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May 20

Maclaurin series

What will the Maclaurin series for $i^{i}$ be? Here $i={\sqrt {-1}}$ —Preceding unsigned comment added by Πrate (talk • contribs) 12:55, 20 May 2010 (UTC)[reply]

i^i has to be defined by choosing some branch for the logarithm. If you put i = exp(i pi/2), you get i^i = exp(-pi/2). So, you could insert -pi/2 in the Maclaurin series for exp(x). Count Iblis (talk) 14:06, 20 May 2010 (UTC)[reply]

Note that i^i is just a (real) number once you've picked your branch of the logarithm. Therefore it doesn't have a Maclauren series expansion in the way you are thinking. It's just a number i.e. a constant function. It's like asking what the Maclauren series of 2 is and writing it out as exp(ln(2)). I think what you're asking is for the Maclauren series of z^i. Am I right? Zunaid 14:16, 21 May 2010 (UTC)[reply]

It's Maclaurin.—Emil J. 15:07, 21 May 2010 (UTC)[reply]

Partial Surface Area of a Sphere

I am looking for a way to find the surface area of a portion of a sphere given two angles and a radius. The two angles would be perpindicular to each other. For example one angle would represent an azimuth and the other would represent elevation, and the radius would be the distance.

I know that arc length (L) can be calcuated by the radius multiplied by the angle (L=rθ).

To get a complete sphere SA=4πr²

The limits for the two angles would be 0 < θ < 2π for azimuth and 0 < φ < π for elevation.

Is there a similar equation for surface area of a sphere as there is for the arc length of a circle? —Preceding unsigned comment added by 138.163.106.71 (talk) 20:07, 20 May 2010 (UTC)[reply]

I don't see how you can define area with only two angles... Unless I'm reading your question wrong? 76.229.164.175 (talk) 21:32, 20 May 2010 (UTC)[reply]

Is what you are describing a "rectangle" on the sphere, - more details please.

As for areas on spheres - yes - Solid angle is one place to start - area = solid angle x radius x radius

also see Steradian

The most common formula for solid angle is that for an area defined by three planes (forming a tetrahedron) - the solid angle is a+b+c-pi where a,b,c are the angles made on the tangential surface to the sphere by the planes.

The simplest solid angle is probably that of a Lune_(mathematics)#Spherical_geometry

I'm guessing you are describing a lune truncated at the top by the elevation angle ? If so you can get the area by subtracting the solid angles of the triangular parts at top and bottom from that of a lune, and then multplying by r^2 to get the area.

Or perhaps you mean a "segment of an orange" with one of the top points cut off.77.86.115.45 (talk) 22:04, 20 May 2010 (UTC)[reply]

Thank you all for the help. What I was trying to describe is a rectangle on a sphere. For example you have an arclenth along a sphere in a vertical plane defined by the radius and some angle φ, and that arc length is swept along another arclength defined by the same radius, but a different angle θ. This appears to be how a solid angle is defined for latitude and longitude rectangles (referencing the above link to solid angles). My only concern is if this works for a sphere, since the earth is actually an oblate spheriod and not a perfect sphere.

This does seem to fit with my original guesses for the surface area, but my limits would need to be adjusted so the angle used in the sine function (elevation or latitude) goes from -π/2 to π/2. The azimuth angle would go from 0 to 2π. I have checked this for the easy measurements that are fractions of a sphere (full sphere, half sphere, quarter, etc.), but I am hoping someone can tell me this works for all angles inside the limits, such as an elevation of π/6 radians (30 degrees) swept around a full 2π circle.

