Always good stuff from Prime Newtons. A great way to start 2023!

Thanks bro....I searched more than 10 videos on this topic but none of them explained it as much simply.....

2wo plus 2wo equals 4our. That pretty much sums it up. Your shirt is awesome!

Thank you for this video. I never knew these forms.

Trig has broken my brain; I keep reading "sin" as "sine."

@PrimeNewtons

6 ай бұрын

😀😀😀😀😀🤣🤣

THANK YOU FOR TEACHING US!!!

So glad to have found this great video again. I forgot to bookmark it when I first came across it, and couldn't remember what it was called, and then just now something like "seven mathematical sins" suddenly popped into my head. Now bookmarked! 👍

I didn’t know this. Wow!

A brilliant methanation 👍👍👍👍👍👍

Woow I love it. THANKS GUY👍🙏

Veramente una bellissima esposizione!!! Complimenti e grazie!

By the way, there are 5 more indeterminate forms that you forgot: log_1(1), log_0(0), log_infinity(infinity), log_0(infinity), log_infinity(infinity). These numbers are called "indeterminate" as they are formed from using the inverse of an annihilation function. An annihilation function is a function that, whatever input you put in, outputs the same output. For example, multiplying by 0 is an annihilation function as all numbers multiplied by 0 equals 0. An inverse function is one that when you take the output of the first function, it would return the input of the previous function. Therefore, if a function that annihilates all real numbers were inverted, and you placed the annihilation result in your function, then does that mean every single number that could possibly be annihilated be produced as outputs?!?! Like I said, any number multiplied by 0 equals 0, so 0/0 is indeterminate. Since 0^-1 is infinity, infinity*0 and infinity/infinity are exact copies of 0/0 in disguise, so they all are indeterminate. In limits, the results are based off of the cardinalities of the 2 numbers, as 0, 1, and infinity (also 4 in googology). All other numbers have 1 cardinal to its ordinal, but those numbers have an infinite number of cardinalities. Infinity-infinity is even more indeterminate than you expect. Consider the natural logarithm function. This function is unique as it is an integral-based function of the integral from 1 to x of 1/x variable x. Since e^(pi*i)=-1, we can conclude that ln(-1)=pi*i. This means that the integral from -1 to 1 of 1/x variable x is -pi*i, as we can reverse the integration bounds by negating the result. Now, let's evaluate the integral of 1/x variable x as a function of area. Since the function 1/x has a vertical asymptote at x=0, we need to split the integral into 2 parts: The integral from -1 to 0, and the integral from 0 to 1. We know that the first integral has infinite area below the x-axis, so it is negative infinity. We also know that the second integral has infinite area above the x-axis, so it is positive infinity. Adding the integrals give the indeterminate form infinity-infinity being equal to complex numbers as well. Since the log base infinity of any nonnegative real number is 0, infinity^0 is indeterminate. 0^0 comes to a problem of being indeterminate due to the arithmetic-geometric transfer. In the arithmetic-geometric transfer, all 0's become 1's and the operation hierarchy moves up by 1 level: Addition to multiplication, multiplication to exponentation, and exponentation to tetration. Therefore, 0^0=1^^1, or 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of one 1, so if 0^0 is indeterminate, 1 tetrated to 1 is also indeterminate, and all numbers are equal to one another. Also, 0 factorial has the same discrepancy as we know that factorials of nonnegative numbers is the product of all numbers less than or equal to the number. Since 0 is an annihilator, 0 isn't included in the product. Therefore, 1 is the lowest number that can use the factorial definition. However, we know that (-1)! is infinity, so 0!=(-1!)*0 (due to the factorial rules), but that is infinity*0, which is also indeterminate. Therefore, 0^0 and 0! live and die together: If 0! is 1, 0^0 is also 1, and if 0^0 is indeterminate, 0! is also indeterminate. Since the infinitieth roots of any number is 1, then 1^infinity is indeterminate. Since 1^x is 1 with x being any number, then log_1(1) is indeterminate. Since 0^x is either 0, 1, or infinity with x being any number, then log_0(0) and log_0(infinity) are both indeterminate. Since infinity^x is either 0 or infinity, then log_infinity(0) and log_infinity(infinity) are both indeterminate.

Great explanation 👌I m also a mathematics teacher

Love from India❤, Way of explaining is awesome

thank you 👏💐

Wow

Excellent one out there sir. Thanks for teaching us. Sir I have a worry, in coordinate geometry the gradient of the x-axis is 0 and the gradient of the y-axis is infinity. The x-axis is perpendicular to the y-axis hence the product of their gradients is -1. Does it mean the product of infinity and 0 is -1? Thanks

isn't 1/0 equal to both positive and negative infinity though?

@enderlord6329

7 ай бұрын

no, the limit of 1/x as x approaches 0 is infinity, 1/x as itself is indeterminate

@thescholarsjourney661

6 ай бұрын

@@enderlord6329 More specifically, the limit of 1/x as x approaches infinity from the right is infinity, but the limit of 1/x as x approaches infinity from the left is *negative* infinity.)

If you write zero over zero as zero to the second power divided by zero to the first power you get zero to the first power and shouldn’t that be indeterminate as well?

@PrimeNewtons

7 ай бұрын

That is the meaning of indeterminate. You could also make 0/0 0^‐1 by that logic

😮

you missed 0^infinite

0^0 is 1

@lawrencejelsma8118

6 ай бұрын

He just proved (1 + 1/x)^x as x goes to infinity ... That is ... (1 + u)^(1/u) as u goes to 0 equals e^1 by a proof that uses L'Hopital's Rule validates (as well as the series representing the function e^x with x=0). Likewise for 0^0 that evaluates to 1 for x^x as x becomes 0 in that function. The question arises can there be another function of f(x)^g(x) where f(some x) = 0^[g(the same x) = 0 also) that doesn't equal 1 proving an f(x)^g(x) at some x where 0^0 is not equal to 1!? Notice he doesn't list 1/(infinity) or any finite number/infinity being 0 because of it's limit. Since he accepts for f(x) = 1/x or f(x) = e^(-x) etc. functions (as do sciences calculations in physics and all engineering) as x becomes sufficiently large proofs: in precision calculations of 1/x and e^(-x) a 0 as a limit acceptable infinite x value equal 0 is okay in math. This extends to x^x as 0^0 limits as equal as (1/u)^(1/u) as u goes to infinity limit of value calculations. In analyzing 0/0 I can agree with disjoint proofs of f(x) = x/x as x goes to 0 that is 1/1 = 1 using L'Hopital's Rule. This becomes a different result that (0)(1/x) produces as x goes to 0. x/x is 1 at x=0 doesn't prove (0)(1/x) is 1 also by assuming 0/0 proves being 1 all the time in limit calculations. (0)/x as x goes to 0 is by L'Hopital's Rule 0/1 = 0.