Advanced Maths

Details: Last Updated: Tuesday, 05 November 2024 02:54; Published: Tuesday, 18 July 2023 17:23; Hits: 347

Advanced Maths

This contains general topics in High School Maths that are not covered anywhere else. Calculus has it's own section.

Induction:

Induction is a vey common technique used to prove theorems that apply to natural numbers n. In absence of induction, such formula are really tedious to proce, but with Induction, they become a piece of cake. Some Maths Olympiad questions involve prove by Induction, so it's good to know this.

Wiki: https://en.wikipedia.org/wiki/Mathematical_induction

Proof by Induction: Given a Statement P(n) true for every natural number n (or for all natural numbers n >= N), we can use induction to prove it. This is done in 2 steps:

First prove that the statement is true for n=0 or n=N (if P(n) is true for n>=N).
Now assume P(n) is true for some n=k. Now, we try to prove that stmt holds for n=(k+1). We find P(n+1) by using P(n) and other maths transformations. We rearrange the terms so that this finally comes in same form as P(n+1). So, this proves that if P(n) holds, then P(n+1) will also hold. Since P(0) is already shown to be true, P(n) is true for all n. NOTE: We can't use formula for P(n+1) while finding P(n+!) since we haven't proved P(n+1) yet.

Sum of 1st N natural numbers: Prove by induction that sum of 1st n natural numbers is n(n+1)/2

This is the most common problem used to prove by induction. Here we have P(n)=n(n+1)/2.

We prove for n=1, i.e P(1)=1(1+1)/2=1 => True

Assume P(n)-n(n+1)/2. So P(k)=k(k+1)/2.

Sum of 1st (k+1) natural numbers = P(k)+k+1 = k(k+1)/2 + (k+1) = (k+1)(k+2)/2 = (k+1)((k+1)+1)/2. This is actually same as P(k+1) as P(k+1)=(k+1)(k+2)/2.

So, assuming Sum of 1st (k) natural numbers = P(k), then Sum of 1st (k+1) natural numbers turns out to be = P(k+1). So, if P(1) is true, then P(2) is true. If P(2) is true then P(3) is true and so on till infinity.

Sum of Squares of 1st N natural numbers: Prove by induction that sum of squares of1st n natural numbers is n(n+1)(2n+1)/6

Here we have P(n)=n(n+1)(2n+1)/6.

We prove for n=1, i.e P(1)=1*2*3/6=1 => True

Assume P(n)-n(n+1)(2n+1)/6. So P(k)=k(k+1)(2k+1)/6.

Sum of squares of 1st (k+1) natural numbers = P(k)+(k+1)^2 = k(k+1)(2k+1)/6 + (k+1)^2 = (k+1)[k(2k+1)/6 + (k+1)] = (k+1)[2k^2+7k+6]/6 = (k+1)[(2k+3)(k+2)]/6. This is actually same as P(k+1) as P(k+1)=(k+1)(k+2)(2k+3)/6. Hence proved.

Similarly for sum of cubes of 1st n numbers can be proved by induction.

Complex Numbers:

First of all, complex numbers are not complex at all. Infact, problems involving complex numbers are usually simple to solve. There's always at least on question on Complex Numbers in US Maths olympiad, and it's relatively easy to solve. So, you should definitely master this.

Secondly, there's no real use of complex numbers in real life. They are a way to solve complicated equations in Physics, electronic, etc by transforming to complex plane. In this maths section, we'll just look at how to work with these.

Wiki => https://en.wikipedia.org/wiki/Complex_number

Imaginary number = √(-1) = i (English letter i)

i^n = ?

i^1 = i
i^2 = -1
i^3 = i^2*i = -i
i^4 = i^2*i^2 = 1
i^n = +1, +i, -1, -i, depending on whether n/4 gives a remainder of 0,1,2,3

