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Actually... To show what I have done:
I found that
And that:
And then plug in that for the first part... but I'm not sure how that would help!????
If i let f(x) = sinx. Find the error E(x, pi/2) in equation:
E(x,a) = f'(a) - [(f(x)-f(a))/(x-a)]
as a function of x. Graph E(x,a). Find a delta if epsilon = .01, .0001, 10^-10.
Honestly I can do the graphing part. I will also be able to do the proof when epsilon = .0001, 10^-10 once I have seen a proof for one part. I am having a hard time with the teacher - he has yet to do an example of how to do one of these proofs... and the book, needless to say, is not much better.
Any help would be great.
I'm working on some tougher derivatives.
1.
and
2.
, x is in the Reals-----------------------------
For #1:
I tried to look at it two separate ways as x>0 and x<0.
I figured if x<0 then:
Does this make sense? It seems too easy?
STUCK PRETTY MUCH COMPLETELY ON #2.
I'm just wondering if there is a good book or place to get self help on Analysis stuff? I learned calc I and II through these books but I'm trying to learn more for Analysis but am not sure where to get help or find a good book about it
AHA!
So, I know how to find that the target value is 4/7 (or .572193) now... (I THINK!)
I want to look at the geometric series:
Which equals 4/7! Thanks Ricky!
Now... to find the value of
so that any partial sum with at least terms is within 0.001 of the target value.I need to:
Find the target of the series:
1 - (3/4) + (9/16) - (27/64) + ... + (-1)^k*(3^k/4^k)
Then i need to find any value of n so that any partial sum with at least n terms is within 0.001 of the target value. Justify the answer.
So here's what I did. I made a program on Mathematica that gave me that the target value of the partial sum is .571429
I also found that at the sum from 0 to the 22nd term is equivalent to 0.572193
I suppose I have my answer but I cannot justify it - except using proof by mathematica program. I am worried though that when I have an exam and am forced do this without mathematica I will not be able to. Is there any advice? Or can someone teach me how to do this?
I need to construct a series that suggests that the sum (of alternating 1s and −1s) is 1/4. Or 2/3.
Here's what was written in class before this:
Here is an example that seems to suggest that
1 − 1 + 1 − 1 + 1 − 1 ± · · · = 1/3
.
Consider the series
S = 1 − x + x3 − x4 + x6 − x7 ± · · · .
Note that
S = (1 − x)(1 + x3 + x6 + · · ·) = (1 − x)*(1/(1 − x^3)) = (1/(1 + x + x^2))
Evaluate at x = 1.
.......
I'm not even sure how the example works completely... let alone how to do one on my own.
Wow. I feel like it was just one of those problems where I went right past something that wouldn't have been that difficult to making the problem very difficult. Thanks Jane. I knew the answer was pi/3 but you've helped and made it completely clear!!
I had a similar problem to do and thanks to you I've already done it! You are a great help!
Question: Prove that the partial sums are always greater than or equal to 1 once we have at least fiver terms. What number does this series appear to approach?
Series: 1 + (1/5) - (1/7) - (1/11) + (1/13) + (1/17) - ...
I found that:
But it is also given in the book that the sum is from k=1 to k with:
[(-1) ^ [Floor((k-1)/2)] ] / [6*(Floor(k/2)) + (-1)^(k-1)] (sorry not being in code - i'll work on it but wanted to get this up)
Any ideas on where to go?
Thanks. I also got number one. Thanks for you're help on 3, it pushed me to finishing one. Hopefully doing this problem on the practice exam I'll be able to do a similar one on the final. You've been a huge help.
phi is the permutation. but on #2 i meant phi(f) = f(1/4)
Let G be the set of all real-valued functions on the interval [0,1]. Define f+g for f,g in G by (f+g)(x)=f(x)+g(x).
1. Prove that G is a group
2. Define phi : G->R by phi(f)=phi(1/4) and prove that phi is a homomorphism
3. Let H={f in G such that f(1/4)=0}. Prove H is a subgroup of G.
4. What is G/H isomorphic to?
All in all - just stuck. Any ideas?
