y=x^x

Sityl · **#21** by **Sityl** » Sep 22, 2010 10:17 pm

logical bob wrote:
Sityl wrote:Could anyone explain this in english?

Only to a limited extent. Maths has notation because it needs it. Could you ask a more specific question?

Well, like, it would take MUCH longer to put it into words, but I'd like to know what cali said.

Example 2 + 2 = 4

If you don't know what those symbols mean, you can explain it like, "If I have two apples and then I get two more apples, then I'll have four apples."

logical bob · **#22** by **logical bob** » Sep 22, 2010 11:39 pm

OK. I don't know how much maths you know, so apologies in advance if I talk down to you by accident.

We're looking for the value of x at the turning point in the curve y = xx. A useful fact is that, at a turning point,

a curve has zero gradient. Cali differentiated the function y = xx i.e. he calculated the gradient of the curve for

any value of x (using rules called the chain and product rules to break the problem down into smaller chunks. It turns out

that the gradient of the curve is (ln x +1)xx.

This uses the exponential function y = ex. The interesting thing about this function is that it is it's own

derivative. At any point on the curve y = ex the gradient of the curve is also ex. e itself is an

irrational number (like pi) - it's about 2.718. y = ln x is the inverse function of y = ex. y = ln x is the power to

which you have to raise e to obtain x. They're inverse because for any value of x, ln ex = eln x = x.

The turning point of the curve occurs when the gradient, (ln x +1)xx = 0. Another handy fact is that two

numbers can only have a product of zero if one or both of them is itself zero. So if (ln x + 1)xx = 0 then either

xx = 0 (but this isn't true for any postitive value of x) or ln x +1 = 0. As the former isn't an option it must be the

latter, meaning the turning point occurs when ln x = -1. We apply the exponential function to both sides and get

eln x = e-1. But y = ex and y = ln x are inverse functions, so eln x = x.

e-1 = 1/e (that's what raising something to the power -1 means) so we have x = 1/e as the value of x at the

turning point.

I hope that makes some sense!

ughaibu · **#23** by **ughaibu** » Sep 23, 2010 12:51 am

logical bob wrote:
Sityl wrote:Could anyone explain this in english?
Only to a limited extent. Maths has notation because it needs it. Could you ask a more specific question?

What maths cant be translated to a natural language?

logical bob · **#24** by **logical bob** » Sep 23, 2010 6:03 am

ughaibu wrote:
logical bob wrote:
Sityl wrote:Could anyone explain this in english?
Only to a limited extent. Maths has notation because it needs it. Could you ask a more specific question?
What maths cant be translated to a natural language?

I'm not saying it can't be translated, but the translation will be far harder to work with because natural language isn't so well suited to the precision and fine distinctions of mathematics. Try doing some maths without using any notation.

chriscase · **#25** by **chriscase** » Sep 23, 2010 6:39 am

The math is not particularly hard if you've taken undergraduate calculus. For someone who has not studied calculus it's bound to be a bit foreign, because the concepts associated with the derivative won't be familiar. Basically, when the OP asked "why does this function have a turning point at this place on its graph," he was asking the kind of question that is generally very hard to answer without calculus, and almost trivial to answer with it.

When we do so-called "analytic geometry" as preparation for calculus, we study specifc conic sections and other functions that can have "turning points", or places where there is a local minimum or maximum on the curve. The canonical example of this is the parabola y=x2, which has an obvious minimum at x=0. When the parabola is inverted, as in the trajectory of a projectile, the parabola can be manipulated to tell us the maximum altitude of the projectile, a result of some practical interest.

While the parabola specifically can be manipulated in such a way as to tell us where the min/max vertex is, other arbitrary curves are not so easy to deal with. Calculus gives us the derivative, which tells us the slope of the tangent to the curve at a given point. Provided that the curve is smooth (differentiable), it becomes a matter of applying some rules and setting the derivative to zero to find the places where the curve levels off and the slope is zero, i.e., "flat".

Calilasseia · **#26** by **Calilasseia** » Sep 23, 2010 10:15 pm

Sityl wrote:Could anyone explain this in english?

The specific properties of xx as described in my exposition are a natural consequence of the laws of logarithms.

If you have a function y = ax, then the inverse function is x = logay. From this elementary fact, a number of laws of logarithms can be deduced from first principles. Once these laws are in place, they govern the behaviour of all functions involving logarithms.

For example, loga(xy) = logax + logay. Which allows you to convert from a multiplication to an addition if you have access to a table of logarithms, or a logarithm function on your calculator. This law is derived as follows:

Let z = logax + logay

Taking antilogarithms to the base a, we have:

az = a(logax + logay)

= alogax.alogay

Since alogax = x, and alogay = y, we have that:

az = xy

Therefore z = loga(xy)

Therefore loga(xy) = logax + logay

Likewise, we have that loga(xy) = y logax. We derive this as follows.

