Some Fun With Maths

[Aethelwulf]

I don't think so. Starting with post #1's

 

[math]\frac{kr}{m} - S + (\frac{5}{ab} - \frac{3}{b^2})= T[/math]

 

I get

 

[math]k = \frac{m(T +S -\frac{5}{ab} +\frac{3}{b^2} )}{r} [/math]

 

Substituting the givens [imath]a=2, b=2, r=20[/imath] and simplifying, that's

 

[math]k = \frac{m(2T +2S -1)}{40}[/math]

 

Evaluating the original equation with sample values for the m, T, S, and k calculated with my result checks out. Using your doesn't. I think you've erred, Aetherwolf.

 

 

No no, you must solve the paranthesis first. Brackets first, always. Then simplify the rest.

[Aethelwulf]

It is a very common rule that what is contained within a paranthesis is solved first in an equation. What happens next in the solving of this equation comes naturally because you end up with a fraction and your unknowns.

[Turtle]

That's not right. the value of [math]a[/math] is what gave other people problems as well. Also, that is not how you simplfy what is inside the paranthesis.

 

The way you divide a number by a fraction is like this

 

[math]6 \frac{2}{\frac{4}{8}} = 6\frac{2}{\frac{8}{4}} = \frac{6 \cdot 2 \cdot 8}{4} = 24[/math]

 

this looks incorrect as written. you must first convert the mixed fraction [math]6 \frac{2}{4}[/math] to the fractional form [math] \frac{26}{4}[/math]. then invert the

denominator, [math] \frac{8}{1}[/math] to [math] \frac{1}{8}[/math] and then multiply the

numerator by the inverted denominator [math] \frac{26}{4}[/math]*[math] \frac{1}{8}[/math] = [math] \frac{26}{32}[/math] = [math] \frac{13}{16}[/math]

 

So when you divide a number by a fraction, you have to flip the fraction around. So let's go back to our example

 

[math]a = \sqrt{\frac{4 \cdot 154}{\frac{22}{7}}}[/math]

 

swap the fraction round

 

[math]a = \sqrt{\frac{4 \cdot 154}{\frac{7}{22}}}[/math]

 

calculate

 

[math]4 \cdot 154 \cdot 7 / 22 = 196[/math]

 

Take the square root of this and it is 14.

 

i agree that to divide a fraction by a fraction, you invert the denominator and multiply by the numerator. "flip the fraction around" is inadequate terminology for the operation imho.

 

so ...

[math]a = \sqrt{\frac{4 \cdot 154}{\frac{22}{7}}}[/math]

 

[math]a = \sqrt{\frac{\frac{616}{1}}{\frac{22}{7}}}[/math]

 

[math]a = \sqrt{\frac{616}{1}}*{\frac{7}{22}}[/math]

 

[math]a = \sqrt{\frac{4312}{22}}[/math]

 

[math]a = \sqrt{196}[/math]

 

[math]a = 14[/math]

 

so, i agree with you on this result.

 

i haven't plugged that in to solve for k, but will check with all due haste.

Edited December 11, 2012 by Turtle

[Turtle]

here's a problem for you wulfy et al. :dog: i haven't worked it out yet myself, but i need it & will get to it soon enough. in the mean time it's fun with maths. :xparty:

 

for the following ordered list, write a generating expression in 1 variable x, indexed for x={1,2,3 ...}.

 

{28,52,84,124,172,228,...}

 

:smart:

[sanctus]

Wulf, read well through my post it changes where you put the main fraction and you are not coherent with it...to obtain your result you have the main fraction elsewhere than in the opening post.

[CraigD]

No no, you must solve the paranthesis first. Brackets first, always. Then simplify the rest.

No, addition is associative, so the equation

[math] \frac{kr}{m} - S + (\frac{5}{ab} - \frac{3}{b^2})= T [/math]

may validly be rewritten

[math] \frac{kr}{m} - S + \frac{5}{ab} - \frac{3}{b^2}= T [/math]

or any other grouping, such as

[math] \frac{kr}{m} +( -S + \frac{5}{ab}) - \frac{3}{b^2}= T [/math]

 

Aethelwolf, your solution (which just means a desired rearrangement),

[math]k = m(T + S) -1[/math]

is simply wrong for the initial equation you gave.

