The Proof is Trivial!

i dont understand this solution. why is it not just n!/ (n-6)!

Using a few approximations, I get it to be close to $\dfrac{\sqrt{21+6e+17\pi+6\log 2}}{2\sqrt 6}$

Is this close at all? I know it isn't correct but it may be close (or not).

Solution 57

Let the integral be $I$. Apply $x\to -x$ to get $I=\displaystyle\int_{-1}^{1} \dfrac{1-x^4\tan x}{1+x^2} dx$ (by the oddity of tan), then add the two forms of $I$ together and divide by 2 to obtain $I = \displaystyle\int_{-1}^{1} \dfrac{1}{1+x^2} dx = \dfrac{\pi}{2}$.

I'm glad the integral has caught the attention of several.



That is not close to what I have - the end result is quite pleasing.

Could someone please explain what is meant by $x\to -x$ - iv seen it on many occasions with integrals on different threads but what does it actually mean (as x approaches -x or sometimes i see as x approaches x-pi/2 with trig functions in integrals) - and how has it been used to solve this integral?

Could someone please explain what is meant by $x\to -x$ - iv seen it on many occasions with integrals on different threads but what does it actually mean (as x approaches -x or sometimes i see as x approaches x-pi/2 with trig functions in integrals) - and how has it been used to solve this integral?

I suspect it might be zero by any chance?...

It's basically making the substitution u=-x then relableling u as x.

I'm pretty sure it should be written $x \mapsto -x$. It means making the substitution x=-u. It's notation which I, for some reason, really don't get on with.

Ok i still dont see how this works: sub in u=-x

$I=\displaystyle\int_{-1}^{1} \dfrac{u^4\tan u -1}{1+u^2} du$

Then if you relabel as x (why- i dont get why you can do this either- they are not the same)

Ok i still dont see how this works: sub in u=-x

$I=\displaystyle\int_{-1}^{1} \dfrac{u^4\tan u -1}{1+u^2} du$

Then if you relabel as x (why- i dont get why you can do this either- they are not the same)
Then you add the integrals and divide by 2- this part i understand but my problem is why are you allowed to do everything before that?

In my increasing annoyance, I told Mathematica to evaluate the integral. It can't make heads or tails of the infinite one, but the integral from 0 to pi/2 is 0.476799233144858312460420290485 (I truncated just before Mathematica became unsure of the value).
Series magic isn't the way forward, probably, because of the incredible oscillation.
The next interval (pi/2 to pi) is 0.06445155429406938922998100618002101588561.
The integral from 0 to 52pi is 0.5771487445508202137340201446320565614660,
while the integral from 0 to 52pi-pi/2 is
0.5768302091014543319600824273038202792160.
The integral from 0 to infinity is therefore probably a bit over 0.57.
The function is even.
Integrals from 0 to successive multiples of pi/2:

This is starting to get infuriating :P

I'm glad the integral has caught the attention of several.

0.5771487445508202137340201446320565614660

This is starting to get infuriating :P

In indefinate integrals the variable you are integrating is irrelivent, the integral of x between 0 and 1 = the integral of u between 0 and 1 = the integral of between 0 and 1, provided you are integrating with respect to that variable.

k but when you change it integral is the wrong way around isnt it? there should be u^4tanu -1 not 1-u^4tanu as farhan did
im very confused

Problem 263 (Haven't actually done this but looks like fun -sorry)

Let $a, b, c, d$ be integers with $a>b>c>d>0$.

Suppose that $\displaystyle ac+bd=(b+d+a-c)(b+d-a+c)$.

Prove that $ac+bd$ is not prime.

Is the definiately typed correctly? I'm not great at number theory but seem to have got to an answer without much touble.

**** it, may as well jump on the bandwagon and add to the struggling masses!

$\gamma$ ?

Problem 263 (Haven't actually done this but looks like fun -sorry)

Let $a, b, c, d$ be integers with $a>b>c>d>0$.

Suppose that $\displaystyle ac+bd=(b+d+a-c)(b+d-a+c)$.

Prove that $ac+bd$ is not prime.

Is the definiately typed correctly? I'm not great at number theory but seem to have got to an answer without much touble.

It certainly is! If you think it's right, go for it