Some Fun With Maths

[Turtle]

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?

 

yes; the problem was to find a function in one variable that generates the given list. the list itself is an artifact of a particular cellular array and it was found by inspection.

 

in the context i am using it, it is a strictly increasing function because x is always positive. outside of my context and in the general sense, 4x²+12x+12 is [only] an increasing function as it graphs as a parabola when x is allowed to take on negative values.

 

was that fun, or what!? :xparty:

[sanctus]

Re-read post 15 then.

 

You always say that [math]\frac{x}{\frac{y}{z}}\equiv \frac{\frac{x}{y}}{z}[/math] just develop it and you see it is wrong:

 

Left side:

[math]\frac{x}{\frac{y}{z}}= x\cdot \frac{z}{y}[/math]

Right side:

 

[math]\frac{\frac{x}{y}}{z}= \frac{x}{y}\cdot \frac{1}{z}[/math]

 

Put some numbers if you do not agree, say x=2,y=2,z=2:

Left side=2

Rightside=0.5

 

And if you say you do not mix stuff:

 

Fraction from your opening post:

 

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

 

Fraction in your post where you say we are wrong (post #9 of the thread):

 

 

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

 

I have to admit that if you can not see this difference and accept that your solution is for a different equation than the one you posted in the OP I see no point in keeping this discussion up, sorry.

[CraigD]

... the givens [imath]a=2, b=2, r=20[/imath] ...

:doh: Oops! I erred on the trick calculation

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

Which evaluates not to 2, but, as Awthelwulf points out, to 14.

 

So my

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

solution should be

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

 

I’ve corrected my previous 2 posts. Apologies for any confusion I caused.

 

I have no idea how you get any of this.

I used simple algebra, as follows:

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

Add [imath] S - \frac{5}{ab} + \frac{3}{b^2}[/imath] to each side of the equation

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

Multiply each side of the equation by [imath]m[/imath]

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

Divide both sides of the equation by [imath]r[/imath]

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

Replace [imath]b[/imath] with [imath]2[/imath]

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

Replace the [imath]r[/imath] with [imath]20[/imath]

[math] k = \frac{m \left(T + S - \frac{5}{2a} + \frac{3}{2^2} \right)}{20} [/math]

Replace a with [imath] a = \sqrt{\frac{\frac{4 \cdot 154}{22}}{7}} = 14 [/imath]

[math] k = \frac{m \left(T + S - \frac{5}{2 \cdot 14} + \frac{3}{2^2} \right)}{20} [/math]

Simplify

[math] k = \frac{m \left(T + S - \frac{5}{28} + \frac{3}{4} \right)}{20} [/math]

[math] k = \frac{m \left(T + S - \frac{5}{28} + \frac{21}{28} \right)}{20} [/math]

[math] k = \frac{m \left(T + S - \frac{4}{7}\right)}{20} [/math]

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

 

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.

But as I showed my (now corrected) post#23, your solution,

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

, while “nice and sweet”, is wrong.

 

I can help you understand your mistake, if you will show the steps you used to reach your incorrect solution, Aethelwulf.

[sanctus]

Wow, step by step then:

 

OP (which evaluates to 2):

[math]\sqrt{\frac{\frac{4\cdot 154}{22}}{7}}=\sqrt{\frac{\frac{616}{22}}{7}}=\sqrt{\frac{28}{7}}=\sqrt{4}=2[/math]

 

Which is the standard algebra approach, i.e. solve the nominator of the main fraction then solve the main fraction then the square-root.

Or if you want to do it with this "swap-fractions around" you get:

[math]\sqrt{\frac{\frac{4\cdot 154}{22}}{7}}=\sqrt{\frac{\frac{4\cdot 154}{22}}{\frac{7}{1}}}=\sqrt{\frac{4\cdot 154}{22}\cdot{\frac{1}{7}}}=\sqrt{\frac{616}{22}\cdot\frac{1}{7}}=\sqrt{28\cdot\frac{1}{7}}=\sqrt{\frac{28}{7}}=\sqrt{4}=2[/math]

 

To get 14 like stated in post #9 and following you start from the following squareroot:

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

which is different from the one in the OP!!!!!!!!!

