#$&*
course Mth163
july 12,2013 2:48 AM
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
021. `query 21
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Question: `qWhat are the possible number of linear and irreducible quadratic factors for a polynomial of degree 6?
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Your solution:
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3 identical factors 3 distinct
1 ident 5 dist
5 indent I dist
2 ident 4 dist etc…
3 pairs of identically different factors
2 pairs of identically different factors 1 ident 1 dist
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confidence rating #$&*:: 2
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Given Solution:
`a** You can have as many as 6 linear and 3 irreducible quadratic factors for a polynomial of degree 6.
For a polynomial of degree 6:
If you have no irreducible quadratic factors then to have degree 6 you will need 6 linear factors.
If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 4 linear factors.
If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 2
linear factors.
If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you can have no linear
factors.
For a polynomial of degree 7:
If you have no irreducible quadratic factors then to have degree 7 you will need 7 linear factors.
If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 5 linear factors to give
you degree 7.
If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 3
linear factors to give you degree 7.
If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you will need 1 linear
factor to give you degree 7. **
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Self-critique (if necessary):
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Self-critique rating:OK
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Question: `qFor a degree 6 polynomial with one irreducible quadratic factor and four linear factors list the possible numbers of repeated and distinct zeros.
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Your solution:
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I used an example (x^2+4) (x-4)(x-3)(x+3)(x+2)
My zeros would then be (-4,0)(-4,0)(4,0)(3,0)(-3,0)(-2,0)
This means that the y int. of this function would be 1152
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confidence rating #$&*: 2
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Given Solution:
`a** there could be 1 root repeated 4 times, 2 roots with 1 repeated 3 times and the other distinct from it, 2 distinct roots each repeated twice, three distinct roots with one of them repeated twice, or four distinct roots. Explanation:
You have one zero for every linear factor, so there will be four zeros.
Since the degree is even the far-left and far-right behaviors will be the same, either both increasing very rapidly toward
+infinity or both decreasing very rapidly toward -infinity.
You can have 4 distinct zeros, which will result in a graph passing straight thru the x axis at each zero, passing one way
(up or down) through one zero and the opposite way (down or up) through the next.
You can have 2 repeated and 2 distinct zeros. At the repeated zero the graph will just touch the x axis as does a
parabola at its vertex. The graph will pass straight through the x axis at the two distinct zeros.
You can have 3 repeated and 1 distinct zero. At the 3 repeated zeros the graph will level off at the instant it passes thru
the x axis, in the same way the y = x^3 graph levels off at x = 0. The graph will pass through the x axis at the one distinct
zero.
You can have two pairs of 2 repeated zeros. At each repeated zero the graph will just touch the x axis as does a
parabola at its vertex. Since there are no single zeros (or any other zeros repeated an odd number of times) the graph
will not pass through the x axis, so will remain either entirely above or below the x axis except at these two points.
You can have four repeated zeros. At the repeated zero the graph will just touch the x axis, much as does a parabola at
its vertex except that just as the y = x^4 function is somewhat flatter near its 'vertex' than the y = x^2 function, the graph
will be flatter near this zero than would be a parabolic graph. **
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Self-critique (if necessary):
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How can a zero be repeated 4 times if the degree is six???
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Self-critique rating:
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(x-1)^4 * (x+2) ( x - 5) is an example of this, as is
(x-1)^4 * (x^2 + x + 12).
When multiplied out both yield degree-6 polynomials, and both have a zero at x = 1 that is repeated 4 times.
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Question: `qDescribe a typical graph for each of these possibilities. Describe by specifying the shape of the graph at each of its zeros, and describe the far-left and far-right belavior of the graph.
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Your solution:
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1 root repeated 4 times-parabola, y=x^4, positive
2 roots with 1 repeated 3 times-y=x^3, positive
2 distinct roots- parabola y=x^2 positive
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confidence rating #$&*: 1
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Given Solution:
The graph will pass through the x axis 4 times. Near each of the 4 zeros the graph will appear to be a straight line, though of course as you get away from the x axis the graph in between the zeros will need to curve in order to return to the axis.
