I’ve recently seen a video where someone did an explanation on why 0 to the power of 0 is defined as 1. He went with X to the power of X and made X increasingly smaller decimals. X = 0.9, X = 0.8, and so on. The results for 0.9 to the power of 0.9 and then 0.8 to the power of 0.8 kept getting smaller until about X = 0.4, where it started to increase again, so 0.3 to the power of 0.3 was a bigger number than 0.4 to 0.4. At what number is the exact “turning point” in this kind of series and why is it in such a weird place (as opposed to at 0.5, i.e halfway to 0)?
In: 5
That turning point is at x = 1 / e
This e is Euler’s number, about 2.71828
*Why* is it 1/e? Because e is everywhere.
To find a local minimum one way is to find where the derivative of the function is zero.
the derivate of x^x is simple to to find by using a^b = e^(a ln (b)) so x^x = e^(x ln (x)) let chang that to e^f(x) and f(x)= x ln (x)
The derivate of f ln(x) = 1/x so the derivate f(x) = x/x + 1 ln(x) =1+ ln(x)
The derivate of e^f(x) = f(x)’ e^f(x)
This mean the derivate of x^x = (1+ln(x)) x^x = x^x + ln(x) x^x
The mean point we look for is where ln(x) =-1
e^ln(x) =x it we use that we get e^ln(x)= e^-1 => x= e^-1 =1/e
So the exact point the change happened at 1/e is approximately 0.367879441
Because of the relationship between the power function and the constant e it is not surprising that the answer includes it.
That turning point is at x = 1 / e
This e is Euler’s number, about 2.71828
*Why* is it 1/e? Because e is everywhere.
That turning point is at x = 1 / e
This e is Euler’s number, about 2.71828
*Why* is it 1/e? Because e is everywhere.
To find a local minimum one way is to find where the derivative of the function is zero.
the derivate of x^x is simple to to find by using a^b = e^(a ln (b)) so x^x = e^(x ln (x)) let chang that to e^f(x) and f(x)= x ln (x)
The derivate of f ln(x) = 1/x so the derivate f(x) = x/x + 1 ln(x) =1+ ln(x)
The derivate of e^f(x) = f(x)’ e^f(x)
This mean the derivate of x^x = (1+ln(x)) x^x = x^x + ln(x) x^x
The mean point we look for is where ln(x) =-1
e^ln(x) =x it we use that we get e^ln(x)= e^-1 => x= e^-1 =1/e
So the exact point the change happened at 1/e is approximately 0.367879441
Because of the relationship between the power function and the constant e it is not surprising that the answer includes it.
To find a local minimum one way is to find where the derivative of the function is zero.
the derivate of x^x is simple to to find by using a^b = e^(a ln (b)) so x^x = e^(x ln (x)) let chang that to e^f(x) and f(x)= x ln (x)
The derivate of f ln(x) = 1/x so the derivate f(x) = x/x + 1 ln(x) =1+ ln(x)
The derivate of e^f(x) = f(x)’ e^f(x)
This mean the derivate of x^x = (1+ln(x)) x^x = x^x + ln(x) x^x
The mean point we look for is where ln(x) =-1
e^ln(x) =x it we use that we get e^ln(x)= e^-1 => x= e^-1 =1/e
So the exact point the change happened at 1/e is approximately 0.367879441
Because of the relationship between the power function and the constant e it is not surprising that the answer includes it.
Writing `x^x` as `exp(x ln(x))` may give you a hint as to what the mystery number is. >!It’s 1/e, where e is Napier’s constant, the 2.718… thing you see *everywhere*.!<
# Why `0^0` is one
– Since `exp(x)` is a well-behaved function, the limit it approaches when stuff happens to its argument is the same as taking the limit of its argument and exponentiating it (i.e. applying the `exp` function).
– This is to say, for the moment we can just look at its argument, `x ln(x)`, see what it approaches, and finally put the exponentiation back to get our answer.
– For anywhere `0 < x < 1`, you’ll find `x` small but positive and `ln(x)` very negative, the latter even tending to negative infinity as x approaches zero. The result of multiplying the two (or rather, what happens to that result as you go near zero) is thus not very intuitive to guess – it can be anything from negative infinity to zero, depending on how quickly the two approaches their limiting values.
– For this we have to use **l’Hôpital’s rule**, which basically says that if you have two things both approaching zero (or both diverging to infinity) at some point, and if the ratio between their respective rates (i.e. derivatives) turns out to be something nice, the ratio between said things as you go to said point approaches whatever the ratio between those rates approaches.
– Here we’re looking at an infinity times zero, but we can rearrange it to get `x ln(x) = ln(x) / (1/x)` – hey presto, we get our infinity divided by infinity. And by taking the derivatives of the numerator and the denominator, we see that the ratio tends to zero.
