# $\color{#D61F06}{\text{Unity}} \ \color{#624F41}{\text{Is}} \ \color{#3D99F6}{\text{Strength}} \ \color{#20A900}{\text{Contest}}$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$ Hello every $-e^{i\pi}$ ,

I am holding a contest named "Unity Is Strength".The aim of this contest is to motivate the community to create beautiful mathematical expressions that equal 1 i.e unity.

What you have to do?

You have to generate an expression that equals $1$ and post it on this note in the following format:

"Here's my entry: <expression> = 1"

Rules And Regulations:

• The expression must be original and it must equal one.

• Every member is allowed only 1 entry. You can update your entry by editing your comment, or deleting it and adding a new one.

• Submission of famous identities is prohibited. For example : Proposing Euler's identity in this form: $-e^{i\pi}=1$ is not allowed.

• Just modifying the known identities/equations to make them equal 1 is prohibited. For example: We all know in $\Delta ABC , \sum \tan A = \prod \tan A$ , so submitting it in this form: $\dfrac{\sum \tan A}{\prod \tan A}=1$ is not allowed.

• Trivial proposals like $2-1=1$ or $\dfrac{a\times 1}{a}=1$ or putting some numbers to force the expression to be equal to $1$ are not expected.

• Irrelevant/abusive comments would be deleted.

• This contest starts from the time this note is posted and ends on 18 October 2015 at 3:30 pm GMT. The entries after this time won't be accepted.

• The entry with the maximum value of $upvotes$ will be declared the winning entry. The member who posted that entry will be the Winner of the Unity Is Strength Contest. So please Do Upvote all the entries you like, and even if you downvote any entries ,it won't affect the score of that/those individuals.

• The results will be announced at any day in the next week of the last date of submission.

The contest starts now. Please like,reshare this note so that it will reach most of the brilliantians!

Best Of Luck! Note by Nihar Mahajan
5 years ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
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Here's my entry :

$\Re\left(\frac{2}{1-e^{i\theta}}\right)=1~~\forall~\theta\in\Bbb R\setminus\{x\mid x=2\pi n~,~n\in\Bbb Z\}$

where $\Re(z)$ denotes the real part of $z\in \Bbb C$.

This is something that I noticed a few months ago while solving a few trigonometry problems on Brilliant. I'm not sure whether this complies with Rule #5 or not but I'm posting it anyway.

- 5 years ago

This is great!

- 5 years ago

Here's my submission.

$\lim _{ n\rightarrow \infty }{ \frac { \sin \left( 2\times 10^{ -n }° \right) -\sin \left( 2\times 10^{ -(n+1) }° \right) }{ \pi \times 10^{ -(n+2) } } } =1$

- 5 years ago

Can you tell your friend's name? Is he/she on brilliant?

- 5 years ago

He is called Isaac Lee. Not very active though.

- 5 years ago

It's very easy though it seems to be tough. It can be achieved by using L'Hospital Rule.

- 5 years ago

There's an even easier method, and that is using the approximation

$\sin { \theta = } \theta$

When $\theta$ is very small.

- 5 years ago

yes!! it's very easy. You are right. But we should convert degrees to radians.

- 5 years ago

Here is my entry

$\dfrac{\left(\displaystyle\int_0^1\ln\left(\frac{1}{x}\right)^{49} \dfrac{x^3}{1-2x^4} \text{dx} \right)\times{2^{101}}}{49! \text{Li}_{50}}=1$

$\text{Li}_{x}$ is the polylogarithm function.

- 5 years ago

Invite Parth too :)

• Kritarth Lohomi : In response to Nihar Mahajan: He is busy..

• Aditya Kumar In response to Kritarth Lohomi: Nice:) Well, this doesn't deserve downvotes.

• Julian Poon In response to Aditya Kumar: Maybe downvotes should not be counted

• Aditya Kumar In response to Julian Poon: Yes but still... it doesn't look good.

• Nihar Mahajan In response to Julian Poon: They are not counted.

• Anik Mandal In response to Kritarth Lohomi: Cool!

• Siddharth Bhatt In response to Kritarth Lohomi: Nice!

- 5 years ago

Here's my entry:

$1=\lim_{x\to0}\sum_{n=1}^{\infty}\sum_{r=0}^{\infty}\frac{(-1)^r}{n+r}\dbinom{n+r}{r}x^{n+r-1}$

- 5 years ago

Nice!

- 5 years ago

$\int_{-1}^{1}\frac{d}{dx}\left(\frac{1.5}{2^{\frac{1}{x}} + 1}\right)dx$

- 5 years ago

Isn't it an improper integral? It is discontinuous at $x=0$ .

- 5 years ago

Yes, it is an improper integral. But that doesn't mean it can't be evaluated.

- 5 years ago

$\color{#FFFFFF}{\text{=1 That's a wise one :P}}$

- 5 years ago

My entry: $\left( \frac { \int _{ 0 }^{ \infty }{ \frac { a^{ 3 } }{ { e }^{ a }-1 } } da }{ \int _{ 0 }^{ \infty }{ { b }^{ 3 }{ e }^{ -b } } db } \right) .\frac { 90 }{ { \pi }^{ 4 } } =1$

- 5 years ago

Lol!! Numerator is Riemann zeta function and denominator is Gamma function.

- 5 years ago

Yes. U r right!