The graph will get increasingly steep at far right and far left, with the slopes at left and right having opposite signs (i.e., if the slope at far right is positive, with the graph moving up and to the right, then at far left the slope must be negative, moving down as it moves to the right; similarly if the slope at far left is positive, the slope at far right will be negative).
The graph might look like the figure below; it might also be 'upside down' compared to the graph in the figure.
ERRONEOUS STUDENT SOLUTION AND INSTRUCTOR CORRECTION
for 4 distincts, the graph curves up above the x axis then it crosses it 4 timeswithe the line retreating from the direction it
came from
for say 2 distincts and 2 repeats, line curves above x axis then it kisses the axis then it crosses it twice, retreating to the
side it came from
for 1 distinct and 3 it curves above x axis then it crosses once and kisses, finnaly heading off to the opposite side it came
from
INSTRUCTOR COMMENTS:
{The graph can't go off in th opposite direction. Since it is a product of four linear factors and any number of quadratic
factors its degree is even so its large-x behavior is the same for large positive as for large negative x. It doesn't kiss at a
degree-3 root, it acts like the y = x^3 polynomial, leveling off just for an instant as it passes through the zero and on to the
other side of the axis. **
problems 3-5
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Self-critique (if necessary):
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When it said 1 root repeated 4 times I assumed it was a quadruple root therefore a parabola
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This description is basically OK for the behavior of the graph near the zero, but technically the term 'parabolic' applies only to y = x^2 and other quadratic polynomials.
y = x^2 defines a parabola, as does any polynomial of degree 2.
The graph of y = x^4 has a lot in common with a parabola, with a similar but somewhat flatter shape, but it isn't a parabola..
You could describe it as a sort of flattened parabola if you like.
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You haven't described the behavior at the other zeros, if there are any, and you haven't described the far-left and far-right behavior of the graph.
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(x-1)^4 “ Near each of the 4 zeros the graph will appear to be a straight line”- can you explain this a little more ???
Was I supposed to put all of them together ???
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The given solution to the original problem addresses all the possibilities. The given solution for this question only addresses the graph if the polynomial has 4 distinct zeros, in which case the graph is approximately linear at each one.
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Question: `qIt doesn't matter if you don't have a graphing utility--you can answer these questions based on what you know about the shape of each power function.
Why does a cubic polynomial, with is shape influenced by the y = x^3 power function, fit the first graph better than a quadratic or a linear polynomial?
What can a cubic polynomial do with this data that a quadratic can't?
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Your solution:
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It has more curves
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confidence rating #$&*:: 2
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Given Solution:
`a** the concavity (i.e., the direction of curvature) of a cubic can change. Linear graphs don't curve, quadratic graphs
can be concave either upward or downward but not both on the same graph. Cubics can change concavity from upward
to downward. **
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This says, more specifically, that the graph an increasing cubic will be increasing at a decreasing rate up to a point, the increasing at an increasing rate beyond that.
You should construct the graph of a decreasing cubic and think about how it would be described.
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Self-critique (if necessary):
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Self-critique rating:
Question: `qOn problem 5, how do the shapes of 4th-degree polynomials and 6th-degree polynomials progressively differ from the shape of a 2d-degree polynomial in such a way as to permit a better and better fit to the graph?
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Your solution:
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The larger the polynomial the more flexible it becomes and the more you can accurately show your graph
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confidence rating #$&*::
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Given Solution:
`a** higher even degrees flatten out more near their 'vertices' **
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Self-critique (if necessary):
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Is that the same for higher odd degrees??
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Higher odd degrees flatten out at the point where their concavity changes, e.g., at the point where they go from concave down (increasing at a decreasing rate) to concave up (increasing at an increasing rate).
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Self-critique rating:
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Question: `qOn problem 5, how do the shapes of 4th-degree polynomials and 6th-degree polynomials progressively differ from the shape of a 2d-degree polynomial in such a way as to permit a better and better fit to the graph?
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Your solution:
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The larger the polynomial the more flexible it becomes and the more you can accurately show your graph
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confidence rating #$&*: 2
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Given Solution:
`aSTUDENT RESPONSE: progressively more flexing, because more curves, and fit graph better the that of a lesser degree
INSTRUCTOR COMMENT: On a degree-2 polynomial there is only one change of direction, which occurs at the
vertex. For degrees 4 and 6, respectively, there can be as many as 3 and 5 changes of direction, respectively. For higher
degrees the graph has more ability to 'wobble around' to follow the data points.
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Self-critique (if necessary):
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Self-critique rating:
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Question: `qWhat is the degree 2 Taylor approximation for f(t) = e^(2t), and what is your approximation to f(.5)? How close is your approximation to the actual value of e^(2t)?
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Your solution:
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1 + 2t + (2t)^2 / 2! = 1 + 2 t + 4 t^2 / 2 = 1 + 2 t + 2 t^2
T(.5) = 1 + 2 * .5 + 2 * .5^2 = 1 + 1 + 2 * .25 = 1 + 1 + .5 = 2.5
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confidence rating #$&*: 3
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Given Solution:
`a** The degree 2 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! = 1 + 2 t + 4 t^2 / 2 = 1 + 2 t + 2 t^2.
Therefore we have
T(.5) = 1 + 2 * .5 + 2 * .5^2 = 1 + 1 + 2 * .25 = 1 + 1 + .5 = 2.5.
The actual value of e^(2t) at t = .5 is
f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx..
The approximation is .218 less than the actual function. **
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Self-critique (if necessary):
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Self-critique rating:OK
Question: `qBy how much does your accuracy improve when you make the same estimate using the degree 3 Taylor approximation?
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Your solution:
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1+(3*.5)+(3*.5^2)/3!=2.875
I guess this is more accurate I do not know now what the answer is supposed to be so how would I know if this is more accurate, this is just a little confusing???
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confidence rating #$&*: 2
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Given Solution:
`a** The degree 3 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! + (2t)^3 / 3! = 1 + 2 t + 4 t^2 / 2 + 8 t^3 / 6 = 1 + 2 t
+ 2 t^2 + 4 t^3 / 3.
Therefore we have
T(.5) = 1 + 2 * .5 + 2 * .5^2 + 4 * .5^3 / 3 = 1 + 1 + 2 * .25 + 4 * .125 / 3 = 1 + 1 + .5 + .167 = 2.667.
The actual value of e^(2t) at t = .5 is
f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx..
The approximation is .051 less than the actual function. This is about 4 times closer than the approximation we obtained
from the degree-2 polynomial. **
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Self-critique (if necessary):
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The degree 3 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! + (2t)^3 / 3! = 1 + 2 t + 4 t^2 / 2 + 8 t^3 / 6 = 1 + 2 t
+ 2 t^2 + 4 t^3 / 3.
Ohh!! So you keep the pervious stuff and keep adding on??
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Right.
The degree tells you how many terms to include. A degree 7 polynomial would continue the pattern until you got to t^7.
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Self-critique rating:
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Question: `qDescribe your graph of the error vs. the degree of the approximation for degree 2, degree 3, degree 4 and degree 5 approximations to e^.5.
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Your solution:
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The higher the factorial the lower error
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confidence rating #$&*: 3
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Given Solution:
`a**
The value of e^.5 to 10 significant figures is 1.648721270. Each approximation should be rounded to a number of significant figures great enough to give at least a 2-significant-figure difference.
The Taylor approximation for e^x is 1 + x + x^2 / 2! + x^3 / 3! + ... + x^n / n! + ...
Thus for degree 2 the approximation is 1 + .5 + .5^2/2 = 1.625
F0r degree 3 the approximation is 1 + .5 + .5^2/2 + .5^3/6 = 1.6458
For degree 4 the approximation is 1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 = 1.64838 approx
For degree 5 the approximation is 1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 + .5^5/120 = 1.648697917
The errors, if calculations are done accurately enough to yield a difference with 3 significant figures, are as follows:
ê^.5 - (1 + .5 + .5^2/2 ) = 0.0237
ê^.5 - (1 + .5 + .5^2/2 + .5^3/6 ) = 0.00288
ê^.5 - (1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 ) = 0.000283
e^.5 - (1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 + .5^5/120) = .000023354
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Self-critique (if necessary):
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Closer to zero is more approx. right??? To try to answer a question I asked earlier in this assignment
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The closer the error is to zero, the closer the result is to the actual value, and the better the approximation.
