January 27, 2017 (edited) 7 minutes ago, Glamdring804 said: This is just painful to look at. Don't tell me someone actually gave you this for a homework? Wat. Metallic hydrogen? Oh man, this is awesome. We just made ourselves a little piece of Jupiter. No, I just enjoy learning. This was something I tried to teach myself. I failed. Now I'm asking a college teacher for help. Plus, I enjoy making teachers actually think about their subjects Edited January 27, 2017 by Silverblade5 0 Quote Link to comment Share on other sites More sharing options...

Glamdring804 Posted January 27, 2017 Report Share Posted January 27, 2017 15 minutes ago, Silverblade5 said: No, I just enjoy learning. This was something I tried to teach myself. I failed. Now I'm asking a college teacher for help. Plus, I enjoy making teachers actually think about their subjects You seem to be doing pretty well. If you're working on partial fractions, you've already gotten half way through Calc 2. Come to think of it, you'd probably really enjoy my honors physics classes. They storming hard, but certainly a lot of fun. 0 Quote Link to comment Share on other sites More sharing options...

Silverblade5 Posted January 27, 2017 Author Report Share Posted January 27, 2017 7 hours ago, Glamdring804 said: You seem to be doing pretty well. If you're working on partial fractions, you've already gotten half way through Calc 2. Come to think of it, you'd probably really enjoy my honors physics classes. They storming hard, but certainly a lot of fun. Probably. I'm taking AP Physics C next year. That has two AP tests at the end of the year. Mechanics and electromagnetism 1 Quote Link to comment Share on other sites More sharing options...

Spoolofwhool Posted January 29, 2017 Report Share Posted January 29, 2017 On 1/27/2017 at 0:48 AM, Kaymyth said: Harvard scientists managed to smoosh hydrogen down into what they believe is a solid, metallic form. They're going to leave it smooshed between diamond tips, run some lasers through it, and see if it breaks. If it doesn't, they'll release the pressure and test to see whether it will remain in metallic form even after the pressure is released. If it does, this is going to mean major advances in superconducting technology. I read somewhere else which says that this has been taken with skepticism in some communities. They're not convinced they actually obtained it, and that they're just misinterpreting something else that's there. Unfortunate if it is an error, but if it isn't, awesome. 0 Quote Link to comment Share on other sites More sharing options...

Kaymyth Posted January 30, 2017 Report Share Posted January 30, 2017 20 hours ago, Spoolofwhool said: I read somewhere else which says that this has been taken with skepticism in some communities. They're not convinced they actually obtained it, and that they're just misinterpreting something else that's there. Unfortunate if it is an error, but if it isn't, awesome. Step One: Do the thing. Step Two: Confirm the thing. Step Three: Repeat the thing. Step Four: Other people repeat the thing. Step Five: Party! Clearly, we have completed Step One and are working on Step Two. 3 Quote Link to comment Share on other sites More sharing options...

Silverblade5 Posted February 8, 2017 Author Report Share Posted February 8, 2017 (edited) So, I did some math recently, and came up with the following information: cos(2 * x) = 2cos^2(x) - 1 cos(2^2 * x) = 2(2cos^2(x) - 1)^2 - 1 cos(2^3 * x) = 2(2(2cos^2(x) - 1)^2 - 1)^2 - 1 cos(2^4 * x) = 2(2(2(2cos^2(x) - 1)^2 - 1)^2 - 1)^2 - 1 Anyway I might be able to condense this into a series representation for cos(2^a * x)? @Chaos@Glamdring804 Since you're my go to people for calc advice Edited February 8, 2017 by Silverblade5 1 Quote Link to comment Share on other sites More sharing options...

