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#1 Re: Euler Avenue » Which Prime below 1 million can be expressed as the sum of most primes » 2012-08-13 02:07:11

It should be the largest number which is the sum of the maximum number of distinct primes.

#3 Re: Help Me ! » Adding and Subtracting Polynomials » 2012-08-12 15:58:22

Hey

The negative sign does not "belong" to x, think of it as -1 * x = -1x = - x.

Does that make sense?

#4 Re: Euler Avenue » Which Prime below 1 million can be expressed as the sum of most primes » 2012-08-11 10:07:22

Agnishom, I will post the answer here but I got it by chance, so I cannot tell you how to code it.

#7 Re: Euler Avenue » Which Prime below 1 million can be expressed as the sum of most primes » 2012-08-10 18:25:04

Btw i just solved it by chance.

Good luck...also i believe after prime 550, the sum come to more than a million, if you start from the 4th prime.

#8 Re: Euler Avenue » Which Prime below 1 million can be expressed as the sum of most primes » 2012-08-10 18:12:17

Hey,

I just did this in excel and I get 1074180, if i sum up from the 4th prime to the 565th prime. dunno

SO post your code and lets see where you went wrong.

Also just to make sure we are starting and ending at the same primes: 4th prime is 7 and 565th prime is 4099.

#9 Re: Help Me ! » 1=2 proof » 2012-08-08 14:37:52

Heres an infinite series "proof" that shows 0 = 1

0 = 0 + 0 + 0 + 0 + · · ·
= (1 − 1) + (1 − 1) + (1 − 1) + (1 − 1) + · · ·
= 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + · · ·
= 1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + · · ·
= 1 + 0 + 0 + 0 + · · ·
= 1


"Legend has it that, around 1703, in letters to contemporary mathematicians, an Italian monk by the name of Guido Grandi (often called Guido Ubaldus) presented this as proof of the existence of God, since it suggested that the universe could have been created out of nothing! What he actually meant isn't clear, but we do know that the brightest minds of the day were unsure how to explain what the problem was. Leibniz at least recognized that the problem was in the ____ line above;"

I left it blank intentionally wink

This is copied from my Calculus prof notes, he showed is this "proof" in class.

#10 Re: Help Me ! » Integration » 2012-08-08 11:52:18

I have been looking at this and I realized that if we just square the actual integral and dont use the fomula given to us, we get to the correct answer.
This does not work:


but this does:

#11 Re: Help Me ! » Integration » 2012-08-08 10:34:05

Hi Bob bundy,

Can you verify my calculations below(just for my understanding):



now from here convert to polar co-ordinates
EDIT: I am not sure what the integration limits for theta would be...if it is 0 to pi/2, then this works out.

let u = r^2, du = 2rdr

since we squared the original equation, we square root the final answer so

This doesnt agree with what wolfram and wikipedia get down. Any idea where I went wrong?

Thanks

#12 Re: Help Me ! » Adding and Subtracting Polynomials » 2012-08-08 07:45:05

Hi SlowlyFading,

The degree of the polynomial is the highest degree in the polynomial.
For example:

has 3 as the highest degree, so the degree of the polynomial is 3.


Adding and subtracting polynomials is just like adding and subtracting numbers. You collect the like terms(the terms which have the same degree) and add or subtract them together.

For example:

First you take care of the negative sign:

Then group the like terms together:

And now add or subtract:

Follow these rules and try the questions you have posted again. Post your answers here, so we can see where you went wrong.

Cheers,
C25

PS Also look at the links bobbym posted

#13 Re: Help Me ! » Integration » 2012-08-08 06:16:10

Hahaha I could not have thought of that on the final exam. Lost 10% right there sad.

Thanks Bob

#14 Re: Help Me ! » Integration » 2012-08-08 04:56:48

Thanks bob bundy and bobbym.

Following that we were asked to use that to solve

and told to use polar coordinates.

I was confused on how to go from the above function to a function of x and y.

#15 Help Me ! » Integration » 2012-08-07 13:26:11

careless25
Replies: 8

Hi,

I just had a calc final and one question stumped me. But i would really love to know the answer. Can anyone help me solve this?
I have to show that:

#16 Re: Help Me ! » Local min/max of multivariable function » 2012-07-18 02:47:57

TheDude,

How do we know that we just have to consider the boundary points? What if there was a graph whos min was not on the  boundary point?

Thanks

#17 Help Me ! » Local min/max of multivariable function » 2012-07-17 13:28:06

careless25
Replies: 6

A circular hot plate given by the relationship x^2 + y^2 <= 4 is heated according to the spatial temperature function T(x, y) = 10-x^2 + 2x - 4y^2. Find the hottest and coldest temperatures on the plate and the points at which they occur.