 

As with any arithmetic equation, this can be show by evaluating the initial and candidate solution equations with the same values for each variable. For example, evaluating one side the initial equation for

[imath]a=14, b=2, r=20, k=3, m=4, S=5[/imath]

gives

[math]\frac{3 \cdot 20}{4} - 5 + (\frac{5}{2 \cdot 14} - \frac{3}{2^2}) = \frac{66}{7}[/math]

so chosing

[imath]T=\frac{66}{7}[/imath]

gives a collection of values for the variables for which the initial equation is true. Any correct rearrangement for the equation must be true for the same values.

 

Candidate solution

[math] k = \frac{m(7T +7S +4)}{140} [/math]

evaluates, via

[math] 3 = \frac{4(7 \cdot \frac{66}{7} +7 \cdot 5 +4)}{140} [/math]

to

[math]3 =3 [/math]

which is true. :thumbs_up We can't say with certainty from this check the solution is correct, but the check does not show it to be incorrect. (Though, because it involved only simple algebra I rarely make mistakes with, I'm confident it's correct)

 

Candidate solution

[math]k = m(T + S) -1[/math]

evaluates via

[math]3 = 4(\frac{66}{7} + 5) -1[/math]

to

[math]3 = \frac{397}{7}[/math]

which is false. :thumbs_do We can say with certainty from this check that the solution is incorrect.

 

You must have made a mistake in you algebra. :( If you show your steps, we can show you where you did. :)

Edited December 12, 2012 by CraigD
Replaces incorrect given a=2 with correct a=14

[Aethelwulf]

No, addition is associative, so the equation

[math] \frac{kr}{m} - S + (\frac{5}{ab} - \frac{3}{b^2})= T [/math]

may validly be rewritten

[math] \frac{kr}{m} - S + \frac{5}{ab} - \frac{3}{b^2}= T [/math]

or any other grouping, such as

[math] \frac{kr}{m} +( -S + \frac{5}{ab}) - \frac{3}{b^2}= T [/math]

 

Aethelwolf, your solution (which just means a desired rearrangement),

is simply wrong for the initial equation you gave.

 

As with any arithmetic equation, this can be show by evaluating the initial and candidate solution equations with the same values for each variable. For example, evaluating one side the initial equation for

[imath]a=2, b=2, r=20, k=3, m=4, S=5[/imath]

gives

[math]\frac{3 \cdot 20}{4} - 5 + (\frac{5}{2 \cdot 2} - \frac{3}{2^2}) = \frac{21}{2}[/math]

so chosing

[imath]T=\frac{21}{2}[/imath]

gives a collection of values for the variables for which the initial equation is true. Any correct rearrangement for the equation must be true for the same values.

 

Candidate solution

[math] k = \frac{m(2T +2S -1)}{40} [/math]

evaluates, via

[math] 3 = \frac{4(2 \cdot \frac{21}{2} +2 \cdot 5 -1)}{40} [/math]

to

[math]3 =3 [/math]

which is true. :thumbs_up We can't say with certainty from this check the solution is correct, but the check does not show it to be incorrect. (Though, because it involved only simple algebra I rarely make mistakes with, I'm confident it's correct)

 

Candidate solution

[math]k = m(T + S) -1[/math]

evaluates via

[math]3 = 4(\frac{21}{2} + 5) -1[/math]

to

[math]3 = 61[/math]

which is false. :thumbs_do We can say with certainty from this check that the solution is incorrect.

 

You must have made a mistake in you algebra. :( If you show your steps, we can show you where you did. :)

 

I have no idea how you get any of this. You are fantastically wrong. If you just simplify the fractions like I showed you how, plugged in the values for the knowns, then end up with a fraction of numbers simplifying to 1, the equation simplifies itself nice and sweet.

 

 

I have no idea, where these calculations of yours are coming from, but you are making overly difficult by sure.

[Aethelwulf]

I EVEN tell you in the OP to simplify what is inside the paranthesis. Simplifying fractions is child play.

