Kumaragupta I

(noun)

Son of Chandragupta II; the emperor of the Gupta Dynasty from c. 415-455 CE.

Related Terms

Examples of Kumaragupta I in the following topics:

  • Decline of the Gupta Empire

    • Yet the succeeding rulers, beginning with Kumaragupta I and then Skandagupta, oversaw the eventual end of the Gupta Empire through military defeats, devalued money and withering leadership.
    • In 415 CE, Chandragupta II was succeeded by his second son, Kumaragupta I, who ruled successfully until 455 CE.
    • As his grandfather and father did before him, Kumaragupta also issued news coins to mark his reign.
    • Upon Kumaragupta’s death in 455 CE, his son, Skandagupta, assumed the throne and ruled until c. 467 CE.
    • A silver coin from the reign of Gupta Emperor Kumaragupta I, c. 415-455 CE.

  • Repeated Scattering with Low Optical Depth

    • $\displaystyle A \equiv \frac{E_f}{E_i} \sim \frac{4}{3} \langle \gamma^2 \rangle = 16 \left ( \frac{kT}{mc^2} \right )^2.$
    • The probability that a photon will scatter as it passes through a medium is simply $\tau_{es}$ if the optical depth is low, and the probability that it will undergo $k$ scatterings $p_k \sim \tau_{es}^k$ and its energy after $k$ scatterings is $E_k=A^k E_i$, so we have
    • $\displaystyle I(E_k) = I(E_i) \exp \left ( \frac{\ln\tau_{es} \ln\frac{E_k}{E_i}}{\ln A} \right ) = I(E_i) \left ( \frac{E_k}{E_i} \right )^{-\alpha}$
    • $\displaystyle P = \int_{E_i}^{A^{1/2}mc^2} I(E_k) dE_k = I(E_i) E_i \left [ \int_1^{A^{1/2} mc^2/E_i} x^{-\alpha} dx \right ].$

  • Eigenvectors and Orthogonal Projections

    • $\left( \mathbf{v}_i \mathbf{v}_i ^T \right) \mathbf{x} = (\mathbf{v}_i ^T \mathbf{x}) \mathbf{v}_i $
    • For the operator $\mathbf{v}_i \mathbf{v}_i ^T $ this is obviously true since $\mathbf{v}_i ^T \mathbf{v}_i = 1$ .
    • $\displaystyle{\sum _ {i=1} ^ m \mathbf{v}_i \mathbf{v}_i ^T = V V^ T = I . }$
    • $\displaystyle{\sum _ {i=1} ^ r \mathbf{v}_i \mathbf{v}_i ^T = V_r V_r ^ T }$
    • $\displaystyle{\sum _ {i=r+1} ^ m \mathbf{v}_i \mathbf{v}_i ^T = V_0 V_0 ^T}$

  • Alexander I's Domestic Reforms

  • Fourth (Or I) Conjugation

  • Fitting a Curve

    • $\displaystyle \begin{aligned} b&= \frac{1}{n} \sum_{i=1}^{n} y_{1} - m \frac{1}{n} \sum_{i=1}^{n} x_{i} \\ &= \left (\bar{y} - m \bar{x} \right) \end{aligned}$
    • Let $\bar{y}$, pronounced $y$-bar, represent the mean (or average) $y$ value of all the data points: $\bar y =\frac{1}{n}\sum_{i=1}^{n} y_{i}$.
    • Respectively $\bar{x}$, pronounced $x$-bar, is the mean (or average) $x$ value of all the data points: $\bar x=\frac{1}{n}\sum_{i=1}^{n} x_{i}$.
    • $\displaystyle \begin{aligned} \sum_{i=1}^{n}x_{i}y_{i}&=0+0+1+4+3+10+15+24\\&=57 \end{aligned} $$\displaystyle \begin{aligned} \sum_{i=1}^{n}x_{i}&=-1+0+1+2+3+4+5+6\\&=20 \end{aligned}$$\displaystyle \begin{aligned} \sum_{i=1}^{n}y_{i}&=0+0+1+2+1+2.5+3+4\\&=13.5 \end{aligned}$

  • Complex Numbers and the Binomial Theorem

    • In what follows, it is useful to keep in mind the powers of the imaginary unit $i$:
    • $(2+3i)^4=2^4+4\cdot 2^3\cdot 3i -6\cdot 2^2\cdot 3^2 -4\cdot 2 \cdot 3^3 i +3^4 $
    • Suppose you wanted to compute $(1+i)^3$.
    • Suppose you wanted to compute $(2+i)^5$.
    • $(2+i)^5 =2^5 + 5\cdot 2^4 i + 10\cdot 2^3 i^2 + 10\cdot 2^2 i^3 + 5\cdot 2 \cdot i^4 + i^5$

  • The paradox of pay

    • I am a pro, I get paid pretty good for playing ball.
    • As I have discovered, teaching is the most rewarding thing I can think of doing.
    • I do not just teach; I am a teacher.
    • I am glad I am paid for my work, but truth be known I would do it for free.
    • I walk away from a class where the students and I have really "lit it up", and I do not even have words to say how good it feels.

  • The Radiative Transfer Equation

    • $\displaystyle I_\nu (s) = I_\nu(s_0) \exp \left [ -\int_{s_0}^s \alpha_\nu (s') ds' \right ]$
    • $\displaystyle I_\nu (\tau_\nu) = I_\nu(s_0)e^{-\tau_\nu} + \int_0^{\tau_\nu} e^{-(\tau_\nu - \tau_\nu')} S_\nu(\tau_\nu') d\tau_\nu'.$
    • $\displaystyle{ I_\nu (\tau_\nu) = I_\nu(s_0)e^{-\tau_\nu} + S_\nu \left ( 1 - e^{-\tau_\nu} \right ) = S_\nu + e^{-\tau_\nu} \left ( I_\nu(0) - S_\nu \right ). }$

  • Orthogonal decomposition of rectangular matrices

    • And since $S$ is symmetric it has orthogonal eigenvectors $\mathbf{w}_i$$\lambda _ i$ with real eigenvalues $\lambda _ i$
    • So we see, once again, that the model eigenvectors $\mathbf{u}_i$ are eigenvectors of $AA^T$ and the data eigenvectors $\mathbf{v}_i$ are eigenvectors of $A^TA$ .
    • Also note that if we change sign of the eigenvalue we see that $(-\mathbf{u}_i,\mathbf{v}_i)$ is an eigenvector too.
    • So if there are $r$ pairs of nonzero eigenvalues $\pm \lambda _i$ then there are $r$ eigenvectors of the form $(\mathbf{u}_i,\mathbf{v}_i)$$\lambda_i$ for the positive $\lambda_i$ and $r$ of the form $(-\mathbf{u}_i,\mathbf{v}_i)$ for the negative $\lambda _i$ .
    • Therefore we have $U^TU = UU^T = I_n$$V^TV = VV^T = I_m$ and $V^TV = VV^T = I_m$$U_r$ .