Here is what I have so far:
g(x,y) = x^2 + y^2 - K where K<= 4

del f = (-2x + 2, -8y)
del g = (2x, 2y)

Applying LaGrange multiplier

1) -2x+2 = lambda(2x)
2) -8y = lambda(2y)
3) x^2 + y^2 = K

solving for eqn 2)
we get y=0 or lamba = -4

sub lambda = -4 in eqn 1
we get x = -1/3

now I am stuck, since there are infinite values of K we have to solve for. I know there is going to be points (-1/3, +/- y) and (+/-x, 0).

Another approach that gives me half the answer is solving for the global max of T(x,y), which coincidentally lies in the constraint given, but then I am at a loss at finding the local min.

Any ideas?

#21 Re: Help Me ! » Graph Theory Questions » 2012-07-15 11:05:26

2.

Proof By induction

Base Case:
we need atleast 3 vertices of degree 2 to have a cycle. Example a triangle.

IH: Assume true
IS:
If graph G with n vertices contains a cycle where each vertex has a degree of 2, then a graph M with n+1 vertices contains the graph G. Therefore graph M with n+1 vertices with degree 2 also contains a cycle.

QED

This is just a rough proof to give you the idea.

#22 Re: Help Me ! » Recursion of Binary Strings » 2012-07-09 03:16:35

hi bobbym,
I think an example might help:

Let B be the set of binary strings dened by
(a) {E,0} contained in B; and (where "E" is an empty string)
(b) for each s in the set B; s1 in set B and s10 in set B:
Show that B = C, where C is the set of all binary strings that do not have 2 consecutive
zeros.

Solution:

First we show that C is contained in B and finally B is contained in C.. Let P(n) be the statement that every binary string
of C of length at most n is in B. We use Mathematical Induction to show that P(n) is true
for all non-negative integers n.
Conditions (a) and (b) imply that P(0) and P(1) are both true. This completes the basis
step.

Induction Hyothesis: For n > 1, assume P(n - 1) is true.

Induction Step:

Let s be any binary string of C of length n. If s ends with a 1, then s = t1 where t is a binary string of C of length n-1.
Since P(n-1) is true, t is in B. By condition (b), s = t1 is in B.
If s ends with a 0, then s = t10, since s has no consecutive zeros, and t is a binary string of C of length n - 2.
Since P(n - 1) is true, t is in B.
By condition (b), s = t10 is in B.
Therefore P(n) is true.


By Mathematical Induction, P(n) is true for all positive integers n. Therefore C is contained in B.

Now we show B is contained in C. Let B(i) be the set of all strings of B that can be obtained by applying operation (b) at most i times.
For example, B(0) = {E,0} and B(1) = {E,0,1,10,01,010}.
Let Q(n) be the statement that B(n) is contained in C.

By Inspection, Q(0) is true. This completes the basis step.

Hypothesis: For n > 0, assume Q(n - 1) is true.
Induction Step:
Clearly B(n) = {E,0} union {b1|b in B(n-1)} union {b10| b in B{n-1)}.
If string b does not have any consecutive zeros, then neither does string b1.
Since B(n-1) is contained in C by our hypothesis, {b1|b in B(n-1)} is contained in C.
If string b does not have any consecutive zeros, then neither does string b10.
Since B(n-1) is contained in C by our induction hypothesis, {b10|b in B(n-1)} is contained in C. Therefore, Bn is contained in C;
that is, Q(n) is true. This completes the inductive step.

Therefore B = C.

#23 Re: Help Me ! » Recursion Problems » 2012-07-08 14:32:37

Hi bobbym and bob bundy,

bobbym: can you share the link to proff by induction?
bob bundy: the OP asked for f(2n) = f(n+1)^2 - f(n-1)^2, your proof proves f(2n) = f(n)^2 + f(n-1)^2. I am guessing there is some algebraic manipulation to get to the identity given by OP?

#24 Re: Help Me ! » Discrete Math Advanced Counting Problem » 2012-07-04 04:15:53

Lets say we have a set A containing 600 elements  = {7,9,11,7,9,11,......,11}
and another way of describing that se would be = {l,r,l,r,l,r,l,r,l.....l} where l and r are right and left shoes respectively. Then if we map them 1 to 1 we have exactly 100 pairs of size 7 shoes, size 9 shoes and size 11 shoes. This means we have a total of 300 correct pairs if arranged this way. Would this work?

#25 Re: Help Me ! » Discrete Math Advanced Counting Problem » 2012-07-04 04:12:42

Yeah it is, I cant find a good way to have a union of those sets.

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