Suppose You Wanted to Know if Log x n Was

8/14/2019 Suppose You Wanted to Know if Log x n Was

1/5

Suppose you wanted to know if Logxn was concave up forsome n.We know that it is concave down everywhere for n 1, but hey;

maybe it is concave up for some n 1. n is probably a good guess.Then you might start by defining f x, n Logxn

In[11]:= fx_, n_ :

Logx^ nIn[12]:= Dfx, n, x

Out[12]=

n Logx1nx

In[13]:= FactorDDfx, n, x, x

Out[13]=

n 1 n Logx Logx2nx2

In[14]:= Solve 0, xOut[14]= x 1nAnd 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 n1

.We can also see that as x ,

f'' x, n .So, for any n, fx, n is concave up on0, n1 and concavedownon n1, .So, even if the graph is concave up for a bit, it will always beconcave down as x .


2/5

In[15]:= Plotfx,2,x,0,14,PlotRange1,8,PlotStyleOrange,PlotLabel"yLogx2"

Out[15]=

2 4 6 8 10 12 14

2

4

6

8

yLogx2

But why stop there? In doing that analysis,

you might notice that f 1, n 0 and f , n n; this is illustrated with a few graphs below.

2 Log[x]^n.nb


3/5

In[31]:= Plotfx,0,fx,.5,fx,1,fx,2,fx,3,fx,4,fx,5,fx,100,x,.5,E.5,PlotStyleRed,Blue,Green,Orange,Black,Brown,Cyan,Purple,PlotRange.1,1.1,PlotLabel"Logx in Blue, Logx100 in Purple"

Out[31]=

1.0 1.5 2.0 2.5 3.0

0.2

0.4

0.6

0.8

1.0

Logx in Blue, Logx100 in Purple

And now, you might be curious as to the area between

Logx and each of those other graphs as n increases : for example,what is the area between Logx and Logx2 on this interval?To

evaluate the definate integral that will give us that area,

first we' ll need to derive an expression for Logxn

x.

Log[x]^n.nb 3


4/5

In[32]:= Logxn xOut[32]= logx

n lognx n 1, logx

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 inte-

gral, I began with an integration by parts.

Logxn x = xLogxn - n Logxn1 x, where I used the substitutionu Logxn, u n Logxn1 x

x, v x and v x.

If we then use integration by parts again, we obtain

xLogxn n Logxn1 nn 1 Logxn2 x, and by now you can probably seewhere this is heading. If we nest it into a summation,

then

Logxn x x k0n 1k n Logxnknk , and we have the explicit value of the indefinite integral for any n > 0.

So, let us return to the problem at hand. We are trying to evaluate

1

Logx Logxn x

which we can now re-write as

A = k01 1k Log1k1k - k0n 1

kn Lognknk -k01 1

k Log11k1k k0n 1kn Log1nknk

A = 1 1 k0n 1k n

nk 1 1n n

A = 1n n 1 k0n 1k n

nk

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

In[18]:= A

n_

:E

k0n 1^k n

nk

1

^n n1

4 Log[x]^n.nb


5/5

In[19]:= TableNAx,x,0,5,1Out[19]= 0.718282, 0., 0.281718, 0.436564, 0.535464, 0.6044As you might predict from the graphs given earlier,

Limn 1

Logxn x 0,

which gives us the formula

Limn

1n n 1 k0

n 1k n n k 1, since 1

Logx x 1.

This sequence converges relatively slowly though,

despite the fact that it is always increasing.

A fact that would be interesting to prove, even though a simple method ofdoing so isn' t obvious to me.

Below is a graph of the sequence and a table of its values :

In[35]:= ListPlotTablen,NAn,n,0,16,JoinedTrue,PlotStyleOrange,PlotRange0,1Tablen,NAn,n,0,16

Out[35]=

0 5 10 15

0.2

0.4

0.6

0.8

1.0

Out[36]= 0, 0.718282, 1, 0., 2, 0.281718, 3, 0.436564, 4, 0.535464, 5, 0.6044,6, 0.655315, 7, 0.69451, 8, 0.725638, 9, 0.750972, 10, 0.771998, 11, 0.789735,12, 0.8049, 13, 0.818017, 14, 0.829483, 15, 0.8396, 16, 0.851563

Log[x]^n.nb 5

Documents

Suppose You Wanted to Know if Log x n Was