q help (1 Viewer)

Anounymouse352 · Apr 2, 2022

Hey, guys any help with following questions would be greatly appreciated

jimmysmith560 · Apr 2, 2022

Would the following working help with the question in the fourth attachment?

cossine · Apr 3, 2022

Anounymouse352 said:
Hey, guys any help with following questions would be greatly appreciated
View attachment 35321
View attachment 35322 View attachment 35323 View attachment 35324 View attachment 35325

For the first picture you could use the taylor series expansion of sin x around x=0

Lith_30 · Apr 3, 2022

For the second attachment

part i:

$\\LHS-RHS=\frac{1}{a}+\frac{1}{b}-\frac{2}{3}\\\\=\frac{3b+3a-2ab}{3ab}\\\\=\frac{3(b+a)-2ab}{3ab}\\\\=\frac{0.5(b+a)(b+a)-2ab}{3ab}\\\\=\frac{\frac{(b^2+2ab+a^2)}{2}-2ab}{3ab}\\\\=\frac{\frac{(b^2+a^2)}{2}-ab}{3ab}\\\\\geq\frac{\sqrt{a^2b^2}-ab}{3ab}\ \text{AM GM inequality}\\\\\geq0$

hence $\frac{1}{a}+\frac{1}{b}\geq\frac{2}{3}$

For part ii it is pretty much the same process, do LHS-RHS, combine all the fractions then use AM GM to make it equal zero.

ExtremelyBoredUser · Apr 3, 2022

Anounymouse352 said:
Hey, guys any help with following questions would be greatly appreciated

View attachment 35323

i)
$(1 + i)^n = (\sqrt{2})^n \times e^{i(\frac{n\pi}{4})}$
$(1 - i)^n = (\sqrt{2})^n \times e^{-i(\frac{n\pi}{4})}$

Summing these together, we get the RHS;

let theta equal npi/4
$RHS = (\sqrt{2})^n \times (e^{i\theta} + e^{-i\theta})$
$RHS = (\sqrt{2})^n \times (e^{i\theta} + e^{-i\theta})$
$RHS = (\sqrt{2})^n \times (\cos(\theta) + i\sin(\theta) + \cos(-\theta) + i\sin(-\theta)$
$RHS = (\sqrt{2})^n \times (\cos(\theta) + i\sin(\theta) + \cos(\theta) - i\sin(\theta)$
$RHS = (\sqrt{2})^n \times (2\cos(\theta)$
$RHS = (\sqrt{2})^n \times ((\sqrt{2})^n \times \cos(\theta)$
$RHS = (\sqrt{2})^{n+2} \times \cos(\frac{n\pi}{4})$

Lith_30 · Apr 3, 2022

Anounymouse352 said:

for part ii)

we'll start with
$\\(1+i)^n+(1-i)^n\ \text{Where n is a multiple of 4}\\\\={n\choose0}+{n\choose1}i-{n\choose2}-{n\choose3}i+{n\choose4}-...+{n\choose{n}}+{n\choose0}+{n\choose1}(-1)i-{n\choose2}-{n\choose3}(-1)^3i+{n\choose4}+...+{n\choose{n}}\ \text{Binomial expansion}\\\\={n\choose0}+{n\choose1}i-{n\choose2}-{n\choose3}i+{n\choose4}-...+{n\choose{n}}+{n\choose0}-{n\choose1}i-{n\choose2}+{n\choose3}i+{n\choose4}+...+{n\choose{n}}$

As you may be able to see all the imaginary components cancel each other out and the real components add onto each other, as it should.

$\\2\left({n\choose0}-{n\choose2}+{n\choose4}-...+{n\choose{n}}\right)=(1+i)^n+(1-i)^n\\\\2\left({n\choose0}-{n\choose2}+{n\choose4}-...+{n\choose{n}}\right)=(\sqrt{2})^{n+2}\cos(\frac{n\pi}{4})\\\\\left({n\choose0}-{n\choose2}+{n\choose4}-...+{n\choose{n}}\right)=\sqrt{2}^n\cos(\frac{n\pi}{4})$

we have the $\sqrt{2}^n$ , but what can we do with the cos function to get what we want??

Since we know that n is a multiple of 4, we can let $n=4k, k\in\mathbb{Z},k\geq0$

hence $\cos(\frac{n\pi}{4})=\cos(k\pi)=1,\ -1,\ 1,\ -1...$ (starting at n=0)

therefore we can say that

$\\\cos(k\pi)=(-1)^k\\\\\cos(\frac{n\pi}{4})=(-1)^{\frac{n}{4}}$

therefore ${n\choose0}-{n\choose2}+{n\choose4}-...+{n\choose{n}}=\sqrt{2}^n(-1)^{\frac{n}{4}}$ as required

q help (1 Viewer)

Anounymouse352

New Member

jimmysmith560

Le Phénix Trilingue

cossine

Well-Known Member

Lith_30

o_o

ExtremelyBoredUser

Bored Uni Student

Lith_30

o_o

Users Who Are Viewing This Thread (Users: 0, Guests: 1)