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Section A4,QH4 lecturenotes from 8/21/08 due to technical difficulty
Edit Section A4,QH4 lecturenotes from 8/21/08 due to technical difficulty here.
Class,
I was out of it today in recitation and all the technical difficulties didn't help. Hopefully we will have all that fixed by next week. I want to clarify some things on number 28 and number 36.
28. estimate ln(1.2) to with in .01
First find the taylor series expansion about zero for ln(1+x) (because ln(x) at x=0 is undefined)
F(x)=ln(1+x)
F’(x)=1/(1+x)=(1+x)^(-1)
F’’(x)=-(1+x)^(-2)
F’’’(x)=2(1+x)^(-3)
F(4)(x)=-6(1+x)^(-4)
Etc.
F(0)=ln(1)=0
F’(0)=1
F’’(0)=-1
F’’’(0)=2
F(4)(0)=-6
F(5)(0)=24
Which gives the taylor series:
x-(x^2/2)+(x^3/3)-(x^4/4)+….
Which can be condensed as (SUM from k=1 to infinity) ((-1)^(k+1)x^k)/(k) for -1x1
Here pick x=.2 so that ln(1+x)=ln(1+.2)=ln(1.2).
Now we want the error term to be .01, so in this case it is actually easier to use THE SERIES ITSELF rather than the lagrange form of the remainder.
Note that the absolute value of the series evaluated when x is between -1 and 1 is always decreasing as terms are being added i.e. magnitude( x^3/3)>magnitude(x^4/4) when x is between -1 and 1, so all you need to do in this case is to solve for n in the following expression:
Setting x to .2 and the expression =.01
((.2)^n/n)=.01, which gives n=3 meaning (x^3/3) is the first term with magnitude .01
Therefore use the approximation P2(x)= x-(x^2/2) to approximate ln(1+.2)
P2(.2)= .2-(.2^2/2)=.18
36. Find the lagrange form of the remainder for f(x)=sqrt(1+x); n=3
The lagrange form of the remainder is Rn(x)=(max(f(n+1)(c)))xn+1/(n+1)!
Just compute derivatives
f(x)=sqrt(1+x)=(1+x)^(1/2)
F’(x)=(1/2)(1+x)^(-1/2)
F’’(x)= (-1/4)(1+x)^(-3/2)
F’’’(x)= (3/8)(1+x)^(-5/2)
F(4)(x)= (-15/16)(1+x)^(-7/2)
Now to find max(f(4)(c)) you would naturally want to use the 1st derivative test on the fourth derivative which is F(5)(x)= (105/32)(1+x)^(-9/2)=0
You get x=-1
All values <-1 of F(5)(x) are undefined and all values >-1 are positive , meaning F(4)(x) is constantly increasing for all x>-1, therefore you cannot assume a specific value for max(f(n+1)(c)) because you do not know x (if you did know x you could find max(f(4)(c)) by setting c=x which would be the endpoint of the interval because you know the 4th derivative of the original function is always increasing). Therefore you must leave the lagrange form of the remainder in the form:
R3(x)=(max(f(4)(c)))x4/(4)!=(max((-15/16)(1+c)^(-7/2)))x4/(4)!
If you have anymore questions feel free to ask.
Peace,
Russ
