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Cześć!

Well, one sure way to get a definitive result is to go through all the possibilities, which are not many, since you just have to choose which operation gets performed before which. 4 operations means there are 4!=24 different orders to do the operations in. So, for example, you could label each operation with a number 1 through 4, then list out all the permutations of the sequence 1,2,3,4 and perform the operations in the order signified by each permutation. E.g., 3241 would mean you'd group them like 5-((2*(6-4))+2).

Hi Bob,

You can superscript the "\prime" to have it look like normal, like so:

Hmm, out Intro to NA professor mentioned that trick with the example of the integral

It's quite brilliant.

Also, yes, if you ran it to a[0], you would get an approximation of ln(9/8). This is also, I think, the only value of a[0] for which the sequence a[n] converges.

bobbym wrote:

Here I suggest you use your own head.

f[x_] := x^6 + x^5 - 5 x^4 - 4 x^3 + 6 x^2 + 3 x - 1

x =0.

x = x - f[x]/f'[x]

Well, of course, Newton's method has a number of conditions to be met so that it would converge.

Also, isolating the zeroes might not be a bad idea.

Agnishom wrote:

Something like this?

I do not understand. The NewtonRaphson function does not take f as an argument in his code.

It uses f as a global variable, assigned a value outside of the subroutine, so, to call it for a function, you'd have to set f to that function before calling NewtonRaphson. I assume it works without much problem, but that is a silly thing to do...

I try to stay away from argumentation of any sort since I've found they are more often fruitless than not.

To say that an ordered pair (x,y) satisfies the equations given in post #1 means that, when we substitute the values, we get a true equality. If you try to reduce the equations the way you did it, you cannot guarantee that the solutions are the same. It'd be like saying that the equation

where represents the function that maps the number x to its principal square root, has the solution 1 because squaring it you get the equation x=1, which the number 1 does satisfy.

That's not really how equations work.

What Nehushtan wrote in post #2 is the correct solution.

Grantingriver wrote:

Therefore x=0 and y=∞.Q.E.F

This is obviously wrong, but I cannot decide if purposefully so...

Well, the first thing is to notice a pattern and conclude that you get maximum values of g when it satisfies the following:

With the initial conditions of and .

From here, we can see that if n is the first number for which g(n)=m, then 3n-1 is the first number for which g(3n-1)=m+1. So, if I call the first number for which , then

Now you can solve this recurrence and see what you get for m=2015.

Hej alla!

I am getting

by two methods, no less.

Why should g(225)=g(3)?

Hej!

g:N->N means that g maps natural numbers into natural numbers.

And

means that the images of n and n+1 differ by one. For example, if g(13)=3, then g(14) can be either 2 or 4.

Hej;

All of the numbers given there are correct (which would, I guess, further mean that it is not and ordering of the areas).

Grantingriver wrote:

Since the rational numbers are equivalent calsses so any members of a certain class can represent that class, but numbers of infinite repetition are not rational numbers ,in fact, this is one of the reasons that lead to the extension of the rational numbers into the real numbers.

Wait what? Am I misunderstanding or do you also believe that 1/3≠0.(3)...?

Also, I see no inconsistencies in defining 0.(9) to be equal to 1, and thus see no argument against such a definition... What's more, defining 0.(9) to be anything other than 1 would be inconsistent, cause then the difference 1-0.(9) would be nonnegative, but also less than 1/n, for every natural number n, which would make it *have* to be 0. So, it's either that, or making it invalid notation.

Grantingriver wrote:

This question is not impossble!! You only have to be more creative to solve this type of problems. For example, your question did not include any information about the bases of the numbers in the list, so if you choose to take the base to be tridecimal or tetradecimal then 71 will be equivelant to 92 and 99 in the decimal base respectively, and this renders the problem to a trivial one since all the numbers in that list using these bases are more then 100 except "71" also this answer can not be rejected since the base of the numbers in the list is not restricted in the question. Therefore the answer is "71"!!!

What if I take the answer C to be in base 8 and the rest in base 20, though? I think the answer might be C!

Hej!

This topic has been discussed many times here and you'll get different opinions on it. Since opinions should not affect absolute truth, the solution is to either take 0.(9)=1 by convention, or just define decimal represntation so that 0.(9) is not a possible decimal representation, which is the variant I support.

Hej, Bernard!

I am failing to see how the "hidden" triangle is different from just the Pascal's triangle multiplied by 1001 and inserted somewhere in-between the numbers of a regular Pascal triangle?

Hej, evene

That is indeed correct, but I feel the approach given on MIF might be a bit more satisfying, as stating that the roots of a quadratic equation are this and that just makes you wonder about how we got that formula, which in turn brings you to a variation of the explanation over there.

Hej, anna!

For v2 permutations, you need to determine which two people will get the same question, which can be done in 10 ways, and then which question each person will get, which can be done in 5*4*3*2 ways. Similar logic for v3.

Okay, so I have discovered that my code for the optional backwards step is not correct, despite giving the correct result (it fails on lower numbers, etc.).

I /have/ written new code for this one, which is the same principle, basically...

I notice Mathematica can't actually expand around negative infinity.

I am not sure what you even tried to do.

See ya around! And thanks!