Glamdring804 Posted February 9, 2017 Report Share Posted February 9, 2017 6 hours ago, Silverblade5 said: So, I did some math recently, and came up with the following information: cos(2 * x) = 2cos^2(x) - 1 cos(2^2 * x) = 2(2cos^2(x) - 1)^2 - 1 cos(2^3 * x) = 2(2(2cos^2(x) - 1)^2 - 1)^2 - 1 cos(2^4 * x) = 2(2(2(2cos^2(x) - 1)^2 - 1)^2 - 1)^2 - 1 Anyway I might be able to condense this into a series representation for cos(2^a * x)? @Chaos@Glamdring804 Since you're my go to people for calc advice Well, I'm not a mathematician, (I'm a physicist, we specialize in doing terrible math because it works) but a series is a summation of terms, and it would be fairly difficult to pull that stuff out of the embedded cosines. @Chaos, you are probably a lot more qualified for this, what do you think? 0 Quote Link to comment Share on other sites More sharing options...

Silverblade5 Posted February 9, 2017 Author Report Share Posted February 9, 2017 2 minutes ago, Glamdring804 said: Well, I'm not a mathematician, (I'm a physicist, we specialize in doing terrible math because it works) but a series is a summation of terms, and it would be fairly difficult to pull that stuff out of the embedded cosines. @Chaos, you are probably a lot more qualified for this, what do you think? What I currently know is that it's a single function, y = 2x^2 -1, where x = cos(x), and the output is constantly being fed into the input a certain number of times. The amount of times the function is repeated is determined by the value of 'a'. 0 Quote Link to comment Share on other sites More sharing options...

Chaos Posted February 9, 2017 Report Share Posted February 9, 2017 I don't have a lot of time, but in the way you wrote it, it is not conducive for a Taylor series approximation in the higher orders as you'd need to expand a lot of stuff out. It'd be useful if you just wanted a first order approximation, though. I will also tell you advice on trig identities: at some point it's way easier to just derive them all in the complex plane. (That's how most are derived, and would be very tedious to do in the reals) 0 Quote Link to comment Share on other sites More sharing options...

Dunkum Posted February 9, 2017 Report Share Posted February 9, 2017 I multiplied the first 2 or 3 terms out and didn't see any particularly easy patterns to turn into sequence notation, but as Chaos notes, the trig functions have interesting forms in the complex plane. it is not impossible that what would be an intractable expression in the reals turns out to be a lot simpler when put in terms of e and i. 0 Quote Link to comment Share on other sites More sharing options...

Pattern Posted February 16, 2017 Report Share Posted February 16, 2017 (edited) Ah well, my crappy mobile post wasn't posted at all - no matter. You might use just a recursive representation, like cos(2^(a+1)*x)=2*cos^2(2^a*x) - 1, but that's probably not what you want. For a series you will also need sin^n(b*x) terms. The Moivre formula comes to mind here, if I remember correctly it is derived from: cos(nx)+i sin(nx)= exp(i*nx)=(exp(i*x))^n=(cos(x)+i*sin(x))^n (*) You can expand this binomial using binomial coefficients which i have no idea how to type in this forum, so let me just define bin(n,k):=n!/(k!*(n-k)!) (*)=Sum(k=0...n) i^k*bin(n,k)*cos^(n-k)(x)*sin^k(x) (In this sum k runs from 0 to n) Collecting real and imaginary terms yields formulae for cos(nx) (real) and sin (nx) (imaginary). i^2=-1 leads to alternating signs. So cos(nx)=cos^n(x) - bin(n,2)*cos^(n-2)(x)*sin^2(x) + bin(n,4)*cos^(n-4)(x)*sin^4(x) - + - +... set n=2^a and you've got a series for your problem. But I am really not sure whether that is as handy as desired... If you are uncomfortable with the sin(x) terms you can still use sin^2(x)+cos^2(x)=1 and replace the sin(x) terms (which all have an even exponent, so no roots involved). Best ingredients for a pen and paper evening ;-) Edited February 16, 2017 by Pattern 0 Quote Link to comment Share on other sites More sharing options...

Silverblade5 Posted February 19, 2017 Author Report Share Posted February 19, 2017 So, class projects for my digital electronics class were recently announced, and I was thinking of creating a soundboard by using timer circuits to alter the frequencies of a signal and feed it through a buzzer to make a sound. All the timer circuits will be in parallel with each other, and will also be connected with a push button switch. What do you people think of this idea? @Glamdring804 since this is basically applied physics. 0 Quote Link to comment Share on other sites More sharing options...

