I've been tidying and revising some of my older posts. Some errors have been corrected, some bad writing made less awful.
However, although it's blog etiquette to notify your readers of when and how you have updated your posts, I'm not doing that because it will take too long and I cannot be bothered.
Do feel free to point out errors that are still obscuring the light of truth...
Monday 2 May 2011
Sunday 1 May 2011
Completing the Square Redux
An amazing site which explains completing the square much better than I ever could:
http://www.mathsisfun.com/algebra/completing-square.html
http://www.mathsisfun.com/algebra/completing-square.html
The Modulus Function
Put very basically, the modulus function finds the absolute value of a number.
It is indicated by a pair of straight brackets around an x or a part of an equation.
It just means that -x when operated on by the modulus function becomes x while x stays as x.
It ignores negative values of x, or of the statement in x, such as (x-2).
It is indicated by a pair of straight brackets around an x or a part of an equation.
It just means that -x when operated on by the modulus function becomes x while x stays as x.
It ignores negative values of x, or of the statement in x, such as (x-2).
Transformation of Graphs, contd
When you are doing composite transformation of graphs, as you will at C3 and presumably C4 level, you do need to remember that sometimes the order of transformation counts. I have to admit to struggling with this (I don't have a tutor right now) so forgive me if I fail to offer a convincing explanation. But it's something like this:
If you are stretching in the x-direction, that comes BEFORE any translation.
If you are stretching in the y-direction, that comes AFTER any translation.
That's what I've picked up from getting lots of answers wrong, anyhow. I think it's to do with the way that a y-stretch affects the whole equation, while an x-stretch affects anything directly attached to the x part of the equation.
Some websites do suggest that you start at the innermost point of the equation and then work out - I don't know if this works for all composite transformations. I doubt it if what I've written above is right.
If you are stretching in the x-direction, that comes BEFORE any translation.
If you are stretching in the y-direction, that comes AFTER any translation.
That's what I've picked up from getting lots of answers wrong, anyhow. I think it's to do with the way that a y-stretch affects the whole equation, while an x-stretch affects anything directly attached to the x part of the equation.
Some websites do suggest that you start at the innermost point of the equation and then work out - I don't know if this works for all composite transformations. I doubt it if what I've written above is right.
Tuesday 26 April 2011
More LaTeX Experiments: Transformation of Graphs
If you take a graph and transform it in various ways, here is what you get:
A reflection in the y-axis turns a graph of y=f(x) into the graph of y= f(-x).
A reflection in the x-axis turns a graph of y=f(x) into the graph of y= -f(x)
- note the difference -
A stretch of factor s in the y- direction turns the graph of y= f(x) into the graph of
y = s f(x).
A stretch of factor m in the x-direction turns the graph of y = f(x) into the graph of \[y = f(\frac{x}{m})\]
A translation by \[ \left( \frac{k}{l} \right) \] turns the graph of y = f(x) into the graph of y = f(x-k) + l
A reflection in the y-axis turns a graph of y=f(x) into the graph of y= f(-x).
A reflection in the x-axis turns a graph of y=f(x) into the graph of y= -f(x)
- note the difference -
A stretch of factor s in the y- direction turns the graph of y= f(x) into the graph of
y = s f(x).
A stretch of factor m in the x-direction turns the graph of y = f(x) into the graph of \[y = f(\frac{x}{m})\]
A translation by \[ \left( \frac{k}{l} \right) \] turns the graph of y = f(x) into the graph of y = f(x-k) + l
Labels:
c2,
c3,
graphs,
transformation of graphs,
translation
Another Test
I'm trying to see if I can use LaTeX to write formulae onto the blog, but I've always been a bit befuddled by it. Instead I've installed a javascript thing from www.watchmath.com and that's supposed to recognise code and turn it into proper maths.
Well here goes.
Let's have a look at the function f(x) = 1/x. By rights in LaTeX that should be \[ \frac{1}{x} \]
Well here goes.
Let's have a look at the function f(x) = 1/x. By rights in LaTeX that should be \[ \frac{1}{x} \]
Test
\[ E(\mathbf{x}) = \sum_{i \in \mathcal{V}} \theta_i(x_i) + \sum_{ij \in \mathcal{E}} \theta_{ij}(x_i, x_j) \]
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