root mean square
(noun)
The square root of the arithmetic mean of the squares.
Examples of root mean square in the following topics:
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Root Mean Square Values
- The root mean square (RMS) voltage or current is the time-averaged voltage or current in an AC system.
- Given the current or voltage as a function of time, we can take the root mean square over time to report the average quantities.
- The root mean square (abbreviated RMS or rms), also known as the quadratic mean, is a statistical measure of the magnitude of a varying quantity.
- It is especially useful when the function alternates between positive and negative values, e.g., sinusoids.The RMS value of a set of values (or a continuous-time function such as a sinusoid) is the square root of the arithmetic mean of the squares of the original values (or the square of the function).
- Relate the root mean square voltage and current in an alternating circut with the peak voltage and current and the average power
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Nuclear Size and Density
- However, the nucleus can be modelled as a sphere of positive charge for the interpretation of electron scattering experiments: because there is no definite boundary to the nucleus, the electrons "see" a range of cross-sections, for which a mean can be taken.
- The qualification of "rms" (for "root mean square") arises because it is the nuclear cross-section, proportional to the square of the radius, which is determining for electron scattering.
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Temperature
- The components of its velocity momentum in the y- and z-directions are not changed, which means there is no force parallel to the wall.
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Phase Angle and Power Factor
- It can be shown that the average power is IrmsVrmscosϕ, where Irms and Vrms are the root mean square (rms) averages of the current and voltage, respectively.
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Damped transient motion
- But the second term may or may not be a sinusoid, depending on whether the square root is positive.
- First if $\frac{\gamma}{2\omega_0} < 1$ , corresponding to small damping, then the argument of the square root is positive and indeed we have a damped sinusoid.
- where, once again, we have arranged things so that the argument of the square root is positive.
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Orthogonal decomposition of rectangular matrices
- The singular values are the square roots of the eigenvalues of $A^TA$ , which are the same as the nonzero eigenvalues of $AA^T$ .
- Keep in mind that the matrices $U$ and $V$ whose columns are the model and data eigenvectors are square (respectively $n \times n$$m \times m$ and $m \times m$ ) and orthogonal.
- It is important to keep the subscript $r$$A$ in mind since the fact that $A$ can be reconstructed from the eigenvectors associated with the nonzero eigenvalues means that the experiment is unable to see the contribution due to the eigenvectors associated with zero eigenvalues.
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Eigenvalues and Eigenvectors
- That means we must choose $\lambda$$\mbox{Det}(A - \lambda I) = 0$ so that $\mbox{Det}(A - \lambda I) = 0$ .
- Can you imagine having to compute the roots of a 10-th order polynomial?
- Therefore $AS = S\Lambda$ which means that $S^{-1} AS = \Lambda$ .
- Not all square matrices possess $n$ linearly independent eigenvectors.
- So the roots are both 0.
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Introduction to Least Squares
- where we have used $x _{\rm ls}$ to denote the least squares value of $x$ .
- In other words the value of $x$ that minimizes the sum of squares of the errors is just the mean of the data.
- In other words, find a linear combination of the columns of $A$ that is as close as possible in a least squares sense to the data.
- Now $A$$A \mathbf{x_{ls}}$ applied to the least squares solution is the approximation to the data from within the column space.
- This is just the sum of the squared errors, but for $n$$m$$m$$m$ simultaneous equations in $m$ unknowns.
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The Speed of a Wave on a String
- The speed of a wave on this kind of string is proportional to the square root of the tension in the string and inversely proportional to the square root of the linear density of the string:$v=\sqrt{\frac{T}{\mu}}$
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Another velocity-dependent force: the Zeeman effect
- Specifically, let's assume that $\frac{qB}{m} << \omega_0$ so the square root reduces to $2 \omega _0$ .
- We take the positive square root since otherwise we would have negative frequencies.