This
article apparently first ran in Popular Mechanics... who knew to look there for math news??
Let the doubters and defenders come forth and critique....
Hello!
(04-23-2024 07:20 PM)rprosperi Wrote: [ -> ]This article apparently first ran in Popular Mechanics... who knew to look there for math news??
Let the doubters and defenders come forth and critique....
The YouTube video linked in the article is four years old and we haven't heard of this "easy" method ever since. So my conclusion is, that Mr. Darwin's principle took care of it ;-)
But really I find it hard to understand what he is explaining there and how his method would work so easily with non-integer factors.
Regards
Max
(04-23-2024 07:20 PM)rprosperi Wrote: [ -> ]This article apparently first ran in Popular Mechanics... who knew to look there for math news??
Here is an HP-42S program using his method:
Code:
00 { 26-Byte Prgm }
01▸LBL "Q"
02 X<> ST Z
03 STO÷ ST Z
04 STO+ ST X
05 +/-
06 ÷
07 STO ST Z
08 X↑2
09 -
10 +/-
11 SQRT
12 RCL+ ST Y
13 X<>Y
14 RCL- ST L
15 END
Same usage of
this quadratic solve and same number of steps. This first attempt takes up one extra byte, though.
(04-23-2024 08:36 PM)Maximilian Hohmann Wrote: [ -> ]Hello!
(04-23-2024 07:20 PM)rprosperi Wrote: [ -> ]This article apparently first ran in Popular Mechanics... who knew to look there for math news??
Let the doubters and defenders come forth and critique....
The YouTube video linked in the article is four years old and we haven't heard of this "easy" method ever since. So my conclusion is, that Mr. Darwin's principle took care of it ;-)
But really I find it hard to understand what he is explaining there and how his method would work so easily with non-integer factors.
Indeed he doesn’t explain it step by step. Hopefully, the following is easier to understand:
x² + b.x + c = 0
x₁ + x₂ = -b
x₁.x₂ = c
u = (x₁ + x₂)/2 = -b/2
=>
x₁ = -b/2 - u (1)
x₂ = -b/2 + u (2)
and
x₁.x₂ = c
(-b/2 - u)(-b/2 + u) = c
or, from the difference of squares formula,
(-b/2)² - u² = c
u² = (-b/2)² - c
u = √((-b/2)² - c)
Then, from (1) and (2) above:
x₁ = -b/2 - √((-b/2)² - c)
x₂ = -b/2 + √((-b/2)² - c)
Regards,
Gerson.
(04-23-2024 11:19 PM)Gerson W. Barbosa Wrote: [ -> ]Same usage of this quadratic solve and same number of steps. This first attempt takes up one extra byte, though.
Second attempt. Same size, same number of steps:
Code:
00 { 25-Byte Prgm }
01▸LBL "Q"
02 X<> ST Z
03 +/-
04 STO÷ ST Z
05 STO+ ST X
06 ÷
07 STO ST Z
08 X↑2
09 +
10 SQRT
11 RCL+ ST Y
12 X<>Y
13 RCL- ST L
14 END
P.S.:
Exactly the same program, actually. But I don’t remember I had previously used this method.
P.P.S.:
HP-41 version:
Code:
01 LBL "Q"
02 X<> Z
03 CHS
04 ST/ Z
05 ST+ X
06 /
07 STO Z
08 X^2
09 +
10 SQRT
11 ST- Y
12 ST+ X
13 RCL Y
14 +
15 END
28 bytes
Notice these preserve the original stack register X.
(04-23-2024 07:20 PM)rprosperi Wrote: [ -> ]This article apparently first ran in Popular Mechanics... who knew to look there for math news??
Let the doubters and defenders come forth and critique....
Ahhh, Popular Mechanics. My step-father had a subscription/prescription. I remember the article from the '60's about spiders on LSD spinning rectangular webs.
I couldn't get it to work on this: x^2+4x+7=0.
(04-25-2024 12:17 AM)ttw Wrote: [ -> ]I couldn't get it to work on this: x^2+4x+7=0.
Complex roots?
There must be something I haven't understood. How is this new? I learned this method at school 40 years ago.
That said, thanks for reminding me.
Hello all,
it works on equations with complex roots, too.
The new way can be reduced to two steps:
\[ 1.\quad u^2 = \frac{b^2}{4} - c \qquad\qquad 2.\quad x_{1,2} = - \frac{b}{2} \pm \sqrt{u^2} \]
The traditional way:
\[ 1.\quad D = b^2 - 4c \qquad\qquad 2.\quad x_{1,2} = \frac{- b \pm \sqrt{D}}{2} \]
For me it seems to be a question of taste which way someone prefers. The only difference is:
\[ D = 4u^2 \]
I'm delighted that this subject has allowed me to rediscover this formula, whose simplicity I admire.
x² − Sx + P = 0
With S = Sum, P = Product, M = Mean, δ = Deviation
M = S/2
δ² = M² − P
x = M ± δ
Code:
Input Sum
ENTER
ENTER
2
÷
ENTER
x²
Input Product
−
√x
− gives x₁
− gives x₂
Would be grateful if someone could direct to instructions for including formula with text.
