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

In summary: How can i get rid of the...remaining square and third power?You could try using the cosine formulas for the square and third power, like cos(mx)^3 and cos(mx)^2.
  • #71
Hi!:smile:
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:-
[tex]sinx=\frac{e^{ix}-e^{-ix}}{2i}[/tex]
 
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  • #72
Hi again! :smile:

Did you find Euler's formula eix = cos x + i sin x in your chapter?
 
  • #73
I like Serena said:
Hi again! :smile:

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

No there was no "Euler's formula" in my book. :frown:
 
  • #74
Pranav-Arora said:
No there was no "Euler's formula" in my book. :frown:

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." :smile:

The formula ehild mentioned follows from this formula.
 
  • #75
I like Serena said:
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." :smile:

The formula ehild mentioned follows from this formula.

I read it on Wikipedia, i think its not that hard as i thought of. :smile:
 
  • #76
Good! Indeed I never found it hard to understand! :)

Do you see how ehild's formula follows from it?
 
  • #77
I like Serena said:
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. :confused:
Also how did ehild get -8?:confused:
 
  • #78
Pranav-Arora said:
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. :confused:
Also how did ehild get -8?:confused:

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.
 
  • #79
I like Serena said:
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.:rolleyes:
Now i need to practice more questions on Complex Numbers.
 
  • #80
Well here's a nice (and fundamental) one: :smile:

Can you find the solutions of z3 = 1?
 
  • #81
I like Serena said:
Well here's a nice (and fundamental) one: :smile:

Can you find the solutions of z3 = 1?

Real solutions or complex solutions?
Real solution z=1
Complex Solution=?
 
  • #82
Complex solutions of course. :smile:

To find them substitute z = e and remember that ϕ is an angle with a period of 2π.
 
  • #83
I like Serena said:
Complex solutions of course. :smile:

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

Do i have to find the angle ϕ?
 
  • #84
Yes.
 
  • #85
Here are my steps:-
[tex](e^{iϕ})^3=1[/tex]
[tex]e^{3iϕ}=e^0[/tex]
[tex]3iϕ=0[/tex]
[tex]ϕ=0[/tex]

Right..?
 
  • #86
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 = ei(ϕ + 2kπ) for any integer k.

Can you repeat what you just did with ei(ϕ + 2kπ)?
 
  • #87
I like Serena said:
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 = ei(ϕ + 2kπ) for any integer k.

Can you repeat what you just did with ei(ϕ + 2kπ)?

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

Btw, i still didn't get why there are more solutions?
 
  • #88
Pranav-Arora said:
Btw, i still didn't get why there are more solutions?

Aren't you forgetting a power or rather factor of 3 here? ;)
 
  • #89
I like Serena said:
Aren't you forgetting a factor 3 here? ;)

What? :confused:
 
  • #90
Pranav-Arora said:
What? :confused:

Try:
(ei(ϕ + 2kπ))3 = 1

Or rather:
i(3ϕ + 2kπ) = 0
 
  • #91
I like Serena said:
Try:
(ei(ϕ + 2kπ))3 = 1

But i am still getting -2kπ.
 
  • #92
Yeah, I just edited my previous post...
 
  • #93
This time i am getting (-2kπ)/3...
But how do you get i(3ϕ+2kπ)=0?
 
  • #94
Look at it this way:

You already had:
ei(3ϕ) = 1

So:
cos(3ϕ) + i sin(3ϕ) = 1

This means that
cos(3ϕ) = 1 and sin(3ϕ) = 0

Can you solve that?
 
  • #95
And here's another way to look at it:

A complex number is represented as in the following picture.
180px-Euler%27s_formula.svg.png


An exponentiation with a power of 3, means a multiplication of the angle with a factor 3.
If you have an angle of 120 degrees, a factor of 3 will make it fully go round.
 
Last edited:
  • #96
I like Serena said:
Look at it this way:

You already had:
ei(3ϕ) = 1

So:
cos(3ϕ) + i sin(3ϕ) = 1

This means that
cos(3ϕ) = 1 and sin(3ϕ) = 0

Can you solve that?

