
</pedantic>
Good question. "Infinity" is not a real number, so the usual understanding of the multiplication operator (which takes two real numbers and produces a real number) does not apply. You can try to extend the real numbers by adding two objects called positive and negative infinity, defined in terms of the limit of an unbounded sequence. So positive infinity is something greater than any finite number (and negative infinity), and vice-versa. Then things like dividing something by infinity, or adding infinity to something, follow naturally from limit operations, and have the values that you'd expect.
Unfortunately there's still no help for "0 * infinity" and other so-called indeterminate forms. This kind of infinity is defined as the limit of any unbounded sequence, but unlike the defined forms like "1 / infinity = 0", the value of "0 * infinity" will depend on the details of the particular sequence. Consider: "x / infinity = 0" is true for any constant, finite x; substituting it into "0 * infinity" gives "(x / infinity) * infinity". The infinities cancel (since we can let them represent the same unbounded sequence), leaving "0 * infinity = x", for any x that you like.
So my preferred answer to the question "What is 0 * infinity" is: mu. It can take on any value, but only because you've thrown away the information about what unbounded sequence, exactly, is represented by this infinity. It means you need to go back and analyze the original problem more carefully. In the context of this particular problem, the question is if a finite bandwidth, divided into an infinite number number of infinitely small channels, gives you infinite bandwidth. Stated precisely this way it's almost obvious. If N is the number of channels and B is the total bandwidth, then the bandwidth of each channel is "B/N". What happens to the total of the channels' bandwidths as N approaches infinity? Well, N goes to infinity and B/N goes to zero, which is where the question "0 * infinity" comes from. But if we keep the information about the original sequences, "N * (B/N)" is just B, and nothing else.
Wrong. The ISS is in constant free fall, far from zero G.
I don't believe there is a difference.
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