**1. Euler's Formula** - \(cos x + isin x = {e}^{ix}\)

**2. Relation between Mean, Mode, Median** - \(Mode = 3*Median - 2*Mean\)

**3. Poisson's Distribution** - \(P(X=r) = {e}^{-\lambda}.\frac{{\lambda}^{r}}{r!}\) where \(\lambda(mean) = np(n = No. of trials, p = probability of success)\)

**4. RMS - Arithmetico-Geometric and Harmonic Mean inequality**

**5. Cauchy-Shwartz's inequality**

**6. Sum of sines with angles in AP**- \(sin\alpha + sin(\alpha + \beta) + ...... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. sin(\alpha + (n-1)\frac{\beta}{2})\)

and similarly

**Sum of cosines with angles in AP** - which gives \(\frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. cos(\alpha + (n-1)\frac{\beta}{2})\)

THANKS a lot in anticipation! And I believe that you have the answers and you must also be in need of some of the formulas. You are free to share them here to clear all your doubts. :)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestTo prove \((i)\) we have \(z=cos(x)+isin(x)\) where \(z\) is a function of \(x\)

Differentiating \(z\) with respect to \(x\) we get :

\(\frac{dz}{dx}=-sin(x)+icos(x)\)

Since \({i}^{2}=-1\) hence we have :

\(\frac{dz}{dx}={i}^{2}sin(x)+icos(x)=i(cos(x)+isin(x))\)

Since \(cos(x)+isin(x)=z\) hence we have :

\(\frac{dz}{dx}=iz\)

Seperating variables and integrating both sides we have :

\( \int { \frac { dz }{ z } } =\int { idx } \)

\(ln(z)=ix+C\)

When \(x=0,z=1\) hence we get \(C=0\)

\(\Rightarrow ln(z)=ix\)

\(\Rightarrow z={e}^{ix}\)

Finally :

\(cos(x)+isin(x)=e^{ix}\)

Log in to reply

Wow! Amazing! This made me become your fan! Thank You! Try other proofs too! And yes can you give me an explanation on why Taylor series makes sense?

Log in to reply

Seriously is my proof that much bad, I wonder about the reason for 4 downvotes.

Sorry if you didn't like the proof.

Log in to reply

I think when x=0 , then at the same time z cannot be equal to one.

Log in to reply

I Love Your way To prove This equation ! Hats Off!! I 'am Amazed How Can any person down-votes it ??

Log in to reply

To prove \(1\) Euler's Formula, simple use the expansion of \(e^x\) , i.e. \[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+.....\]

\(\implies e^{ix}=1+x.i+\frac{x^2.i^2}{2!}+\frac{x^3.i^3}{3!}+\frac{x^4.i^4}{4!}+\frac{x^5.i^5}{5!}+.....\)

\( \quad \quad \quad \quad =1+x.i-\frac{x^2}{2!}-i.\frac{x^3}{3!}+\frac{x^4}{4!}+i.\frac{x^5}{5!}+........\)

\( \quad \quad \quad \quad =(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.....)+i.(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......)\)

\(\quad \quad \quad \quad =cosx +i.sinx\)

Because we know that\[cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.....\] \[sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......\]Hence proved that \[\boxed{cosx+isinx=e^{ix}}\]

Log in to reply

Can you tell me why does Taylor series make sense? How to find them? *if you don't mind

Log in to reply

5) and 6) are simple, i don't know about rest

\( [\sum_{i = 1}^{n} a_{i}b_{i}]^{2} \leq [\sum_{i = 1}^{n} a_{i}^{2}]^{2}[\sum_{i = 1}^{n} b_{i}^{2}]^{2}\)

and the equality holds if \( \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = .....=\frac{a_{n}}{b_{n}}\)

Let\( A = [\sum_{i = 1}^{n} a_{i}^{2}]^{2} , B =[\sum_{i = 1}^{n} b_{i}^{2}]^{2} , C = [\sum_{i = 1}^{n} a_{i}b_{i}]^{2}\)

then \( C^{2} \leq AB\)

If B = 0 then \(b_{i} = 0\) for i = 1 , 2, . , n .Hence C = 0 and is true .Therefore it is sufficient to consider the case \( B\neq 0 \) This implies B > 0 .

\( 0 \leq \sum_{i = 1}^{n} (Ba_{i} - Cb_{i})^{2} = \sum_{i = 1}^{n} ( B^{2}a_{i}^{2} - 2BCa_{i}b_{i} + C^{2}b_{i}^{2}\)

distributing the sum,

\( = B(AB - C^{2})\)

Sinc eB> 0 we get \( AB - C^{2} > 0\) which is the inequality \(C^{2} \leq AB\). Moreover , equality holds if \(\sum_{i = 1}^{n} (Ba_{i} - Cb_{i})^{2} = 0\)

This is equivalent to

\( \frac{a_{i}}{b_{i}} = \frac{C}{B}\) for i = 1 , 2 , .... , n

Simple application and beautiful application

If a, b , c and d are positive real numbers such that \(c^{2} + d^{2} = (a^{2} + b^{2})^{3}\),

prove that \( \frac{a^{3}}{c} + \frac{b^{3}}{d}\) if ad = bc .

