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 29, 2009

### Holidaze

Posted by Matt Haggard at 6:19 PM 4 comments

## 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...

:)

Posted by Matt Haggard at 5:06 PM 4 comments

## 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 = 9Squares can be represented with squares. Go figure. Here's a 3x3:

000We'll define

000

000

AThen let's define:_{1}= x^{2}= 9 (area of square)

x = 3 (side of square)

AThe new square has four sections:_{2}= (x + Δx)^{2}= 10 (area of new square)

x + Δx = ? (side of new square - this is the answer to the original question)

1113And 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:

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

A_{section 0}= x^{2}= 9

A_{section 1}= x(Δx)

A_{section 2}= x(Δx)

A_{section 3}= (Δx)^{2}

111If we set the total area of that shape to 10, then we can easily find Δx:

0002

0002

0002

A_{2}= x^{2}+ 2x(Δx)

... (math) ...

Δx = (A_{2}- x^{2}) / 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 = 8,3872. Take the difference of N and x

100^{2}= 10,000 (too big)

50^{2}= 2,500 (too small)

90^{2}= 8,100 (perfect)

x = 90

^{2}(and call it d)

d = N - x3. This equation:^{2}

= 8387 - 8100

= 287

(x + a)tells us the difference effected in a square by incrementing the root by a. For example:^{2}= x^{2}+ 2xa + a^{2}

2So choose a to get as close as you can to d without going over (and call that value b):^{2}= 4

(2 + 1)^{2}= 2^{2}+ 2(2)(1) + 1^{2}

= 2^{2}+ 5 (so adding 1 to the root increases the square by 5)

= 9 = 3^{2}

(2 + 5)^{2}= 2^{2}+ 2(2)(5) + 5^{2}

= 2^{2}+ 45 (so adding 5 to the root increases the square by 45)

= 49

2xa + aAdd your chosen a to make a new x:^{2}

2(90)(1) + 1^{2}= 181

2(90)(2) + 2^{2}= 364 (too big)

b = 181

a = 1

x = 914. If you want more precision (x is already within 1 of the answer), subtract b from d:

d = 287 - 181then divide that number by 2x:

= 106

Add that result to x and you've got a pretty precise answer:

106 / 2x = 106 / 2(91)

= 106 / 182

= 53 / 91

≈ 5/9

≈ .55

√8387 ≈ 91.55 (estimate)

= 91.58 (real answer)

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

On that one, if I knew that 11

√127 = ?

10^{2}= 100

127 - 100 = 27

27 / (2*10) = 1 + 7/20

7/20 = .35

√127 ≈ 11.35 (estimate; not bad)

= 11.26 (actual answer)

^{2}= 121, then

11^{2}= 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)

Posted by Matt Haggard at 4:11 PM 3 comments