The Proof is Trivial!

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Because the

u

is just a dummy variable - you'll get exactly the same result whether you integrate

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

I=\displaystyle\int_{-1}^{1} \dfrac{x^4\tan x -1}{1+x^2} dx

original question says
or

I=\displaystyle\int_{-1}^{1} \dfrac{1+ x^4\tan x }[br]{1+x^2} dx

Then after some substitution

I=\displaystyle\int_{-1}^{1} \dfrac{1- x^4\tan x}{1+x^2} dx

But because you said variable irrelevant with definite

You can add.. left with integral of 1/x^2 +1 wich gives result of pi/2

But if you sub in u=-x though should get

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

if you replace the u you get

I=\displaystyle\int_{-1}^{1} \dfrac{x^4\tan x -1}{1+x^2} dx

How can you add this now- the substitution has got you nowhere( from this perspective) I know I am wrong but i just dont get this

Reply 1881

james22

Original post by nahomyemane778

original question says

I=\displaystyle\int_{-1}^{1} \dfrac{1+ x^4\tan x }[br]{1+x^2} dx

Then after some substitution

I=\displaystyle\int_{-1}^{1} \dfrac{1- x^4\tan x}{1+x^2} dx

But because you said variable irrelevant with definite

You can add.. left with integral of 1/x^2 +1 wich gives result of pi/2

But if you sub in u=-x though should get

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

if you replace the u you get

I=\displaystyle\int_{-1}^{1} \dfrac{x^4\tan x -1}{1+x^2} dx

How can you add this now- the substitution has got you nowhere( from this perspective) I know I am wrong but i just dont get this

The original I has a 1+stuff on the top, the equivalent, but rearanged I, has a 1-stuff on the top. When you add these together the stuff cancels out and you are left with 1/something.

Reply 1882

Arcane1729

Original post by james22

The original I has a 1+stuff on the top, the equivalent, but rearanged I, has a 1-stuff on the top. When you add these together the stuff cancels out and you are left with 1/something.

but when you substitute you dont get 1-stuff you get stuff-1 on top
because you did u=-x so du/dx=-1 so you must multiply by -1 and you
get stuff-1 not 1-stuff so when you add the numerator does not simplify

(edited 10 years ago)

Reply 1883

Smaug123

Original post by nahomyemane778

You have forgotten that during the course of the substitution, you also changed the limits of integration from {-1,1} to {1,-1}. It takes an extra minus sign to flip them back round.
EDIT: Oh, sorry, that makes three minus signs - I confess that I misunderstood your question when you asked (I thought it was just asking why you could relabel u as x), so I didn't check the integral. It is true that the integral of

\dfrac{1-x^4 \tan (x)}{x^2+1}

from -1 to 1 is

\dfrac{\pi}{2}

, while the integral of

\dfrac{u^4 \tan (u)-1}{u^2+1}

from -1 to 1 is

-\dfrac{\pi}{2}

(edited 10 years ago)

Reply 1884

Arcane1729

Original post by Smaug123

You have forgotten that during the course of the substitution, you also changed the limits of integration from {-1,1} to {1,-1}. It takes an extra minus sign to flip them back round.

Oh my goodness me. :facepalm2:

I finally understand thank you! :adore:

.
What a genius trick then- how do people think of this kind of stuff?

Reply 1885

Smaug123

Original post by nahomyemane778

Oh my goodness me. :facepalm2:

I finally understand thank you! :adore:

.
What a genius trick then- how do people think of this kind of stuff?

Practice, and by noticing that almost none of the function cares about whether x is negative or positive. Also divine inspiration, I think… but it's a fairly common pattern: "make a substitution to get the negative of the original", it happens in a variety of trig contexts.

Reply 1886

Jkn

Original post by james22

Assume, for a contradiction, that

ac+bd

is a prime. Using the inequality we can immediately deduce that

b+d+a-c \neq 1

Therefore

b+d-a+c=1

(otherwise we have found 2 factors, contradicting the fact that

ac+bd

is prime).

The equation then reduces to

ac+bd=b+d+a-c

which we can rearange then make use of the inequalities given to show that

0=a(c-1)+b(d-1)+c-d) \geq 0+0+c-d=c-d>0

which is a contradiction as required.

Omg, sorry dude, it turns out there was a typo, no idea how I misread it over and over again! :facepalm:

I was driving myself mad trying to find an error :lol:

Reply 1887

james22

Original post by Jkn

Omg, sorry dude, it turns out there was a typo, no idea how I misread it over and over again! :facepalm:

I was driving myself mad trying to find an error :lol:

No problem, I thought it was a bit easy.

Reply 1888

Jkn

Original post by james22

No problem, I thought it was a bit easy.

Mm, what happened is that I misread it, solved it quickly (like you did) but then got really confused considering how absolutely insane this question is supposed to be. Kind of annoyed I looked at the solution though, would have been nice to see if I would have known where to start :tongue:

Reply 1889

FireGarden

Problem 264**

Find all

x : \sin(x) = 13

Reply 1890

bananarama2

Solution 264

\sin(x)=\frac{\sinh (ix)}{i}=13

ix=arsinh (13i)=\ln (13i+\sqrt{-169+1} )=ln (13+2\sqrt{42}) +ln(i)

x=\frac{\pi}{2} +2 n \pi-i \ln (13+2\sqrt{42})

Where n is an integer.

Reply 1891

FireGarden

Let's see who can think of some nice shortcuts.

Problem 265**

Is the following matrix singular?

Unparseable latex formula:

[br]\[ \left( \begin{array}{cccc}[br]54401 & 57668 & 15982 & 103790 \\[br]33223 & 26563 & 23165 & 71489 \\[br]36799 & 37189 & 16596 & 46152 \\[br]21689 & 55538 & 79922 & 51277\end{array} \right)\] [br]

Reply 1892

Smaug123

Original post by FireGarden

Let's see who can think of some nice shortcuts.

Problem 265**

Is the following matrix singular?

Unparseable latex formula:

Spoiler

Reply 1893

james22

Original post by FireGarden

Let's see who can think of some nice shortcuts.

Problem 265**

Is the following matrix singular?

Unparseable latex formula:

No, as it has determinent

-1559852323213094031 \neq 0

Reply 1894

bananarama2

Original post by james22

No, as it has determinent

-1559852323213094031 \neq 0

*Engineers solution :biggrin:

Reply 1895

james22

Original post by bananarama2

*Engineers solution :biggrin:

I disagree, I tihnk a sign of a true mathematician is one who can reduce a problem to one that he knows he can solve. In this case I reduced the problem to finding the determinent with the assistance of a computer. Why make things more difficult than needed?

Reply 1896

Hasufel

Solution 265

If you divide every row in the matrix by the leading number in the corresponding row, (1st row divided by leading number in the first row, etc) then, every number in the first column is 1 (EDIT)

This is relatively easy to put into row echelon form. (in reduced row-echelon form, it`s the 4x4 identity matrix!)

Take the trace of the resulting upper diagonal matrix, which is non-zero, so the matrix is non-singular.

(OR: perform row operations until the first column is (1,0,0,0) then use co-factor expansion along the first column)

(edited 10 years ago)

Reply 1897

FireGarden

Original post by Smaug123

Spoiler

The very solution i had in mind :wink:

Reply 1898

FireGarden

Now I'm going to be a dick head and post something I think not even LotF will evaluate without a lot of head scratching. The solution I have is only using knowledge from A level.. but I doubt it'll be used. It has a nice answer, too.

Problem 266***

Evaluate