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March 30

when the power of number is not a whole number

Hi friends. I know how to calculate a number to the power of two, which means the number times itself, N*N. What do I enter in a calculator to calculate a number to the power of 2.5? Thank you. 70.67.193.176 (talk) 15:41, 30 March 2022 (UTC)[reply]

Exponentiation (raising to a power) is not primarily about repeated multiplication. Only in the very limited case of having positive integer exponents do the two functions align; we tend to teach exponentiation as repeated multiplication, because it makes the function seem like a simpler way to do repeated multiplication, but as I said, this only works for integers. The better way to think about exponentiation is as a type of function where the rate of change of the function is equal to a constant times the function itself, which is to say that d(a^x)/dx = c*a^x. That definition is sufficiently narrow to include only the exponential function itself, and broad enough to encompass all forms of exponentiation, including using all real number, complex numbers, even matrices and tensors and even other functions as exponents. This definition also includes the facile definition of "repeated multiplication" for the positive integers. However, getting into this more fundamental definition of exponentiation really probably needs us to do more calculus than to answer your question directly. I instead direct you to the EXCELLENT work of YouTuber 3blue1brown, who does a great job of explaining exponentiation and its core definition in many of his videos; he addresses it in his "Lockdown Math" series, and also in his "Introduction to Calculus" and any video he does on Euler's number and exponentiation in general. He's usually the first place I recommend people go when they are struggling with hard math concepts.

To more directly answer your question in direct terms, fractional exponents mean the root of a number, so a^0.5 = a^1/2 = square root of a, and similarly a^1/3 = cube root of a. You didn't ask about negative numbers, but negative exponents just mean the multiplicative inverse of the number, so a^-x = 1/a^x. There are a few ways to solve a^2.5; the easiest may be to consider it as a^5/2, which is the square root of a⁵. --Jayron 32 16:06, 30 March 2022 (UTC)[reply]

Answering the calculator part of the question: Many calculators perform this function with a button labeled x^y. Use for your example: a number → x^y → 2.5 → = . By the way, 3b1b is an outstanding channel!:-) DB1729 (talk) 16:23, 30 March 2022 (UTC)[reply]

Thank you both, both methods got me the number. 70.67.193.176 (talk) 19:47, 30 March 2022 (UTC)[reply]

In school we were trained to use logarithm tables, which were base 10. To find the value of

x\times y

, the drill was:

Find (using the table) the value of $\log x$ ; let's call it $\xi .$
Find (using the table) the value of $\log y$ ; let's call it $\eta .$
Compute $\zeta =\xi +\eta .$
Find (using the table) the value of $\operatorname {antilog} \zeta .$

To find the value of

x^{y}

, you could do this:

Find (using the table) the value of $\log x$ ; let's call it $\xi .$
Compute $\zeta =y\times \xi .$
Find (using the table) the value of $\operatorname {antilog} \zeta .$

--Lambiam 23:51, 30 March 2022 (UTC)[reply]

Wow, Lambiam, you must be as old as me! —Tamfang (talk) 02:19, 2 April 2022 (UTC)[reply]

Circles on Regular Polygon sides overlaps.

If I have a square and add a circle centered on the side and whose diameter is the entire side, the overlap between the circles is *relatively* easy to calculate since the other crossing point of the circles is in the middle of the square. (two circles sharing a corner will share an area equal to the two quarter circles minus the quarter of the square). Is there *any* hope of doing this with other Regular Polygons (or even a general formula?) (I know the Triangle has to deal with a three way overlap, so that case can be excluded, I think).Naraht (talk) 17:22, 30 March 2022 (UTC)[reply]

The area of overlap between two circles is twice the area of a circular segment:

A={\tfrac {R^{2}}{2}}\left(\theta -\sin \theta \right).

I get R = s/2 where s is the side of the polygon, and θ = 2π/n, making the area of the overlap:

2A={\tfrac {s^{2}}{4}}\left({\tfrac {2\pi }{n}}-\sin {\tfrac {2\pi }{n}}\right).

To get the total overlap, assuming n≥4, multiply this by n. --RDBury (talk) 19:14, 30 March 2022 (UTC)[reply]