My work so far:
G is a group means that it is assosciative, closed, has an inverse and identity. (It's associative because addition is)
Need some help in neutral geometry.
The first one is to prove that parallelograms exist.
(the others depend on this proof so hopefully getting this will help me do those) I started by using a saccheri quadrilateral and using triangle congruences to prove it but i get stuck.
Thanks -
1. Given 3 non-collinear points, there is one and only one circle containing all three points.
2. If triangles ABC and DEF have right angles at A and D and have AB=DE and BC=EF show the triangles are congruent.
For 1: The three points are on the circle. I'm not sure where to start. We've really only done work with triangles and lines in the class so far.
For 2: I now have angle-side-side which is not a valid form of proof. I've tried proving by contradiction with no success. I'm not sure what step to take now.
ideas?
1. Prove no line can intersect a circle more than twice.
2. Assume that any straight line drawn from a point on the circle to a point inside will intersect the circle at least one more time. Prove that it will intersect the circle exactly once more.
for 1... I did it using Pasch's postulate but now the prof. wants us to do it again with a different method and I just haven't figured it. I have 3 pages of drawings and attempts but it's due tomorrow now... and the past week I've yet to figure it out???
and for 2 i feel like i need to use the fact that "no 3 lines can intercept a line from a point such that all lines have the same length" [which i've proved in this same problem set].
Ideas?
Do the odd permutations of S_n form a group? Explain.
I KNOW that they dont for the reason that the group of odd permutations does not include the identity. I'm not sure how to prove this in general.
Thanks.
Let Q be the group of rational numbers under addition and let Q* be the group of nonzero rational numbers under multiplication. In Q, list the elements in <1/2>. In Q*, list the elements in <1/2>. Find order of each element in Q and in Q*.
I know that....
for any element, a, from a group, G, we let <a> denote the set {a^n l n in Z}. I've finished other problems dealing with this but this problem is fooling me.
Thanks.
1. What are the symmetries of a nonsquare rectangle?
2. What are the symmetries of a parallelogram that is neither a rectangle or a rhombus.
3. What are the symmetries of a rhombus that is not a rectangle?
My thoughts:
1. 4: The identity, 180 degree rotation, vertical reflection and horizontal refletion?
2. 4: The identity, 180 degree rotation, 2 reflections from opposite angles.
3. 4: same as above?
it doesn't seem right because 2 and 3 seems like they shouldn't be the same? THOUGHTS? Thanks.
The Question: If S is a subset of {1,2,...,n} having size 2n+1 prove S contains 3 consecutive numbers. Show that this is best possible by exhibiting a set of size 2n for which the conclusion is false.
My Answer: Suppose S has no consecutive elements, and label its elements k(0), k(1), ..., k(2n - 1), k(2n) in order, so that
k(0) < k(1) < ... < k(2n - 1) < k(2n).
Since these aren't consecutive, we have that k(i) - k(i - 1) ≥ 2 for i = 1, 2, ..., 2n. Thus
k(2n) - k(0) ≥ [k(2n) - (k(2n - 1)] + [k(2n - 1) - k(2n -2)] + ... + (k(2) - k(1)) + (k(1) - k(0)) ≥ 2n * 2 = 4n.
Since the smallest k(0) can be is 1 and k(2n) can be no larger than 3n, it follows that 3n ≥ k(2n) ≥ k(0) + 4n = 4n + 1, which is contradiction. Hence S must contain some consecutive elements.
I need to use the pigeonhole principle in my answer (says the teacher). I'm not sure how I can use it in this case. I thought i did, but.... Any ideas??
Thanks everyone, I got #3; and got it correct on the test!
#1 I got wrong, I attempted to not directly copy anyones work here, but instead tried to use advice from different posts and I was wrong and have to try and fix it now, but I appreciate all the help.
I didn't actually understand #2 really until WHATISMATH posted. I appreciate it
Thanks for the help everyone.