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Question: `qWhat are your degree four approximations for e^.2, e^.4, e^.6 e^.8 and e^1? Describe the graph of the approximation error vs. x.
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Your solution:
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. e^(.2)= 1 + .2 + .2^2 / 2! + .2^3 / 3! + .2^4 / 4!=1.23133
e^(.4)= 1 + .4 + .4^2 / 2! + .4^3 / 3! + .4^4 / 4!=1.53067
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Evaluated correctly this would be a little over 1.49.
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e^(.6)= =1.096
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1 + .6 + .6^2 / 2! + .6^3 / 3! + .6^4 / 4!is not 1.096.
All these terms are positive, and 1 + .6 is already equal to 1.6.
When you add in the rest of these numbers you get a little over 1.82.
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e^(.8)= 1 + .8+ .8^2 / 2! + .8^3 / 3! + .8^4 / 4!=2.36533
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This should be a little over 2.22.
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e^(1)= 1 + 1 + 1^2 / 2! + 1^3 / 3! + 1^4 / 4!=2.91667
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This should be a little under 2.71.
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due to the information on my graph the outlier or error was .6
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Self-critique rating:
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You were also asked to compare with the correct values e^.2, e^.4, e^.6, e^.8 and e^1.
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Question: `qThe following are the approximations and errors:
The degree-4 approximation to the exponential function e^x is 1 + x + x^2 / 2! + x^3 / 3! + x^4 / 4!. We evaluate this expression for x = .2, .4, 6., .8, 1.0:
For x = 0.2 the degree 4 approximation is 1.2214
This differs from e^.2 = 1.221402758 by 2.75816*10^(-06) = .000002758
For x = 0.4 the degree 4 approximation is 1.49173333
This differs from e^.4 = 1.491824698 by 9.13643*10^(-05) = .00009136
For x = 0.6 the degree 4 approximation is 1.8214
This differs from e^.6 = 1.8221188 by 0.0007188
For x = 0.8 the degree 4 approximation is 2.2224
This differs from e^.8 = 2.225540928 by 0.003140928
For x = 1 the degree 4 approximation is 2.70833333
This differs from e^1 = 2.718281828 by 0.009948495
The errors can be written to 2 significant figures as .0000027, .000091, .000071, .0031 and .0099.
A graph of approximation error vs. x increases exponentially, with over a 10-fold increase with every increment of 0.2.
This graph is shown below:
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Self-critique (if necessary):
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Wait am I trying to find the error in the pattern of the e^x factorial sequences???
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That is correct.
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Question: `qWhat is the function which gives the quadratic approximation to the natural log function?
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Your solution:
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Ln(x)=(x-1)-(x-1)^2/2+2(x-1)^3/6-3(x-1)/24+4(x-1)^3/120
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Good, but the quadratic approximation would be only
Ln(x)=(x-1)-(x-1)^2/2
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The natural log is defined in terms of powers of (x-1)
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confidence rating #$&*: 3
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Given Solution:
`a** The function is P2(x) = (x-1) - (x-1)^2/2.
A table of values of ln(x), P2(x) and P2(x) - ln(x):
x ln(x) P2(x) P2(x) - ln(x)
.6 -0.5108256237 -0.48 0.03082562376
.8 -0.2231435513 -0.22 0.003143551314
1.2 0.1823215567 0.18 -0.002321556793
1.4 0.3364722366 0.32 -0.01647223662
At x = 1 we have ln(x) = ln(1) = 0, and P2(x) = P2(1) = (1-1) - (1-1)^2 / 2 = 0. There is no difference in values at x =
1.
As we move away from x = 1 the approximation becomes less and less accurate. **
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Self-critique (if necessary):
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Self-critique rating:
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Self-critique rating:OK
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Question: `qWhat are your degree four approximations for e^.2, e^.4, e^.6 e^.8 and e^1? Describe the graph of the approximation error vs. x.
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Self-critique (if necessary):
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Question: `qWhat is the error in the degree 2 approximation to ln(x) for x = .6, .8, 1.2 and 1.4? Why does the approximation get better as x approaches 1?
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Your solution:
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.03, .00314, .00232, .016472
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confidence rating #$&*:
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Given Solution:
`a** The respective errors are .03, .00314, .00232, .016472.
