Any math wizzes in the house?

[Cream Soda]

Looking for help explaining how to do the following problems

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i43.tinypic.com/viijxl.jpg
i41.tinypic.com/bevq78.jpg

[Jodwahh]

Here is the answer:

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http://i38.photobucket.com/albums/e1...ahh/viijxl.jpg

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I suck at maths

[0ni0n]

I think the sum of these perimeters is 60.

All the triangles are equilateral, so the perimeter of the largest triangle ABC is 30.

Since AD = DE = AE = 5 (D and E are the mid points of AB and AC respectively), the perimeter of triangle DEF is 15.

By the same token, the next triangle inside of DEF is half of 15 or a fourth of 30, which is 7.5

So we derive this equation adding up all the triangles....


30x1 + 30x(1/2) + 30x(1/4) + 30x(1/8 ) + 30x(1/16) + ........

>> 30 x (1+1/2+1/4+1/8+1/16+......)

since (1/2 + 1/4 + 1/8 + 1/16 +..........) = 1, it's actually forever converging to 1, and the wiki's definition of 1 is 0.999999999999.... so basically it's 1.

>> 30 x 2 = 60

[Drex]

Brute force way for part 2 wasn't too bad, so I did it that way. Part B honestly should've been first in the question, since you can solve part A using part B.

1st term: x
2nd term: y
3rd term: x+y
4th term: x+2y
5th term: 2x+3y
6th term: 3x+5y
7th term: 5x+8y
8th term: 8x+13y
9th term: 13x+21y
10th term: 21x+34y

If you add those ten terms, you get 55x+88y.

The 7th term is 5x+8y. 11(5x)+11(8y) yields 55x+88y.

For part A, just plug in any 2 numbers for x and y and generate your list.

I imagine there's some sort of trick or formula, but there's the long way to do it.

edit - looking at it like this, there actually is a pretty easy way to generate a long list - the x-term is just the previous y-term, and the y-term is just the sum of the previous x-term and the current x-term.

[Daydreamer]

I think the sum of these perimeters is 60.

All the triangles are equilateral, so the perimeter of the largest triangle ABC is 30.

Since AD = DE = AE = 5 (D and E are the mid points of AB and AC respectively), the perimeter of triangle DEF is 15.

By the same token, the next triangle inside of DEF is half of 15 or a fourth of 30, which is 7.5

So we derive this equation adding up all the triangles....


30x1 + 30x(1/2) + 30x(1/4) + 30x(1/8 ) + 30x(1/16) + ........

>> 30 x (1+1/2+1/4+1/8+1/16+......)

since (1/2 + 1/4 + 1/8 + 1/16 +..........) = 1, it's actually forever converging to 1, and the wiki's definition of 1 is 0.999999999999.... so basically it's 1.

>> 30 x 2 = 60

Pretty sure this is correct. If I remember right the actual formula for a geometric sequence is a/(1-r) where a is the first term and r is what each following term is multiplied by.

So we have 30/(1-1/2) = 60

[Cream Soda]

Here is the answer:

Spoiler: show

http://i38.photobucket.com/albums/e1...ahh/viijxl.jpg

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I suck at maths

LMFAO


and thanks everyone else

[Diamicles]

LOL @ the Triforce

There's a big difference in being a math whiz and being good at finite algebra!! Shit, I have a degree in physics with a minor in math, and I never had to do anything that resembled this.

[BarthelloSylph]

Infinite series is Calculus II. Use the fact that you can represent a function with A/(1-R) equals the geometric infinite series SUMMATION(N>>infinity) R^N. Just find a way to represent the equation as R.

[Diamicles]

Infinite series is Calculus II. Use the fact that you can represent a function with A/(1-R) equals the geometric infinite series SUMMATION(N>>infinity) R^N. Just find a way to represent the equation as R.

Could have fooled me. I took all the classes through differential equations, maybe I just don't remember doing these at all >.>

[0ni0n]

It's easy to prove (1/2 + 1/4 + 1/8 + 1/16 + ...) = 1,
since
1/2 + 1/4 = 3/4
1/2 + 1/4 + 1/8 = 7/8
1/2 + 1/4 + 1/8 + 1/16 = 15/16

you probably already see the pattern and extrapolate that in an infinite series, the solution would be (infinity-1)/infinity.

infinity - 1 = infinity
(infinity-1)/infinity = infinity/infinity = 1

[Kyten]

Lol @ the Triforce response.

[Trajan]

It's easy to prove (1/2 + 1/4 + 1/8 + 1/16 + ...) = 1,
since
1/2 + 1/4 = 3/4
1/2 + 1/4 + 1/8 = 7/8
1/2 + 1/4 + 1/8 + 1/16 = 15/16

you probably already see the pattern and extrapolate that in an infinite series, the solution would be (infinity-1)/infinity.

infinity - 1 = infinity
(infinity-1)/infinity = infinity/infinity = 1

Infinity/infinity =/= 1 ... it's an indeterminate form, and one of the reasons why you learn L'Hospital's Rule

[Jodwahh]

I'm sticking with my tri-force answer.

[antihero]

http://img300.imageshack.us/img300/8...ceyiswrong.jpg

[Cream Soda]

Hey frens, got a few more for you. Due tomorrow and I was absent this week cause it stormed and I don't have a car xD so I'm a bit behind and would be awesome if any could help. I'm good with number 1 and 2 though it would be nice to have something to check against. The help I need though is with 3 and 4 definitely. Thanks in advance

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http://i44.tinypic.com/f4icnq.jpg

[Demosthenes11]

taylor series is stuff I would recommend learning >.>
You should sit down and learn it tonight. These problems are fairly basic too, so I won't give you answers

[Ferien]

Fuck me now I remember why I stopped taking math after Multivariable.

[Octavious]

taylor series is stuff I would recommend learning >.>
You should sit down and learn it tonight. These problems are fairly basic too, so I won't give you answers

I can second this. I've used various properties of Taylor's Series multiple times in engineering, especially in my CS Numerical Analysis. D:

[Trajan]

Hey frens, got a few more for you. Due tomorrow and I was absent this week cause it stormed and I don't have a car xD so I'm a bit behind and would be awesome if any could help. I'm good with number 1 and 2 though it would be nice to have something to check against. The help I need though is with 3 and 4 definitely. Thanks in advance

Spoiler: show

http://i44.tinypic.com/f4icnq.jpg

I have to leave for work now, so i didn't get the chance to work on any of the problems, however I suggest going to Physics Help and Math Help - Physics Forums
and click on the "homework and coursework questions" and post your question there. I am not sure if you will get a response tonight but there are a lot of smart people there who could help you in the future.