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Post by Sandmaster on Feb 28, 2009 14:30:37 GMT -5
That won't help, but I think there is an equation for how large an object would appear using the altitude (distance) and the leg (size, in a way)
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Post by General Veers on Feb 28, 2009 14:49:13 GMT -5
That's what I was doing.
w=base= double the short leg
a=altitude=distance from back of eye
An isosceles triangle can be split into two back-to-back right triangles.
I GOT IT!
Assuming that t is the angle from the altitude to h (the hypotenuse) of the right triangle that makes up an isosceles triangle,
sint=w/h cost=a/h tant=w/a
There it is! That is what I need!
Now my limit will be in terms of t (I will substitute w/a for tant).
limt ► infinitytant
The above expression, the limit of the tangent of t as t approaches infinity, will provide the value that w'/a' will approach.
I might need a graphing calculator for this...hold on.
Darn.
limt ► infinitytant is undefined.
If you look at a graph of f(x)=tanx, the graph shows oscillating values. That graph does not approach any single value as it goes on to infinity.
The world will never know what an infinitely large mass an infinite distance away will look like or seem to look like...
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Post by Sandmaster on Feb 28, 2009 15:02:40 GMT -5
this could take a while...
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Post by General Veers on Feb 28, 2009 15:04:45 GMT -5
The limit is undefined. See above post.
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Post by Sandmaster on Feb 28, 2009 15:06:00 GMT -5
So i was right!
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Post by General Veers on Feb 28, 2009 15:10:11 GMT -5
The question is, what exactly does that mean if the apparent size of an object is undefined?
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Post by Sandmaster on Feb 28, 2009 18:34:20 GMT -5
It means that infinity has the attributes of 0 (try dividing infinity by infinity)
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Post by The Dark Master on Mar 3, 2009 14:29:31 GMT -5
So infinitiy's partner is 0? That just came as final proof as what the image of perfection is.
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Post by General Veers on Mar 3, 2009 19:47:06 GMT -5
The limit of x as x approaches infinity does not exist (because it approaches infinity, which is not any fixed value).
The limit of x-1 as x approaches infinity is 0.
Yes, they are pretty much a good pair, 0 and the concept of infinity...
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Post by General Veers on Mar 3, 2009 22:07:03 GMT -5
Edit: Post removed. Thank you.
DO NOT SPAM OR POST USELESS COMMENTS
Please ensure that you actually ask questions concerning math principles, postulates, theorems, misconceptions, etc., or else that you answer such questions.
You may also bring up new principles or neat observations, such as what vaconcovat did with the palindrome numbers, or request discoveries such as what Sandmaster did with the "infinite size - infinite distance" problem.
You can also correct any mistakes that you see.
Thank you.
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Post by General Veers on Mar 4, 2009 0:17:59 GMT -5
I see that you edited the post above my reminder. Thank you, but it still isn't contributing to the thread. I am trying to keep out as much spam as possible. Please cooperate. Thank you for your patience and understanding.
Edit: Post removed. Thank you.
Sequences (Condensed)
A sequence is a list of terms whose value depends upon where it is in the list. The first term of a sequence is referred to as a1, the second term as a2, and so on. Some nth term, somewhere in the list, is called an. A sequence can have a finite amount of terms or an infinite amount of terms.
For example...
1,2,3,4,5,6,7,8,9,10,11
That is a sequence of numbers that follow the rule an=1+1(n-1)=1-1+n=n. a1=1 a2=2 a3=3 a4=4 a5=5 a6=6 a7=7 a8=8 a9=9 a10=10 a11=11
1,2,4,8,16,32,64,128,256,512,1024
That is a sequence of numbers that follow the rule an=1*2n-1=2n-1 a1=1 a2=2 a3=4 a4=8 a5=16 a6=32 a7=64 a8=128 a9=256 a10=512 a11=1024
The first sequence was an arithmetic sequence: subtracting any one term by the term before it would provide the same difference no matter which two consecutive terms you choose. Arithmetic series follow the pattern an=a1+(n-1)d, where n is which term in the sequence you want to find and d is the common difference found by subtracting any two consecutive terms.
The second sequence was a geometric sequence: dividing any one term by the term before it would provide the same ratio no matter which two consecutive terms you choose. Geometric series follow the pattern an=a1rn-1, where n is which term in the sequence you want to find and r is the common ratio found by dividing any two consecutive terms.