There is no error at x = 1, since both the function and the approximation give us 0. As we move away from 1 the
approximation becomes less and less accurate. **
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Self-critique (if necessary):
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Self-critique rating:
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Question: `qproblem 12. What does the 1/x graph do than no quadratic function can do?
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Your solution:
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It has two asymptotes
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confidence rating #$&*: 3
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Given Solution:
`a** The y = 1/x graph has vertical asymptotes at the y axis and horizontal asymptotes at the x axis. The parabolas we
get from quadratic functions do have neither vertical nor horizontal asymptotes. **
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Self-critique (if necessary):
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vertical asymptotes at the y axis and horizontal asymptotes at the x axis of 1/x
while parabolas do not have asymptotes.
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Self-critique rating:3
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Question: `qWhat are the errors in the quadratic approximation to 1/x at x = .6, .8, 1, 1.2, and 1.4?
Describe a graph of the approximation error vs. x.
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Your solution:
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P2(.6)=1-(.6-1)+(.6-1)^2=1.56
P2(.8)=1.24
P2(1)=1
P2(1.2)=.84
P2(1.4)=.76
The differences are .32, .24, .16, .08
Error is .24, which is the interval between P2(.8)=1.24
P2(1)=1
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confidence rating #$&*:
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Given Solution:
`a** The quadratic approximation to 1/x is the second-degree Taylor polynomial
P2(x) = 1 - (x - 1) + (x - 1)^2.
A table of values of 1/x, P2(x) and P2(x) - 1/x:
x 1/x P2(x) P2(x) - 1/x
.6 1.666... 1.56 -.1066...
.8 1.25 1.24 -0.01
1.2 0.8333.... 0.84 0.006666
1.4 0.714285... 0.76 0.457
A graph of approximation error vs. x decreases at a decreasing rate to 0 at x = 1, then increases at an increasing rate for x
> 1.
This shows how the accuracy of the approximation decreases as we move away from x = 1.
The graph of approximation error vs. x gets greater as we move away from x = 1.**
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Self-critique (if necessary):
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Does this maean as we move further away from one our error becomes greater???
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The further our value of x is from 1, the greater our error.
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Self-critique rating:3
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: What is the maximum number of zeros possible for a polynomial of degree 5 which contains one irreducible quadratic factor?
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Your solution:
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You need to try to answer this.
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If you review the worksheets and the Query, you should develop some idea of how this question can be answered.
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You might find it helpful to make up such a polynomial. You can do this by just writing down an irreducible quadratic as one of your factor, and then inserting whatever additional factors are required.
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confidence rating #$&*:
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Question: What are the zeros of each of the following polynomials, and what is the multiplicity of each zero?
y = (x - 3) * (x + 2) * (x^2 + 4 x + 3)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 4)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 5)
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Your solution:
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(3,0)(-2,0) (-3,0)(-1,0) so for the first function we have multiplicities of one for all of our zeros
(3,0)(2,0)(-2,0) for the second function all of our multiplicities are one
(3,0)(-2,0) each with a multiplicity of one as for (x^2 + 4 x + 5) is irreducible therefore cannot have zeros or multiplicity
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At least one of those quadratic expressions factors into two linear factors, and at least one doe not.
This will affect at least some of your answers.
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confidence rating #$&*: 2
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Question: What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 5?
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Your solution:
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2 identical 3 distinct
3 ident 2 dis
2 pairs of identical factors and one distinct
4 identical and 1 distinct
3 identical and the other two ident to once another but now others
5 identical
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That's a good list, but there are more possibilities.
For example you could have 5 distinct linear factors.
And none of your possibilities includes an irreducible quadratic factor.
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confidence rating #$&*: 2
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Question: What are the degree 2 and degree 3 Taylor approximations of the function f(t) = e^(t/2)?
The approximate value of e^(t / 2) for t = 0.2 is 1.1052, accurate to five significant figures.
By how much does the value of the degree-2 Taylor polynomial, evaluated at t = 0.2, differ from this value?
By how much does the value of the degree-3 Taylor polynomial, evaluated at t = 0.2, differ from this value?