- 5 years ago

Here's my entry, $\displaystyle{ \lim_{x \to 0}} \log_{tan^{2} x} \left(tan^{2} 2x \right) = 1$

$\quad\displaystyle\lim_{x\to0}\log_{\tan^2x}(\tan^22x)$

$=\displaystyle\lim_{x\to0}\frac{\log(\tan^22x)}{\log{\tan^2x}}$

$=\displaystyle\lim_{t\to0}\frac{\log\left(\frac{2t}{1-t^2}\right)^2}{\log{t^2}}$

$=\displaystyle\lim_{t\to0}\frac{\log\left(\frac{2t}{1-t^2}\right)}{\log{t}}$

$=\displaystyle\lim_{t\to0}\frac{\frac{2(1-t^2)-2t(-2t)}{(1-t^2)^2}/\left(\frac{2t}{1-t^2}\right)}{1/t}$

$=\displaystyle\lim_{t\to0}t\frac{2+2t^2}{(1-t^2)^2}/\left(\frac{2t}{1-t^2}\right)$

$=\displaystyle\lim_{t\to0}t\frac{2+2t^2}{1-t^2}/(2t)$

$=\displaystyle\lim_{t\to0}\frac{2+2t^2}{2(1-t^2)}$

$=\displaystyle\lim_{t\to0}\frac{1+t^2}{1-t^2}$

$=\displaystyle\frac11$

$=1$

- 5 years ago

Well yeah you could do that. But wouldnt it be easier to apply L'Hopital rule right after applying change of base theorem?

Right.

$\quad\displaystyle\lim_{x\to0}\log_{\tan^2x}(\tan^22x)$

$=\displaystyle\lim_{x\to0}\frac{\log(\tan^22x)}{\log{\tan^2x}}$

$=\displaystyle\lim_{x\to0}\frac{\log(\tan2x)}{\log{\tan x}}$

$=\displaystyle\lim_{x\to0}\frac{2\sec^2x/(\tan2x)}{\sec^2x/(\tan x)}$

$=\displaystyle\lim_{x\to0}\frac{2/(\tan2x)}{1/(\tan x)}$

$=\displaystyle\lim_{x\to0}\frac{2\tan x}{\tan2x}$

$=\displaystyle\lim_{x\to0}\frac{2\sec^2x}{2\sec^2x}$

$=\displaystyle1$

- 5 years ago

Yeah that was my solution too.Good job.

Thanks!

- 5 years ago

My submission: $\Large\int^1_0\frac{\ln(\sin^{-1}[\int^3_e\Gamma(x)dx]+e)}{e^{\delta(\int^\pi_0[\cot x]dx)}}dx$
where $[.]$ denotes the floor function/ greatest integer function, $\delta(.)$ denotes the Delta Function and $\Gamma(.)$ denotes the Gamma Function. All the best!

- 5 years ago

my late entry @Nihar Mahajan $\prod_{k=2}^\infty \left(\zeta(k)\sum_{n=1}^\infty \dfrac{\mu(n)}{n^k}\right) =1$

- 4 years, 9 months ago

No problem. Though I am unable to actually evaluate the LHS , the equation looks interesting!

- 4 years, 9 months ago

@Sravanth Chebrolu Sorry to say , but I have confirmed that your proposal is not original with some people. Your proposal is not accepted. I am deleting it. Don't worry , you can post another , you have whole week in hand! :)

- 5 years ago

I just think that this would be fun if it implies the rule (but I don't think it is): $\displaystyle \ce{^16_8A} \ce{^20_10B} = 1$

- 5 years ago

I don't really understand your notation. Mind if you elaborate on the notation used? Thanks a lot!

- 5 years ago

Even I don't understand your notation.Wait , are you using the notation for atomic mass , atomic number of an element? If so , then it is not allowed.

- 5 years ago

Is the equation in your entry somehow related to radioactivity? Because I see a nice correlation between A and B (according to your notation and how it's written). B reduces to A after occurrence of one $\alpha$-decay (followed by a $\gamma$-decay, as usual to stabilize energy levels).

Then again, looking at the numbers, this might be something else because B doesn't satisfy the "n/p ratio > 1.5" criteria for radioactivity to occur.

- 5 years ago

This was just a silly equation which won't make any sense here. So A is Oxygen, denoted by O and B is Neon, denoted by Ne. When we connect both, we get ONe = 1

- 5 years ago

Lol... I like that

- 5 years ago

Ah.. ok lol

- 5 years ago

Here's my entry

$\large{\frac { 7 }{ 26 } \left[ \frac { d }{ dx } { \left[ \tan ^{ -1 }{ \left( \frac { 2{ x }^{ 4 } }{ 3{ x }^{ 2 }-7x+1 } \right) } +\sin ^{ -1 }{ \left( \frac { 26x+4 }{ 7 } \right) } \right] }_{ x=0 } \right] =1}$

- 5 years ago

The contest is already over and I am in process of posting the results. Don't delete this , I would mention this as a "late entry".

- 5 years ago

Here is my entry $-(\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}})(\large \lim_{n\to\infty}(\frac {n^2(n sin(\frac {1}{n})-2)}{1+3n^2}))=1$

- 5 years ago

You mean $n\to\infty$ right? And use "\left(...\right)" and also "\displaystyle" to make things neater :)

- 5 years ago

This is actually a bit forced and not really original. The nested radical is actually a problem by Ramanujan which is well known to equal $3$. So, you just multiply it with $\frac {-1}3$ to get $-1$, the $\frac {-1}3$ coming very trivially from the limit. The negative sign at the very left results in getting $(-1)\times (-1)=1$.

Seeing that your profile picture is a picture of Ramanujan, this post looks quite apt, haha.

- 5 years ago

Lol I have not copied exactly the same I have made edits in it :-p

- 5 years ago

Sorry , but your submission includes the famous Ramanujan's nested radicals which is not allowed as stated in rule number 3. Don't delete this , let it stay :)

- 5 years ago

I have sent you some others as message on fb

- 5 years ago