– Raising any positive number to the zeroth power results in one; and since `exp(x) = e^x`, our limit is also one at `x = 0`
# Why do we have a minimum
– `exp(x)`, `x` and `ln(x)` are all **monotonically increasing** functions in x, meaning that their value always increases as you increase their argument. They are also continuous over the entire positive real line, i.e. for all `x > 0`. Plus they are also **differentiable** over said range, implying that their n-th derivatives are also all continuous, no matter how big n is.
– Hence, asking whether our function attains a minimum is the same as asking whether `f(x) := x ln(x)`, the argument to the outermost exponential function, does so.
– For `x > 1`, it is obvious that f(x) increases since both `x` and `ln(x)` are positive and increasing.
– Previously, we got that the limit of f(x) as you approaches zero from above (i.e. the positive side) is also zero. We also note that f(x) is zero at one since `ln(1) = 0`.
– We’ve also established that for `0 < x < 1` f(x) is negative. This means that for very small values of x, f(x) must be decreasing, going from (the limit) zero to a negative value.
– So at the two ends of the interval `0 < x < 1`, f(x) is respectively decreasing and increasing. This means that its derivatives are respectively negative and positive.
– Since the (first) derivative of f(x) is continuous, there must have been at least one point in the interval that it attains the value zero, flipping from the negative into the positive. This is the result of the **intermediate value theorem**, and such is a stationary point of f(x).
– Noting that the point is smaller than its surroundings, we see that it is indeed a minimum.
# Why the minimum is what it is
It is trivial to get the minimum at `1/e` with calculus.* While this does sound unfulfilling, you may feel better looking at our form of f(x), which involves a natural logarithm (i.e. logarithm to base `e`); it is then not surprising to see that `e` is involved in the answer to “where the minimum of f(x) is”.
* EDIT: explicitly: the derivative `d/dx(x ln(x)) = d(x)/dx ln(x) + x d(ln(x))/dx = ln(x) + 1`. At the minimum it must be zero, so we need `ln(x) = -1` and thus `x = 1/e`.
Writing `x^x` as `exp(x ln(x))` may give you a hint as to what the mystery number is. >!It’s 1/e, where e is Napier’s constant, the 2.718… thing you see *everywhere*.!<
# Why `0^0` is one
– Since `exp(x)` is a well-behaved function, the limit it approaches when stuff happens to its argument is the same as taking the limit of its argument and exponentiating it (i.e. applying the `exp` function).
– This is to say, for the moment we can just look at its argument, `x ln(x)`, see what it approaches, and finally put the exponentiation back to get our answer.
– For anywhere `0 < x < 1`, you’ll find `x` small but positive and `ln(x)` very negative, the latter even tending to negative infinity as x approaches zero. The result of multiplying the two (or rather, what happens to that result as you go near zero) is thus not very intuitive to guess – it can be anything from negative infinity to zero, depending on how quickly the two approaches their limiting values.
– For this we have to use **l’Hôpital’s rule**, which basically says that if you have two things both approaching zero (or both diverging to infinity) at some point, and if the ratio between their respective rates (i.e. derivatives) turns out to be something nice, the ratio between said things as you go to said point approaches whatever the ratio between those rates approaches.
– Here we’re looking at an infinity times zero, but we can rearrange it to get `x ln(x) = ln(x) / (1/x)` – hey presto, we get our infinity divided by infinity. And by taking the derivatives of the numerator and the denominator, we see that the ratio tends to zero.
– Raising any positive number to the zeroth power results in one; and since `exp(x) = e^x`, our limit is also one at `x = 0`
# Why do we have a minimum
– `exp(x)`, `x` and `ln(x)` are all **monotonically increasing** functions in x, meaning that their value always increases as you increase their argument. They are also continuous over the entire positive real line, i.e. for all `x > 0`. Plus they are also **differentiable** over said range, implying that their n-th derivatives are also all continuous, no matter how big n is.
– Hence, asking whether our function attains a minimum is the same as asking whether `f(x) := x ln(x)`, the argument to the outermost exponential function, does so.
– For `x > 1`, it is obvious that f(x) increases since both `x` and `ln(x)` are positive and increasing.
– Previously, we got that the limit of f(x) as you approaches zero from above (i.e. the positive side) is also zero. We also note that f(x) is zero at one since `ln(1) = 0`.
– We’ve also established that for `0 < x < 1` f(x) is negative. This means that for very small values of x, f(x) must be decreasing, going from (the limit) zero to a negative value.
– So at the two ends of the interval `0 < x < 1`, f(x) is respectively decreasing and increasing. This means that its derivatives are respectively negative and positive.
– Since the (first) derivative of f(x) is continuous, there must have been at least one point in the interval that it attains the value zero, flipping from the negative into the positive. This is the result of the **intermediate value theorem**, and such is a stationary point of f(x).
– Noting that the point is smaller than its surroundings, we see that it is indeed a minimum.
# Why the minimum is what it is
It is trivial to get the minimum at `1/e` with calculus.* While this does sound unfulfilling, you may feel better looking at our form of f(x), which involves a natural logarithm (i.e. logarithm to base `e`); it is then not surprising to see that `e` is involved in the answer to “where the minimum of f(x) is”.
* EDIT: explicitly: the derivative `d/dx(x ln(x)) = d(x)/dx ln(x) + x d(ln(x))/dx = ln(x) + 1`. At the minimum it must be zero, so we need `ln(x) = -1` and thus `x = 1/e`.
Writing `x^x` as `exp(x ln(x))` may give you a hint as to what the mystery number is. >!It’s 1/e, where e is Napier’s constant, the 2.718… thing you see *everywhere*.!<
# Why `0^0` is one
– Since `exp(x)` is a well-behaved function, the limit it approaches when stuff happens to its argument is the same as taking the limit of its argument and exponentiating it (i.e. applying the `exp` function).
– This is to say, for the moment we can just look at its argument, `x ln(x)`, see what it approaches, and finally put the exponentiation back to get our answer.
– For anywhere `0 < x < 1`, you’ll find `x` small but positive and `ln(x)` very negative, the latter even tending to negative infinity as x approaches zero. The result of multiplying the two (or rather, what happens to that result as you go near zero) is thus not very intuitive to guess – it can be anything from negative infinity to zero, depending on how quickly the two approaches their limiting values.
– For this we have to use **l’Hôpital’s rule**, which basically says that if you have two things both approaching zero (or both diverging to infinity) at some point, and if the ratio between their respective rates (i.e. derivatives) turns out to be something nice, the ratio between said things as you go to said point approaches whatever the ratio between those rates approaches.
– Here we’re looking at an infinity times zero, but we can rearrange it to get `x ln(x) = ln(x) / (1/x)` – hey presto, we get our infinity divided by infinity. And by taking the derivatives of the numerator and the denominator, we see that the ratio tends to zero.
– Raising any positive number to the zeroth power results in one; and since `exp(x) = e^x`, our limit is also one at `x = 0`
# Why do we have a minimum
– `exp(x)`, `x` and `ln(x)` are all **monotonically increasing** functions in x, meaning that their value always increases as you increase their argument. They are also continuous over the entire positive real line, i.e. for all `x > 0`. Plus they are also **differentiable** over said range, implying that their n-th derivatives are also all continuous, no matter how big n is.
– Hence, asking whether our function attains a minimum is the same as asking whether `f(x) := x ln(x)`, the argument to the outermost exponential function, does so.
– For `x > 1`, it is obvious that f(x) increases since both `x` and `ln(x)` are positive and increasing.
– Previously, we got that the limit of f(x) as you approaches zero from above (i.e. the positive side) is also zero. We also note that f(x) is zero at one since `ln(1) = 0`.
– We’ve also established that for `0 < x < 1` f(x) is negative. This means that for very small values of x, f(x) must be decreasing, going from (the limit) zero to a negative value.
– So at the two ends of the interval `0 < x < 1`, f(x) is respectively decreasing and increasing. This means that its derivatives are respectively negative and positive.
– Since the (first) derivative of f(x) is continuous, there must have been at least one point in the interval that it attains the value zero, flipping from the negative into the positive. This is the result of the **intermediate value theorem**, and such is a stationary point of f(x).
– Noting that the point is smaller than its surroundings, we see that it is indeed a minimum.
# Why the minimum is what it is
It is trivial to get the minimum at `1/e` with calculus.* While this does sound unfulfilling, you may feel better looking at our form of f(x), which involves a natural logarithm (i.e. logarithm to base `e`); it is then not surprising to see that `e` is involved in the answer to “where the minimum of f(x) is”.
* EDIT: explicitly: the derivative `d/dx(x ln(x)) = d(x)/dx ln(x) + x d(ln(x))/dx = ln(x) + 1`. At the minimum it must be zero, so we need `ln(x) = -1` and thus `x = 1/e`.
> I’ve recently seen a video where someone did an explanation on why 0 to the power of 0 is defined as 1
That video is bad. Why? Because defining(!) 0^0 = 1 has nothing to do with limits. At all. You could just as well take the limit of 0^x for x>0 as x approaches 0, but then you would get 0 instead of 1. There is hardly any limit reason where 0^0 = 1 simplifies anything.
The true reason is way simpler: it is a convention that works extremely well, while any other choice makes things uglier (but not “wrong”). We write down many expressions in ways where it would be very tedious to mention 0 explicitly, such as polynomials or sums.
Also, there is a combinatorial reason that is actually related: m^n is the number of ways to color n items, with one of m colors each, no restrictions. So how many ways are there to color 0 things with 0 colors? Well, 1: do exactly nothing, it is already okay as it is. Nothing to color, after all. Having more colors as options changes nothing, hence m^0 is still 1. But if you have one or more objects but no colors, there is no way to accomplish the task, hence 0^n = 0 for n>0.
Latest Answers