6

[EarlyMon]

Well I disagree with your disagreement! So there!

edit: to explain myself a little better, no, i'm not saying f(x) = 2*x. In fact, I'm saying that's the problem-- OTHER people are saying that, but it's not true. f(x) is a function, that is f of x (f is a function of x), meaning the variable f is applied to the variable x. In our equation, we would apply 2 to 3, which in effect is multiplying it... but it's not the same as saying f(x) = f*x (though isolated, those equations are redundant). The problem is that USUALLY creating a function simply means multiplying it, so we've been trained to think that 2(3) = 2*3. But it isn't.

Um. f(x) is literally function of x - and does not mean the variable f is applied to x.

How about functions of the form -

where the last sum is simply the first sum rewritten using the definitions ξn = n/T, and Δξ = (n + 1)/T − n/T = 1/T.

f(x) is a function, that is f of x (f is a function of x), meaning the variable f is applied to the variable x.

No. In the above example, there is no variable f that is applied to variable x.

The f in f(x) is simply a placeholder - for the operations that are applied to x.

we've been trained to think that 2(3) = 2*3. But it isn't.

We've been trained to think that way for a reason. Commutation - it comes from commutation notation and means 2*3.

Honest.

So, by direct evaluation, the equation must become 6/2*3 - as had been said before.

 

[330D]

Um. f(x) is literally function of x - and does not mean the variable f is applied to x.

How about functions of the form -

where the last sum is simply the first sum rewritten using the definitions ξn = n/T, and Δξ = (n + 1)/T − n/T = 1/T.No. In the above example, there is no variable f that is applied to variable x.

The f in f(x) is simply a placeholder - for the operations that are applied to x.

We've been trained to think that way for a reason. Commutation - it comes from commutation notation and means 2*3.

Honest.

So, by direct evaluation, the equation must become 6/2*3 - as had been said before.

This post makes me realize how much I dislike math. (That is math, right?)

 

[EarlyMon]

It's a common approximation(*) of the Fourier transform, used to estimate frequency components in time series data.

It was an Easter egg for Master Po, who may have missed implied time when I said rhythm.

(*)The even more common one is to use Euler's and just directly evaluate from there, but it was just easier to copy from wikipedia than spell out the eq'n. A great deal of what I've done and published started with that eq'n. (Stated for Po's benefit, just to beat him to that punch. )

Ok Bob - once again 1,1,1,2,__ or 1,1,2,1,3,1,4,__

I guess time is running out.

 

[sonofaresiii]

Um. f(x) is literally function of x - and does not mean the variable f is applied to x.

How about functions of the form -

where the last sum is simply the first sum rewritten using the definitions ξn = n/T, and Δξ = (n + 1)/T − n/T = 1/T.No. In the above example, there is no variable f that is applied to variable x.

The f in f(x) is simply a placeholder - for the operations that are applied to x.

We've been trained to think that way for a reason. Commutation - it comes from commutation notation and means 2*3.

Honest.

So, by direct evaluation, the equation must become 6/2*3 - as had been said before.

well, fine.

 

[Member243850]

I GOT IT I GOT I GOT!! xD

I remeber now!!!

It is a quadratic sequence!

I should have known this! I was the only one in my class to get it out of every body!!!

Mr Lars was very proud of me!

It is comming back to me now!

T1 = ax^2 + bx + c Okay???? You get that???

GOOD

NOW

T1 = 1(1)^2 + 1(b) + c

but T1 = 1

SO

1 = 1 + b + c

b = - c

NOW

T5 = 1(2)^2 + 2b + c

but T5 = 2

therefore

2 = 4 + 2b + c

therefore

2b = -2 -c

b = -1 - c/2

OKAY now we try simultanous equations!!!

We know that:

b = -1 - c/2

AND

we know that:

b = -c

SO therfore

- c = - 1 - c/2 (that is c divided by 2)

Therefore

-2c = -2 - c

-c = -2

THERFORE

c = 2

NOW we go back to the beginning xP

1 = 1 + b+ c

THERFORE

b = -2

NOW we just do the rest

T6 = 1(6) squared -2 = 2


THE ANSWER IS

ONE

 

[alostpacket]

42

 

[alostpacket]

It's funny how people used a mnemonic to justify an answer.

That's like saying Math is the way it is because I tied this string around my finger

Multiplication and division are equal im PEMDAS it's actually P E MD AS. It could have just as easily be written PEDMSA.

The only argument for the answer being 1 is if you interpret the 2 being a function of the parenthesis. Since it's genreally considered shorthand for multiplication and not a function the answer is 9. (Unless your talking computer science or a different type of math like calculus where x and * dont mean the same thing anymore). Basically what EarlyMon said.

Still, my answer is 42.

 

[alostpacket]

Also, here's the answer Google gave me

If we claim to have fellowship with him and yet walk in the darkness, we lie and do not live out the truth. But if we walk in the light, as he is in.

 

[Gnomad]

I figured I better get my vote in the poll since the correct answer is down by three votes.

 

[alostpacket]

Well if I write 6/2(1+2) in Java for Android I get:

6/2(1+2) 
Syntax error on token "2", * expected after this token.

So, that settles......something

 

[DaSchmarotzer]

Calculations:

6

 

[Bomfy]

All I know is that the answer I selected will undoubtedly be the wrong answer.

