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[PDF]Shot Noise in Photodiodes and Extraction of the ... - 北京大学
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我们将在后⾯实际对这个−1/2 幂次律性质进⾏检验. 2.4 系统噪声的测量. ⾸先，我们对PCI4474 数据采集卡的背景噪声进⾏测量. 我们使⽤ BNC 短. 接头将数据采集 ...移动载荷作用下漂浮薄板水弹性响应的渐近分析--《上海大学 ...
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由 张辉 著作 - 2013
当载荷移动速度不等于临界速度时,瞬态波阻和波形均随时间衰减到稳态;而当移动速度等于临界速度时,瞬态波阻和波形均随时间以1/2幂次增加,无法达到稳态. 3)针对 ...音乐快递：布朗运动01：朗之万方程其中势函数在平衡态附近 ...
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2011年5月30日 - 这里的1/2幂次出现在高分子构象统计等许多涉及随机运动的理论中。 离散的无规行走问题本身早已经发展成一个活跃的研究领域。最简单的等步长 ...布朗運動理論一百年_HPNET的主頁_hpnet的和訊博客
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2009年5月31日 - 這裏的1/2冪次出現在高分子構象統計等許多涉及隨機運動的理論中。 離散的無規行走問題本身早已經發展成一個活躍的研究領域。最簡單的等步長 [PDF]布朗運動、郎之萬方程式、與布朗動力學
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但現代對布朗運動的理論描述常採用較易瞭解的朗之萬(Langevin)理論，其特點是我們可以將 ... 本文將就朗之萬方程 ... 在無外力作用下，一維空間中的布朗運動可寫成，. 2 .... 速度。藉由這種方式，研究者能夠直接研究布朗粒子. 間的相互作用，及如何受外加力場的影響；利用電腦 ... 方程式(5)對時間t 積分，Ermak 等人得到簡單的計算.朗之萬方程(Langevin's equation) 描述布朗運動
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朗之萬方程(Langevin's equation) 描述布朗運動 ... 當不存在其它外力時F'(t)=0，朗[PDF]布朗运动理论一百年1)
经典和量子耗散系统的随机模拟方法_百度百科
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... 与涨落耗散、朗之万方程的数值模拟及其策略、主方程的蒙特卡罗模拟、反常扩散的 ... 简谐噪声的频域带宽5．7 简谐速度噪声5．7．1 简谐速度噪声的关联函数5．7．朗之万方程- 维基百科，自由的百科全书
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在统计物理中，朗之万方程（保罗·朗之万，1908年）是一个描述自由度子集的时间 ... 作用在粒子上的力写成正比于粒子的速度（斯托克斯定律）的粘滞力，和一个表示 ...朗之万方程,langevin equation,音标,读音,翻译,英文例句,英语 ...
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由颗粒运动的朗之万方程出发,对流体脉动速度采用扩维方法,把色噪声转化为白噪声,再降 ... 该解法从自由状态空间中的广义朗之万梯度方程出发,利用常数变易法导出了与广义朗之万方程等价的广义的第二类非线性、随机性Valterra积分方程,采用逐次 ...音乐快递：布朗运动01：朗之万方程其中势函数在平衡态附近 ...
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2011年5月30日 - 我们先不把随机力F(t)具体化，直接对线性的朗之万方程求积分： .... 扩散项的系数矩阵，而Yi是支撑起随机过程的空间中的“场”量，例如坐标或速度。
Why Is an Exponent of 1/2 the Same as a Square Root?
Date: 05/05/2004 at 09:05:30 From: Will Subject: Algebra 2 Why is raising a number to the 1/2 power the same as taking the square root of the number? For example, 36^(1/2) = 6. It's not confusing, Iam just wondering why that is the only number that can do that. It seems pretty interesting to me, so could you please answer?
Date: 05/05/2004 at 14:03:02 From: Doctor Terrel Subject: Re: Algebra 2 Hi Will - To show you why the "1/2 power" of a number means the same as "square root" of the number, I need to give a little background first. Let's try this... The square root of 9 is 3 because 3^2 = 3 * 3 = 9. The square root of 64 is 8 because 8^2 = 8 * 8 = 64. In summary, the square root of some number N is a value that when multiplied by itself (or squared) produces the given number N. Or in other words, r will be the square root of N if r^2 or r * r = N. Are you familiar with the basic laws of exponents? Let's apply some of them to this question. We'll start with: x^a * x^b = x^(a + b) This law says that when you multiply like bases, you keep the base and add the exponents, as shown here: x^2 * x^3 = (x*x)*(x*x*x) = x*x*x*x*x = x^5, which is x^(2 + 3) Let's try using that law with the 1/2 power: x^(1/2) * x^(1/2) = x^(1/2 + 1/2) = x^1 = x But look at what just happened! We multiplied x^(1/2) by itself (or squared it) and we got x. According to our earlier summary/definition of the square root, that means that x^(1/2) must be the square root of x. Can you see that? Here's one more way to look at it using exponent laws. There is a law that says that a power raised to a power is the product of the powers. In other words: (x^2)^3 = x^(2*3) or x^6 This is actually an extension of the exponent addition rule we already looked at, since (x^2)^3 = x^2 * x^2 * x^2 = x^(2 + 2 + 2) = x^6 Let's suppose for a moment that we don't know how to write a square root as an exponent, and we'll try to figure out what would work. We know that when we square the square root of x we will get x, as we defined above. So if we let n be this unknown exponent that represents the square root of x, we know that (x^n)^2 = x Applying our "power to a power" rule, we can rewrite that equation as x^(2n) = x^1 Since the bases are the same on each side of the equation (x), and the two quantities are equal, the exponents must also be the same: 2n = 1 n = 1/2 Ahah! The mystery exponent n that represents the square root turns out to be 1/2. You can see why it works: [x^(1/2)]^2 = x^[(1/2) * 2] = x^1 = x Going back to your example: [36^(1/2)] * [36^(1/2)] = [36^(1/2)]^2 = 36^[(1/2) * 2] = 36^1 = 36 Since multiplying 36^(1/2) by itself (or squaring it) gave 36, 36^(1/2) must in fact be the square root of 36. Hope this helps. Good luck. Write again if need be...
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