Ended up attaching a pdf.
Regards - RR
(04-26-2024 04:18 AM)Rolief_Rechner Wrote: [ -> ]Would be grateful if someone could direct to instructions for including formula with text.
\(
\begin{align}
x_{1,2} = - \frac{b}{2a} \pm \sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}
\end{align}
\)
Which is the same as the well-known
quadratic formula:
\(
\begin{align}
x_{1,2} = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}
\end{align}
\)
Just
![[Image: postbit_quote.gif]](https://www.hpmuseum.org/forum/images/english/postbit_quote.gif)
this post to see how to use
LaTeX.
PS: You missed a factor 2 in the denominator.
Thomas Klemm wrote:
...
PS: You missed a factor 2 in the denominator.
- - - -
Enjoy reading your (and others) post here.
Have learned much. It is nice knowing RPN users are not alone.
However, this must be politely pointed out.
Those who are used to textbook formula make this error.
A consolation is you are not the first to do this.
Have attached pdf of short Mathcad worksheet for illustration.
best regards - RR
(04-26-2024 09:25 PM)Rolief_Rechner Wrote: [ -> ]A consolation is you are not the first to do this.
I'm not sure I understand you correctly, but your
QuadForm.pdf attachment uses the following incorrect formula:
\(
\begin{align}
x_{1,2} = - \frac{b}{a} \pm \sqrt{\left(\frac{b}{a}\right)^2 - \frac{c}{a}}
\end{align}
\)
This led to my comment.
Example
\(
\begin{align}
a &:= 1 \\
b &:= -5 \\
c &:= 6 \\
\\
x_{1,2} &= 5 \pm \sqrt{19} \\
\\
x_1 &\approx 0.641101 \\
x_2 &\approx 9.3589 \\
\end{align}
\)
\(\Rightarrow\)
Incorrect
BTW: The proper way to quote a post is using the
button:
(04-26-2024 06:01 AM)Thomas Klemm Wrote: [ -> ]PS: You missed a factor 2 in the denominator.
(04-27-2024 10:50 AM)Thomas Klemm Wrote: [ -> ] (04-26-2024 09:25 PM)Rolief_Rechner Wrote: [ -> ]A consolation is you are not the first to do this.
I'm not sure I understand you correctly, but your QuadForm.pdf attachment uses the following incorrect formula:
\(
\begin{align}
x_{1,2} = - \frac{b}{a} \pm \sqrt{\left(\frac{b}{a}\right)^2 - \frac{c}{a}}
\end{align}
\)
This led to my comment.
Example
\(
\begin{align}
a &:= 1 \\
b &:= -5 \\
c &:= 6 \\
\\
x_{1,2} &= 5 \pm \sqrt{19} \\
\\
x_1 &\approx 0.641101 \\
x_2 &\approx 9.3589 \\
\end{align}
\)
\(\Rightarrow\) Incorrect
BTW: The proper way to quote a post is using the button:
(04-26-2024 06:01 AM)Thomas Klemm Wrote: [ -> ]PS: You missed a factor 2 in the denominator.
The formula is derived by completing the square.
It was checked with Wolfram Alpha, attached.
If you still insist this is wrong, remedial education is suggested.
(04-27-2024 08:21 PM)Rolief_Rechner Wrote: [ -> ]The formula is derived by completing the square.
It was checked with Wolfram Alpha, attached.
This is where it get wrong. (x + y)^2 = x^2 +
2 xy + y^2
a*x^2 + b*x + c = 0
x^2 + (b/a)*x + (c/a) = 0
x^2 + (b/a)*x + (b/(2a))^2 = (b/(2a))^2 - (c/a)
(x + b/(
2a))^2 = (b/(2a))^2 - (c/a)
x = -b/(
2a) ± √((b/(2a))^2 - (c/a))
(04-27-2024 09:02 PM)Albert Chan Wrote: [ -> ] (04-27-2024 08:21 PM)Rolief_Rechner Wrote: [ -> ]The formula is derived by completing the square.
It was checked with Wolfram Alpha, attached.
This is where it get wrong. (x + y)^2 = x^2 + 2 xy + y^2
a*x^2 + b*x + c = 0
x^2 + (b/a)*x + (c/a) = 0
x^2 + (b/a)*x + (b/(2a))^2 = (b/(2a))^2 - (c/a)
(x + b/(2a))^2 = (b/(2a))^2 - (c/a)
x = -b/(2a) ± √((b/(2a))^2 - (c/a))
That is what I ended up with also.
It is simpler than what most textbooks have.
Once I stumbled upon this, loved it.
Regards - RR