What is the value of n in the identity sin^3xsin3x=\sum^n_{m=0}{}^nC_mcosmx?

[ehild]

Hi, tiny-tim, is not there a sign error in your derivation?

Just to show off my way:

sinx=\frac{e^{ix}-e^{-ix}}{2i},
cosx=\frac{e^{ix}+e^{-ix}}{i},

sin(x)^3 sin(3x)=(\frac {e^{ix}-e^{-ix}}{2i})^3 \frac{e^{3ix}-e^{-3ix}}{2i}=

=\frac {e^{3ix}-3 e^{ix}+3 e^{-ix}-e^{-3ix}}{-8i}\frac{e^{3ix}-e^{-3ix}}{2i}=
1/16(e^{6ix}-3 e^{4ix}+3 e^{2ix}-1-1+3 e^{-2ix}-3 e^{-4ix}+e^{-6ix})=
1/8(cos(6x)-3cos(4x)+3cos(2x)-1)

ehild

 

[I like Serena]

@tiny-tim: Nice improvement on the calculation btw. It's way less work!

Just to show off my way:

Thank you ehild, this is a nice solution, and shows a couple of things.

First, I think that Pranav-Arora is very well able to follow this one and do it himself.
It only requires a basic grasp on complex numbers, and beyond that it's just exponentiation rules, which Pranav-Arora's has already practiced quite a bit.

And it also shows that in my humble opinion calculating with sines, cosines and angles is kind of obsolete.
I found that calculating with imaginary e-powers (or vectors in geometric problems) is much easier. :shy:

I'm interested in examples where you would really want to use sines, cosines and angles to do your calculations...
Does any of you have any?

 

[ehild]

And it also shows that in my humble opinion calculating with sines, cosines and angles is kind of obsolete.
I found that calculating with imaginary e-powers (or vectors in geometric problems) is much easier. :shy:

I'm interested in examples where you would really want to use sines, cosines and angles to do your calculations...
Does any of you have any?

What about determining the height of a tower from the angle of view at a known distance?

ehild

 

[I like Serena]

What about determining the height of a tower from the angle of view at a known distance?

ehild

Yes, it's still useful for immediate applications of the definitions of sines, cosines and tangents based on an angle.
What I'm looking for is actually calculating with it.
Usually I convert an angle as soon as I can to a vector, then calculate with it, and possibly convert it back into an angle to present a result.
To me it feels a bit like converting inches to metric before calculating, and converting back to present a result.

And even here...
If you measure an angle, you use a tangent to convert distance to height.
But what you actually measure, is a distance on a measuring device that is marked as an angle, but it might as well be marked as a distance at a specific short distance, or as a tangent ratio.
Multiply by the ratio in distance and you get the height without ever using an angle or the tangent function.

 

[tiny-tim]

hi ehild!

yes that's nice ​

and it's certainly shorter for k = 3 (but not so good for large k)

 

[icystrike]

Please try to use euler's forumla and u will realize that it is made of some even m (2,4,6).

So n=6

 

[I like Serena]

and it's certainly shorter for k = 3 (but not so good for large k)

Let's see.

For k is odd we have:
\begin{eqnarray}<br /> \sin^kx \sin kx<br /> &amp;=&amp; \left(\frac {e^{ix} - e^{-ix}} {2i} \right)^k \left(\frac {e^{ikx} - e^{-ikx}} {2i} \right) \\<br /> &amp;=&amp; \frac 1 {(2i)^{k+1}} \left(\sum_{m=0}^k \binom k m (-1)^m e^{i(k-2m)x} \right) (e^{ikx} - e^{-ikx}) \\<br /> &amp;=&amp; \frac 1 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m (e^{i(2k-2m)x} + e^{-i(2k-2m)x}) \\<br /> &amp;=&amp; \frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m \cos(2k-2m)x \\<br /> &amp;=&amp; (-1)^{\frac {k+1} 2} \frac 1 {2^k} \left(\binom k 0 \cos 2kx - \binom k 1 \cos(2k-2)x + \binom k 2 \cos(2k-4)x - ... + \binom k {k-1} \cos 2x - 1 \right)<br /> \end{eqnarray}

As you can see, I've also corrected the sign error.

 

[tiny-tim]

mmm … let's compare length …

sin^kxsinkx = sin^kx Im((cosx + isinx)^k)

= Im((cosxsinx + isin²x)^k)

= (1/2^k) Im((±sin2x + i(1 - cos2x))^k)

= (1/2^k) Im(i^k(1 - cos2x ± isin2x)^k)

(I lost an i² here originally )

= (1/2^k) Im(i^k(1 - e^2ix)^k)

= (1/2^k) Im(i^k ∑ ^kC_m (-1^m) e^2imx)

= (1/2^k)(-1)^(k-1)/2 ∑ ^kC_m (-1^m) cos2mx for k odd

= (1/2^k)(-1)^k/2 ∑ ^kC_m (-1^m) sin2mx for k even​

ooh, yes, yours is 2 lines shorter

\frac 1 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m (e^{i(2k-2m)x} + e^{-i(2k-2m)x})
\frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m \cos(2k-2m)x

oh, that's how it works! nice!