Ricky - I didn't have another example to ask the question for. Sorry, but I still dont understand anyone's answers.
2. Lattice path - as my book says - A lattice path in the plan is a path joining together integer points via steps of unit legth rightward and upward. Typically the rightward is given a value of 1 and the upward is given a value of 0. If given a path from (0,0) to (0,1) to (1,1) to (2,1) to (2,2) I understand the path will be (1,0,0,1) but after that I'm not sure what good they are? What's their use?
Cardinatlity - I guess what the point is?
Some things I'm confused about:
1. How to show that the union of countable sets is also countable. For example, let A1, A2, ... be a sequence of sets, each countable. How to prove the union is a countable set.
2. What the use of lattice paths is. I understand how to make/get one... by I'm not sure how, on the test for example, it could be used in a problem.
3. How to count something in 2 ways
4. Cardinality.
and
5. Binary coding / n-tuples. For example, our teacher emailed us today with a question he says we should be able to answer easily: The 51st State of the union is going to be the State of Tom. I will be
issuing license plates for cars using 3 letters from the alphabet. How many plates can I make? In essence we are counting ?-ary ?-tuples. Finally, what is the probability that a randomly made plate will spell Tom?
ANY ADVICE/HELP WOULD BE WONDERFUL.
sorry, i dont know how to make equations?!
1. Need to Use Pascal's Formula and induction on n to prove the Binomial Theorem:
(DONT UNDERSTAND THE PART IN BOLD) Correct otherwise??
Base Case: (a+b)^0 = 1 = \sum_{k=0}^0 { 0 \choose k } a^{0-k}b^k.
For the inductive step, assume the theorem holds when the exponent is m. Then for n = m + 1
(a+b)^{m+1} = a(a+b)^m + b(a+b)^m \,
= a \sum_{k=0}^m { m \choose k } a^{m-k} b^k + b \sum_{j=0}^m { m \choose j } a^{m-j} b^j by the inductive hypothesis
= \sum_{k=0}^m { m \choose k } a^{m-k+1} b^k + \sum_{j=0}^m { m \choose j } a^{m-j} b^{j+1} by multiplying through by a and b
= a^{m+1} + \sum_{k=1}^m { m \choose k } a^{m-k+1} b^k + \sum_{j=0}^m { m \choose j } a^{m-j} b^{j+1} by pulling out the k = 0 term
= a^{m+1} + \sum_{k=1}^m { m \choose k } a^{m-k+1} b^k + \sum_{k=1}^{m+1} { m \choose k-1 }a^{m-k+1}b^{k} by letting j = k − 1
= a^{m+1} + \sum_{k=1}^m { m \choose k } a^{m-k+1}b^k + \sum_{k=1}^{m} { m \choose k-1 }a^{m+1-k}b^{k} + b^{m+1} by pulling out the k = m + 1 term from the RHS
= a^{m+1} + b^{m+1} + \sum_{k=1}^m \left[ { m \choose k } + { m \choose k-1 } \right] a^{m+1-k}b^k by combining the sums
= a^{m+1} + b^{m+1} + \sum_{k=1}^m { m+1 \choose k } a^{m+1-k}b^k from Pascal's rule
= \sum_{k=0}^{m+1} { m+1 \choose k } a^{m+1-k}b^k by adding in the m + 1 terms.
2. Summing the cubes?
a)Prove directly that m^3=6(m choose 3)+6(m choose 2)+m
b)Use part (a) to prove that the sum(from i=1 to n) i^3 = [(n(n+1)/2]^2.(without using induction)
c)Prove part (a) by counting a set in two ways (hint: Count the ordered triples that can be formed from [m].)
Really just not sure where to go here... maybe some hint may help? I think I can do it... not sure where to start?
Yes you are right pi man... I made a mistake... In my notes i did write 3^k... Made an accident here. But I see what you are saying with:
∑(n choose k)*(3^n) :: (sum from k=0 to n)
But I think that it will be
∑(n choose 2k)*(3^n) :: (sum from k=0 to n/2)
and then divide that by 6^n to get .5?