blocks|key|2231388|text|我不知道Matlab计算sin(x)的确切方法--但是你可以通过使用幂级数来计算它来研究它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2231389|sin+x+=+x+-+(x%5E3)/3!+%2B+(x%5E5)/5!+-+(x%5E7)/7!+%2B+(x%5E9)/9!+...|code-block|syntax|javascript|2231390|将其转换为一些Matlab代码，我们通过以下方式表示它：|2231391|clc
x+=+pi;+++++%25++or+x+=+pi/2
res+=+x;
factor+=+-1;
for+ii=3:2:19
++res+=+res+%2B+factor*power(x,ii)/factorial(ii);
++factor+=+factor*-1;++
++fprintf+(+'iteration+%252i++sin(x)=%251.16f\n',+(ii-1)/2,+res+);
end
res|2231392|在x=pi和x=pi/2上运行这段代码，你可以看到x=pi/2收敛到正确的结果(在每秒误差内)相当快(9次迭代)-而x=pi的情况不会在相同的时间范围内收敛。|2231393|值得注意的是，在9次迭代时，在阶乘(19)中计算的最后一个阶乘。在这个序列中要计算的下一个阶乘是21。这是最后一个阶乘，由于双精度，它可以100%25准确地表示(参见help+factorial)。|2231394|所以我认为，对于pi/2，数学解收敛到1，比pi的情况快两倍精度。事实上，由于数学上的限制和可以存储在双精度结果中的准确性，圆周率的情况不能完全收敛。|2231395|已经说过了，sin(pi)在eps中，所以您应该将这一事实用于您的目的。|2231396|我已经复制了下面得到的结果(R2015b)：|2231397|Results+for+PI/2
iteration++1++sin(x)=0.9248322292886504
iteration++2++sin(x)=1.0045248555348174
iteration++3++sin(x)=0.9998431013994987
iteration++4++sin(x)=1.0000035425842861
iteration++5++sin(x)=0.9999999437410510
iteration++6++sin(x)=1.0000000006627803
iteration++7++sin(x)=0.9999999999939768
iteration++8++sin(x)=1.0000000000000437
iteration++9++sin(x)=1.0000000000000000
Final+Result:+1.0000000000000000


Results+for+PI
iteration++1++sin(x)=-2.0261201264601763
iteration++2++sin(x)=0.5240439134171688
iteration++3++sin(x)=-0.0752206159036231
iteration++4++sin(x)=0.0069252707075051
iteration++5++sin(x)=-0.0004451602382092
iteration++6++sin(x)=0.0000211425675584
iteration++7++sin(x)=-0.0000007727858894
iteration++8++sin(x)=0.0000000224195107
iteration++9++sin(x)=-0.0000000005289183
Final+Result:+-0.0000000005289183|2231398|entityMap^0|C|6|0|0|0|0|1|4|6|6|P|6|1M|4|0|29|E|0|0|6|7|E|3|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1H|8|@$9|1I|A|1J|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|1K|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1L|8|@$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]]|D|@]|E|$]]|$1|W|3|X|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1R|8|@]|D|@]|E|$I|J]]|$1|10|3|-4|5|6|7|1S|8|@]|D|@]|E|$]]]|11|$]]

I dont know the exact way that Matlab calculates <code>sin(x)</code> - but you can investigate this by calculating it using the power series, i.e.

<pre><code>sin x = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + (x^9)/9! ...
</code></pre>

Turning this into some Matlab code we represent it by:

<pre><code>clc
x = pi; % or x = pi/2
res = x;
factor = -1;
for ii=3:2:19
 res = res + factor*power(x,ii)/factorial(ii);
 factor = factor*-1; 
 fprintf ( 'iteration %2i sin(x)=%1.16f\n', (ii-1)/2, res );
end
res
</code></pre>

Running this code for both <code>x=pi</code> and <code>x=pi/2</code> you can see that the <code>x=pi/2</code> converges on the correct result (within eps error) quite quickly (9 iterations) - while the <code>x=pi</code> case doesn't converge in the same time frame. 

Its useful to note that at 9 iterations the last factorial that is being calculated in factorial(19). The next factorial that would be calculated in this sequence is 21. This is the last factorial that can be represented with 100% accuracy due to double precision (see <code>help factorial</code>).

So I think whats happening is that for pi/2 the mathematical solution converges on 1 to within double precision quicker that the pi case. In fact the pi case cant converge completely due to limitations in the maths and the accuracy that can be stored in a double precision result.

Having said all that the <code>sin(pi)</code> is within <code>eps</code> so you should use that fact for you purposes.