(and for k even, there's a sign-change somewhere)

though we can then use ^kC_m = ^kC_k-m to go straight to …

\frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^{k-m} \cos2mx

 

[ehild]

I just wanted to show this, but you both beat me while I watched Poirot on TV

ehild

 

[ehild]

If you measure an angle, you use a tangent to convert distance to height.
But what you actually measure, is a distance on a measuring device that is marked as an angle, but it might as well be marked as a distance at a specific short distance, or as a tangent ratio..

As far as I know the angle of view is measure of the length of an arc between the arms of a compass, when those arms point to the desired directions. And the tangent of that angle is used to get distance or height.

ehild

 

[Saitama]

Hi!
I had read the complex numbers chapter from my textbook and i think i now have a basic idea of them. So may i know how ehild has got this relationship:-
sinx=\frac{e^{ix}-e^{-ix}}{2i}

 

[I like Serena]

Hi again!

Did you find Euler's formula e^ix = cos x + i sin x in your chapter?

 

[Saitama]

Hi again!

Did you find Euler's formula e^ix = cos x + i sin x in your chapter?

No there was no "Euler's formula" in my book.

 

[I like Serena]

No there was no "Euler's formula" in my book.

That's a problem then, since this identity is the most important one in the entire field of the complex numbers.

Richard Feynman called Euler's formula "our jewel" and "one of the most remarkable, almost astounding, formulas in all of mathematics."

The formula ehild mentioned follows from this formula.

 

[Saitama]

That's a problem then, since this identity is the most important one in the entire field of the complex numbers.

Richard Feynman called Euler's formula "our jewel" and "one of the most remarkable, almost astounding, formulas in all of mathematics."

The formula ehild mentioned follows from this formula.

I read it on Wikipedia, i think its not that hard as i thought of.

 

[I like Serena]

Good! Indeed I never found it hard to understand! :)

Do you see how ehild's formula follows from it?

 

[Saitama]

Good! Indeed I never found it hard to understand! :)

Do you see how ehild's formula follows from it?

I was checking out the ehild's solution, i got confused at one step when ehild multiplied -8 with 2 in the denominator. In the next step there was no minus sign.
Also how did ehild get -8?

 

[I like Serena]

I was checking out the ehild's solution, i got confused at one step when ehild multiplied -8 with 2 in the denominator. In the next step there was no minus sign.
Also how did ehild get -8?

The other thing you need to know about complex numbers is the definition of i:
i² = -1

So (2i)³ = 2³ i² i = 8 (-1) i = -8i.

 

[Saitama]

The other thing you need to know about complex numbers is the definition of i:
i² = -1

So (2i)³ = 2³ i² i = 8 (-1) i = -8i.

Oh sorry! I am really a fool to ask this question.
Now i need to practice more questions on Complex Numbers.

 

[I like Serena]

Well here's a nice (and fundamental) one:

Can you find the solutions of z³ = 1?

 

[Saitama]

Well here's a nice (and fundamental) one:

Can you find the solutions of z³ = 1?

Real solutions or complex solutions?
Real solution z=1
Complex Solution=?

 

[I like Serena]

Complex solutions of course.

To find them substitute z = e^iϕ and remember that ϕ is an angle with a period of 2π.

 

[Saitama]

Complex solutions of course.

To find them substitute z = e^iϕ and remember that ϕ is an angle with a period of 2π.

Do i have to find the angle ϕ?

 

[I like Serena]

Yes.

 

[Saitama]

Here are my steps:-
(e^{iϕ})^3=1
e^{3iϕ}=e^0
3iϕ=0
ϕ=0

Right..?

 

[I like Serena]

Almost. :)

As I said, you have to take into account that ϕ is an angle with a period of 2π.
It means that there are more solutions than just this one.

So e^iϕ = e^{i(ϕ + 2kπ)} for any integer k.

Can you repeat what you just did with e^{i(ϕ + 2kπ)}?

 

[Saitama]

Almost. :)

As I said, you have to take into account that ϕ is an angle with a period of 2π.
It means that there are more solutions than just this one.

So e^iϕ = e^{i(ϕ + 2kπ)} for any integer k.

Can you repeat what you just did with e^{i(ϕ + 2kπ)}?

Ok, i repeat the steps:-
i(ϕ+2kπ)=0
ϕ=-2kπ

Btw, i still didn't get why there are more solutions?

 

[I like Serena]

Btw, i still didn't get why there are more solutions?

Aren't you forgetting a power or rather factor of 3 here? ;)

 

[Saitama]

Aren't you forgetting a factor 3 here? ;)

What?

 

[I like Serena]

What?

Try:
(e^{i(ϕ + 2kπ)})³ = 1

Or rather:
i(3ϕ + 2kπ) = 0