Optional moderately challenging question: Accurate to within +- 0.1, what is the smallest value of t for which the degree-3 Taylor polynomial differs from the value of the exponential function by more than your answer to the first question (i.e., by more than the value of the degree-2 polynomial evaluated at t = 0.2 differed from that of the given function at t = 0.2)?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
1+.2+.2^2/2=1.22
1+.2+.2^2/2+.2^3/6=1.22133
The difference is -.00133
@&
Your approximations are for e^t, not for e^(t/2).
The difference should be between the approximation and the actual value of the function; e.g., between 1.22 and e^(.2 / 2).
*@
@&
The Taylor expansion for e^t is
1 + t + t^2 / 2 + t^3 / 3! + t^4 / 4! + ... etc.
What therefore would be the first three terms of the Taylor expansion for each of the following?
e^z
(hint: this starts out 1 + z + z^2 / 2)
e^q
e^(aardvark)
e^(t-x)
e^(t / 2)
*@
&&&
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
*********************************************
Question: What is the maximum number of zeros possible for a polynomial of degree 5 which contains one irreducible quadratic factor?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
@&
You need to try to answer this.
*@
@&
If you review the worksheets and the Query, you should develop some idea of how this question can be answered.
*@
@&
You might find it helpful to make up such a polynomial. You can do this by just writing down an irreducible quadratic as one of your factor, and then inserting whatever additional factors are required.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the zeros of each of the following polynomials, and what is the multiplicity of each zero?
y = (x - 3) * (x + 2) * (x^2 + 4 x + 3)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 4)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 5)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
(3,0)(-2,0) (-3,0)(-1,0) so for the first function we have multiplicities of one for all of our zeros
(3,0)(2,0)(-2,0) for the second function all of our multiplicities are one
(3,0)(-2,0) each with a multiplicity of one as for (x^2 + 4 x + 5) is irreducible therefore cannot have zeros or multiplicity
@&
At least one of those quadratic expressions factors into two linear factors, and at least one doe not.
This will affect at least some of your answers.
*@
&&&
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 5?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
2 identical 3 distinct
3 ident 2 dis
2 pairs of identical factors and one distinct
4 identical and 1 distinct
3 identical and the other two ident to once another but now others
5 identical
@&
That's a good list, but there are more possibilities.
For example you could have 5 distinct linear factors.
And none of your possibilities includes an irreducible quadratic factor.
*@
&&&
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the degree 2 and degree 3 Taylor approximations of the function f(t) = e^(t/2)?
The approximate value of e^(t / 2) for t = 0.2 is 1.1052, accurate to five significant figures.
By how much does the value of the degree-2 Taylor polynomial, evaluated at t = 0.2, differ from this value?
By how much does the value of the degree-3 Taylor polynomial, evaluated at t = 0.2, differ from this value?
Optional moderately challenging question: Accurate to within +- 0.1, what is the smallest value of t for which the degree-3 Taylor polynomial differs from the value of the exponential function by more than your answer to the first question (i.e., by more than the value of the degree-2 polynomial evaluated at t = 0.2 differed from that of the given function at t = 0.2)?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
1+.2+.2^2/2=1.22
1+.2+.2^2/2+.2^3/6=1.22133
The difference is -.00133
@&
Your approximations are for e^t, not for e^(t/2).
The difference should be between the approximation and the actual value of the function; e.g., between 1.22 and e^(.2 / 2).
*@
@&
The Taylor expansion for e^t is
1 + t + t^2 / 2 + t^3 / 3! + t^4 / 4! + ... etc.
What therefore would be the first three terms of the Taylor expansion for each of the following?
e^z
(hint: this starts out 1 + z + z^2 / 2)
e^q
e^(aardvark)
e^(t-x)
e^(t / 2)
*@
&&&
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
*********************************************
Question: What is the maximum number of zeros possible for a polynomial of degree 5 which contains one irreducible quadratic factor?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
@&
You need to try to answer this.
*@
@&
If you review the worksheets and the Query, you should develop some idea of how this question can be answered.
