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Need derviation of some formulas! Help!

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. :)

Note by Kartik Sharma
2 years, 8 months ago

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To 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 :

\(cos(x)+isin(x)=e^{ix}\) Ronak Agarwal · 2 years, 8 months ago

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@Ronak Agarwal I Love Your way To prove This equation ! Hats Off!! I 'am Amazed How Can any person down-votes it ?? Deepanshu Gupta · 2 years, 8 months ago

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

Sorry if you didn't like the proof. Ronak Agarwal · 2 years, 8 months ago

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@Ronak Agarwal I think when x=0 , then at the same time z cannot be equal to one. Sneha Sharma · 2 years, 7 months ago

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@Ronak Agarwal 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? Kartik Sharma · 2 years, 8 months ago

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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 \[\boxed{cosx+isinx=e^{ix}}\] Sandeep Bhardwaj · 2 years, 8 months ago

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@Sandeep Bhardwaj Can you tell me why does Taylor series make sense? How to find them? *if you don't mind Kartik Sharma · 2 years, 8 months ago

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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 Megh Choksi · 2 years, 8 months ago

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@Megh Choksi Nice Thanks! I solved that problem! Maybe I would give my solution too! Kartik Sharma · 2 years, 8 months ago

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Comment deleted Nov 12, 2014

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@Megh Choksi Thanks! Nice! Try other ones too! Kartik Sharma · 2 years, 8 months ago

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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 ! Sandeep Bhardwaj · 2 years, 8 months ago

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@Sandeep Bhardwaj Thanks for sharing! You are quite good at Latex Kartik Sharma · 2 years, 8 months ago

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7. Proof for the integrating factor - \(I = {e}^{\int{p(x)} dx}\) where \(\frac{dy}{dx} + p(x)y = Q\) is the differential equation. Kartik Sharma · 2 years, 8 months ago

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If you have some more of which you need derivations of, you can share them here! Kartik Sharma · 2 years, 8 months ago

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@Kartik Sharma and more.. (I can't write names of all of 'em). Kartik Sharma · 2 years, 8 months ago

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You sir, can sometimes be so overt that it's covert. Saurabh Chauhan · 2 years, 8 months ago

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@Saurabh Chauhan ART OF IGNORING! (I also know it) Kartik Sharma · 2 years, 8 months ago

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@Kartik Sharma HUH! I bet you're gonna read this. Saurabh Chauhan · 2 years, 8 months ago

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AoPS (Art of Problem Solving) has the answer to your question. That's where I learnt a lot of my formulae. Sharky Kesa · 2 years, 8 months ago

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@Sharky Kesa Yeah! But it would be better if I get them here. Kartik Sharma · 2 years, 8 months ago

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@Kartik Sharma Maybe, there should be wiki pages for these. Sharky Kesa · 2 years, 8 months ago

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@Sharky Kesa YEAH! Agree! Kartik Sharma · 2 years, 8 months ago

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