[Blindmath] Basic mathematical concept question

Sat Apr 21 08:33:09 UTC 2012

Hello all,
Thanks for all explanations.
Now I have soem other concepts I studied from high school but I don't know
wwhy is that.

 The function is increasing when the derivative interval of that function is
positive, and decresing when interval of derivative is negative.

The root of second derivative is the inflection point of the draf of the
cube function.

Some tregonometric algorithms.

sin(a + b) =sin(a) cos(b) + sin(b) cos(a)

cos(a)cos(b) = 1/2(cos(a - b) + (a + b))
Could you please help me explain why?
Thanks you very muchh,
Nam
 ----- Original Message ----- 
From: "David Tseng" <davidct1209 at gmail.com>
To: "Blind Math list for those interested in mathematics"
<blindmath at nfbnet.org>
Sent: Sunday, March 25, 2012 4:17 AM
Subject: Re: [Blindmath] Basic mathematical concept question

>I personally don't think it's the mechanics that gives blind math
> students trouble. It's rather the intuition concerning a topic or
> concept as it's usually conveyed in a highly visual manner.
>
> Your example, for instance, is a linear transformation of a matrix of
> pixels, in which you take differences between successive columns. The
> intuition behind such operations is expressed as vectors in R^n. The
> actual "picture" of the R^3 vector space, as an example isn't ever
> really mentioned in the kind of explicit detail that I feel new blind
> students need. Having the very tangible notion that you're "smoothing"
> out a particular axis/dimension of the vectors is more
> powerful/long-lasting than say taking differences.
>
> On 3/24/12, Richard Baldwin <baldwin at dickbaldwin.com> wrote:
>> Now for the practical.
>>
>> Many real-life "curves" cannot be analytically defined in precise
>> mathematical terms without using an infinite series. Even for those that
>> can be analytically defined, you might find this information useful.
>>
>> Consider, for example, the set of values that define the red color values
>> in a single row of pixels in a digital image. It is highly unlikely that
>> you can, with any reasonable number of terms, fit a mathematical
>> expression
>> to to that shape. However, using a computer, you can approximate the
>> first
>> derivative of that shape by successively subtracting the red value of one
>> pixel from the red value of the next pixel from the left end of the row
>> to
>> the right end of the row and saving those difference values along the
>> way.
>>
>> If you plot those values, you will be plotting an estimate of the first
>> derivative of the curve.
>>
>> Viewed from a completely different viewpoint, you will have applied a
>> high-pass frequency filter to the curve. This filter will suppress the
>> low
>> frequency values and accentuates the high frequency values. In general,
>> you
>> can expect the new function to appear to be more "ragged" than the first.
>>
>> If the set of values to which the process is applied happen to be samples
>> of audio data, and you listen to the "before" and "after" versions, you
>> might be able to audibly detect the change in the frequency spectrum.
>>
>> Perform that process for all three color values for every row of pixels
>> in
>> the image, normalize the results to guarantee that all of the new values
>> fall in the range from 0 to 255 inclusive, display those new values as an
>> image, and you will have produced a new image in which the features tend
>> to
>> be sharper than in the original image on a horizontal basis. Since you
>> didn't do the same thing vertically, you won't experience the increased
>> sharpness along the vertical axis of the image.
>>
>> Math is a wonderful thing. I continue to be amazed as to how mathematical
>> concepts in one area tie into mathematical concepts in an apparently
>> different area if you keep an open mind for such connections.
>>
>> Now back to the practical. When I took differential calculus about 100
>> years ago, there was no such thing as a personal computer and barely such
>> a
>> thing as a mainframe computer. However, with the availability of personal
>> computers, you now have the opportunity to estimate and graph the
>> derivative of a given function and to compare that graph with a graph of
>> the derivative function that you produce analytically. If they don't
>> match
>> -- it's back to the analytic drawing board.
>>
>> Here is the procedure:
>>
>> 1. Evaluate the original function at a set of equally and closely spaced
>> points and save those values as a sequence of sampled data values.
>>
>> 2. Perform the difference operation that I described above on those
>> sampled
>> data values, save, and graph those results. (Be aware that you cannot
>> perform the difference operation on the last data value.)
>>
>> 3. Graph your analytically derived differential function and compare the
>> two.
>>
>> Hope this helps.
>> Dick Baldwin
>>
>> On Sat, Mar 24, 2012 at 2:30 PM, David Tseng <davidct1209 at gmail.com>
>> wrote:
>>
>>> With regard to accessible materials I would start with seeking a
>>> non-lossy medium for your textbook. When I went through my
>>> undergraduate work in mathematics, I employed a variety of techniques
>>> to accomplish this. If you can find it, audio recordings of Calculus
>>> books (Stewart comes to mind) are available in the U.S. Despite the
>>> very sequential (tape) nature of this method, humans reading the book
>>> render the text precisely and insert appropriate descriptions for
>>> figures/graphs when appropriate. This becomes invaluable as you move
>>> through Calculus in its differential, integral, and multi variable
>>> variants.
>>>
>>> There are also now Daisy equivalents of the above. So, you can
>>> download an audio book that has chapter/section indecies.
>>>
>>> Alternatively, you can try to have relevant materials hand-annotated
>>> by whatever resources your college may offer.
>>>
>>> Finally, wikipedia, though it reads more like a reference than say a
>>> tutorial, is quite accessible. Images have alt tags that expose the
>>> raw LaTeX of the expression under discussion.
>>>
>>> Hth,
>>> David
>>>
>>> On 3/24/12, Andrew Stacey <andrew.stacey at math.ntnu.no> wrote:
>>> > That was a nearly perfect explanation.
>>> >
>>> > Just one small technicality.  The properties that Richard refers to as
>>> > "continuous" and "discontinuous" are actually "differentiable" and
>>> > "non-differentiable".  The rough-and-ready description of a function
>>> being
>>> > continuous is that its graph can be drawn without taking the pencil
>>> > off
>>> the
>>> > paper.  The actual definition is a little more complicated, but that's
>>> the
>>> > basic idea.  A continuous function has no jumps, no gaps.  It is
>>> > a differentiable function that has a well-defined slope everywhere.
>>> >
>>> > A differentiable function is automatically continuous, but not vice
>>> versa.
>>> >
>>> > Some examples:
>>> >
>>> > 1. A discontinuous function: define f(t) = 0 if t < 0 and f(t) = 1
>>> > otherwise.
>>> > This has a jump at 0 so is discontinuous.
>>> >
>>> > 2. A continuous function that isn't differentiable: f(t) = 0 if t < 0
>>> > and
>>> > f(t)
>>> > = t otherwise.  This is continuous, but it has no well-defined slope
>>> > at
>>> > 0 since as we approach 0 from the left the slope appears to be 0
>>> > (flat)
>>> > whereas if we approach it from the right then the slope appears to be
>>> > 1.
>>> >
>>> > Most functions that you encounter are differentiable, in fact probably
>>> even
>>> > better than that.  But properly defined, most functions that exist are
>>> not
>>> > differentiable anywhere.  It's just that since differentiable
>>> > functions
>>> are
>>> > so
>>> > easy to study (compared to arbitrary functions), we use them if at all
>>> > possible.
>>> >
>>> > Andrew Stacey
>>> >
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>>>
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>>
>>
>>
>> --
>> Richard G. Baldwin (Dick Baldwin)
>> Home of Baldwin's on-line Java Tutorials
>> http://www.DickBaldwin.com
>>
>> Professor of Computer Information Technology
>> Austin Community College
>> (512) 223-4758
>> mailto:Baldwin at DickBaldwin.com
>> http://www.austincc.edu/baldwin/
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>
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