Suppose You Wanted To Know If Log X N Was

Suppose you wanted to know if Log@xDn was concave up for some n.We know that it is concave down everywhere for n = 1, but hey; maybe it is concave up for some n > 1. Hn = ã is probably a good guess.L Then you might start by defining f Hx, nL = Log@xDn In[11]:=

f@x_, n_D := Log@xD ^ n

In[12]:=

D@f@x, nD, xD n Log@xD-1+n

Out[12]=

x In[13]:=

Factor@D@D@f@x, nD, xD, xDD n H- 1 + n - Log@xDL Log@xD-2+n

Out[13]=

In[14]:= Out[14]=

x2 Solve@% Š 0, xD 99x ® ã-1+n ==

And you might take a couple of derivatives and set the second derivative equal to zero and solve for x to determine what the points of inflection are.And so, the only point ofinflectionarex = ãn-1 .We can also see that as x ® ¥, f '' Hx, nL ® - ¥.So, for any n, fHx, nL is concave up onI0, ãn-1 M and concave

downon Iãn-1 , ¥M.So, even if the graph is concave up for a bit, it will always be concave down as x ® ¥.

2

Log[x]^n.nb

In[15]:=

PlotAf@x,2D,8x,0,14<,PlotRange®8-1,8<, PlotStyle®Orange,PlotLabel®"y=Log@xD2 "E

y=Log@xD2 8

6

Out[15]=

4

2

2

4

6

8

10

12

14

But why stop there ? In doing that analysis, you might notice that f H1, nL = 0 and f Hã, nL = n; this is illustrated with a few graphs below.

Log[x]^n.nb

In[31]:=

PlotA8f@x,0D,f@x,.5D,f@x,1D,f@x,2D,f@x,3D,f@x,4D,f@x,5D,f@x,100D<,8x,.5,E+.5<, PlotStyle®8Red,Blue,Green,Orange,Black,Brown,Cyan,Purple<,PlotRange®8-.1,1.1<, PlotLabel®"Log@xD in Blue, Log@xD100 in Purple"E

Log@xD in Blue, Log@xD100 in Purple 1.0

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0.6 Out[31]=

0.4

0.2

1.0

1.5

2.0

2.5

And now, you might be curious as to the area between Log@xD and each of those other graphs as n increases : for example, what is the area between Log@xD and Log@xD2 on this interval ? To evaluate the definate integral that will give us that area, first we ' ll need to derive an expression for à Log@xDn â x.

3.0

3

4

Log[x]^n.nb

In[32]:=

Out[32]=

n à Log@xD â x

H-logHxLL-n logn HxL GHn + 1, -logHxLL

Well, that' s not very helpful. It is a further generalized version of the expression I came up with, which works only for the integers; but to see what we want to see, all we need is the integers. To do this integral, I began with an integration by parts. Ù Log@xDn âx = xLog@xDn - Ù n Log@xDn-1 âx, where I used the substitution u = Log@xDn , âu =

n Log@xDn-1 âx , x

âv = âx and v = x.

If we then use integration by parts again, we obtain xLog@xDn - n Log@xDn-1 + Ù nHn - 1L Log@xDn-2 âx, and by now you can probably see where then

this

is

Ù Log@xDn âx = x Únk=0

heading.

If

H-1Lk n! Log@xDn-k Hn-kL!

we

nest

it

into

a

summation,

, and we have the explicit value of the indef-

inite integral for any n > 0. So, let us return to the problem at hand. We are trying to evaluate Ù1 HLog@xD - Log@xDn L âx ã

which we can now re-write as A = Jã Ú1k=0

H-1Lk Log@ãD1-k H1-kL!

A = JãH1 - 1L - ã Únk=0

-ã Únk=0

H-1Lk n! N Hn-kL!

A = H-1Ln n! + 1 - ã Únk=0

H-1Lk n! Log@ãDn-k Hn-kL!

N-JÚ1k=0

H-1Lk Log@1D1-k H1-kL!

- Únk=0

-H-1 - H-1Ln n!L

H-1Lk n! Hn-kL!

And we have an explicit function in terms of n for the area. n

In[18]:=

A@n_D:=-Eâ k=0

H-1L^k n! Hn-kL!

+H-1L^n n!+1

H-1Lk n! Log@1Dn-k Hn-kL!

N

Log[x]^n.nb

In[19]:=

Table@N@A@xDD,8x,0,5,1
Out[19]=

As you might predict from the graphs given earlier, Lim à Log@xDn â x = 0, ã

n®¥

1

which gives us the formula H- 1Ln n ! + 1 - ã â n

Lim n®¥

k=0

H- 1Lk n ! Hn - kL !

= 1, since à Log@xD â x = 1. ã

1

This sequence converges relatively slowly though, despite the fact that it is always increasing. HA fact that would be interesting to prove, even though a simple method of doing so isn ' t obvious to me.L Below is a graph of the sequence and a table of its values : In[35]:=

ListPlot@Table@8n,N@A@nDD<,8n,0,16
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0.6 Out[35]=

0.4

0.2

0

Out[36]=

5

10

15

880, - 0.718282<, 81, 0.<, 82, 0.281718<, 83, 0.436564<, 84, 0.535464<, 85, 0.6044<, 86, 0.655315<, 87, 0.69451<, 88, 0.725638<, 89, 0.750972<, 810, 0.771998<, 811, 0.789735<, 812, 0.8049<, 813, 0.818017<, 814, 0.829483<, 815, 0.8396<, 816, 0.851563<<

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