You are Here: Home >< Maths

# The Proof is Trivial! watch

1. (Original post by Star-girl)
Solution 56

By the t-substitution , the limits get transformed into and since the limits are the same, the integral evaluates to .
I get
2. (Original post by Felix Felicis)
I get
Seems more plausible than 0. My solution is probably dodgy then... but why?
3. (Original post by Star-girl)
Solution 56

By the t-substitution , the limits get transformed into and since the limits are the same, the integral evaluates to .

Seems a bit too quick - maybe this is a little dodgy...
The integrand is positive over the interval, so it's impossible for the integral to be 0.
4. (Original post by Star-girl)
Seems more plausible than 0. My solution is probably dodgy then... but why?
Maybe because of the asymptotes of tan between the limits?
5. (Original post by und)
The integrand is positive over the interval, so it's impossible for the integral to be 0.
(Original post by Felix Felicis)
Maybe because of the asymptotes of tan between the limits?
Ha! What a noobish error I made... I think it's sleepy-time for me...
6. Solution 56

Solution 56

Let

7. (Original post by MsDanderson)
hey! IF YOU'RE A FEMALE, could you fill this out please? it literally only takes 2mins do it -> http://www.surveymonkey.com/s/B3MVL58

I need 100 done in the next 24 hours or I'll fail my project Please help! I'm so grateful
You come looking for females... in a maths thread?
8. (Original post by und)
You come looking for females... in a maths thread?
I only just read that it's for females...I already filled it in Ah well
9. (Original post by MsDanderson)
it's okay! i appreciate it anyway!
You might want to discard my results anyway...my answers may skew your data for a survey targeted at females to put it lightly
10. (Original post by Star-girl)
Solution 56

By the t-substitution , the limits get transformed into and since the limits are the same, the integral evaluates to .

Seems a bit too quick - maybe this is a little dodgy...

EDIT: Ignore this rubbish.
Can't just sub in t=tan(x/2) as it is not continuous across the range of integration.
11. Problem 59**

Prove that there are no integers for which .
12. (Original post by dknt)
Well I guess I'll provide another physicsy mathsy problem

Problem 58 **/***

A double pendulum consists of a mass m2 suspended by a rod of length l2 from a mass m1, which is itself suspended by a rod of length l1 from a fixed pivot, as shown below.

Attachment 208929

Show that the equations of motion for small displacements can be written as

where,

and
Spoiler:
Show
Would you use the small displacement approximation before or after taking the lagrangian to find the equations of motion?
13. (Original post by cpdavis)
Spoiler:
Show
Would you use the small displacement approximation before or after taking the lagrangian to find the equations of motion?
Spoiler:
Show
Far easier to do it right at the beginning when finding the potentials and kinetic energies. Of course, you could do it after and you should still get the same result, however it makes applying Lagrange's equation and therefore the equations of motion much harder! Of course for a challenge try finding the EOMs for arbitrary displacements
14. (Original post by dknt)
Spoiler:
Show
Far easier to do it right at the beginning when finding the potentials and kinetic energies. Of course, you could do it after and you should still get the same result, however it makes applying Lagrange's equation and therefore the equations of motion much harder! Of course for a challenge try finding the EOMs for arbitrary displacements
Spoiler:
Show
Yeah I realised that it would be easier to do it at an earlier stage I remember coming across a problem like this but we assumes that the lengths and masses were the same. Now working through this solution, good practice for my dynamics course
15. (Original post by bananarama2)
I just lost the will to live with that question after I completed a load of school work (binomial expansion). I generally go for energy, but just didn't in that instance, nice one Star-girl (I had to think about that name.)
For some reason I didn't see this yesterday. Ah yes... school work... what I should have been doing this holiday...

Thanks (and haha).
16. (Original post by und)
You come looking for females... in a maths thread?

(Original post by james22)
Can't just sub in t=tan(x/2) as it is not continuous across the range of integration.
Yeah - I realised afterwards, hence the edits...
17. Solution 59

More general:

Denote . This is called binary quadratic form. Its discriminant is defined as .
We say that represents given number properly if there are integers such that and .

We have the following proposition:
Let and be given integers. Then there exists a binary quadratic form of discriminant that represents properly if and only if is solvable.

Let and . Since , we can choose integers such that , , , and . Notice that . Now there are integers such that and . It remains to cite the Chinese remainder theorem.
Conversely, let . Then the form has discriminant and represents properly.

Next, I claim that there is not binary quadratic form of discriminant which represents properly. Since is square-free, we cannot have improper representations.
The congruence is obviously solvable. Consider the congruence . Now . Hence . Therefore the congruence has no solutions and my claim holds true.
18. Solution 58

Ignoring none linear terms.

Ignoring non linear terms.

We have

and

So;

where,

and
19. (Original post by bananarama2)
Solution 58

Ignoring none linear terms.

Ignoring non linear terms.

We have

and

So;

where,

and
Beaten was planning to type this later
20. (Original post by cpdavis)
Beaten was planning to type this later
Sorry Although I think I may have done you a favor, it took and hour and a half to type that.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 22, 2018
Today on TSR

### The most controversial member on TSR?

Who do you think it is...

### Is confrontation required?

Discussions on TSR

• Latest
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE