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reinout-g

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 #1
avatar+2353 
0
Bertie:

Melody says that things have gone quiet so I am taking the opportunity to share a problem.
I first encountered it as a twelve year old and have just rediscovered it in an old notebook.
I remember that it took me (seemingly) an age to come up with a solution that didn't just
involve trial and error, and I'm wondering now if there are any other methods of solution
different to the one I found.

A man receives a cheque and goes to his local bank to cash it.
The cashier at the bank is having an exceedingly bad day and mistakenly mixes up the
dollars and cents, (pounds and pence in England), giving the man y dollars and x cents
rather than x dollars and y cents. The man doesn't notice the mistake and returns home,
on the way spending 5 cents of the money. (He had no other money with him when he left home.)
Only when he gets home does he check, and he finds that he now has exactly double
the amount on the cheque.
The question is how much was the cheque for ?



Took me a while to figure out, but I've found a way without guessing.

First, let's start by making an equation which suits the problem.


y+1/100x - 5/100 is the amount the man has left at the end of the day.
This is twice the amount of the check and we know the check had a value of x+1/100y
Therefore;

2(x+1/100y) = y + 1/100x - 5/100

We can rewrite this to;

98y-199x= 5

Define x' = -x

then we have

98y+199x' = 5

What we have here is called a linear diophantine equation (since both x and y are integers) and lucky for us it is solvable.

What we want to do is apply the euclidean algorithm which seeks the greatest common divisor.



This iterates into;

199= 98 * 2 + 3
98 = 3 * 32 + 2
3 = 2*1 + 1
2 = 1*2 + 0

where the red number indicated the GCD.

Sadly though, we won't need the GCD :



We do need the steps we made to get there. Pay attention, this is where it is all about to unfold

we rewrite the 3th equation into

1 = 3-2*1

using the second equation we can rewrite this to

1 = 3-1*(98-32*3)

This can be rewritten to

1 = 33*3-98

We can rewrite this using the first equation into

1 = 33*(199-2*98) - 98

Which can be rewritten to

1 = 33* 199 - 66 * 98 - 98

or

1= 33*199 - 67 * 98

so,

5 = 165*199 - 335*98

So a solution to our problem would be (x',y) = (165,-335)

However, since we're looking for values of x' and y within the bounds -100<x'<0 and 0<y<100 (since x'= -x) we want to alter this equation to show values (if they exist) within our bounds.


I can do this in the following way;

5 = (165-98)*199-(335-199)*98

5 = 67*199-136*98

Since i basically add and substract 98*199 the equation is still correct.

Doing this again gives

5 = (67-98)*199-(136-199)*98

And now we have

5 = (-31)*199 +63*98

We therefore know that (x',y) = (-31,63)

Since I defined x' = -x, (x,y) = (31,63)

Hence the amount on the check was $31,63, the man received $63,31 and found out he had $63,26.

Since $63,26 = 2*$31,63, he does indeed have twice the amount on the check after spending 5 cents.






http://mathworld.wolfram.com/DiophantineEquation.html for more information on the algorithm
Apr 5, 2014
 #3
avatar+2353 
+5
Okay, so f(x) = (2x+1) 9(3x-1) 7.

Now we have two functions so we want to apply the product rule.

The product rule works for functions in the following way f(x) = g(x) * h(x)

in this case g(x) = (2x+1) 9 and h(x) = (3x-1) 7

Since we know the product rule states f'(x) = g'(x)*h(x) + g(x) * h'(x)

we have to calculate the derivatives of g(x) and h(x).

For both we need the chain rule which states f(x) = g(h(x)) => f'(x) = g'(h(x))*h'(x).

That equation might be an appaling sight so let me attempt to state it to you in simpler words.

The derivative of a function 2 within a function 1 is the derivative of the function 1 leaving the function 2 as it is multiplied by the derivative of the function 2

Since this still might seem a little fuzzy let me give you an example with the functions we have g(x) = (2x+1) 9

We could again rewrite this to g(x) = (u(x)) 9. where u(x) = 2x+1 (so u'(x) = 2) then g'(x) = 9*u(x) 8*u'(x) = 9*(2x+1) 8*2 = 18(2x+1) 8

Now, let me continue with the rest of the original question.

We just saw that g'(x) = 18(2x+1) 8

Similarly we can calculate (again using the chain rule) h'(x) = 7*(3x-1) 6*3 = 21*(3x-1) 6

Then for f'(x), we can fill in the blanks.

f'(x) = g'(x)h(x) + g(x)h'(x) = 18(2x+1) 8* (3x-1) 7 + (2x+1) 9*21(3x-1) 6

We can simplify this to 18(2x+1) 8*(3x-1) 6*(3x-1) + (2x+1) * (2x+1) 8 * 21(3x-1) 6

I did this so both terms have (2x+1) 8*(3x-1) 6 in common.

This means I can now simply the equation to (18(3x-1)+21(2x+1)) * (2x+1) 8*(3x-1) 6

Note that: 18(3x-1)+21(2x+1) = 54x-18+42x+21 = 96x+3 = 3(32x+1)

So we calculated that f'(x) = 3(32x+1)(2x+1) 8(3x-1) 6

f"(x) works in quite a similar way, where we can again use f(x) = g(x)h(x) only this time g(x) also is a multiplication of two functions so for g'(x) you'd also need to use the product rule.

Must be a case of productruleception

See whether you can do that one yourself.

Reinout

edit: just saw Alan did the same thing in the mean time
Apr 5, 2014
 #1
avatar+2353 
+5
Stu:

Online test questions, you don't have to give me the answer as I do want to learn the process and be able to do it, but I suspect that the way the test is set up, that investigation and getting help is part of the process to encourage learning of difficult material. So answers are not necessary but mighty helpful since, I can see what is wrong, thus feel free to put new variables in. Note there is a few questions here, challenging for me, and since I can not follow the derivative rules to well, or if I do the answers are not put it in the correct form etc. The last answers I got were input right so thank you and that shows that if the process is right the system will accept the answers.

1
http://snag.gy/w8UHZ.jpg

2
http://snag.gy/FrQ15.jpg

3
http://snag.gy/sCKWg.jpg

4
http://snag.gy/eE7hY.jpg
I really would like an explanation of this question and answer, or if there was a another answer to arrive at through a different method and what that method would be. thanks.

I appreciate the help.




Okay, let's see what I can do here,
I think I'll give you important hints for the first three and show you the whole solution to the fourth one to give you an idea.

Also I think it might help you to look at this site; http://www.mathsisfun.com/calculus/derivatives-rules.html
This website has lot's of simple explanations on various mathematical subjects and this section pretty much summarizes the rules for derivatives.



Now, for the first one, you should look into the product rule. The idea is that f(x) = g(x)*h(x) then f'(x) = g'(x)*h(x) + g(x)*h'(x)

So for example if g(x) = x 2 and h(x) = e x so f(x) = x 2*e x

then f'(x) = 2xe x + x 2e x = (2x+x 2)e x



For the second one, look at the derivatives of trigonometric functions (and learn them by heart). So, f(x) = sin(x) => f'(x) = cos(x) and g(x) = cos(x) => g'(x) = -sin(x)

Also have a look at the chain rule to see what to do with cos(pi*x).



For the third one,

Same as the second one, look at the derivatives of trigonometric functions and the chain rule. That should do the trick

I'll do the fourth one in a seperate post.
Apr 5, 2014