Using above inequality we get ,

\((a^{2} + b^{2})^{2} = (\sqrt{\frac{a^{3}}{c}}\sqrt{ac} + \sqrt{\frac{b^{3}}{d}}\sqrt{bd} )^{2}\)

\( \leq ( \frac{a^{3}}{c} + \frac{b^{3}}{d})(ac + bd)\) , where the equality holds

Thus \( \leq ( \frac{a^{3}}{c} + \frac{b^{3}}{d})(ac + bd) \geq ( a^{2} + b^{2})^{3}\)

\( = ( a^{2} + b^{2})^{1/2} ( a^{2} + b^{2})^{3/2}\)

\( = ( a^{2} + b^{2})^{1/2}( c^{2} + d^{2})^{1/2} \geq ac + bd\)

again using the inequality , the equality holds in the last step ( above) if \( \frac{a}{c} = \frac{b}{d}\)

Combining both we get \(\frac{a^{3}}{c} + \frac{b^{3}}{d} \geq 1 \) and the equality holds if ad = bc.

6) One of my favorite

let s be equal to the given series , \( \alpha =a , \beta = b\)

\( 2Xsin(b/2) = 2sinasin(b/2) + 2sin(a + b)sin(b/2) ............\)

\( = cos( a - b/2) - cos( a + b/2) + cos(a + b/2) - cos( a + 3b/2) .... + cos[(a + (n - 3/2)b)] - cos[a + (n - 1/2)b]\)

\( = cos(a - b/2) - cos(a + (n - 1/2)b]\)

\( = 2sin[a +((n-1)/2)b]sin(nb/2)\)

\( X = \huge{\boxed{\frac{sin[a + \frac{n - 1}{2}b]sin\frac{nb}{2}}{sin\frac{b}{2}}}}\)

Similarly for the cosine series it can be done , try it ,its interesting and enjoyable.

Lovely Application of this can be found in THIS QUESTION

Enjoy@Kartik Sharma

Log in to reply

Nice Thanks! I solved that problem! Maybe I would give my solution too!

Log in to reply

I am giving proof of \(6th\) formula : \[S(let)=sin\alpha + sin(\alpha + \beta) + ...... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. sin(\alpha + (n-1)\frac{\beta}{2})\]

We know that \[\boxed{2.sinA.sinB=cos(A-B)-cos(A+B)}\] \(\implies 2.sin\alpha.sin\frac{\beta}{2}=cos(\alpha-\frac{\beta}{2})-cos(\alpha+\frac{\beta}{2})\)

\(\implies 2.sin(\alpha+\beta).sin\frac{\beta}{2}=cos(\alpha+\frac{\beta}{2})-cos(\alpha+\frac{3\beta}{2})\)

\(\implies 2.sin(\alpha+2\beta).sin\frac{\beta}{2}=cos(\alpha+2\beta-\frac{\beta}{2})-cos(\alpha+2.\beta+\frac{\beta}{2})\)

\(\bullet\)

\(\bullet\)

\(\bullet\)

\(\bullet\)

\(\bullet\)

\(\implies 2.sin(\alpha+(n-1)\beta).sin\frac{\beta}{2}=cos(\alpha+(n-1)\beta-\frac{\beta}{2})-cos(\alpha+(n-1).\beta+\frac{\beta}{2})\)

By adding, we get\(2.sin\frac{\beta}{2}[sin\alpha+sin(\alpha+\beta)+.....+sin(\alpha+(n-1)\beta)]=cos(\frac{\alpha-\beta}{2})-cos(\alpha+(\frac{n-1}{2}).\beta)\)

\(\implies 2.sin\frac{\beta}{2} \times S=2sin(\alpha+(\frac{n-1}{2}).\beta).sin\frac{n.\beta}{2}\) \[\implies S=sin\alpha + sin(\alpha + \beta) + ...... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. sin(\alpha + (n-1)\frac{\beta}{2})\]

In the similar way,we can prove\[cos\alpha + cos(\alpha + \beta) + ......+cos(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. cos(\alpha + (n-1)\frac{\beta}{2})\]To prove this, we will use the formula \(\boxed{2.cosA.cosB=sin(A+B)-sin(A-B)}\)

enjoy !

Log in to reply

Thanks for sharing! You are quite good at Latex

Log in to reply

@Calvin Lin @Michael Mendrin @Sandeep Bhardwaj @Sanjeet Raria @Sharky Kesa @Krishna Ar @Satvik Golechha @Finn Hulse @Daniel Liu @megh choksi

Log in to reply

and more.. (I can't write names of all of 'em).

Log in to reply

If you have some more of which you need derivations of, you can share them here!

Log in to reply

7. Proof for the integrating factor- \(I = {e}^{\int{p(x)} dx}\) where \(\frac{dy}{dx} + p(x)y = Q\) is the differential equation.Log in to reply

You sir, can sometimes be so overt that it's covert.

Log in to reply

ART OF IGNORING! (I also know it)

Log in to reply

HUH! I bet you're gonna read this.

Log in to reply

AoPS (Art of Problem Solving) has the answer to your question. That's where I learnt a lot of my formulae.

Log in to reply

Yeah! But it would be better if I get them here.

Log in to reply

Maybe, there should be wiki pages for these.

Log in to reply

Log in to reply