User:Egm6321.f09.team7.benedict

Problem Statement
Pg. 37-1 []

Show that the following relation is true:

$$x\frac12ln(x\frac{1+x}{1-x})-1 = xtanh^{-1}(x)-1$$

Problem Solution
We can recall from HW 6 Problem 13 the proof of $$\frac12ln(\frac{1+x}{1-x}) = tanh^{-1}(x)$$

From here a few easy steps give us the proof we are looking for.

First recognize that the relation from the last homework above is part of our original relation if we isolate it we have:

$$x(\frac12ln(\frac{1+x}{1-x}))-1 = x(tanh^{-1}(x))-1$$

we can cross out the relation from the previous homework and are left with:

$$x\cancel{(\frac12ln(\frac{1+x}{1-x}))}-1 = x\cancel{(tanh^{-1}(x))}-1$$ = $$x(1)-1=x(1)-1$$

or

x-1 = x-1

proving the relation is true

Problem Statement
Pg. 37-1 []

From the transparencies equation (2) is defined as:

$$Q_n(x) = \underline{P}_n(x)tanh^{-1}(x)-2 \sum\nolimits_{j=1,3,5}^{J} \frac{2n-2j+1}{(2n-j+1)j}\underline{P}_{n-j}(x)$$

Using (2)show the relationship of the odd or evenness of Q_n to "n"

Problem Solution
First we need to make some definitions within the equation number (2)

The function $$tanh^{-1}(x)$$ is an odd function both from the definition and also as explained on the transparency

Next the previous homework proved that the polynomial $$\underline{P}_n(x)$$ is an even function when n is even

It follows that the function $$\underline{P}_{n-j}(x)$$ is odd for even n since n-j would be odd (j always being odd)

we recognize that the order of the functions is $$tanh^{-1}(x)$$ multiplied by $$\underline{P}_n(x)$$ then subtracted by $$\underline{P}_{n-j}(x)$$ defining an odd function as "O" and an even function as "E" we will have the following two possibilities:

for an even n:

E * O - O leads to O - O which finally gives us O

for odd n:

O * O - E leads to E - E which finally gives us E

(The relationships for odd and even functions were explained in the previous homework if further information is needed)

So the final result for Qn is:

n=odd Qn=even, n=even Qn=odd

Problem Statement
Pg. 37-1 []

Plot the polynmial $$/underline{P}_{n}$$ for n 0 through 4 and $$Q_{n}$$ for n 0 through 4

Problem Statement
Pg. 37-2 []

Show that  = 0

Problem Solution
recall from lectures 32 and 33 the definition of a scalar product. For these two functions which are symmetric function in the range of -1 to 1 their scalar product will be zero if they are opposite (if one is odd and one is even).

Now recall from the second problem in this homework that for n=odd Qn=even or n=even Qn=odd

Also from the previous homework n=odd Pn=odd  n=even Pn=even

So no matter what if Q and P have the same "n" they will be opposite

Therefore:  by definition always = 0