ϕ=0.
I want to know why you take z=e? :confused:
 
  • #97
Pranav-Arora said:
ϕ=0.

Err... no.

The solution of cos(3ϕ) = 1
is 3ϕ=0 mod 2π,
which comes out as ϕ = 0 mod (2/3)π.

My bad, I thought you already knew this.
But now that I think about it, I recall that you did not know the "mod" notation yet.
So are you aware of the periodicity of the cosine function?
That it has a period 2π?
Pranav-Arora said:
I want to know why you take z=e? :confused:

Actually, I didn't do that quite right.
It should be:
z = r e

Any complex number z has 2 representations:
z = x + i y
z = r e

Are you aware of the first representation?
The second form follows from the first combined with Euler's formula.
 
Last edited:
  • #98
I like Serena said:
Err... no.

The solution of cos(3ϕ) = 1
is 3ϕ=0 mod 2π,
which comes out as ϕ = 0 mod (2/3)π.

My bad, I thought you already knew this.
But now that I think about it, I recall that you did not know the "mod" notation yet.
So are you aware of the periodicity of the cosine function?
That is has a period 2π?

Yes, i know about the periodicity of cosine function which is 2π.:smile:

I like Serena said:
Actually, I didn't do that quite right.
It should be:
z = r e

Any complex number z has 2 representations:
z = x + i y
z = r e

Are you aware of the first representation?
The second form follows from the first combined with Euler's formula.

I know about the first representation. :smile:
But i don't know about the second one. :frown:
 
  • #99
Pranav-Arora said:
Yes, i know about the periodicity of cosine function which is 2π.:smile:

So do you get that cos(3ϕ) = 1 has 3 solutions?


Pranav-Arora said:
I know about the first representation. :smile:
But i don't know about the second one. :frown:

Are you familiar with so called polar coordinates?
That is:
x = r cos(ϕ)
y = r sin(ϕ)
 
  • #100
I like Serena said:
So do you get that cos(3ϕ) = 1 has 3 solutions?
Yep, i found out three solutions using graph. :smile:
Is there any other way to find the number of solutions?

I like Serena said:
Are you familiar with so called polar coordinates?
That is:
x = r cos(ϕ)
y = r sin(ϕ)

No, i don't know about polar coordinates. :frown:
 
  • #101
Pranav-Arora said:
No, i don't know about polar coordinates. :frown:

Well, I guess we're chunking off a bit more than I originally thought.
But then, you seem so knowledgeable already! :smile:

Well, as long as you want to learn, that's fine by me.

Here's a picture that shows which r and phi I'm talking about (in the picture they use theta instead of phi though).

250px-Polar_to_cartesian.svg.png


If you have a point (x, y) it has a distance to the origin, which we call "r".
As you can see in the picture, you can define a rectangular triangle with an angle.
If you apply the definition of the cosine and the sine, you should be able to see that
x = r cos angle
y = r sin angle

That's it! :smile:
This is what we call polar coordinates, which is a different way to identify points in a plane.


Pranav-Arora said:
Yep, i found out three solutions using graph. :smile:
Is there any other way to find the number of solutions?

Erm... what are you thinking of?

The key is that any angle has a period of 2pi.
 
  • #102
I like Serena said:
Well, I guess we're chunking off a bit more than I originally thought.
But then, you seem so knowledgeable already! :smile:

Well, as long as you want to learn, that's fine by me.

Here's a picture that shows which r and phi I'm talking about (in the picture they use theta instead of phi though).

250px-Polar_to_cartesian.svg.png


If you have a point (x, y) it has a distance to the origin, which we call "r".
As you can see in the picture, you can define a rectangular triangle with an angle.
If you apply the definition of the cosine and the sine, you should be able to see that
x = r cos angle
y = r sin angle

That's it! :smile:
This is what we call polar coordinates, which is a different way to identify points in a plane.

Thank you for your explanation I Like Serena! :smile:
So now let's get back to the question i.e. z=re. :wink:

I like Serena said:
Erm... what are you thinking of?

The key is that any angle has a period of 2pi.

