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I tried to use the "spoiler" BB Code to hide the answer, but it didn't work.
[font=Courier]Maybe I'll use the hide tag later.[/font]
|-a1a2a3=-15
|a1a2+a2a3+a3a1=13
|a1+a2+a3=-3 <------shouldn't that be positive 3 ?
Also, thanks for the explanation, I see now.
Awesome work with the "pre". Braille test sheet works great. Only drawback I can see is when you go to edit it, all of the carriage returns are missing in the editor so the lines are all wrapped around together.
Can't complain though, it reads fine when you use the sheet.
I found Gadwin PrintScreen, free also.
Maybe this will help some. It's way over my head.
http://cgd.best.vwh.net/home/flt/flt08.htm
or same linky
Nil 3.125%
because 30 / (100+300+150+200+180+30) = 0.03125
Move decimal over two to the right to make a percent. .03125 is then 3.125% or 3 and 1/8 percent.
I editted the Braille worksheet, but I noticed that multiple spaces are truncated to one space, like in HTML, so
I put in the "pre" thing, but no use.
As a test I put in 1 2 3 4 5 6 7 8 9, but it ended up 1 2 3 4 5 6 7 8 9
I found this a very useful page on this subject.
http://uzweb.uz.ac.zw/science/maths/zimaths/62/dioph.htm
or
My favorite part is at the end of the page:
(i) The equation 3x + 6y = 22 has no solution since (3, 6) = 3 does not divide 22.
(ii) The equation 7x + 11y = 13 has solution x = -39, y = 26. For
11 = 1·7 + 4, 7 = 1·4 + 3, 4 = 1·3 + 1.
Thus (7, 11) = 1, which divides 13. Further, working from the last equation back to the first,1 = 4 - 3 = 4 -(7-4) = 2·4 - 7 = 2·11 - 3·7.
Hence7·(-3) + 11(2) = 1, 7·(-39) + 11(26) = 13.
The other solutions are given byx = -39 + 11r, y = 26 - 7r
where r is any integer.
Also equation # 10 at their site has an incorrect sign, I think.
Is the Nil Fault an actual fault or a service call that was fixed some other way?
Do we count the Nil ones?
In second question, what does "mode" mean?
At wolfram's site, equation # 6 has a "+ 1" in it (Euclidian thing).
I wonder why it always works out to a one, like the example they have?
How did you get |a1a2+a2a3+a3a1=13 ?? and what does the | symbol mean?
Also I was wondering if you considered that if the function is rotated,
some of the new vertical values are directly above one another, thus
it is not a one-to-one function.
I don't understand a lot of this rotation, but here are some questions.
Before we attack the integral as asked by krassi,
I am most intrigued by these 2 lines.
I wouldn't have figured this part out.
Very nice.
lny = xlnx now differentiate both sides
y'/y = lnx + x/x
I understand the product rule on the right side of equation,
but the y'/y surprised me because I probably would have
just thought of 1/y, but I guess because we are differentiating
with respect to x this happens? Please enlighten me to the
difference with lny going to y'/y, but if lnx was on right side of
equation, we don't write down x'/x, just 1/x. I think I am
missing something key and basic.
Differentiating x^x is the following according to an online differentiator
x x ^ (x - 1) + ln x x ^ x
But I was wondering how you arrive at this?
I think you could divide the area up into parts as a start.
x^3 + 3x^2 + 13x - 15
the factor theorm, what is that?
x (x^2 + 3x + 13) = 15
Guess 1, too big
Guess .9, .9 (.81 + .27 + 13) = 15
Guess .95, too big
Guess .92, 15.28
Guess .91, 15.07
Guess .9067, 14.999 Close enough.
So factor out (x - .9067)
x - .9067 x^3 + 3x^2 + 13x - 15
quotient: x^2 + (3 + .9067)x + (13 + .9067(3 - .9067)) There is a remainder so imaginary pair??
dividing calculations: x^3 - .9067x^2 - .9067(3 - .9067)x - .9067(13 + .9067(3 - .9067))
krassi_holmz's mathimatica got: .906756576Because it didn't divide almost evenly, instead of getting -15 on end, it multiplies to about -13.5,
I guess this means the other roots aren't real numbers?? I'm just guessing from what
krassi said.
I followed your logic until you said "But we have
at least 2 squares greater than n
at least 2 squares greater than n-1".
What does this mean? Can you give numerical example?
ax + by = c
by = -ax + c
y = (-a/b)x + (c/b)
So y = at + ?1? and
x = bt + ?2?
-------
Put into original equation ax + by = c
a(bt + ?2?) + b(at + ?1?) = c
abt + a?2? + bat + b?1? = c
2bat + a?2? + b?1? = c
---------
y=7x + .3
y=7t + ?
x=t + ?
No integer solutions for this 7x + .3
---------
No luck yet.sorry, don't really understand how you do this.
can show me how you get this formula (-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + (-3x0+3x1)t + x0 ??
from x0(1-3t+3t^2-t^3) + 3x1 (t-2t^2+t^3) + 3x2(t^2-t^3) + x3t^3
Method used is similar to this example:
(A+B)C + (A+B)D
See how I am skipping a step and not writing it down.
This is the equivalent.
(C+D)A + (C+D)B
The step I didn't ever write down is because you just organize by
jumping around and choosing the A's first, and the B's second.
But it would be:
AC + BC + AD + BD
Then get the A's on the left, and B's on the right.
Holy Toledo, those lists of 32 and 40 use every number 1 to 32 and 1 to 40!!!!! That's incredible.
Were they computed by hand or by computer? Do you know?
What about length 16? I didn't see that one? What is the post # for it?
I found a method to possibly get close to an answer.
a = (-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + (-3x0+3x1)t + x0
Get t on left side, and put a on right, leave t^3 and t^2 on right side.
-(-3x0+3x1)t = (-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + x0 - a
Divide by -(-3x0+3x1)
t = (-1/(-3x0+3x1)) ((-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + x0 - a )
Make a guess at t and plug it into the t^3 and t^2 and solve right side of equation.
Take answer and put it back in again.
Repeat until you appear to be getting closer and closer to last result.
I don't know if this will work, but I just read this on a web page about solving a cubic.a = x0 * (1-t)³ + x1 * 3 * t * (1-t)² + x2 * 3 * t² * (1-t) +x3 * t³
1-2t+t^2(1-t) 1-2t+t^2 3x2(t^2-t^3) x3t^3
3x1 (t-2t^2+t^3) | |
1-2t+t^2 | | |
-t +2t^2 -t^3 | | |
-------------- v v v
x0(1-3t+3t^2-t^3)
(-x0+3x1-3x2+x3)t^3 + (3x0-6x1+3x2)t^2 + (-3x0+3x1)t + x0
I don't know what to do next to this cubic equation.