Sunday, November 29, 2009

Holidaze

Yesterday was Saturday. So was the day before. And so was the day before that. Tomorrow is Monday. Today, when I first remembered that tomorrow is Monday, I couldn't remember what projects I'm working on and thought, "I wonder if I still know how to program..." Time will tell.

One of my family's best traditions is Practice Thanksgiving. Here's a good way to imagine it (remember these from elementary school?):

Practice Thanksgiving : Thanksgiving :: Rehearsal : Performance

After this year's two practices, the real thing went very well. We elected for the "healthy" yams instead of the healthy yams, the rolls were butter-sodden and fluffy...

Aside: I'm pretty sure I'm going to leave this earth clutching my failing heart as it struggles to pump blood through my dry, sealed arteries. Either that, or the butter and bacon grease lubricating my veins will ease my heart's burden to the tune of 40 extra years of life. End aside.

...I made a banana cream pie with a pecan, graham cracker crust (ingredients: 1 pkg graham crackers, 3/4 c. or so pecans, 1 Tbsp sugar, 1 stick butter). Definitely my favorite pie. The turkey, potatoes, stuffing, cranberry sauce (made fresh by Mom), gravy and green beans all tasted great. And it was fun to spend the day with my family.

No pic today. But the square root of 689 is about 26.3

Sunday, November 15, 2009

Arranged Marriage

I just signed into Family Search to follow through with a request made by my stake. On my pedigree chart there's a link below my name that says "Add or find wife." I clicked on it hopefully, but then it asks me for her name. How is that supposed to help me find a wife?

I'm thinking I might just fill it in anyway...

:)

Sunday, November 8, 2009

Squarish Roots

I love math and I love solving clever math problems. This guy is pretty cool. Well, he's a nerd, but he's a pretty cool nerd. Also, Donald encounters square roots here.

Anyway, for the longest time I've wondered how to quickly find the ballpark square root of a number. The square root of 9 is easy, but what about 10, or 11, or 8,387? Chace once said that his dad knew of a way that was pretty quick, but we couldn't figure it out. Or maybe Chace figured it out and never told me :)

Well, I finally found a way! So.... here we go:

How I figured it out:
What's the square root of 10? It's probably just bigger than 3 because

3 * 3 = 9
Squares can be represented with squares. Go figure. Here's a 3x3:
000
000
000
We'll define
A1 = x2 = 9 (area of square)
x = 3 (side of square)
Then let's define:
A2 = (x + Δx)2 = 10 (area of new square)
x + Δx = ? (side of new square - this is the answer to the original question)
The new square has four sections:
1113
0002
0002
0002

section 0: area encompassing the original 3x3 square
section 1: additional area on top
section 2: additional area on the side
section 3: additional area in the top right corner

Asection 0 = x2 = 9
Asection 1 = x(Δx)
Asection 2 = x(Δx)
Asection 3 = (Δx)2
And here's the cool part! (You didn't think there'd be a cool part, did you?) Δx is less than 1, because otherwise x + Δx = 4 (or more) which squares to 16. And since I just want the ballpark answer, I can ignore the area of section 3. That leads to:
111
0002
0002
0002
If we set the total area of that shape to 10, then we can easily find Δx:
A2 = x2 + 2x(Δx)
... (math) ...
Δx = (A2 - x2) / 2x
= (10 - 9) / 2(3)
= 1 / 6
= 0.16

√10 ≈ x + Δx
≈ 3.16

Which is close to the real answer:
3.16227766


Steps for any number:
Given any number (let's call it N):
1. Choose a number, x, whose square is just less than N
N = 8,387
1002 = 10,000 (too big)
502 = 2,500 (too small)
902 = 8,100 (perfect)
x = 90
2. Take the difference of N and x2 (and call it d)
d = N - x2
= 8387 - 8100
= 287
3. This equation:
(x + a)2 = x2 + 2xa + a2
tells us the difference effected in a square by incrementing the root by a. For example:
22 = 4

(2 + 1)2 = 22 + 2(2)(1) + 12
= 22 + 5 (so adding 1 to the root increases the square by 5)
= 9 = 32

(2 + 5)2 = 22 + 2(2)(5) + 52
= 22 + 45 (so adding 5 to the root increases the square by 45)
= 49
So choose a to get as close as you can to d without going over (and call that value b):
2xa + a2
2(90)(1) + 12 = 181
2(90)(2) + 22 = 364 (too big)

b = 181
a = 1
Add your chosen a to make a new x:
x = 91
4. If you want more precision (x is already within 1 of the answer), subtract b from d:
d = 287 - 181
= 106
then divide that number by 2x:

106 / 2x = 106 / 2(91)
= 106 / 182
= 53 / 91
≈ 5/9
≈ .55
Add that result to x and you've got a pretty precise answer:

√8387 ≈ 91.55 (estimate)
= 91.58 (real answer)


So here's one done really quickly with even less precision:

√127 = ?
102 = 100
127 - 100 = 27
27 / (2*10) = 1 + 7/20
7/20 = .35
√127 ≈ 11.35 (estimate; not bad)
= 11.26 (actual answer)
On that one, if I knew that 112 = 121, then

112 = 121
127 - 121 = 6
6 / (2*11) = 6 / 22
= 3 / 11
= 3 * (1 / 11)
≈ 3 * .09
≈ .27
√127 ≈ 11.27 (even better)
= 11.26 (actual answer)