May I tell you that is not a real number.
That is a shock to many. I hope you are not among them.
Therefore neither nor has any meaning.
If they are meaningless, then there is no comparison. Is there?
I am using a TI-89 calculator. The highest level of math I've taken is high school calculus. I was playing around with infinity, and entered cos(infinity), which returned sin(infinity). Why is this? I don't understand limits incredibly well, but I know that the range of the sine and cosine functions fluctuates between 1 and -1. I guess you could also say that their domain is the same, as -infinity to infinity? It is approaching infinity along the x axis, but does not approach anything along the y-axis?
Can anyone shed more light on this? Thanks.
As previously said, infinity is not a number so it cant be treated as one. Its like saying, what value on the real number line corresponds to 5*bedroom? The question doesnt make sense, so anything the calculator may spit out is worthless information.
well, in general, f(∞) may not be meaningless. in the case of real (or even rational) numbers, we can set:
, IF this limit exists (well, technically, we should also have some other restrictions on f, but
normally continuity will suffice).
both the sine and cosine function oscillate. however, the "phase shift" between sine and cosine remains constant, no matter how large x becomes:
as functions they do not "approach each other". since calculators are necessarily finite and have a limit to how large a number they can store,
(which typically is actually an integer (my guess is 1 less than some power of 2), converted to scientific notation when too large to display),
my guess is that "default values" for f(x), when x exceeds the maximum value storeable, are pre-programmed in. most of these are probably not to be trusted.
(For consistency 1/0 might return +inf rather than nan etc.)
Now if your calculator is programmed correctly and it allows you to enter this nonsense, you should get
But note the operative word here is nonsense.
Can one just define cos(\infinity) to be 0 via the following limit process? \int_0^\infty sin(x) dx=1-cos(\infty)=1, where cos(\infty)=0. The integral was done by adding a factor e^(-\alpha x) to the integrand and letting \alpha-> 0. Not sure how to type equations, so I wrote it in Tex format.
[tex] x = \neg A\cdot \neg B\cdot C + \neg A\cdot B\cdot \neg C + \neg A\cdot B\cdot C + A\cdot \neg B\cdot \neg C [/tex] gives
Click on the “go advanced” tab. On the toolbar you will see clicking on that give the LaTeX wraps, [tex] [/tex]. The code goes between them.
Define and then and .
That makes your idea impossible.
as x goes to infinity, cos(x) takes on EVERY VALUE between -1 and 1.
this is a bit easier to understand if you look at the behavior of cos(1/x) near 0 (this switches the "far ends" of the real line to a bounded area we can look at in detail). there's no number we can pick in the interval [-1,1] that will make cos(1/x) continuous at 0 (it's oscillating too quickly).