The Proof is Trivial!

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Yes, but the way GCSEs are taught means they wouldn't have a clue what a function was even if they do stuff related to them....

Ah ok. See in France we study these things rather than dropping coins off the Eiffel tower. :teehee:

Reply 1141

Zakee

Original post by Felix Felicis

Some more (*) problems for people who want them:

Problem 172*

Is

\displaystyle\prod_{r=0}^{100} r!

a perfect square? If not, could we remove one of the factorials to make it a perfect square?

Solution 172*

No, it is not a perfect square. How? Well, I can compute 100! in my mind. (like this guy: http://www.youtube.com/watch?v=8b2jOcKI798)

Taking the premise of 100! is not a perfect square.
A perfect square for factorials can be defined as below:

a! \times a! = (a!)^2

100! \times 99! \times 98! ... \times 1!

Note that:

100! = 100 \times 99!

100! \times 99!

100 \times 99! \times 99!

a! \times a! = (a!)^2

, that means:

99! \times 99! = (99!)^2

which is a perfect square.

We can continue this procedure mechanically by pairing up the rest of the factorials so we have:

(100 \times (99!)^2) \times (98 \times (97!)^2) \times (96 \times (95!)^2)... \times (2 \times (1!)^2)

Rearranging these (there are no minus terms or chicanery occurring here)

(100 \times 98 \times 96 \times 94... \times 2) \times ((99!)^2 \times (97!)^2 \times (95!)^2 \times... (1!)^2)

Using the LHS, we have:

2(1) (\times 2(2) \times 2(3) \times 2(4) ... \times 2(50))

So LHS of the equation is simply:

Unparseable latex formula:

2^5^0 \times 50!

We know that

Unparseable latex formula:

\sqrt {2^5^0}

is a perfect square (I think this is what you could call trivial), by removing

50!

, we can actually end up with the initial result forming a perfect square, such that:

\displaystyle\prod_{r=0}^{100} r! - (50!)

x^2

(edited 10 years ago)

Reply 1142

Zakee

Original post by Lord of the Flies

Ah ok. See in France we study these things rather than dropping coins off the Eiffel tower. :teehee:

PRSOM.

In all fairness though, Lagrange would be quite content with his question. :wink:

(edited 10 years ago)

Reply 1143

bananarama2

Original post by Lord of the Flies

Ah ok. See in France we study these things rather than dropping coins off the Eiffel tower. :teehee:

That question was off you first year dynamics and relativity course smart alec :wink:

Reply 1144

Felix Felicis

Original post by Jkn

Lightweight, you can get it form applying integration by parts to the other one by integrating

\ln(x)

(and differentiating the denominator) and then evaluating a fairly trivial other part with a tanx substitution :wink:

Yeah but that took me like an hour and half to think of on the problem LotF set earlier :lol:

Reply 1145

bananarama2

Original post by DJMayes

Hyperbolics are more fun. :tongue:

(What answer do you get with that sub though?)

The same

Reply 1146

Jkn

Original post by DJMayes

Let

x^3 = sinh^2u

3x^2 \ dx = 2sinhucoshu \ du

\Rightarrow I = \frac{2}{3} \int sinh^3ucosh^2u \ dx

= \frac{2}{3} \int sinhu(cosh^2u-1)cosh^2u \ dx

= \frac{2}{3} \int sinhucosh^4u - sinhucosh^2u \ dx

= = \frac{2}{3} [ \frac{1}{5}cosh^5u -\frac{1}{3}cosh^3u ]

x = 0 so u = 0, coshu = 1, x = 2 so coshu = 3

Finally plugging those in I get 1192/45, although this isn't really a nice answer...

Hopefully now you will follow me to the algebraic promised land...

Solution 178 (2)

\displaystyle \int_0^2 ((1+x^3)x^2-x^2)\sqrt{1+x^3} dx = \left[\frac{2}{15} (1+x^3)^{\frac{5}{2}} -\frac{2}{9} (1+x^3)^{\frac{3}{2}} \right]_0^2 =\frac{1192}{45} \ \square

Reply 1147

Zakee

Give me 2-3 months, and I'll be able to do those integrals which look so palatable. :frown:

Reply 1148

Jkn

Original post by bananarama2

That question was off you first year dynamics and relativity course smart alec :wink:

Ooo I'm interested again :tongue:

I got an answer but I think it was wrong.. I might type it up later

Spoiler

Original post by Felix Felicis

Yeah but that took me like an hour and half to think of on the problem LotF set earlier :lol:

When was it set? :eek:

Reply 1149

bananarama2

Original post by Zakee

PRSOM.

In all fairness though, Lagrange would be quite content with his question. :wink:

I like the blase way physics questions are given. The sort of "a washing machine can be powered by a solar umbrella, calculate the rate of loss of mass of the sun." type questions :tongue:

Reply 1150

Felix Felicis

Original post by Jkn

Ooo I'm interested again :tongue:

I got an answer but I think it was wrong.. I might type it up later

Spoiler

When was it set? :eek:

Problem 132 xD Also, a few other integrals were set earlier as well but hey ho, nothing wrong with other people attempting them :biggrin:

Reply 1151

username937760

Original post by Zakee

Solution 172*

No, it is not a perfect square. How? Well, I can compute 100! in my mind.

I like all of your solution except this bit - even a computation of the number doesn't prove it isn't not a perfect square. I think a better way of doing this would be to count the number of a specific prime in your solution. For example, the number of 47's. From 47 to 100, we get 54 lots of 47. Also, we have another 47 included in each number between 94 and 100, giving another 7 lots of 47. This gives an odd number of 47's, and so it isn't a perfect square as you cannot divide the amounts of 47 equally amongst the two square factors.

(edited 10 years ago)

Reply 1152

bananarama2

Original post by Jkn

Ooo I'm interested again :tongue:

I got an answer but I think it was wrong.. I might type it up later

Spoiler

When was it set? :eek:

I did it differently to the lecture notes, but got the same answer. I didn't use dodgy limited calculus :wink:

If you just want the direction you just need two lines of working. But you need the moment of inertia.

Reply 1153

Zakee

Original post by bananarama2

I like the blase way physics questions are given. The sort of "a washing machine can be powered by a solar umbrella, calculate the rate of loss of mass of the sun." type questions :tongue:

My favorite is:

"Jimmy is spinning around on the base of cylindrical chamber such that he is in contact with the base and well as the cylindrical wall. He is just on the point of lifting and the cylinder has a radius of 10m, and his mass is 60kg. Given an arbitrary compact gauge group, does a non-trivial quantum Yang-Mills theory with a finite mass gap exist?"

Reply 1154

Zakee

Original post by DJMayes

That is true. I was sort of joking about the computation part that I had calculated 100! x 99! x 98! x 97!... (If I did, then well, I might as well quit school now and just end up replacing the TI-84 plus).

However, I didn't think of doing what you did as I just made an assumption (which I'm aware is scandalous). I'm actually fond of viewing it the way that you've shown Cheers man. :smile:

I guess it's just my innate laziness which prevents me from being rigorous enough -sighs-. I should probably quit Mathematics and end up doing poetry. :nopity: