计算对数

发布于 2024-11-01 09:22:23 字数 485 浏览 7 评论 0原文

我正在尝试编写一个方法，该方法接受基数k和值n到小数点后两位，然后计算n的对数基数k，而不使用任何Java的Math.log 方法。这是我到目前为止所得到的：

public static double log(double k, double n) {
    double value = 0.0;

    for(double i = 1; i > .001; i /= 10) {
        while(!(Math.pow(k, value) >= n )) {
            value += i;
        }
    }

    return value;
}

当我尝试计算 5.0625 的以 4 为底的对数时，问题出现了，它返回 2.0，但应该返回 1.5。

我不知道为什么这不起作用。任何帮助表示赞赏。

不，这不是作业，这是我为了好玩而试图解决的问题集的一部分。

原文

I'm trying to write a method that takes in a base k and a value n to 2 decimal places, then computes the log base k of n without using any of Java's Math.log methods. Here's what I have so far:

public static double log(double k, double n) {
    double value = 0.0;

    for(double i = 1; i > .001; i /= 10) {
        while(!(Math.pow(k, value) >= n )) {
            value += i;
        }
    }

    return value;
}

The problem comes up when I try computing log base 4 of 5.0625, which returns 2.0, but should return 1.5.

I have no idea why this isn't working. Any help is appreciated.

No this is not homework, it's part of a problem set that I'm trying to solve for fun.

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吃兔兔 2024-11-08 09:22:23

您添加金额 i 的次数太频繁了。因此，您很快就会达到比实际值大的值，并且 while 循环将永远不会再次进入。

从该值中减去一次 i 就可以了：

for(double i = 1; i > .001; i /= 10) {
    while(!(Math.pow(k, value) > n )) {
        value += i;
    }
    value -= i;
}

You're adding the amount i once too ofter. Thus you'll quite soon reach a value larger than the actual value and the while loop will never be entered again.

Subtract i once from the value and you'll be fine:

for(double i = 1; i > .001; i /= 10) {
    while(!(Math.pow(k, value) > n )) {
        value += i;
    }
    value -= i;
}

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浮生未歇 2024-11-08 09:22:23

单步执行纸上的代码：

Iteration: i=1 value = 0.0, calculated power = 1
Iteration: i=1 value = 1.0, calculated power = 4
Iteration: i=1 value = 2.0, calculated power = 16

现在，您的值是 2.0。但代码中的任何一点都无法向另一个方向纠正。您需要检查过冲和下冲情况。

Step through the code on paper:

Iteration: i=1 value = 0.0, calculated power = 1
Iteration: i=1 value = 1.0, calculated power = 4
Iteration: i=1 value = 2.0, calculated power = 16

Now at this point, your value is 2.0. But at no point in the code to you have a way to correct back in the other direction. You need to check for both overshoot and undershoot cases.

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晨敛清荷 2024-11-08 09:22:23

这个循环

    while(!(Math.pow(k, value) >= n )) {
        value += i;
    }

走得太远了。它仅在超过正确值后停止。所以在计算个位的时候，1不够，所以就到2.0，而后面的所有测试都表明至少是够了，所以就到此结束。

This loop

    while(!(Math.pow(k, value) >= n )) {
        value += i;
    }

goes too far. It only stops after the correct value has been surpassed. So when calculating the ones place, 1 isn't enough, so it goes to 2.0, and all subsequent tests show that it is at least enough, so that's where it ends.

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