Glamdring804 Posted February 20, 2017 Report Share Posted February 20, 2017 18 hours ago, Silverblade5 said: So, class projects for my digital electronics class were recently announced, and I was thinking of creating a soundboard by using timer circuits to alter the frequencies of a signal and feed it through a buzzer to make a sound. All the timer circuits will be in parallel with each other, and will also be connected with a push button switch. What do you people think of this idea? @Glamdring804 since this is basically applied physics. Well...That's actually more of electrical engineering. If you showed me the circuit diagram, I could tell you the equivalent resistance, the current drawn from the source, etc. I haven't yet taken advanced enough courses to really evaluate your idea from a physics perspective. But, I can tell you that it sounds pretty cool to me. 0 Quote Link to comment Share on other sites More sharing options...

Silverblade5 Posted May 5, 2017 Author Report Share Posted May 5, 2017 @Chaos, @Glamdring804, I think you two might enjoy this. It always bothered me that when looking at a table of trig identities, there wasn't a sum to product identity for sin(a) + cos(b). It's actually a really simple one, but the derivation involves the culmination of almost all the other ones, so it's really elegant in that regard. Below is the identity, and below that will be the derivation. sin(a) + cos(b) = ((1 + sin(a+b))(1 + sin(a-b)))^.5 Derivation: Spoiler sin(a) + cos(b) sin(a) + cos((a+b) - a) sin(a) + cos(a)cos(a+b) + sin(a)sin(a+b) cos(a)cos(a+b) + sin(a)(1 + sin(a+b)) let u = cos(a+b) and let v = 1 + sin(a+b) ucos(a) + vsin(a) = rcos(c-a) ucos(a) + vsin(a) = rcos(c)cos(a) + rsin(c)sin(a) Solve for c and r in terms of u and v u = rcos(c) v = rsin(c) r = u/cos(c) v = usin(c)/cos(c) v = utan(c) c = arctan(v/u) r = u/cos(arctan(v/u)) cos(arctan(v/u)) = u/(u^2 + v^2)^.5 r = (u^2 + v^2)^.5 With back-substituting, we get r = (cos^2(a+b) + (1 + sin(a+b))^2)^.5 r = (cos^2(a+b) + sin^2(a+b) + 2sin(a+b) + 1)^.5 r = (2 + 2sin(a+b))^.5 r = (2(1 + sin(a+b)))^.5 Now lets define c. c = arctan(v/u) c = arctan((1 + sin(a+b) ) / (0 + cos(a+b))) c = arctan((sin(pi/2) + sin(a+b)) / (cos(pi/2) + cos(a+b))) c = arctan((2sin((pi/2 + a + b)/2)cos((pi/2 - a - b)/2)) / (2cos((pi/2 - a - b)/2)cos((pi/2 + a + b)/2))) Factor out 2cos((pi/2 - a - b)/2), and you get c = arctan(sin((pi/2 + a + b)/2) / cos((pi/2 + a + b)/2)) c = arctan(tan((pi/2 + a + b)/2)) c = (pi/2 + a + b)/2 If we plug all this into the formula ucos(a) + vsin(a) = rcos(c-a), we get cos(a)cos(a+b) + sin(a)(1 + sin(a+b)) = (2(1 + sin(a+b)))^.5 * cos((pi/2 + a + b)/2 - a) Now lets simplify the bit that came from cos(c-a). cos((pi/2 + a + b)/2 - a) = cos((pi/2 - a + b)/2) = ((1 + cos(pi/2 - a + b))/2)^.5 = ((1 + cos(pi/2 - a)cos(b) - sin(pi/2 - a)sin(b))/2)^.5 = ((1 + sin(a)cos(b) - cos(a)sin(b))/2)^.5 = ((1 + sin(a-b))/2)^.5 (2(1 + sin(a+b)))^.5 * ((1 + sin(a-b))/2)^.5 = ((1 + sin(a+b))(1 + sin(a-b)))^.5 Therefore, sin(a) + cos(b) = ((1 + sin(a+b))(1 + sin(a-b)))^.5 3 Quote Link to comment Share on other sites More sharing options...

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