<p>QUESTION #1 </p>
<p>f(x)= 1 / (1-2x)</p>
<p>a.) write the first four terms and the general term of the taylor series expansion of f(x) about x=0
b.)what is the interval of convergence of the series you found in part a?
c) find the value of f at x = -1/4</p>
<p>QUESTION #2</p>
<p>A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per squar meter for the sides, what is the cost of the least expensive tank?</p>
<p>Your solutions to these two Questions would be very much appreciated</p>
<p>1/(1-u) = 1 + u + u^2 + u^3 and now because u = 2x, you just plug that in to get
1/(1-2x) = 1 + 2x + (2x)^2 + (2x)^3....(2x)^n
The interval of converge is -.5<x<.5 , but I am not that sure on this
at x=-1/4, I believe you can just plug that into the original equation
1/(1 - 2(1/4)) = 1/.5 = 2. You should come real close to this value by doing the first around 6 terms. After the first 4 its 1.875, after 5 it is 1.9375.</p>
<p>Similiarly
Tan inverse derivative is 1 / (1+u^2) = 1/(1-(-u^2)) which you write as
1 - u^2 + u^4 - u^6 +... (sign changes because the extra - sign) and then because that is the derivative, you can integrate this formula to approximate tan inverse of u so
x - u^3/3 + u^5/5 -u^7/7 and you let u=x or whatever.</p>
<p>For the second one, you need to get two equations
Volume
36 = L<em>w</em>h = L<em>4</em>h so 9 = L*h</p>
<p>Surface Area
2sides<em>5per</em>h(4=width + L) + $10<em>4width</em>L== min </p>
<p>Combine the two equations by plug in for h as h=9/L
10<em>9/L(4+L) + 4L
360/L + 90 + 4L take the derivative, to find rel min and max, you might need to check both answers or multiple answers.
-360/L^2 + 40 = 0
360/40=L^2 so L= root9 = 3
9=L</em>h so h=3</p>
<p>Variables known
Width = 4 constant
length = 3
height=3</p>
<p>Total Cost
2sides<em>5per</em>h(4=width + L) + $10<em>4width</em>L== min
$10<em>3(4+3) + $10</em>4*3 =$330.
You can check this by changing h to say 5, and making L=9/5, then the total cost =362.
Making h to 2 and L = 9/2 cost =350.</p>
<p>thanks for your help matt but i have a question regarding the second problem</p>
<p>what do you mean by "2sides<em>5per</em>h(4=width + L) + $10<em>4width</em>L== min "</p>
<p>and i got lost in how you combined the equations? and also dont you need to use the second derivative test to check for minimum and maximum?</p>
<p>what do you mean by "2sides<em>5per</em>h(4=width + L) + $10<em>4width</em>L== min "</p>
<p>and i got lost in how you combined the equations? and also dont you need to use the second derivative test to check for minimum and maximum?</p>
<p>2 sides for each because if you have a box, two sides of the box are bounded by height and by either the length or the width
5 per = $per square unit
h= the height, an unknown
4=width means that just put a 4 there, and it represents the width
and the second half follows this
I combined equations by plugging in for h and you know that 9=Lh so h=9/L. Then you can just plug that into where you see the h on the first one.</p>
<p>You need the first derivative to test for the min, max, and the second deriv to see if it is a min or max. I did it by instead checking the points to the left and to the right of L=3. But yea, you should just use the 2nd derivative, but since you only got one answer and the endpoints are like infinity, and 9/infinity, if thats even possible, it wasnt really necessary to check the answer.</p>
<p>hey matt i think u messed up on question 1 part c. f(x) = -1/4 , u plugged in 1/4
That also means your following sentence is incorrect
"You should come real close to this value by doing the first around 6 terms. After the first 4 its 1.875, after 5 it is 1.9375."</p>
<p>also for part c how many terms aare adeuate for approximating f(-1/4) with an error not exceeding one per cent??</p>