NOT ALL SEQUENCES ARE ARITHMETIC OR GEOMETRIC!
For example, the sequence whose nth term is 1/n! is neither an arithmetic or geometric sequence.
Series (Condensed)
A series is the sum of the terms in a sequence.
For example...
S11=1+2+3+4+5+6+7+8+9+10+11=66
This is the sum of the arithmetic sequence example from a1 to a11. Another way to solve for the sum is with the formula Sn=(n/2)(a1+an).
S11=1+2+4+8+16+32+64+128+256+512+1024=2047
This is the sum of the geometric sequence example from a1 to a11. Another way to solve for the sum is with the formula Sn=a1(1-rn)/(1-r).
NOT ALL SERIES ARE ARITHMETIC OR GEOMETRIC!
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Post by Qwerty on Mar 4, 2009 0:36:50 GMT -5
Okay...
Any way to say it in English?
Indeed...
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Post by General Veers on Mar 4, 2009 0:41:09 GMT -5
I edited it some, although not by much. I hope this helps?
Sequences: list of terms that follow a pattern
Series: sum of terms in a sequence.
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Post by Qwerty on Mar 4, 2009 0:44:24 GMT -5
Yup, that clears it up.
Indeed, it does...
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Post by The Dark Master on Mar 4, 2009 14:06:04 GMT -5
0x0=infinity Infinity x infinity=infinity
infinity-infinty=0 0 x infinity= 0
Likeness? Could this mean 0=infinity? You may think: 'The man's mad, 0 is nothing and infinity is, well...Infitity!' Howver, that is in physics, as maths thingymabobies, they are very similar.... I'm not really a dab hand at maths, but it's worth a shot.....
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Post by General Veers on Mar 4, 2009 15:45:51 GMT -5
Any value that is not undefined times zero is zero. E.g. 0n=0, when n is a defined value.
Zero times an undefined number technically cannot be determined. (0/x)(x/0)=0x/0x=0/0, which is an indeterminate form.
Zero times infinity...well, the limit of 0n as n approaches infinity would indeed be zero, but you couldn't perform that operation without limits.
Practically any operation with the concept of infinity used as a value would be indeterminate, although limits can be used to find one of several possible values for any of the indeterminate forms. For example, (5*infinity5-2*infinity4+10*infinity3+infinity2-7*infinity+8)/(infinity6+6*infinity5+15*infinity4+20*infinity3+15*infinity2+6*infinity+1) would be indeterminate as is, but by taking the limit of that function (with x's instead of infinities) as x approached infinity and using L'Hopital's Rule, you would get the following:
limitx ► infinity(5x5-2x4+10x3+x2-7x+8)/(x6+6x5+15x4+20x3+15x2+6x+1) limitx ► infinity(25x4-8x3+30x2+2x-7)/(6x5+30x4+60x3+60x2+30x+6) limitx ► infinity(100x3-24x2+60x+2)/(30x4+120x3+180x2+120x+30) limitx ► infinity(300x2-48x+60)/(120x3+360x2+360x+120) limitx ► infinity(600x-48)/(360x2+720x+360) limitx ► infinity(600)/(720x+720) 0 (The limit of a constant over a variable as that variable approaches infinity is equivalent to zero.)
You might not understand L'Hopital's Rule until you take differential calculus, since L'Hopital's Rule uses derivatives to make one of two specific indeterminate forms (infinity/infinity and 0/0) determinable.
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Post by Sandmaster on Mar 9, 2009 17:10:13 GMT -5
What about phi (1.168...or something)?
It's awsome
Also you get closer to it every time you divide the larger by the smaller of two consecutive numbers in the Fibbonacci sequence!
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Post by General Veers on Mar 9, 2009 19:57:34 GMT -5
Phi, the golden ratio?
It's exact value is (1 + 51/2)/2
It is approximately 1.61803398875...
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Post by Sandmaster on Mar 10, 2009 14:48:28 GMT -5
yeah, phi is awesome
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Post by General Veers on Mar 10, 2009 16:45:18 GMT -5
I never did realize that limn ► infinity(an/an-1)=phi when an is a term in the Fibonacci sequence.
Thank you.
Did you know that sum(1/n!,n,0,infinity)=e?
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