[Aethelwulf]

here's a problem for you wulfy et al. :dog: i haven't worked it out yet myself, but i need it & will get to it soon enough. in the mean time it's fun with maths. :xparty:

 

for the following ordered list, write a generating expression in 1 variable x, indexed for x={1,2,3 ...}.

 

{28,52,84,124,172,228,...}

 

:smart:

 

 

A generating function with x increasing as (1,2,3...n), so somthing like

 

[math]g(x) = \sum_{x = (1,2,3..n. )}^{\infty} a_n \cdot x[/math]

 

Which isn't your ordinary generating function since x is not squared. If you can find an easy solution to this, good for you. Right now I don't see one.

[Aethelwulf]

When you have a fraction of the form

 

[math](\frac{5}{ab} - \frac{3}{b^2})[/math]

 

And in an initial problem asked to simplify it, you really should simplify, not plug in numbers like Craig keeps doing, and leaving it in this form. To simplify it, one gets

 

[math](\frac{5}{ab} - \frac{3}{b^2}) = \frac{5b - 3a}{ab^2}[/math]

 

This is not a ''desired'' form as such, it is really the basics of simplifying fractions. Algebra even.

[Aethelwulf]

this looks incorrect as written. you must first convert the mixed fraction [math]6 \frac{2}{4}[/math] to the fractional form [math] \frac{26}{4}[/math]. then invert the

denominator, [math] \frac{8}{1}[/math] to [math] \frac{1}{8}[/math] and then multiply the

numerator by the inverted denominator [math] \frac{26}{4}[/math]*[math] \frac{1}{8}[/math] = [math] \frac{26}{32}[/math] = [math] \frac{13}{16}[/math]

 

 

 

 

 

You made a mistake here. There is nothing ''mixed'' about this fraction, we are dividing a whole number by a fraction, the same rule as before.

[Aethelwulf]

Look at my example again:

 

[math]6 \frac{2}{\frac{4}{8}} = 6\frac{2}{\frac{8}{4}} = \frac{6 \cdot 2 \cdot 8}{4} = 24[/math]

 

check this example with the original OP example

 

[math]a = \sqrt{\frac{4 \cdot 154}{\frac{22}{7}}}[/math]

 

They are identical, you can't say me solving the latter here is correct but then claim the other one is wrong.

[Turtle]

A generating function with x increasing as (1,2,3...n), so somthing like

 

[math]g(x) = \sum_{x = (1,2,3..n. )}^{\infty} a_n \cdot x[/math]

 

Which isn't your ordinary generating function since x is not squared. If you can find an easy solution to this, good for you. Right now I don't see one.

 

did you write "since x is not squared" because you didn't square x, or did you think i meant you weren't allowed to square x? do whatever you like to x by all means. :smart:

Edited December 12, 2012 by Turtle

[Turtle]

Look at my example again:

 

[math]6 \frac{2}{\frac{4}{8}} = 6\frac{2}{\frac{8}{4}} = \frac{6 \cdot 2 \cdot 8}{4} = 24[/math]

 

check this example with the original OP example

 

[math]a = \sqrt{\frac{4 \cdot 154}{\frac{22}{7}}}[/math]

 

They are identical, you can't say me solving the latter here is correct but then claim the other one is wrong.

 

whatever you intend, [math]6 \frac{2}{\frac{4}{8}} = 6\frac{2}{\frac{8}{4}} = \frac{6 \cdot 2 \cdot 8}{4} = 24[/math] does not read well; it is ambiguous in the first form whether 6 is a multiplier or part of a mixed fraction. don't worry yourself further by any means. :shrug:

Edited December 12, 2012 by Turtle

[Aethelwulf]

did you write "since x is not squared" because you didn't square x, or did you think i meant you weren't allowed to square x? do whatever you like to x by all means. :smart:

 

 

Have you solved it yet?

[Turtle]

Have you solved it yet?

 

just did. you?

 

Edit: d'oh! :doh: forgot to post it. no peeky 'til you have a solution! :naughty:

 

 

4x²+12x+12

 

Edited December 12, 2012 by Turtle

[Aethelwulf]

I don't think I have understood your problem.

 

You state that the way to solve it is by stating

 

4x2+12x+12

 

to get these numbers

 

{28,52,84,124,172,228,...}

 

Your problem suggests that it is an increasing function?