*@
@&
You might find it helpful to make up such a polynomial. You can do this by just writing down an irreducible quadratic as one of your factor, and then inserting whatever additional factors are required.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the zeros of each of the following polynomials, and what is the multiplicity of each zero?
y = (x - 3) * (x + 2) * (x^2 + 4 x + 3)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 4)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 5)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
(3,0)(-2,0) (-3,0)(-1,0) so for the first function we have multiplicities of one for all of our zeros
(3,0)(2,0)(-2,0) for the second function all of our multiplicities are one
(3,0)(-2,0) each with a multiplicity of one as for (x^2 + 4 x + 5) is irreducible therefore cannot have zeros or multiplicity
@&
At least one of those quadratic expressions factors into two linear factors, and at least one doe not.
This will affect at least some of your answers.
*@
&&&
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 5?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
2 identical 3 distinct
3 ident 2 dis
2 pairs of identical factors and one distinct
4 identical and 1 distinct
3 identical and the other two ident to once another but now others
5 identical
@&
That's a good list, but there are more possibilities.
For example you could have 5 distinct linear factors.
And none of your possibilities includes an irreducible quadratic factor.
*@
&&&
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the degree 2 and degree 3 Taylor approximations of the function f(t) = e^(t/2)?
The approximate value of e^(t / 2) for t = 0.2 is 1.1052, accurate to five significant figures.
By how much does the value of the degree-2 Taylor polynomial, evaluated at t = 0.2, differ from this value?
By how much does the value of the degree-3 Taylor polynomial, evaluated at t = 0.2, differ from this value?
Optional moderately challenging question: Accurate to within +- 0.1, what is the smallest value of t for which the degree-3 Taylor polynomial differs from the value of the exponential function by more than your answer to the first question (i.e., by more than the value of the degree-2 polynomial evaluated at t = 0.2 differed from that of the given function at t = 0.2)?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
1+.2+.2^2/2=1.22
1+.2+.2^2/2+.2^3/6=1.22133
The difference is -.00133
@&
Your approximations are for e^t, not for e^(t/2).
The difference should be between the approximation and the actual value of the function; e.g., between 1.22 and e^(.2 / 2).
*@
@&
The Taylor expansion for e^t is
1 + t + t^2 / 2 + t^3 / 3! + t^4 / 4! + ... etc.
What therefore would be the first three terms of the Taylor expansion for each of the following?
e^z
(hint: this starts out 1 + z + z^2 / 2)
e^q
e^(aardvark)
e^(t-x)
e^(t / 2)
*@
&&&
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!#*&!
@&
You're not doing badly, but you need a little more work on some of these questions.
You should at the very least do your best on revisions of the last 4 questions.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
*********************************************
Question: What is the maximum number of zeros possible for a polynomial of degree 5 which contains one irreducible quadratic factor?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
@&
You need to try to answer this.
*@
@&
If you review the worksheets and the Query, you should develop some idea of how this question can be answered.
*@
@&
You might find it helpful to make up such a polynomial. You can do this by just writing down an irreducible quadratic as one of your factor, and then inserting whatever additional factors are required.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the zeros of each of the following polynomials, and what is the multiplicity of each zero?
y = (x - 3) * (x + 2) * (x^2 + 4 x + 3)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 4)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 5)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
(3,0)(-2,0) (-3,0)(-1,0) so for the first function we have multiplicities of one for all of our zeros
(3,0)(2,0)(-2,0) for the second function all of our multiplicities are one
(3,0)(-2,0) each with a multiplicity of one as for (x^2 + 4 x + 5) is irreducible therefore cannot have zeros or multiplicity
@&
At least one of those quadratic expressions factors into two linear factors, and at least one doe not.
This will affect at least some of your answers.
*@
&&&
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 5?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
2 identical 3 distinct
3 ident 2 dis
2 pairs of identical factors and one distinct
4 identical and 1 distinct
3 identical and the other two ident to once another but now others
5 identical
@&
That's a good list, but there are more possibilities.
For example you could have 5 distinct linear factors.
And none of your possibilities includes an irreducible quadratic factor.
*@
&&&
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the degree 2 and degree 3 Taylor approximations of the function f(t) = e^(t/2)?
The approximate value of e^(t / 2) for t = 0.2 is 1.1052, accurate to five significant figures.
By how much does the value of the degree-2 Taylor polynomial, evaluated at t = 0.2, differ from this value?
