# 2 minus infinity

New comments cannot be posted and votes cannot be cast, Press J to jump to the feed. Obviously, changing the letter denoting the variable will not change the value of the integral (we are still integrating the same function), so Recall that we have arrived at the following equation: I^2 = ∫_{-∞}^∞ e^{-y^2} \l(∫_{-∞}^∞ e^{-x^2}\,dx\r)\,dy\,. The comment that prompted the question was: In some ways infinity is related to zero - they're both more states of mind than actual things. I^2 = ∫_{-∞}^∞ ∫_{-∞}^∞ e^{-y^2}e^{-x^2}\,dx\,dy = ∫_{-∞}^∞ ∫_{-∞}^∞ e^{-(x^2+y^2)}\,dx\,dy\,. (we have used the formula $|(x,y)| = √{x^2+y^2}$, which allowed us to write $e^{-(x^2+y^2)} = e^{-|(x,y)|^2} = e^{-|\mathbf z|^2}$). At first, you may think that infinity divided by infinity equals one. I^2 = 2\pi ∫_0^∞ \frac{1}{2}e^{-s}\,ds = \pi\l[-e^{-s}\r]_0^∞ = \pi\,. Notice the term $r$ that wasn’t there in the ordinary Cartesian coordinates. Like other things that don't exist (the square root of -1 for example), they are useful in maths. Riemann/Lebesgue integration works in the same way in two dimensions (or any number of dimensions, for that matter); only now we are calculating volumes, not areas. Now, $∫_{-∞}^∞ e^{-x^2}\,dx$ is just a number (a constant), so we can put it inside the second integral: If you get a copy, you can learn new things and support this website at the same time—why don’t you check them out? Now we only need to calculate the two-dimensional integral $∫_{ℝ^2} e^{-|\mathbf{z}|^2}\,d\mathbf{z}$, and we’re done. Tip: See my list of the Most Common Mistakes in English. I = ∫_{-∞}^∞ e^{-x^2}\,dx\,. And you can't add to infinity. I understand infinity isn't a number but a concept- I'm just looking for an explanation as to how infinity minus 2 is logical while infinity minus 1 is not. $$ I^2 = ∫_0^∞ ∫_0^{2\pi} e^{-r^2}r\,dφ\,dr = ∫_0^∞ e^{-r^2}r ∫_0^{2\pi} 1\,dφ\,dr = 2\pi ∫_0^∞ e^{-r^2}r\,dr\,. I am going to prove what infinity minus infinity really equals, and I think you will be surprised by the answer. Under certain conditions that are satisfied in our case (such as when the function is positive and continuous), this area is equal to the definite integral calculated via primitive functions (antiderivatives), so we don’t really have to worry about the interpretation of the integrals above. 1.00) for Nintendo emulator online Free, Famicom games. $$ Is this correct? where the notation $∫_{ℝ^2} f(\mathbf{z})\,d\mathbf{z}$ simply means that we are calculating the volume under the graph of $f$ using the two-dimensional Riemann or Lebesgue integral. \text{volume under }g = ∫_0^∞ ∫_0^{2\pi} g(r,φ)\,r\,dφ\,dr\,. $$ $$ to receive a weekly summary of new articles, Follow me to get updates and engage in a discussion, You can use the image on another website, provided that you. Thanks to the $r$ we now have in the integral, we can use a simple substitution $s = r^2$: I am going to prove what infinity divided by infinity really equals, and … Infinity minus two is a possible concept, but only useful in abstract calculations. First, I am going to define this axiom (assumption) that infinity subtracted from infinity is equal to zero: ∞ - ∞ = 0; Next, I am going to add the number one to both sides of the equation. $$ Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. It is as much used in common speech (well, the adjectival form eternal) as is infinity (infinite) - but very few really mean them or understand what they really are. I don't know of any context where infinity minus 1 wouldn't make sense, but infinity minus 2 would. The way you calculate volume under the graph of a function $g$ in polar coordinates is as follows: 1.00) (NES) classic game online in your browser, Play Rockman 4 – Minus Infinity (Ver. What does Infinity Divided by Infinity Equal? Fubini’s theorem tells us that a two-dimensional integral can be split into two one-dimensional ones (if the integral is finite): $$ As it is,.as others have said, it seems inexplicable. It will teach you how to avoid mistakes with commas, prepositions, irregular verbs, and much more. However, the mathematicians haven't managed to work this one into their stuff yet. However, from the point of view of the integral with respect to $x$, $e^{-y^2}$ is also just a number, so we can put it inside, too: However, the mathematicians haven't managed to work this one into their stuff yet. Another similar concept is eternity. I^2 = ∫_{ℝ^2} e^{-|\mathbf{z}|^2}\,d\mathbf{z}\,. $$ $$ How is this all helping? $$ And that’s the end of our journey—we can only conclude that By using our Services or clicking I agree, you agree to our use of cookies. You can't divide by zero. Cookies help us deliver our Services. I recently read that infinity minus two is a possible concept, while infinity minus 1 is still infinity. In mathematics, the affinely extended real number system is obtained from the real number system ℝ by adding two elements: + ∞ and − ∞ (read as positive infinity and negative infinity respectively), where the infinities are treated as actual numbers. Zero is the absence of anything, while infinity is the top limit (or bottom limit if negative) of everything. $$ Let’s continue our calculation of $I^2$. Until you define rigorously what is meant by infinity minus a natural number, both statements are equally meaningless. $$ Another similar concept is eternity. Infinity minus two is a possible concept, but only useful in abstract calculations. Press question mark to learn the rest of the keyboard shortcuts. $$ Two times infinity is infinity. The volume under the graph of $g$ is approximately the sum of the values $g(r,φ)$ multiplied by the area of the corresponding rectangle, which is $r\Delta φ\Delta r$. $$ I = ∫_{-∞}^∞ e^{-y^2}\,dy\,. $$ "Infinity" can refer to many different concepts in many different contexts. Using Fubini’s theorem, we can convert the two one-dimensional integrals into just one two-dimensional integral: I = √\pi\,. If you provide the source where you read this, it might be possible for people to figure out the context. Infinity plus one is infinity. By the way, I have written several educational ebooks. I^2 = ∫_{-∞}^∞ e^{-x^2}\,dx ∫_{-∞}^∞ e^{-y^2}\,dy\,. $$ ∫_{ℝ^2} f(\mathbf{z})\,d\mathbf{z} = ∫_ℝ \l(∫_ℝ f(x,y)\,dx\r)\,dy = ∫_{-∞}^∞ ∫_{-∞}^∞ f(x,y)\,dx\,dy\,, Now the fun part begins, but we need a little bit of theory of two-dimensional integration.

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