Rom

\(i^{4k+2}+i^{4k+4}=0 \\ i^{4k+1}+i^{4k+3}=0\)

\(\text{we can rewrite the sum as}\\ (i + i^3) + (i^2 + i^4) + (i^5 + i^7) + (i^6 + i^8) + \dots + (i^{98}+i^{100}) +(i^{97}+ i^{99}) - i^{100} = -i^{100} \\ -i^{100} =- i^{4(25)} = -1\)

if you start with one layer and double the number of layers with each fold, and noting that

\(2^{10}=1024\)

it's clear that 10 folds is 1024 layers

if you wanted some number of layers that was enormous say N you would find

\(folds = \left \lceil \log_2(N) \right \rceil\)

where

\(\lceil x \rceil = \text{ the least integer }k,~\ni k > x\)

\(\dfrac{w+z}{1+w z}= \dfrac{(w+z)(1+w z)^*}{(1+w z)(1+w z)^*} = \\ \dfrac{w+z + w(wz)^* + z(wz)^*}{|1+wz|^2}\)

The denominator is real so we just need to show the numerator is also real

\(\text{Noting that }(wz)^* = w^* z^* \text{ we have}\\ w+z+w(wz)^*+z(wz)^* = \\ w + z + w w^* z^* + z z^* w^* = \\ w + z + 1\cdot z^* + 1 \cdot w^* = \\ (w+w^*)+(z+z^*) = 2Re(w) + 2Re(z) = \\ 2Re(w+z) \in \mathbb{R} \)

\(\text{Let's say I can finish the job in }n \text{ hours} \\ \text{thus I work at a rate of }\dfrac{1~job}{n~hr} \\ \text{You finish the job in (}n-2) \text{ hours}\\ \text{and thus work at a rate of }\dfrac{1~job}{(n-2)~hr}\)

\(\text{When working together the rates combine.}\\ \text{I work 1 hour at a rate of }\dfrac{1~job}{n~hr} \\ \text{and then work 3 hours at a rate of } \left(\dfrac{1}{n}+\dfrac{1}{n-2}\right) \dfrac {job}{hr} \text{completing the job}\)

\(1~job = \left(\dfrac {1~job}{n~hr} \cdot 1~hr+\left(\dfrac 1 n+\dfrac{1}{n-2} \right)\dfrac{job}{hr} \cdot 3~hr\right)\)

dropping units for a moment

\(1 = \dfrac 1 n +\dfrac{2n-2}{n(n-2)}\cdot 3\)

\(n^2 - 2n = 7n-8 \\ n^2 -9n+8 = 0 \\ (n-1)(n-8) = 0 \\ n=1,~8\)

\(\text{n can't be 1 as that would make your partner finish the job in (-1) hours}\\ \text{so it must be that }n=8~ hrs\)

Ok.. if we have a resistance of R ohms in a piece of perfect wire we will hve R/2 ohms in half that length of wire.

If we pass 12 volts through and measure 100 mA of current we have 12/(0.1) = 120 Ohms of resistance in the wire.

If we cut it in half we will only have 60 Ohms of resistance in the wire.

So if we pass 24 volts through it we will measure 24v/60ohs = 0.4 Amps = 400 mA of current

I can also quickly see this by noting that we halve the resistance and double the voltage.

2/(1/2) = 4, so we increase the original current by a factor of 4.

I'D SAY THEY ARE 25...... AND 140!!!!!!!

the word is modulus

\(|Z| = \left | \dfrac 1 Z \right | = | 1 - Z |\)

first off you should know that

 \(\left | \dfrac 1 Z \right | = \dfrac {1}{| Z |} \\ \\ \text{so } |Z| = \left|\dfrac 1 Z \right| \Rightarrow |Z|=1\)

\(|1-Z|=1 \\ \\ \sqrt{(1-Z)(1-Z^*)} = 1 \\ \\ (1-Z)(1-Z^*)=1 \\ \\ 1 -Z -Z^* + Z Z^* = 1\)

\(\text{but }Z Z^* = |Z|^2 = 1 \text{ so} \\ 1-Z-Z^* + 1 = 1 \\ Z+Z^* = 1 \\ 2 Re(Z) = 1 \\ Re(Z) = \dfrac 1 2\)

\(1 = 1^2 = |Z|^2 = Re(Z)^2 + Im(Z)^2 = \left(\dfrac 1 2\right)^2 + Im(Z)^2 \\ \\ Im(Z)^2 = \dfrac 3 4 \\ \\ Im(Z) = \pm \dfrac{\sqrt{3}}{2} \\ \\ Z = \dfrac 1 2 (1 \pm i\sqrt{3})\)

\(\text{For a quadratic equation }a x^2 + b x + c = 0 \\ \\ \text{a non-negative discriminant, i.e. }\\ D=b^2 - 4 a c \\ \text{indicates real roots}\)

\(\text{the discriminant in this case is } \\ D=64+16 = 80 > 0 \\ \text{and thus we have real roots}\)

\(f(0)=2 \Rightarrow c=2 \\ \\ \text{Then we have }a+b+2=-1 \Rightarrow b=-3-a\)

\(\text{ so we have }f(x) = a x^2 -(3+a)x + 2\)

\(\text{This has zeros at }\\ x = \dfrac{(3+a)\pm \sqrt{(3+a)^2-8a}}{2a}\)

\(\text{subtracting we have}\\ \dfrac{\sqrt{(3+a)^2-8a}}{a}=\dfrac{\sqrt{9-2a+a^2}}{a} = 2\sqrt{2}\)

\(\dfrac{9-2a+a^2}{a^2}= 8 \\ \\ 7a^2+2a-9=0 \\ \\ a=\dfrac{-2\pm \sqrt{4+252}}{14}=\dfrac{-2\pm 16}{14} = 1,~-\dfrac 9 7\)

\(f(x) = x^2 -4x+2 \text{ or }-\dfrac 9 7 x^2 - \dfrac{12}{7}+2\)

I think this reads

\(x \geq -1,~\text{ Prove using induction that }(1+x)^n \geq 1+nx\)

\(P_1 = (1+x)^1 \geq 1+(1)x \\ \\ P_1 = (1+x) \geq (1+x) = TRUE\)

\(\text{Assume }P_n \text{ is TRUE} \\ \\ (1+x)^{n+1} = (1+x)^n(1+x) \geq (1+n x)(1+x) =\\ 1+(n+1)x+nx^2 \geq 1+(n+1)x \\ \\ \text{Thus }P_n \Rightarrow P_{n+1} \bigtriangleup\)