میدانیم pi و e متعالی هستند. بنابراین
(x-pi)(x-e)=x^2- (e+pi)x + epi
نمیتواند ضرایب گویا داشته باشد. پس حداقل یکی از epi و e+pi گنگ هستند. اما آیا pi+e گویاست؟
@infinitymath
(x-pi)(x-e)=x^2- (e+pi)x + epi
نمیتواند ضرایب گویا داشته باشد. پس حداقل یکی از epi و e+pi گنگ هستند. اما آیا pi+e گویاست؟
@infinitymath
Infinity ...
... it's not big ...
... it's not huge ...
... it's not tremendously large ...
... it's not extremely humongously enormous ...
... it's ...
Endless!
Infinity has no end
Infinity is the idea of something that has no end.
In our world we don't have anything like it. So we imagine traveling on and on, trying hard to get there, but that is not actually infinity.
So don't think like that (it just hurts your brain!). Just think "endless", or "boundless".
If there is no reason something should stop, then it is infinite.
Infinity does not grow
Infinity is not "getting larger", it is already fully formed.
Sometimes people (including me) say it "goes on and on" which sounds like it is growing somehow. But infinity does not do anything, it just is.
Infinity is not a real number
Infinity is not a real number, it is an idea. An idea of something without an end.
Infinity cannot be measured.
Even these faraway galaxies can't compete with infinity.
Infinity is Simple
Yes! It is actually simpler than things which do have an end. Because when something has an end, we have to define where that end is.
@infinitymath
... it's not big ...
... it's not huge ...
... it's not tremendously large ...
... it's not extremely humongously enormous ...
... it's ...
Endless!
Infinity has no end
Infinity is the idea of something that has no end.
In our world we don't have anything like it. So we imagine traveling on and on, trying hard to get there, but that is not actually infinity.
So don't think like that (it just hurts your brain!). Just think "endless", or "boundless".
If there is no reason something should stop, then it is infinite.
Infinity does not grow
Infinity is not "getting larger", it is already fully formed.
Sometimes people (including me) say it "goes on and on" which sounds like it is growing somehow. But infinity does not do anything, it just is.
Infinity is not a real number
Infinity is not a real number, it is an idea. An idea of something without an end.
Infinity cannot be measured.
Even these faraway galaxies can't compete with infinity.
Infinity is Simple
Yes! It is actually simpler than things which do have an end. Because when something has an end, we have to define where that end is.
@infinitymath
Above image shows all circles of the form
(x-A(k))2+(y-B(k))2=(R(k))2,
for k=-10000, -9999, ... , 9999, 10000, where
A(k)=(3k/20000)+(cos(37πk/10000))6sin((k/10000)7(3π/5))+(9/7)(cos(37πk/10000))16(cos(πk/20000))12sin(πk/10000),
B(k)=(-5/4)(cos(37πk/10000))6cos((k/10000)7(3π/5))(1+😔cos(πk/20000)cos(3πk/20000))8)+(2/3)(cos(3πk/200000)cos(9πk/200000)cos(9πk/100000))12,
R(k)=(1/32)+(1/15)(sin(37πk/10000))2((sin(πk/10000))2+(3/2)(cos(πk/20000))18).
@infinitymath
(x-A(k))2+(y-B(k))2=(R(k))2,
for k=-10000, -9999, ... , 9999, 10000, where
A(k)=(3k/20000)+(cos(37πk/10000))6sin((k/10000)7(3π/5))+(9/7)(cos(37πk/10000))16(cos(πk/20000))12sin(πk/10000),
B(k)=(-5/4)(cos(37πk/10000))6cos((k/10000)7(3π/5))(1+😔cos(πk/20000)cos(3πk/20000))8)+(2/3)(cos(3πk/200000)cos(9πk/200000)cos(9πk/100000))12,
R(k)=(1/32)+(1/15)(sin(37πk/10000))2((sin(πk/10000))2+(3/2)(cos(πk/20000))18).
@infinitymath
@infinitymath
Visit http://blogs.ams.org/visualinsight/2016/01/15/cairo-tiling/
to see how To construct the Cairo tiling
@infinitymath
Visit http://blogs.ams.org/visualinsight/2016/01/15/cairo-tiling/
to see how To construct the Cairo tiling
@infinitymath
Cantor Cube K^3 👆👆👆👆👆👆
Cantor cube is a cantor space and it is homeomorphic to cantor set in reals line K. In fact K^n is homeomorphic to K, for all n.
The above mentioned theorem on uniqueness of the cantor space is due to Brouwer.
Cantor cube is a cantor space and it is homeomorphic to cantor set in reals line K. In fact K^n is homeomorphic to K, for all n.
The above mentioned theorem on uniqueness of the cantor space is due to Brouwer.