Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 Re: Help Me ! » Recurrent relation » 2006-01-02 23:00:06

Bit late, but thanks anyway. wink

#2 Re: Help Me ! » Rounding .5 up or down? » 2006-01-02 22:55:58

I was always taught that .5 rounded up, and I've never heard anyone disagree with that up to now. hmm

#3 Re: Help Me ! » Recurrent relation » 2005-12-21 09:30:24

God wrote:

I really didn't get what you were trying to do with that recursive equation sad or else I'd help

Solve it?

It's a second order linear homogemous recurrent relation with constant coefficients.

I already found a general solution to them (the CE), but it's missing two constants, which can be filled in. The bottom half is me attempting to do that, and failing. hmm

#4 Help Me ! » Recurrent relation » 2005-12-21 02:56:47

dublet
Replies: 5




Characteristic equation:






Check:

hmm

Okay, I'm doing something wrong here. hmm

Simple way won't work, using ABC formula:





So the worked out CE is:


Anyone care to calculate c1 and c2? big_smile







Is it getting ugly yet? hmm

Using the approximations:





Help?

#7 Re: Help Me ! » Graphs Please help » 2005-12-17 14:48:49

kcleigh2 wrote:

lol anything specific lol all of it really lol you know when you just cant seem to get you head around something no matter how you look at it. Well this is one of them things lol . i am so rubish at maths it is unbelieveable lol sorry for being a pain.

A graph is a plot of a mathematical function. If you have the function f(x) = x, then calling the function for 0..100 will give back 0..100. This makes for a linear function.
If you have to draw a graph from 0 to 100 values of x, you just make the first and second point (because it's a linear function), which has the begin point of (0, f(0)) = (0, 0) and the end point (100, f(100)) = (100, 100).

Now you know the range of both axes, which has to be from 0 to 100, if you use a centimeter for each 10, you can make it 10 by 10 centimetres big. You then put the begin and end point in the area, and draw a direct line.

If the function is f(x) = 2 and you need to make the graph contain all solutions from 0 to 50, then the begin points are (0, 2) and (50, 2). Again you place the two points, and draw a direct line.

You do know how a function works, don't you?

Do you also know about coordinates?

Cartesiancoordinates2D.JPG

Wikipedia wrote:

In the two-dimensional Cartesian coordinate system, a point P in the xy-plane is represent by a tuple of two components (x,y).

    * x is the signed distance from the y-axis to the point P, and
    * y is the signed distance from the x-axis to the point P.

#8 Re: Help Me ! » Graphs Please help » 2005-12-17 08:53:08

Simpler? hmm

Is there anything in specific you don't understand?

http://en.wikipedia.org/wiki/Graph_of_a_function

#9 Re: Help Me ! » Graphs Please help » 2005-12-16 13:33:36

John E. Franklin wrote:

£ and p.  How do you pronounce these?  Pound? Cent?  I am an ignorant New Englander over the Atlantic.

Pound and penny. smile

#10 Re: Help Me ! » Graphs Please help » 2005-12-16 13:24:57

First to draw the graph, you have to figure you the formula.

It says that the bill is £18 as starting point, and then for every unit, it's 40p more.

So the function is

, where x is the number of units, and f(x) gives you the cost in pennies. The result of this, you can call y, thus y = f(x). Now you know both the x and y axes of your graph.

After that you have to find the range of both axes. you know that for x it is between 0 and a maximum of 250 units. Now you can calculate the range for y, which then goes from f(0) to f(250).

Once you have calculated the results of both of those, you need to lay it out on paper, and you can then put two points on the graph: (0, f(0)) and (250, f(250)) (written as (x, y) coordinate pairs). If you have drawn your graph accurately, you should now easily be able to read the results.

For the second part, you have to make another function: g(x), which calculates the cost for water company B. This is essentially the same as formula f(x), except with different values. First, since there is no starting charge, there's nothing to add, so the function is just

.

Now that you know this, you can also calculate g(0) and g(250), in the same way as for the f(x) function. Then plot the values on the graph, and make a direct line between the two.

I hope this is enough help, any further, and I'll be giving you the solution. wink

#12 Re: Help Me ! » Solving infinite series » 2005-12-16 09:08:28

mathsyperson wrote:

There is a standard formula that we can use to help us:

With an infinite series, wouldn't this amount to the following?


And as such, for example:

Or am I missing something? hmm

#13 Re: Help Me ! » Solving infinite series » 2005-12-16 07:55:10

mathsyperson wrote:

I plotted the first few hundred points of a_n, and it seems that the function diverges away from 0 in the negative direction. So, if you were to sum the infinite series, you'd just get - ∞.

The

, as you suggested, indeed.

I think you might have made a mistake with the function, though. As it is, it involves dividing by 0 when n = 1, so that indicates that something might be wrong. Plus, -∞ isn't a very satisfying answer.

This is no mistake, it's exactly what I got for one of my math exams.

The thing which I want to know is how you solve a summation, as all the books I have don't really show me a structured way to do it. hmm

Another summation

Do you just plot a few points to see where it goes, and then you see if it goes to infinity, 1, 0, -1 or -infinity, or is there a more precise method (which is easy to understand wink)?

#14 Help Me ! » Solving infinite series » 2005-12-16 03:43:58

dublet
Replies: 24

Given the following:


How would you solve the following summation?

#15 Re: Help Me ! » Help Me!! » 2005-12-16 02:45:35

Nestea wrote:

If you went to the bank for a 15 year, $150,000 mortgage at an interest rate of 6.75% pa, compounding quarterly and you wanted to give payments semi-annually how much would each of your payments have to be over the 15 years?

Answer:$8,057.37

You can solve this by a recurrence relation.



Out of this recurrence, you can formulate the following formula

Now for f(15) this is 399585.399585.31053
Following that, the total payments formula can easily be described as

Which leads to p(15) = f(15) - 150000 = 399585.399585.31053 - 150000 = 249585.31053.

Dividing that over 15 * 2 is 30 payments would be 8319.51035 per payment.

The difference is in the fact that my formula doesn't account for the decreasing of the overall amount by your payment. Left as exercise for the reader. wink

Board footer

Powered by FluxBB