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April 4

a question.

how to solve equation like,sin(x)=exp(x)?AND can we put x=log[sin(x)]?thank youHusseinshimaljasimdini (talk) 11:18, 4 April 2008 (UTC)[reply]

Yes, you can rewrite the equation as

x=\log(\sin(x))

, as long as you add the constraint that

sin(x)>0

. From the graphs of sin(x) and exp(x) it is clear that there are an infinite number of solutions, all less than 0. I very much doubt that there is a closed form solution - I think the best you could hope for would be a power series. Gandalf61 (talk) 13:02, 4 April 2008 (UTC)[reply]

A better method might be to write sin as

{\frac {1}{2i}}\left(e^{ix}-e^{-ix}\right)

, which would at least give solutions without resorting to logs of trig functions. That being said, it's still pretty much impossible to solve analytically I'd think. -mattbuck (Talk) 21:07, 4 April 2008 (UTC)[reply]

Approximate exp(x)−sin(x) by p(x)=1+x²/2+x³/3+x⁴/24+x⁶/720+x⁷/2520 and solve the algebraic equation p(x)=0 by the Durand-Kerner method. Bo Jacoby (talk) 15:51, 5 April 2008 (UTC).[reply]

If you're going to approximate, it's much better to use Newton's method. Use negative multiples of π as seeds. -- Meni Rosenfeld (talk) 16:54, 5 April 2008 (UTC)[reply]

I respectfully disagree. Newton's method is likely not to converge. Smart choises of seed are needed. Nonreal complex roots are often not found. Bo Jacoby (talk) 17:54, 5 April 2008 (UTC).[reply]

So? I wasn't talking about the general case (though I could), but rather for this particular problem. If we are only interested in real roots, I have already given the needed smart seeds. The Durand-Kerner may be good for polynomials, but this is not a polynomial, and approximating it as one is likely to cause problems and should only be used as a last resort. In this case, using the polynomial you have given, the first root is -3.01303. The actual root is -3.18306. Just guessing the solution to be -π is already better. -- Meni Rosenfeld (talk) 18:21, 5 April 2008 (UTC)[reply]

The formulation "how to solve equation like,sin(x)=exp(x)" is referring to the general case. Bo Jacoby (talk) 07:08, 6 April 2008 (UTC).[reply]

That's true. Your suggestion does generalize a bit better than "the roots are clearly close to negative multiples of π", but I maintain my position that it is only applicable if we are willing to ignore roots distant from the point of expansion, to find a few spurious roots, and to either have low accuracy or refine it with a general iterative method (such as Newton's). -- Meni Rosenfeld (talk) 14:38, 6 April 2008 (UTC)[reply]

For any entire function f(x), and for any precision ε>0, and for any radius R>0, there exists a degree N>0 such that the Taylor polynomial approximation p_n(x) to f(x) for any n>N and for any |x|<R satifies |f(x)−p_n(x)|<ε. So a general method to find zeroes of a transcendent entire function f(x) within a circle |x|<R is to compute the zeroes of polynomial approximations. Newton's method finds one root at a time, and it is hard to know if all roots inside some circle have been found. That's why I prefer the Durand-Kerner method which computes all the roots of a polynomial. Bo Jacoby (talk) 15:22, 6 April 2008 (UTC).[reply]

Series having a nice expression for partial sums

I'm looking for a convergent series with positive terms, other than a geometric series or a telescoping series, which converges to a nice value and whose partial sums are also nicely expressible. Does anyone have any examples of series like this? —Bkell (talk) 16:59, 4 April 2008 (UTC)[reply]

After some thought I realize that any series which converges to a nice value and whose partial sums are nicely expressible can be written as a telescoping series, so maybe my question doesn't make sense. I'll have to think some more about what I really mean. —Bkell (talk) 20:07, 4 April 2008 (UTC)[reply]

I think I know what you mean. You are probably looking for an example of an infinite series which you can show converges using only the definition; i.e., the sequence of partial sums converges. Hence, you want a nice expression for that sequence, so that it is "easy" to see that the series converges. Am I right? –King Bee ^{(τ • γ)} 20:20, 4 April 2008 (UTC)[reply]

Converting an infinite series' decimals to fractions

Hi. Now, before you freak about its easiness, let me explain. I know how to do some infinite series of decimals. For example, 0.33333... is 1/3, 0.090909... is 1/11, 0.166666... is 1/6, 0.11111... is 1/9, 0.22222... is 2/9, 0.99999... is 1, and so on. Other ones, like 0.363636... are calculatable by dividing 1 by that number (1/2.75 = 4/11), and other ones like 0.181818..., 0.833333..., and 1.55555... are easy because they're based on the above numbers. However, is there a quick way, other than rounding, truncating, estimating, and making the denominator infinity, to make a number that produces a very long series of decimals when you divide it into 1, and is a repeat of a prime number, into a fraction? For example, 0.23232323... . Or, can it be possible for a repeating series to be irrational, or can only non-repeating series like 3.1415926535897932384626433832... be irrational? I'm not asking for the answer to a specific question, just if there is a way to do this accurately, or is rounding/truncating better? Thanks. ~A H 1^(TCU) 21:14, 4 April 2008 (UTC)[reply]

As long as there's a repeating unit, it can be done. Just multiply by powers of 10 until the first repetition starts just after the decimal, then multiply by 10 again until the second repetition starts after the decimal. You can then subtract one from the other, leaving only integer values.

Take as an example your 0.23232323... let's call that X. 100X = 23.232323.... So 100X - X = 23.23... - 0.23... = 23 = 99X. Thus, X = 23/99.

Thus any repeated decimal can be written as a fraction, and so only non-repeating decimals are irrational. -mattbuck (Talk) 21:47, 4 April 2008 (UTC)[reply]

Also, any fraction has a repeating decimal expansion. That is, the numbers with a repeating decimal expansion (including repeating zeroes) are exactly the rational numbers. Phaunt (talk) 12:52, 5 April 2008 (UTC)[reply]

It should also be mentioned that the repeating decimal expansions aren't necessarily unique (the OP uses 0.999...=1 as an example, similar equalities exist for any terminating decimal). --Tango (talk) 14:36, 5 April 2008 (UTC)[reply]

Not for any terminating decimal but only for 9 and 0 because 0.9999...=1.0000...; 1/3=0.33333... is unique. Bo Jacoby (talk) 15:25, 5 April 2008 (UTC).[reply]

0.333... isn't terminating. Terminating means repeating 0's, and you can reduce the last non-zero digit by one and replace the repeating 0's with repeating 9's. If you have repeating 9's, you can, of course, do the same in reverse. --Tango (talk) 15:36, 5 April 2008 (UTC)[reply]

My favorite way to do this is to treat the fraction as the sum of an infinite Geometric Series, which off the top of my head, I don't remember exactly, but the idea is this: .11111111 is 1*.1 + 1*.01 + 1*.001, etc. --Arcoain (talk) 00:36, 8 April 2008 (UTC)[reply]