 

Developing you get

[math]\sqrt{\frac{4\cdot 154}{\frac{22}{7}}}=\sqrt{\frac{616}{\frac{22}{7}}}=\sqrt{616\cdot \frac{7}{22}}=\sqrt{\frac{4312}{22}}=\sqrt{196}=14[/math]

 

This should prove 2 things

1) It is important where the main fraction is, it gives different results

2) If the orginal equation is like the OP-one then the square-root evaluates to 2, if in the OP wulf did a typo and he means it should be like in post #9 then he is right.

[Aethelwulf]

Wow, step by step then:

 

OP (which evaluates to 2):

[math]\sqrt{\frac{\frac{4\cdot 154}{22}}{7}}=\sqrt{\frac{\frac{616}{22}}{7}}=\sqrt{\frac{28}{7}}=\sqrt{4}=2[/math]

 

Which is the standard algebra approach, i.e. solve the nominator of the main fraction then solve the main fraction then the square-root.

Or if you want to do it with this "swap-fractions around" you get:

[math]\sqrt{\frac{\frac{4\cdot 154}{22}}{7}}=\sqrt{\frac{\frac{4\cdot 154}{22}}{\frac{7}{1}}}=\sqrt{\frac{4\cdot 154}{22}\cdot{\frac{1}{7}}}=\sqrt{\frac{616}{22}\cdot\frac{1}{7}}=\sqrt{28\cdot\frac{1}{7}}=\sqrt{\frac{28}{7}}=\sqrt{4}=2[/math]

 

To get 14 like stated in post #9 and following you start from the following squareroot:

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

which is different from the one in the OP!!!!!!!!!

 

Developing you get

[math]\sqrt{\frac{4\cdot 154}{\frac{22}{7}}}=\sqrt{\frac{616}{\frac{22}{7}}}=\sqrt{616\cdot \frac{7}{22}}=\sqrt{\frac{4312}{22}}=\sqrt{196}=14[/math]

 

This should prove 2 things

1) It is important where the main fraction is, it gives different results

2) If the orginal equation is like the OP-one then the square-root evaluates to 2, if in the OP wulf did a typo and he means it should be like in post #9 then he is right.

 

If it makes you feel better, calling it a typo, then very well.

 

I have clearly stated time and time again now, it's a number divided by a fraction.

[Aethelwulf]

:doh: Oops! I erred on the trick calculation

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

Which evaluates not to 2, but, as Awthelwulf points out, to 14.

 

So my

 

solution should be

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

 

I’ve corrected my previous 2 posts. Apologies for any confusion I caused.

 

 

I used simple algebra, as follows:

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

Add [imath] S - \frac{5}{ab} + \frac{3}{b^2}[/imath] to each side of the equation

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

Multiply each side of the equation by [imath]m[/imath]

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

Divide both sides of the equation by [imath]r[/imath]

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

Replace [imath]b[/imath] with [imath]2[/imath]

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

Replace the [imath]r[/imath] with [imath]20[/imath]

[math] k = \frac{m \left(T + S - \frac{5}{2a} + \frac{3}{2^2} \right)}{20} [/math]

Replace a with [imath] a = \sqrt{\frac{\frac{4 \cdot 154}{22}}{7}} = 14 [/imath]

[math] k = \frac{m \left(T + S - \frac{5}{2 \cdot 14} + \frac{3}{2^2} \right)}{20} [/math]

Simplify

[math] k = \frac{m \left(T + S - \frac{5}{28} + \frac{3}{4} \right)}{20} [/math]

[math] k = \frac{m \left(T + S - \frac{5}{28} + \frac{21}{28} \right)}{20} [/math]

[math] k = \frac{m \left(T + S - \frac{4}{7}\right)}{20} [/math]

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

 

 

But as I showed my (now corrected) post#23, your solution,

, while “nice and sweet”, is wrong.

 

I can help you understand your mistake, if you will show the steps you used to reach your incorrect solution, Aethelwulf.

 

 

Craig

 

 

Of course your solution is wrong, I have shown it to be wrong twice now, you are also making no comment on the fact that the original problem asked you to simplify what was inside the paranthesis, may you tell me why each time around you end up with an equation so mangled in the end, it looks nothing like the right solution? Maybe tell me why you are simply refusing to simplify the fractions like I have shown you how?

[Aethelwulf]

I don't want to listen to these deluded attempts at stating I am wrong about something - when I know clearly I am right in the process of simplifying the equation. Standard algebra, posted in the task OP, was to tell you to simplify what was inside the brackets. Each time you do not do it, your equations at attempting to solve the problem get worse and worse.

