Random math questions

[Deleted User]

Can you expand the induction step please?

If we subbed in n= k+1 into n^3 < 2^n, how would we get to what you wrote?

[brightsky]

2^n > n^3 --> note that this only holds for n greater or equal to 10.

[Deleted User]

Okay so how would I prove by induction that this is true for n>= 10 though?

[brightsky]

need to prove P(n): 2^n > n^3, for n>=10
check P(10): 2^10 > 10^3 or 1024 > 1000 is true
now assume P(k) is true, where k >= 10. we now need to prove that if this assumption is made, then P(k+1) is also true.
P(k): 2^k > k^3
P(k+1): 2^(k+1) > (k+1)^3 = k^3 + 3k^2 + 3k + 1
LHS = 2^(k+1)
=2*2^k
>2*k^3 (from inductive hypothesis)
= k^3 + k^3
>k^3 + 3k^2 + 3k + 1 (prove later that k^3 > 3k^2 + 3k + 1 for k>= 10)
=(k+1)^3
if all's well, then P(k+1) is also true, which means, according to the principle of maths induction, P(n) is true.

now we need to prove that k^3 > 3k^2 + 3k + 1, for k>= 10
now since k>= 10
k^3 = k*k^2 >= 10k^2 = 3k^2 + 3k^2 + 4k^2 which is obviously greater than 3k^2 + 3k + 1

QED

[kamil9876]

Here is a slightly quicker way to do the induction step, which is essentially what I did in the previous post. One can show that for we have . To prove this you can verify this holds for and then notice that it is a decreasing function of hence the expression remains less than . Hence can be proven via:

[#1procrastinator]

How would you show that when you rotate a square, the angles the edges make with the horizontal and vertical are equal? (i'm just assuming it's true :p)

[brightsky]

draw a square around the square you are rotating. you will get 4 congruent triangles.

[#1procrastinator]

is there a non-visual proof?

[kamil9876]

How would you show that when you rotate a square, the angles the edges make with the horizontal and vertical are equal? (i'm just assuming it's true :p)

Can you be a bit more specific? Is the rotation by any angle? Which horizontal and vertical are we talking about? And which angle specifically?

[#1procrastinator]

Can you be a bit more specific? Is the rotation by any angle? Which horizontal and vertical are we talking about? And which angle specifically?

i suppose so. by horizontal and vertical lines i mean the edges of the square you draw around it after rotation

[Deleted User]

How do I do d^4/dt^4 (e^(-t) cos(2t))? Thank you

[brightsky]

d^4/dt^4 (e^(-t) cos(2t))
= d^4/dt^4 (e^(-t) Re(e^(2it))
= Re [d^4/dt^4 (e^(-t)*e^(2it))]
= Re [d^4/dt^4 (e^(t(-1 + 2i)))]
= Re [(-1 + 2i)^4*e^(-t)*e^(2it)]
= e^(-t) Re [(-7+24i) (cos(2t) + i*sin(2t))]
= e^(-t) [-7cos(2t)-24sin(2t)]

[Jeggz]

How do I do d^4/dt^4 (e^(-t) cos(2t))? Thank you

Edit: Majorly Beaten 

[Deleted User]

Thanks guys. Also, can someone explain to me how

Re(1+sqrt(3)i)^51 can be written as -2^51

and

Im(1+i)^75 can be written as 2^37?

[polar]

Thanks guys. Also, can someone explain to me how

Re(1+sqrt(3)i)^51 can be written as -2^51

and

Im(1+i)^75 can be written as 2^37?

hint: use de moivre's theorem