By how much does the value of the degree-3 Taylor polynomial, evaluated at t = 0.2, differ from this value?
Optional moderately challenging question: Accurate to within +- 0.1, what is the smallest value of t for which the degree-3 Taylor polynomial differs from the value of the exponential function by more than your answer to the first question (i.e., by more than the value of the degree-2 polynomial evaluated at t = 0.2 differed from that of the given function at t = 0.2)?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
1+.2+.2^2/2=1.22
1+.2+.2^2/2+.2^3/6=1.22133
The difference is -.00133
@&
Your approximations are for e^t, not for e^(t/2).
The difference should be between the approximation and the actual value of the function; e.g., between 1.22 and e^(.2 / 2).
*@
@&
The Taylor expansion for e^t is
1 + t + t^2 / 2 + t^3 / 3! + t^4 / 4! + ... etc.
What therefore would be the first three terms of the Taylor expansion for each of the following?
e^z
(hint: this starts out 1 + z + z^2 / 2)
e^q
e^(aardvark)
e^(t-x)
e^(t / 2)
*@
&&&
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!#*&!
@&
You're not doing badly, but you need a little more work on some of these questions.
You should at the very least do your best on revisions of the last 4 questions.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@
#*&!
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
*********************************************
Question: What is the maximum number of zeros possible for a polynomial of degree 5 which contains one irreducible quadratic factor?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
@&
You need to try to answer this.
*@
@&
If you review the worksheets and the Query, you should develop some idea of how this question can be answered.
*@
@&
You might find it helpful to make up such a polynomial. You can do this by just writing down an irreducible quadratic as one of your factor, and then inserting whatever additional factors are required.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the zeros of each of the following polynomials, and what is the multiplicity of each zero?
y = (x - 3) * (x + 2) * (x^2 + 4 x + 3)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 4)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 5)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
(3,0)(-2,0) (-3,0)(-1,0) so for the first function we have multiplicities of one for all of our zeros
(3,0)(2,0)(-2,0) for the second function all of our multiplicities are one
(3,0)(-2,0) each with a multiplicity of one as for (x^2 + 4 x + 5) is irreducible therefore cannot have zeros or multiplicity
@&
At least one of those quadratic expressions factors into two linear factors, and at least one doe not.
This will affect at least some of your answers.
*@
&&&
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 5?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
2 identical 3 distinct
3 ident 2 dis
2 pairs of identical factors and one distinct
4 identical and 1 distinct
3 identical and the other two ident to once another but now others
5 identical
@&
That's a good list, but there are more possibilities.
For example you could have 5 distinct linear factors.
And none of your possibilities includes an irreducible quadratic factor.
*@
&&&
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: What are the degree 2 and degree 3 Taylor approximations of the function f(t) = e^(t/2)?
The approximate value of e^(t / 2) for t = 0.2 is 1.1052, accurate to five significant figures.
By how much does the value of the degree-2 Taylor polynomial, evaluated at t = 0.2, differ from this value?
By how much does the value of the degree-3 Taylor polynomial, evaluated at t = 0.2, differ from this value?
Optional moderately challenging question: Accurate to within +- 0.1, what is the smallest value of t for which the degree-3 Taylor polynomial differs from the value of the exponential function by more than your answer to the first question (i.e., by more than the value of the degree-2 polynomial evaluated at t = 0.2 differed from that of the given function at t = 0.2)?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
&&&
1+.2+.2^2/2=1.22
1+.2+.2^2/2+.2^3/6=1.22133
The difference is -.00133
@&
Your approximations are for e^t, not for e^(t/2).
The difference should be between the approximation and the actual value of the function; e.g., between 1.22 and e^(.2 / 2).
*@
@&
The Taylor expansion for e^t is
1 + t + t^2 / 2 + t^3 / 3! + t^4 / 4! + ... etc.
What therefore would be the first three terms of the Taylor expansion for each of the following?
e^z
(hint: this starts out 1 + z + z^2 / 2)
e^q
e^(aardvark)
e^(t-x)
e^(t / 2)
*@
&&&
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!#*&!
@&
You're not doing badly, but you need a little more work on some of these questions.
You should at the very least do your best on revisions of the last 4 questions.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@
#*&!#*&!