M-line

Examples of M-line in the following topics:

• Slope

  • Putting the equation of a line into this form gives you the slope ($m$) of a line, and its $y$-intercept ($b$).
  • We will now discuss the interpretation of $m$, and how to calculate$m$ for a given line.
  • The slope is positive ($m > 0$).
  • A line is decreasing if it goes down from left to right and the slope is negative ($m < 0$) .
  • Mathematically, the slope m of the line is $\displaystyle m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$, given two points$(x_1, y_1)$and$(x_2, y_2)$ that fall on the line.

• Slope and Intercept

  • In the regression line equation the constant $m$ is the slope of the line and $b$ is the $y$-intercept.
  • A simple example is the equation for the regression line which follows:
  • The constant $$$m$ is slope of the line and $b$ is the $y$-intercept -- the value where the line cross the $y$ axis.
  • So, $m$and$b$ are the coefficients of the equation.
  • An equation where y is the dependent variable, x is the independent variable, m is the slope, and b is the intercept.

• Parallel and Perpendicular Lines

  • Recall that the slope-intercept form of an equation is: $y=mx+b$ and the point-slope form of an equation is: $y-y_{1}=m(x-x_{1})$, both contain information about the slope, namely the constant$m$.
  • If two lines, say $f(x)=mx+b$ and $g(x)=nx+c$, are parallel, then $n$ must equal $m$.
  • For example, in the graph below, $f(x)=2x+3$ and $g(x)=2x-1$ are parallel since they have the same slope, $m=2$.
  • For example given two lines, $f(x)=m_{1}x+b_{1}$and$g(x)=m_{2}x+b_{2}$,$f(x)\perp g(x)$ states that the two lines are perpendicular to each other.
  • This means that if the slope of one line is $m$m, then the slope of its perpendicular line is $\frac{-1}{m}$.

• Equations of Lines and Planes

  • A line is described by a point on the line and its angle of inclination, or slope.
  • A line in three dimensional space is given by a point, $P_0 (x_o,y_o,z_o)$, or a plane, $M$, and its direction.
  • This direction is described by a vector, $\mathbf{v}$, which is parallel to plane and $P$ is the arbitrary point on plane $M$.
  • Now, we can use all this information to form the equation of a line on plane $M$.
  • where $t$ represents the location of vector $\mathbf{r}$ on plane $M$.

• Derivatives and Rates of Change

  • Historically, the primary motivation for the study of differentiation was the tangent line problem, which is the task of, for a given curve, finding the slope of the straight line that is tangent to that curve at a given point.
  • Thus, to solve the tangent line problem, we need to find the slope of a line that is "touching" a given curve at a given point, or, in modern language, that has the same slope.
  • In this case, $y = f(x) = m x + b$, for real numbers m and b, and the slope m is given by:
  • $y + \Delta y = f(x+ \Delta x) = m (x + \Delta x) + b = m x + b + m \Delta x = y + m\Delta x$
  • This gives an exact value for the slope of a straight line.

• Derivatives and the Shape of the Graph

  • In this case, $y=f(x) = m \cdot x+b$, for real numbers$m$and$b$, and the slope m is given by $\frac{\Delta y}{\Delta x}$, where the symbol $\Delta$ (the uppercase form of the Greek letter Delta) is an abbreviation for "change in. "
  • This gives an exact value for the slope of a straight line.
  • At each point, the derivative of is the slope of a line that is tangent to the curve.
  • The line is always tangent to the blue curve; its slope is the derivative.
  • Note derivative is positive where a green line appears, negative where a red line appears, and zero where a black line appears.

• Green's Theorem

  • Green's theorem gives relationship between a line integral around closed curve $C$ and a double integral over plane region $D$ bounded by $C$.
  • Green's theorem gives the relationship between a line integral around a simple closed curve $C$ and a double integral over the plane region $D$ bounded by $C$.
  • $\displaystyle{\oint_{C} (L\, \mathrm{d}x + M\, \mathrm{d}y) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, \mathrm{d}x\, \mathrm{d}y}$
  • Green's theorem can be used to compute area by line integral.
  • Provided we choose $L$ and $M$such that:$\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1$, then the area is given by$A=\oint_{C} x$.

• Alternating Series

  • If $m$is odd and$S_m - S_n < a_{m}$ via the following calculation:
  • $\begin{aligned} S_m - S_n & = \sum_{k=0}^m(-1)^k\,a_k\,-\,\sum_{k=0}^n\,(-1)^k\,a_k\ \\& = \sum_{k=m+1}^n\,(-1)^k\,a_k \\ & =a_{m+1}-a_{m+2}+a_{m+3}-a_{m+4}+\cdots+a_n\\ & =\displaystyle a_{m+1}-(a_{m+2}-a_{m+3}) -\cdots-a_n \le a_{m+1}\le a_{m} \\& \quad [a_{n} \text{ decreasing}].
  • Thus, we have the final inequality $S_m - S_n \le a_{m}$.
  • Similarly, it can be shown that, since $a_m$converges to$0$,$S_m - S_n$converges to$0$for$m, n \rightarrow \infty$.
  • The first fourteen partial sums of the alternating harmonic series (black line segments) shown converging to the natural logarithm of 2 (red line).

• Dynamics of UCM

  • Newton's universal law of gravitation states that every particle attracts every other particle with a force along a line joining them.
  • Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times.
  • Gravity supplies the centripetal force to mass $m$.
  • (b) For any closed gravitational orbit, $m$follows an elliptical path with$M$ at one focus.
  • The mass $m$moves fastest when it is closest to$M$.

• The Born-Oppenheimer Approximation

  • For diatomic molecules there is still a rotational symmetry about the line connecting the two nuclei.
  • $\displaystyle E_\textrm{elec} \sim \frac{p^2}{m} \sim \frac{\hbar^2}{m a^2} \sim \alpha^2 m c^2 \sim 1~\textrm{eV}$
  • $\displaystyle E_\textrm{vib} \sim \hbar \omega = \hbar \left (\frac{k}{M} \right )^{1/2} = \hbar \left ( \frac{\hbar^2}{m M a^4} \right )^{1/2} = \left ( \frac{m}{M} \right )^{1/2} \frac{\hbar^2}{m a^2} \sim 0.01-0.1~\textrm{eV}.$
  • $\displaystyle E_\textrm{rot} \sim \frac{\hbar^2 L \left (L+1\right) }{2 I} = \frac{\hbar^2 L \left (L+1\right )}{M a^2} = \frac{m}{M} \frac{\hbar^2}{m a^2} L \left (L+1\right )\sim 10^{-3}~\textrm{eV}.$