[Aethelwulf]

yes; the problem was to find a function in one variable that generates the given list. the list itself is an artifact of a particular cellular array and it was found by inspection.

 

in the context i am using it, it is a strictly increasing function because x is always positive. outside of my context and in the general sense, 4x²+12x+12 is [only] an increasing function as it graphs as a parabola when x is allowed to take on negative values.

 

was that fun, or what!? :xparty:

 

 

I still don't see how it generated the value 28... can you give some examples? I could be being a bit dense here, but I'd like to see how you derive everything, like I have done with my own OP example.

[Aethelwulf]

Ok, assuming everyone is comfortable with the answer for the equation in post 1, (with the exception curiously of Craig), I begin the next problem. No one has to solve it, but try and brush up on algebra skills. It is probably only slightly harder than the first, but not by much.

 

 

 

[math]a = \frac{d}{dx}[\frac{1}{5}(3x - 2)^5][/math]

 

[math]b = \frac{\frac{3d}{d-2}}{\frac{9d^2}{7d - 14}}[/math]

 

You need to differentiate the first equation, the second equation requires some algebra. When you have done this, rewrite your equation in the form

 

[math]a + b = e[/math]

 

and then solve for [math]d[/math] (the simplest part)

[CraigD]

Of course your solution is wrong…

No, yours is. I prove this in post #23 that your solution is wrong, and show all the algebra demonstrating mine is right in post #37.

 

My undergraduate degree is Math, and I tutored it for 5 and taught remedial college algebra for a year. Please don’t continue insisting you are right after being shown otherwise – it is preventing you from learning algebra, and is a violation of our site rules :naughty:

 

Wow, step by step then:

 

OP (which evaluates to 2):

[math]\sqrt{\frac{\frac{4\cdot 154}{22}}{7}}=\sqrt{\frac{\frac{616}{22}}{7}}=\sqrt{\frac{28}{7}}=\sqrt{4}=2[/math]

 

…

 

[math]\sqrt{\frac{4\cdot 154}{\frac{22}{7}}}=\sqrt{\frac{616}{\frac{22}{7}}}=\sqrt{616\cdot \frac{7}{22}}=\sqrt{\frac{4312}{22}}=\sqrt{196}=14[/math]

 

This should prove 2 things ...

It’s also a keen demonstration that division is not associative, ; and that explicit association using parenthesis can never hurt, eg:

 

[math]\sqrt{\frac{ \left( \frac{4\cdot 154} {22} \right) }{7}} [/math]

 

[math]\sqrt{ \frac{ \left( 4\cdot 154 \right) }{ \left( \frac{22}{7} \right) }} [/math]

 

;)

[Aethelwulf]

Craig I don't know what you are going on about, I showed the solution

 

4*154/22/7

 

next step

 

616*7/22

 

=14

 

There is really nothing more to this.

 

It is a number divided by a fraction, not a fraction divided by a number. It cannot be explained any more clearer.

 

Also, none of that is consequential towards your own calculations. You don't simplify the fractions straight away is where you err.

[CraigD]

Craig I don't know what you are going on about, I showed the solution

 

4*154/22/7

 

next step

 

616*7/22

 

=14

The value of [imath]a[/imath] isn't important.

 

If it's 14, a solution for k is: [math] k = \frac{m \left(7T + 7S + 4 \right)}{140} [/math]

 

If it's 2, a solution is: [math] k = \frac{m \left(2T + 2S -1 \right)}{40} [/math]

 

in general, assuming no given values, it's: [math] k = \frac{m \left(T + S - \frac{5}{ab} + \frac{3}{b^2} \right)}{r} [/math]

 

In no case, with the other given values, however, is it [math]k = m(T + S) -1[/math]

 

If you gave different given values for [imath]a[/imath] [imath]b[/imath], and [imath]r[/imath] (such as [imath]a=\sqrt2[/imath], [imath]a=\sqrt2[/imath], [imath]r=1[/imath], or in general, [imath]r=1[/imath], [imath]a=\frac{5b}{b^2+3}[/imath] ), that solution would be correct, but it isn't for the values given in post #1.

 

What I was trying to show in post #23 is that you don't need to do anything but pick some values for the other variables that satisfy the original equation, and evaluate each side of your proposed solution to see it's wrong. This isn't an algebraic manipulation, just an arithmetic check. It can't prove a solution is right, but can show it's wrong.

Edited December 14, 2012 by CraigD
Added a general formula for a

[Turtle]

... 4x²+12x+12 ...