 

[EarlyMon]

(a+b)(a-b) = b(a-b)____________(factorization)
____(a+b) = b_________________(simplification)

The simplification step required dividing both sides by zero.

 

[DaSchmarotzer]

The simplification step required dividing both sides by zero.

Of course, but you'd be surprised at the number of educated people who will be stumped. Maybe not the ones used to work specifically with Mathematics though.

For example, the CEO of a consulting engineering firm that I know didn't find it.

 

[Benjaguy]

Welcome to one of the easiest algebraic equations in the world!... or is it? MUAHAHAHAHA!

Can you solve this pesky little problem while also giving damning evidence that the other 2 can not possibly work?

YOU MUST CHOOSE ONE AND ONLY ONE!

Calculations:

6

 

[EarlyMon]

Of course, but you'd be surprised at the number of educated people who will be stumped. Maybe not the ones used to work specifically with Mathematics though.

For example, the CEO of a consulting engineering firm that I know didn't find it.

And you didn't take the opportunity to show him that dividing by zero will double a number and is therefore perfectly valid?!?

Get on his payroll - you'll be set for life in the first month!

 

[novox77]

I answered '1.'

It was a Malcolm Gladwell "Blink" answer. I didn't ponder my decision at all. Doesn't make my answer correct, but it does prevent me from getting sucked into a self debate. Here's my logic:

We make the equation a bit more algebraic without changing the operators and parens:

c/2(a+b) where a=1, b=2, and c=6

If you assume c/2 comes after evaluating (a+b), you are essentially putting (a+b) in the numerator of your division:

c/2 * (a+b)/1 --> c(a+b)/2

So if the equation was written as c(a+b)/2 ((a+b) is in the numerator), your final answer is 9. But the equation was definitely not written in this way. So 9 is wrong.

As the original equation was written, (a+b) is clearly in the denominator because I proved above it can't be in the numerator without a drastic re-write of the equation. In other words, c/2(a+b) can only imply that (a+b) is in the denominator without re-writing the equation. Therefore, as the original equation was written, the answer is the following:

eval parens first: (a+b) = 3
3 is part of the denominator, so 6/2(3) = 6/6 = 1

 

[alostpacket]

If you assume c/2 comes after evaluating (a+b), you are essentially putting (a+b) in the numerator of your division:

Yes

c/2 * (a+b)/1 --> c(a+b)/2

EDIT: quoted wrong part should have used this:

c/2(a+b) can only imply that (a+b) is in the denominator without re-writing the equation.

As soon as you've done this part of your calculation you've assumed the order of operations incorrectly.



Think about it this way, 6 over 2 times 3. This is still written correctly, right?


6
__ (3) = 18/2 = 9
2


You would never multiply the (1+2) to the denominator, that would be like buying an iPhone.

 

[adi19956]

WolframAlpha says it's 9? it's 9. http://www.wolframalpha.com/input/?i=6/2(1+2)

 

[novox77]

Yes



As soon as you've done this part of your calculation you've assumed the order of operations incorrectly.

Think about it this way, 6 over 2 times 3. This is still written correctly, right?


6
__ (3) = 18/2 = 9
2


You would never multiply the (1+2) to the denominator, that would be like buying an iPhone.

you totally misread what I posted. Cause the part you quoted and said was incorrect is exactly how you would arrive at "9."

 

[alostpacket]

you totally misread what I posted. Cause the part you quoted and said was incorrect is exactly how you would arrive at "9."

Ah yer right sorry, but it was a mistake of quoting the wrong part of your post, not misreading what you said.

Your proof is still wrong though because it rests on the "because it was written this way, I can perform the order of operations this way"


This is what I should have put in the second quote:

c/2(a+b) can only imply that (a+b) is in the denominator without re-writing the equation.

That's where you make an assumption and rewrite the equation yourself.

You're basically saying because there are parenthesis, you can do the multiplication of 2 * (a+b) before c/2 even though that's not how the order of operations work. After the parenthesis, you're supposed to go back to division/multiplication and left->right. x is just another way of writing x*y, it doesnt change the rules

 

[jroc]

The more I read this thread....the more it looks like its time for me to go back to school...lol.

But it is an interesting read. If I didnt post in here, this was gonna be one of my "subscribe to and just read" threads.

And where in the world is Vihzel!?!?!?

 

[sonofaresiii]

I'm sticking with 1. Not because it's right, but because I don't like being wrong. What's that? You still say I'm wrong? LALALA I CAN'T HEAR YOU.

I will go invent a time machine and punch the Egyptians in the head until they re-write math to make me right.

How will I invent a time machine? Well, with my new-math that I invent, it'll be easy!

 

[novox77]

You're basically saying because there are parenthesis, you can do the multiplication of 2 * (a+b) before c/2 even though that's not how the order of operations work. After the parenthesis, you're supposed to go back to division/multiplication and left->right. x is just another way of writing x*y, it doesnt change the rules

Well there's debate on how order of operations work as it pertains to M and D. But I was trying to eliminate that debate entirely by focusing on the OP's statement in one of his posts:

You have to make a decision simply based on the equation given in the title

And in the title, it was written as: 6

 

[Vihzel]

Oh I'm here. I am just enjoying reading all of these fine posts.