To find the solutions of cos3x=1, what i did was that i draw a graph of cos 3x, then draw a line parallel to x-axis at y=1 on the graph of cos 3x. The points where both the graphs intersect are the solutions for cos 3x=1.
This method is graphical but i want to know the other way to find the number of solutions for cos3x=1, i don't want to use graphical method.
 
  • #103
Pranav-Arora said:
Thank you for your explanation I Like Serena! :smile:
So now let's get back to the question i.e. z=re. :wink:

What do you get if you substitute Euler's formula?



Pranav-Arora said:
To find the solutions of cos3x=1, what i did was that i draw a graph of cos 3x, then draw a line parallel to x-axis at y=1 on the graph of cos 3x. The points where both the graphs intersect are the solutions for cos 3x=1.
This method is graphical but i want to know the other way to find the number of solutions for cos3x=1, i don't want to use graphical method.

Well, the algebraic method is to use the "mod 2pi" notation, or the "+2 k pi" notation, meaning that a cosine will be the same if you use an angle that is a multiple of 2pi bigger or smaller.

And another graphical method, is to look at the polar coordinates and consider what the angle should be to come out at (1, 0).

What do you know already of this?
 
  • #104
I like Serena said:
What do you get if you substitute Euler's formula?
Substituting Euler's formula:-
z=r(cosx+isinx)
What's this "r"?

I like Serena said:
Well, the algebraic method is to use the "mod 2pi" notation, or the "+2 k pi" notation, meaning that a cosine will be the same if you use an angle that is a multiple of 2pi bigger or smaller.

And another graphical method, is to look at the polar coordinates and consider what the angle should be to come out at (1, 0).

What do you know already of this?

I think i should not go to algebraic method.
75px-Puzzled.svg.png
 
  • #105
Pranav-Arora said:
Substituting Euler's formula:-
z=r(cosx+isinx)
What's this "r"?

The same r as in the polar coordinates.
It is the distance of point (X, Y) to the origin, where z = X + i Y.
Btw, I used capital letters to distinguish them from the "x" in your equation which is actually phi.

Taking it one step further, you have:
z = r cosx + i r sinx = X + i Y

Do you see now?


Pranav-Arora said:
I think i should not go to algebraic method.

Nice icon! :smile:

So you have not learned yet how to solve cos x = c?
 
<h2>1. What is the meaning of n in the identity sin^3xsin3x=\sum^n_{m=0}{}^nC_mcosmx?</h2><p>In this identity, n represents the number of terms in the summation on the right side of the equation. It is the highest degree of the cosine function that appears in the summation.</p><h2>2. How do you determine the value of n in this identity?</h2><p>The value of n can be determined by looking at the highest degree of the cosine function in the summation. This will be the value of n.</p><h2>3. What is the purpose of n in this identity?</h2><p>The value of n helps to determine the number of terms in the summation and the degree of the cosine function that appears in the summation. It allows for a more general representation of the identity.</p><h2>4. Can n be any value in this identity?</h2><p>No, n must be a non-negative integer. This is because the summation starts at m=0 and goes up to n, so n must be a non-negative value to ensure all terms are included.</p><h2>5. How does changing the value of n affect the identity?</h2><p>Changing the value of n will change the number of terms in the summation and the degree of the cosine function that appears. This will result in a different representation of the identity, but it will still hold true as long as n is a non-negative integer.</p>

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

In this identity, n represents the number of terms in the summation on the right side of the equation. It is the highest degree of the cosine function that appears in the summation.

2. How do you determine the value of n in this identity?

The value of n can be determined by looking at the highest degree of the cosine function in the summation. This will be the value of n.

3. What is the purpose of n in this identity?

The value of n helps to determine the number of terms in the summation and the degree of the cosine function that appears in the summation. It allows for a more general representation of the identity.

4. Can n be any value in this identity?

No, n must be a non-negative integer. This is because the summation starts at m=0 and goes up to n, so n must be a non-negative value to ensure all terms are included.

5. How does changing the value of n affect the identity?

Changing the value of n will change the number of terms in the summation and the degree of the cosine function that appears. This will result in a different representation of the identity, but it will still hold true as long as n is a non-negative integer.

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