Complex Numbers

[Vee3020]

Calculate the sum:
S = cosϕ + acos 2ϕ + a ^ 2 cos 3ϕ +… + a ^ (n-1) cos nϕ

I think this question is related to Euler's number? Like, if I substitute cosϕ as e^iϕ , then it will be a series which will have cosϕ + acos 2ϕ + a ^ 2 cos 3ϕ +… + a ^ (n-1) cos nϕ as it's real part. Is the logic correct?

I need step wise solution to this since I'm stuck.

 

[Vee3020]

I posted a question here yesterday since I was stuck on it. But unfortunately, no one has replied to me or even gave me a hint about solving it. I have even include the thought that I put into solving the question but no one has taken notice of it.

It was kind of urgent for me. Questions which got posted after m

 

[Deleted member 4993]

Calculate the sum:
S = cosϕ + acos 2ϕ + a ^ 2 cos 3ϕ +… + a ^ (n-1) cos nϕ

I think this question is related to Euler's number? Like, if I substitute cosϕ as e^iϕ , then it will be a series which will have cosϕ + acos 2ϕ + a ^ 2 cos 3ϕ +… + a ^ (n-1) cos nϕ as it's real part. Is the logic correct?

I need step wise solution to this since I'm stuck.

You are thinking in the correct way - sort of. Let us define a new series:

B = e^(iΦ) [ 1 + a*e^(iΦ) + {a*e^(iΦ)}^2 + {a*e^(iΦ)}^3 ......]

Now you have a geometric series. The real part of "B" will be the solution you are seeking.

Continue......

 

[Vee3020]

Thankyou so much Sir! Can you also tell me what the solution will be? I'm studying this topic on my own so I'm clearing up my concepts. If you don't mind, can I get a step wise solution. Please?

 

[Vee3020]

Calculate the sum:
S = cosϕ + acos 2ϕ + a ^ 2 cos 3ϕ +… + a ^ (n-1) cos nϕ

I think this question is related to Euler's number? Like, if I substitute cosϕ as e^iϕ , then it will be a series which will have cosϕ + acos 2ϕ + a ^ 2 cos 3ϕ +… + a ^ (n-1) cos nϕ as it's real part. Is the logic correct?

I need step wise solution to this since I'm stuck.

Thankyou so much sir! But can you please explain me the solution step wise? And the answer as well. I'm learning this on my own and I'm still working to clear my concepts. So I don't know how to apply the method even though I can tell which method to use to solve it. Can you please include the answer as well? Please?

 

[Deleted member 4993]

Thankyou so much sir! But can you please explain me the solution step wise? And the answer as well. I'm learning this on my own and I'm still working to clear my concepts. So I don't know how to apply the method even though I can tell which method to use to solve it. Can you please include the answer as well? Please?

We will help you find the answer - but you will have to FIND it.

What is the sum of following:

x + x*r + x*r^2 + x*r^3 + ... + x*r^n = ?

 

[Vee3020]

We will help you find the answer - but you will have to FIND it.

What is the sum of following:

x + x*r + x*r^2 + x*r^3 + ... + x*r^n = ?

That would be : x * (1 - r^n+1 / 1-r )

Right?

 

[topsquark]

That would be : x * (1 - r^n+1 / 1-r )

Right?

Parentheses!
x*(1 - r^(n + 1))/(1 - r)

I'll write what you did in LaTeX to make it clearer:
[math]x * \left ( 1 - r^n +\dfrac{1}{1} - r \right )[/math]
-Dan

 

[Vee3020]

Parentheses!
x*(1 - r^(n + 1))/(1 - r)

I'll write what you did in LaTeX to make it clearer:
[math]x * \left ( 1 - r^n +\dfrac{1}{1} - r \right )[/math]
-Dan

Yes! Thankyou!

 

[Deleted member 4993]

That would be : x * (1 - r^(n+1)) / (1-r )

You have:

B = e^(iΦ) [ 1 + a*e^(iΦ) + {a*e^(iΦ)}^2 + {a*e^(iΦ)}^3 ......]

B * e^(-iΦ) = 1 + a*e^(iΦ) + {a*e^(iΦ)}^2 + {a*e^(iΦ)}^3 ......