Let z = y logax

Taking antilogarithms, we have that:

az = ay logax

From an elementary rule governing exponents, (uv)w = uvw

Therefore az = (alogax)y

Now alogax is simply x, therefore az = xy

Therefore loga(az) = z = loga(xy)

Therefore y logax = loga(xy)

Once you have these rules in place (along with a couple of others), properties of functions involving logarithms can be deduced using these rules.

Sityl · **#27** by **Sityl** » Sep 23, 2010 10:19 pm

logical bob wrote:OK. I don't know how much maths you know, so apologies in advance if I talk down to you by accident.

We're looking for the value of x at the turning point in the curve y = xx. A useful fact is that, at a turning point,

a curve has zero gradient. Cali differentiated the function y = xx i.e. he calculated the gradient of the curve for

any value of x (using rules called the chain and product rules to break the problem down into smaller chunks. It turns out

that the gradient of the curve is (ln x +1)xx.

This uses the exponential function y = ex. The interesting thing about this function is that it is it's own

derivative. At any point on the curve y = ex the gradient of the curve is also ex. e itself is an

irrational number (like pi) - it's about 2.718. y = ln x is the inverse function of y = ex. y = ln x is the power to

which you have to raise e to obtain x. They're inverse because for any value of x, ln ex = eln x = x.

The turning point of the curve occurs when the gradient, (ln x +1)xx = 0. Another handy fact is that two

numbers can only have a product of zero if one or both of them is itself zero. So if (ln x + 1)xx = 0 then either

xx = 0 (but this isn't true for any postitive value of x) or ln x +1 = 0. As the former isn't an option it must be the

latter, meaning the turning point occurs when ln x = -1. We apply the exponential function to both sides and get

eln x = e-1. But y = ex and y = ln x are inverse functions, so eln x = x.

e-1 = 1/e (that's what raising something to the power -1 means) so we have x = 1/e as the value of x at the

turning point.

I hope that makes some sense!

Brilliant! No, you didn't talk down, there was maybe one or two parts you talked too up, but on the whole it was understandable and I have an idea of what cali was talking about!

Thanks! :cheers:

Sityl · **#28** by **Sityl** » Sep 23, 2010 10:22 pm

chriscase wrote:The math is not particularly hard if you've taken undergraduate calculus. For someone who has not studied calculus it's bound to be a bit foreign, because the concepts associated with the derivative won't be familiar. Basically, when the OP asked "why does this function have a turning point at this place on its graph," he was asking the kind of question that is generally very hard to answer without calculus, and almost trivial to answer with it.

When we do so-called "analytic geometry" as preparation for calculus, we study specifc conic sections and other functions that can have "turning points", or places where there is a local minimum or maximum on the curve. The canonical example of this is the parabola y=x2, which has an obvious minimum at x=0. When the parabola is inverted, as in the trajectory of a projectile, the parabola can be manipulated to tell us the maximum altitude of the projectile, a result of some practical interest.

While the parabola specifically can be manipulated in such a way as to tell us where the min/max vertex is, other arbitrary curves are not so easy to deal with. Calculus gives us the derivative, which tells us the slope of the tangent to the curve at a given point. Provided that the curve is smooth (differentiable), it becomes a matter of applying some rules and setting the derivative to zero to find the places where the curve levels off and the slope is zero, i.e., "flat".

Can't graphs exist that have multiple points at wich there is no gradient? Or is slope =/= gradient?

Calilasseia · **#29** by **Calilasseia** » Sep 23, 2010 10:48 pm

Sityl wrote:
chriscase wrote:The math is not particularly hard if you've taken undergraduate calculus. For someone who has not studied calculus it's bound to be a bit foreign, because the concepts associated with the derivative won't be familiar. Basically, when the OP asked "why does this function have a turning point at this place on its graph," he was asking the kind of question that is generally very hard to answer without calculus, and almost trivial to answer with it.

When we do so-called "analytic geometry" as preparation for calculus, we study specifc conic sections and other functions that can have "turning points", or places where there is a local minimum or maximum on the curve. The canonical example of this is the parabola y=x2, which has an obvious minimum at x=0. When the parabola is inverted, as in the trajectory of a projectile, the parabola can be manipulated to tell us the maximum altitude of the projectile, a result of some practical interest.

While the parabola specifically can be manipulated in such a way as to tell us where the min/max vertex is, other arbitrary curves are not so easy to deal with. Calculus gives us the derivative, which tells us the slope of the tangent to the curve at a given point. Provided that the curve is smooth (differentiable), it becomes a matter of applying some rules and setting the derivative to zero to find the places where the curve levels off and the slope is zero, i.e., "flat".

Can't graphs exist that have multiple points at wich there is no gradient? Or is slope =/= gradient?