Once again thank you all for your help. 138.163.106.72 (talk) 15:10, 21 May 2010 (UTC)[reply]

I'm not sure if this is what you mean, but given a pair of latitudes φ₁ and φ₂ and a pair of longitudes θ₁ and θ₂, the area that falls between them is r²(sinφ₂ - sinφ₁)(θ₂ - θ₁). Rckrone (talk) 17:16, 21 May 2010 (UTC)[reply]

Applications of calculus in lower maths

What are some was that calculus can be applied to figure out lower-maths problems? By lower maths, I mean algebra, geometry, trig, and precalculus, not addition and subtraction, of course :) These ways should be quicker or about the same speed as doing it conventionally--for example it would not make much sense to use the derivative formula to find the slope of a linear function! 76.229.164.175 (talk) 21:25, 20 May 2010 (UTC)[reply]

If you know only trigonometry, you could wonder for what value of x the function sin(x) + sin(3x) has its largest value. The answer can be found quickly by using calculus. That's just one example. Michael Hardy (talk) 23:41, 20 May 2010 (UTC)[reply]

Agreed, finding the minimum or maximum of a function is perhaps the simplest common application of calculus. And, going the other way, there's also finding the area under a curve. The relationships between position, velocity, and acceleration (and a curve's equation, slope, and curvature) are common physical manifestations of the derivative/integral relationships. StuRat (talk) 02:00, 21 May 2010 (UTC)[reply]

I can think of the following examples. A fast way to find a partial fraction expansion is to perform Laurent expansions around the poles and then to add up the singular parts of all the expansions.

Sometimes an algebra problem can be efficiently solved using series expansions, see e.g. here

Division of numbers or polynomials is done most efficiently using Newton Raphson instead of long division.

Combinatoric problems can sometimes be efficiently solved using formal power series. Count Iblis (talk) 02:37, 21 May 2010 (UTC)[reply]

Definition of the (units of time)...that make up the components of one second.

A second is a unit of time. There are 60 seconds(unit of time) in one minute. What is the name or number of the components of one second of time? I realize there are parts ie: 10ths 100ths 1000ths (etc) of a second.! When asked to add 9.6 and 8.4 and 7.8 and 1.9 seconds...and then... how do you add the fractional parts,and what do you carry over? ie 6+4+8+9=27. do you say .7 and carry over the 20 as an equal to one second.? does .27 equal 2.7 seconds.?

At what point does the ttl fractional units make up one second.? is .10 equal to one second?

Thank You —Preceding unsigned comment added by 67.142.130.40 (talk) 23:05, 20 May 2010 (UTC)[reply]

To evaluate 9.6 + 8.4 + 7.8 + 1.9 you can proceed just as if you had 96 + 84 + 78 + 19 except that the decimal point would remain just before the last digit. That doesn't depend on whether they're seconds or liters or dollars or anything else.

I had to cogitate a bit before figuring out what your last question means. Here's my guess as to what you meant: Suppose you have, for example, 6.2 + 9.8, and you go add the 2 in 6.2 to the 8 in 9.8. You have 2 + 8 = 10. That's not .10. Rather, you put the 0 in the place right after the decimal point and carry the 1, getting 1 + 6 + 9 = 16, so the answer is 16.0 seconds, which is the same as 16 seconds. Michael Hardy (talk) 23:47, 20 May 2010 (UTC)[reply]

I don't think that was what he was saying, Michael. He might have thought that when computing seconds from minutes you would use base 60! Now, it's ridiculous to use that high of a base, but... 76.229.164.175 (talk) 02:19, 21 May 2010 (UTC)[reply]

Also, there isn't any common unit of time smaller than a second in the English system, although a moment is sometimes humorously used to mean 1/60th of a second. In metric, we can go to milliseconds (1/1000th) or even microseconds (a millionth). StuRat (talk) 01:51, 21 May 2010 (UTC)[reply]

Historically, a "minute" was the first minute (= small) subdivision of an hour, a second was the second one, and so on. That's where the name "second" comes from. So a "third" would be 1/60 second, a "fourth" would be 1/3,600 second, etc. But by the time people actually started needing time units smaller than a second, they were more likely to use milliseconds. (Degrees of arc could be divided the same way, too. An angle of 12°34'56.75" could be written 12°34'56''45''', where 1''' was a third of arc and equal to 1/60 second of arc. Again, this was not normally ever needed.) --Anonymous, 23:33 UTC, May 21, 2010.