Complex number is any number with 2 parts: a real number (Re) added to an imaginary number (Im). Represented by Z. Z=a+b*i => Here a and b are both real, where a is the real part of Z and b is the imaginary part of Z. We usually plot complex numebrs on x,y ais, where x axis is the real part of Z, and Y axis is the Im part of Z. So, now we can plot coordinates of X not only in cartesian coordinates as (a,b), but also in polar coordinates (r,theta). Th polar form is much more helpful in solving complex arithmetic. Magnitude of Z (or r in polar coord) is defined as distance from origin r = √(a^2+b^2) and angle theta is tan^-1 (b/a). Magnitude of Z is also noted as |Z|

Addition/subtraction = Z1+Z2 = (a1+b1*i) + (a2+b2*i) = (a1+a2) + (b1+b2)*i => (real parts add/sub and Im parts add/sub)

Multiplication: Z1*Z2 = (a1+b1*i) * (a2+b2*i) => Use algebraic multiplication to get (a1*a2 - b1*b2) + (a1*b2 + a2*b1)*i

Division: Z1/Z2 = (a1+b1*i) / (a2+b2*i) => Multiply both top and bottom with (a1-b2*i) to get rid of complex number in denominator. [(a1+b1*i) *(a2-b2*i)] / [(a2+b2*i)*(a2-b2*i)] = (a1*a2 + b1*b2) + (a1*b2 - a2*b1)*i / √(a2^2 + b2^2)

Z = rCosθ + i* rSinθ (in polar form for any complex number). It's also written as Z = rCisθ or r∠θ

It's much easier to multiply and divide polar numbers by using polar form.

Z1*Z2 = ( r1Cosθ1 + i* r1Sinθ1 ) * ( r2Cosθ2 + i* r2Sinθ2 ) = r1*r2 [ ( Cosθ1.Cosθ2 - Sinθ1.Sinθ2) + i (Sinθ1.Cosθ2 + Cosθ1.Sinθ2) = r1.r2 [ Cos(θ1+θ2) + i Sin(θ1+θ2) ] => In polar form, magnitude multiply while angles add.

Similarly Z1/Z2 = ( r1Cosθ1 + i* r1Sinθ1 ) / ( r2Cosθ2 + i* r2Sinθ2 ) => Multiply both top and bottom by ( r2Cosθ2 - i* r2Sinθ2 ) to get rid of complex number in denominator => r1/r2 [ ( Cosθ1 + i* Sinθ1 ) * ( Cosθ2 - i* Sinθ2 ) ] / ( 1 )] = r1/r2 [ ( Cosθ1.Cosθ2 + Sinθ1.Sinθ2) + i (Sinθ1.Cosθ2 - Cosθ1.Sinθ2) = r1/r2 [ Cos(θ1-θ2) + i Sin(θ1-θ2) ] => In polar form, magnitude divide while angles subtract.

We can extend above observation to n complex numbers, where angles add/sub and magituddes mult/div.

i.e Za1.Za2....Zan / (Zb1.Zb2... Zbn) = ra1.ra2...ran/(rb1.rb2...rbn) . Cis [ θa1+θa2...+θan - (θb1+θb2+...+θbn) ]

De Moivre's Theorem:

It's a special case of above, where power n of any complex number is Z^n = r^n Cis nθ

Link => https://en.wikipedia.org/wiki/De_Moivre%27s_formula

To find n-th roots of Z, we can use the above formula, where Z^(1/n) = r^(1/n) Cis [ (θ+2πk)/n] where k is an integer from 0 to (n-1). As the fundamental theorem of Algebra says that there has to be n roots, we get n solutions. The term 2πk can be added to the angle of any complex number, but it remains the same number. But when we divide by n, then the angle 2π divided by n gives new angles within the 360 degrees yielding unique points.

Though De Moivre's theorem as written above is not true for non integer values of n, it was proved by Euler that it holds for even complex number n.

If $z = ρ e^{i θ}$ then $z^{p} = ρ^{p} e^{i θ p}$ .where p is any real or imaginary number.

Proof is from Euler’s Theorem (which is very elegantly proved from Maclaurin Series), which says: $e^{i θ} = \cos θ + i \sin θ$

So, letting $r$ be any real/complex number: $(\cos θ + i \sin θ)^{r} = (e^{i θ})^{r} = e^{i (r θ)} = \cos (r θ) + i \sin (r θ)$

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Advanced Maths