I still don't see how it generated the value 28... can you give some examples? I could be being a bit dense here, but I'd like to see how you derive everything, like I have done with my own OP example.

 

examples? sure.

 

4x²+12x+12=

solve for x={1,2,3...}

 

let x= 1:

4*1²+12*1+12=?

4*1+12+12=?

4+12+12=28

 

let x=2:

4*2²+12*2+12=?

4*4+24+12=?

16+24+12=52

 

let x=3:

4*3²+12*3+12=?

4*9+36+12=?

36+36+12=84

...

 

the given set: {28,52,84,124,172,228,...}

they match; problem solved.

 

here's another set i haven't worked up the expression for yet. the expression will be quadratic like the last. ax²+bx+c and it will be fun to find it.

 

#2 {82, 124, 174, 232, 298, 372 ...}

 

edit: append solution spoiler.

 

this one was a little tricky fun. :xparty: 4x²+30x+48

 

Edited December 14, 2012 by Turtle

[sanctus]

If it makes you feel better, calling it a typo, then very well.

 

I have clearly stated time and time again now, it's a number divided by a fraction.

 

I can agree with that :-). Then the square root gives what you say. Next time just check how you wrote down the equation before saying everyone is wrong ;-). Because you do agree that as it was in the OP it gives 2, no?

[sanctus]

Craig I don't know what you are going on about, I showed the solution

 

4*154/22/7

 

next step

 

616*7/22

 

=14

 

There is really nothing more to this.

 

It is a number divided by a fraction, not a fraction divided by a number. It cannot be explained any more clearer.

 

Also, none of that is consequential towards your own calculations. You don't simplify the fractions straight away is where you err.

 

No sorry, how you wirte it it is not conform with what you mean it be, if you want it to be a number divided by a fraction you should write it as:

4*154/(22/7)

 

And a fraction divided by a number is:

(4*154/22)/7

 

which can be simplified to

4*154/22/7

 

but you never have (in standard writing, you might solve things like this on your paper and know it to mean what you want it to):

4*154/22/7 NEVER= 4*154/(22/7)

 

This is all I meant about the main fraction location.

 

In other words, no one is ever guessing that you mean number divided by a fraction if you write 4*154/22/7, since what we learn at school is that we can solve it just sequentially...

 

In other words, I do not criticize your math-skills, but only that you fail to see that you are not following the convention on how to write fractions.

[sanctus]

You both wouldn't have needed to think anything if it had been calculated right. The rule for dividing whole numbers by fractions is well-known.

 

http://www.cimt.plymouth.ac.uk/resources/help/miscon5.pdf

 

An example from this site is

 

[math]3/1/4[/math]

 

The answer they give is 12. Use the rule I gave, remember to swap the fraction about

 

[math]3/4/1[/math]

 

is

 

[math]3 \cdot 4/1 = 12[/math]

 

It is perfectly consistent and nothing wrong with any kind of representation. I did warn you it could be tricky.

 

Ok you are coherent with your convention. But what they actually write on this pdf you linked to is not 3/1/4=12, BUT 3/(1/4)=12 or in latex:

[math]3/\frac{1}{4}=\frac{3}{\frac{1}{4}}=12[/math]

 

3/1/4 is just (in the standard convention, not yours;-)) 3 divided by divided by 4 =3/4. Since writing 3/1/4 there is NO WAY of knowing that you mean 3/4 being a fraction.

 

 

BTW, I agree with Craig, your solution is wrong and it has been shown repetedly now ;-), so far I was only concentrating on the square-root.

[sanctus]

Ok, assuming everyone is comfortable with the answer for the equation in post 1, (with the exception curiously of Craig), I begin the next problem. No one has to solve it, but try and brush up on algebra skills. It is probably only slightly harder than the first, but not by much.

 

 

 

[math]a = \frac{d}{dx}[\frac{1}{5}(3x - 2)^5][/math]

 

[math]b = \frac{\frac{3d}{d-2}}{\frac{9d^2}{7d - 14}}[/math]

 

You need to differentiate the first equation, the second equation requires some algebra. When you have done this, rewrite your equation in the form

 

[math]a + b = e[/math]

 

and then solve for [math]d[/math] (the simplest part)

 

I already know, that with your conventions we will not agree...but:

a=3*(3x-2)^4

b=7/(3*d)

 

d=7/(3e - 9*(3x-2)^4)