If you mean by the above, functions whose graphs have multiple points at which the gradient is zero, yes, there are an infinite number of these. y=sin(nx) and y=cos(nx) are two families of curves which have an infinite number of such points. Many polynomials of degree 3 have two such points, many polynomials of degree 4 have 3 such points, and so on. Indeed, in the general case, a polynomial of degree n will have n-1 such turning points, provided that all of the roots of the polynomial in question are distinct. For example, y=x3-6x2+11x-6 should have 3 turning points, because it has three distinct roots (it can be factorised as (x1-)(x-2)(x-3), and the roots of y=0 are x=1, x=2 and x=3). On the other hand, y=x3-5x2+8x-4 will probably NOT have two turning points, because it factorises as (x-1)(x-2)2, and therefore it has two identical roots at x=2.

Note that in general, a polynomial of degree n (i.e., one whose highest power is xn) will have n roots. This property of polynomials is known formally as the Remainder Theorem, and is one of the fundamental theorems of both real and complex analysis. In the most general case, one has to include complex number roots in order for the theorem to hold, but there are a large number of cases where polynomials of degree n have n real roots. In order to have n real roots, a polynomial has to have n-1 turning points, as a graph of a selection of such polynomials will verify quickly upon inspection.

However, for functions other than polynomials, determining the existence and nature of turning points in the general case requires the use of calculus, if one wishes to do this analytically (i.e., without plotting a graph, and using algebraic tools alone).

Calilasseia · **#30** by **Calilasseia** » Sep 24, 2010 11:49 am

Correction to above: a polynomial of degree n has to have n-1 turning points if it has n distinct real roots.

I've just compiled a nice Excel spreadsheet that illustrates some salient points about polynomials and turning points. However, in my above examples with cubic polynomials, I made an error. It transpires that (x-1)(x-2)2 does have two turning points, it's just that the section of the curve between the two turning points has only a slight gradient. However, if one moves to quintic polynomials (i.e., polynomials whose highest power of x is x5), then the relationship between distinct roots and turning points is much easier to visualise. For example, if you plot the following polynomial:

y=(x-1)(x-2)(x-3)(x-4)(x-5)

between the values x=0.5 and x=5.5, then you obtain a very nice curve that intersects the x-axis five times (at x=1, x=2, x=3, x=4 and x=5 respectively) and has four turning points. However, if you plot the following polynomial:

y=(x-1)2(x-5)3

between the same x-values, then you end up with a curve that only has three turning points (and one of those is a point of inflexion, not a transition from positive to negative gradient or vice versa).

chriscase · **#31** by **chriscase** » Oct 05, 2010 12:49 am

Isn't the relevant question how many distinct zero soutions there are for the derivative. I'm not sure if there are any examples of polynomials that have zeros of degree greater than one, but whose derivative does not. Or maybe an example of a polynomial that has distinct zeros but whose derivative does not...

Preno · **#32** by **Preno** » Oct 05, 2010 4:38 pm

chriscase wrote:Isn't the relevant question how many distinct zero soutions there are for the derivative. I'm not sure if there are any examples of polynomials that have zeros of degree greater than one, but whose derivative does not.

Trivially, x2 has a zero of degree greater than one and its derivative has a zero of degree one.

Or maybe an example of a polynomial that has distinct zeros but whose derivative does not...

x2-1.

chriscase · **#33** by **chriscase** » Oct 05, 2010 4:46 pm

Yes, I think the examples I am thinking of would need to be of higher degree than two. The stement I am looking at is:

"a polynomial of degree n has to have n-1 turning points if it has n distinct real roots."

This seems somewhat imprecise, since what we are looking for are distinct zeros for the derivative, not the original polynomial. Unless we can prove that, if a polynomial has n distinct real roots, its derivative must have n-1 distinct real roots. I'm not sure if this is the case.

twistor59 · **#34** by **twistor59** » Oct 05, 2010 5:02 pm

if I take a couple of distinct roots, doesn't Rolle's theorem imply that there's a point between them with zero derivative ? So I can take successive pairs of roots and know that there's a point between each with zero derivative.

Or something like that.

chriscase · **#35** by **chriscase** » Oct 05, 2010 5:41 pm

twistor59 wrote:if I take a couple of distinct roots, doesn't Rolle's theorem imply that there's a point between them with zero derivative ? So I can take successive pairs of roots and know that there's a point between each with zero derivative.

Or something like that.

Sounds right to me. Thanks :-)

Broiled Jogger · **#36** by **Broiled Jogger** » Oct 11, 2010 11:38 pm

0.4 ^ 0.4 is equal to the natural logarithm of 2 to five decimal places, if anyone cares. I certainly don't care, but someone might.

Someone · **#37** by **Someone** » Jan 05, 2011 4:11 am

If anyone wants to go deeper on a somewhat related subject, http://en.wikipedia.org/wiki/Lambert_W_Function partially deals with the infinite tower of exponentiation x^(x^(x^....., which I don't know how to show prettily here.

Someone · **#38** by **Someone** » Jan 05, 2011 4:22 am

Also, while giving wikipedia references, the following two will correct an error in naming by an earlier poster.
http://en.wikipedia.org/wiki/Polynomial ... er_theorem
http://en.wikipedia.org/wiki/Fundamenta ... of_algebra

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