I've copied the results I get below (R2015b):

<pre><code>Results for PI/2
iteration 1 sin(x)=0.9248322292886504
iteration 2 sin(x)=1.0045248555348174
iteration 3 sin(x)=0.9998431013994987
iteration 4 sin(x)=1.0000035425842861
iteration 5 sin(x)=0.9999999437410510
iteration 6 sin(x)=1.0000000006627803
iteration 7 sin(x)=0.9999999999939768
iteration 8 sin(x)=1.0000000000000437
iteration 9 sin(x)=1.0000000000000000
Final Result: 1.0000000000000000


Results for PI
iteration 1 sin(x)=-2.0261201264601763
iteration 2 sin(x)=0.5240439134171688
iteration 3 sin(x)=-0.0752206159036231
iteration 4 sin(x)=0.0069252707075051
iteration 5 sin(x)=-0.0004451602382092
iteration 6 sin(x)=0.0000211425675584
iteration 7 sin(x)=-0.0000007727858894
iteration 8 sin(x)=0.0000000224195107
iteration 9 sin(x)=-0.0000000005289183
Final Result: -0.0000000005289183
</code></pre>

blocks|key|2578664|text|原因是，与0相比，sin(pi)=0.0，所以每个小错误，无论多么小，都是巨大的，因此是可见的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2578665|不同的是，对于sin(pi/2)=1：如果算法产生的误差小于eps+(在2.220446e-16附近)，您将不会看到这个错误，因为1%2Beps=1。|2578666|误差部分是输入不精确(pi值不精确)的结果，部分是计算过程中舍入的结果。人们必须深入研究代码才能得到正确的结果。|2578667|另一个重要因素是函数本身。考虑误差传播，通过查看pi和pi/2的泰勒级数，我们可以看到：|2578668|sin(pi%2Bdx)=sin(pi)%2Bcos(pi)dx%2Bo(dx%5E2)=-dx%2Bo(dx%5E2)
sin(pi/2%2Bdx)=sin(pi/2)%2Bcos(pi/2)dx%2Bo(dx%5E2)=1%2Bo(dx%5E2)|code-block|syntax|javascript|2578669|很明显:如果dx是关于eps的，那么由于输入不精确而导致的错误将是关于eps*eps的，因此与1相比是不可见的。|2578670|entityMap^0|5|1|9|B|0|7|B|U|3|10|C|1T|7|0|B|2|0|O|2|R|4|0|0|6|2|B|3|Z|7|1B|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|L|3|M|5|N|7|1G|8|@]|D|@]|E|$O|P]]|$1|Q|3|R|5|6|7|1H|8|@$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1Q|8|@]|D|@]|E|$]]]|T|$]]

The reason is, that <code>sin(pi)=0.0</code>, so every small error no matter how small is is huge compared to <code>0</code> and thus is visible.

Differently, for <code>sin(pi/2)=1</code>: if the algorithm produces an error smaller than <code>eps</code> (around <code>2.220446e-16</code>), you would not see this error because <code>1+eps=1</code>. 

The error is partly the result of the imprecise input (<code>pi</code> value is not exact) and partly the result of the round-off during the calculation. One has to look deep into the code to get it right.

Another important factor is the function itself. Considering the error propagation by looking at the Taylor's series for <code>pi</code> and <code>pi/2</code> we can see:

<pre><code>sin(pi+dx)=sin(pi)+cos(pi)dx+o(dx^2)=-dx+o(dx^2)
sin(pi/2+dx)=sin(pi/2)+cos(pi/2)dx+o(dx^2)=1+o(dx^2)
</code></pre>

It is clear: if <code>dx</code> is about <code>eps</code>, than the error due to the imprecise input will be about <code>eps*eps</code> and thus not visible as compared to <code>1</code>.

I have a problem in calculation with <code>matlab</code>.
I know that "<code>pi</code>" is a floating number and is not exact. So, in matlab <code>sin(pi)</code> is not exactly zero.
My question is if "<code>pi</code>" is not exact then why <code>sin(pi/2)</code> is exactly equal 1.

<code>sin(pi)</code> --> is not exact beacause of <code>pi</code>.
but
<code>sin(pi/2)</code> is exactly equal 1 

I am wonder and confused!

why in matlab sin(pi) is not exact but sin(pi/2) is exact?

我在计算matlab时遇到了一个问题。我知道"pi“是一个浮点数，并不精确。所以，在matlab中，sin(pi)不完全是零。我的问题是，如果"pi“不精确，那么为什么sin(pi/2)恰好等于1。由于pi的原因，sin(pi) -->不是完全正确的。但是sin(pi/2)恰好等于1我既好奇又困惑！

问为什么在matlab中sin(pi)不是精确的，而sin(pi/2)是精确的？
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问为什么在matlab中sin(pi)不是精确的，而sin(pi/2)是精确的？
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