Math Is Fun Forum

Oops, that does not work. If tanAtanB = 1, then the denominator in the formula for tan(A+B) is 0.

Sorry, back to the drawing board …

269) Roger Federer

Roger Federer (German pronunciation: [ˈrɔdʒər ˈfeːdərər]; born 8 August 1981) is a Swiss professional tennis player who is currently ranked world No. 2 in men's singles tennis by the Association of Tennis Professionals (ATP). Federer has won 20 Grand Slam singles titles and has held the world No. 1 spot in the ATP rankings for 302 weeks, including 237 consecutive weeks. After turning professional in 1998, he was continuously ranked in the top ten from October 2002 to November 2016. He re-entered the top ten following his victory at the 2017 Australian Open.
https://en.wikipedia.org/wiki/Roger_Federer

Roger Federer won his sixth Australian Open and 20th Grand Slam title with a five-set victory over Marin Čilić.
The Swiss lost five games in a row as he dropped the fourth set but recovered to win 6-2 6-7 (5-7) 6-3 3-6 6-1.
Federer, 36, becomes only the fourth player after Margaret Court, Serena Williams and Steffi Graf to win 20 or more major singles titles.
"It's a dream come true and the fairytale continues," said Federer, who has won three of the last five majors.
http://www.bbc.co.uk/sport/tennis/42851064

Has there ever been, or will there ever be, a more fantastic tennis player than Rogerer Federer, winner of 20 Grand Slam singles title (and counting)?
http://speakforum.forumotion.com/t43-roger-federer

Your question makes no sense to me.  Branches of mathematics (or any other subject) are neither “sequential” nor complete: they don’t form any kind of order,  nor do they have a beginning and end. Well, maybe there is beginning, couched in the obscurities of prehistoric time, but definitely no end. New branches are cropping up all the time – for example, chaos theory and the study of fractals were unheard of before the 1960s. There are also topics (such as non-standard analysis and rational trigonometry) that are not universally accepted by mathematicians as valid branches of mathematics.

But if you think your question makes sense, then go and google it yourself – don’t be lazy.

A wave with a high frequency has a short wavelength, so its peaks and troughs are bunched up close together. One with a low frequency will have a longer wavelength, i.e. its peaks and troughs are more spread out.

It still works, provided the modulus of the common ratio is less than 1.

So if n ≡ 2 or 3 (mod 4) it is never possible to have all lights on at the end. I will show that you can have them all on if n ≡ 0 or 1 (mod 4).

For n = 4, the process is simple:

0000
1000
1110
0000
1111

using bob bundy’s notation. Let us give this an easy-to-remember name and call it the “four-switch process”.

Now label the switches 1, 2, …, n. Suppose n is a multiple of 4. The first four people perform the “four-switch process” to the first four switches. The next four people then toggle the first four switches and perform the “four-switch process” to switches 5 to 8. Then switches 5–8 will all be on; since switches 1–4 are already on, toggling them an even number of times will leave them on. Hence after the first 8 people, switches 1–8 will be on. Then we can continue the process, persons 9–12 toggling switches 1–8 and doing the “four-switch process” to swtiches 9–12, and so on. Hence when n ≡ 0 (mod 4) it is always possible to leave all the lights on.

Bob was scratching his head over the case n = 5; let me put him out of his misery.

00000
10000
01000
11110
00000
11111

The first person turns the first switch on, persons 2–5 toggle the first switch and perform the four-switch process on switches 2–5. So if n ≡ 1 (mod 4) proceed as follows: persons 6–9 toggle the first 5 switches and do the four-switch process on switches 6–9, persons 10–13 toggle the first 9 switches and do the four-switch process on switches 10–13, and so on – leaving all lights on at the end.

So there you have it:

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I think it looks great!

I'm not as au fait with Fourier series (and analysis topics in general) as I am with other, more algebraic areas of mathematics, so it's great to learn something new.

This YouTube video may be Quite Interesting to you:

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[*]https://www.youtube.com/watch?v=BE657F9sqyQ[/*]
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http://www.mathsisfun.com/puzzles/12-da … ution.html

It’s a partridge IN a pear tree, not partridge AND a pear tree. The tree provides a setting for the bird to be in and is not counted as a separate gift.

The solution is involves a sum of triangular numbers

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where

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