What are x , r and n this case? What do you get for summation?

 

[Vee3020]

You have:

B = e^(iΦ) [ 1 + a*e^(iΦ) + {a*e^(iΦ)}^2 + {a*e^(iΦ)}^3 ......]

B * e^(-iΦ) = 1 + a*e^(iΦ) + {a*e^(iΦ)}^2 + {a*e^(iΦ)}^3 ......

What are x , r and n this case? What do you get for summation?

After substituting I got the following result :
1 - (a * e^iΦ)^n+1
___________________
1 - (a * e^iΦ)

Is that correct?

Or we have to consider n+1 differently? Since nth term will be now n-1 and we have to add 1 to it thus making it (n-1)+1 = n?

 

[pka]

Calculate the sum: S = cosϕ + acos 2ϕ + a ^ 2 cos 3ϕ +… + a ^ (n-1) cos nϕ
I think this question is related to Euler's number? Like, if I substitute cosϕ as e^iϕ , then it will be a series which will have cosϕ + acos 2ϕ + a ^ 2 cos 3ϕ +… + a ^ (n-1) cos nϕ as it's real part. Is the logic correct? I need step wise solution to this since I'm stuck.

I have been extremely reluctant to reply to reply not knowing your level of knowledge of complex number theory.
The real part, \(\Re{(r\exp(i\theta))}=r\cos(\theta)\).
Thus \(\Re \left( {\sum\limits_{k = 1}^n {{a^{k - 1}}\exp \left( {k\phi i} \right)} } \right) = \sum\limits_{k = 1}^n {{a^{k - 1}}\cos \left( {k\phi } \right)} = \cos (\phi ) + a\cos (2\phi ) + \cdots + {a^{n - 1}}\cos (n\phi )=~?\)

 

[Deleted member 4993]

After substituting I got the following result :
1 - (a * e^iΦ)^n+1
___________________
1 - (a * e^iΦ)

Is that correct?

Or we have to consider n+1 differently? Since nth term will be now n-1 and we have to add 1 to it thus making it (n-1)+1 = n?

The numerical value of 'n' (# of terms in the series) remains the same.

 

[Vee3020]

The numerical value of 'n' (# of terms in the series) remains the same.

So is it correct?

 

[Vee3020]

I have been extremely reluctant to reply to reply not knowing your level of knowledge of complex number theory.
The real part, \(\Re{(r\exp(i\theta))}=r\cos(\theta)\).
Thus \(\Re \left( {\sum\limits_{k = 1}^n {{a^{k - 1}}\exp \left( {k\phi i} \right)} } \right) = \sum\limits_{k = 1}^n {{a^{k - 1}}\cos \left( {k\phi } \right)} = \cos (\phi ) + a\cos (2\phi ) + \cdots + {a^{n - 1}}\cos (n\phi )=~?\)

I know that formula. But I think it's not that simple. By rewriting the series using Euler's number, I think we have to write the summation of that new series in cosΦ + isinΦ format. And the cosΦ part, whatever it may be, will be the sum of real part. I guess?

 

[Vee3020]

The numerical value of 'n' (# of terms in the series) remains the same.

So the equation is correct so far?

 

[Deleted member 4993]

So the equation is correct so far?

Yes... Continue..

 

[Vee3020]

Yes... Continue..

That leaves me with :

1 - (a^n+1* e^n+1iΦ)
___________________
1 - (a * e^iΦ)


Sir please tell me what to do next? Should I substitute e^n+1iΦ as cos(n+1)Φ + isin(n+1)Φ ?

 

[Cubist]

I think it's n, not (n+1)

[MATH] B e^{-?\phi}= \frac{1-a^n \mathrm{e}^{? n\phi}}{1-a \mathrm{e}^{? \phi}}[/MATH]
Next step, can you make the denominator real?

 

[Deleted member 4993]

I think it's n, not (n+1)

[MATH] B e^{-?\phi}= \frac{1-a^n \mathrm{e}^{? n\phi}}{1-a \mathrm{e}^{? \phi}}[/MATH]
Next step, can you make the denominator real?

Calculate the sum:
S = cosϕ + acos 2ϕ + a ^ 2 cos 3ϕ +… + a ^ (n-1) cos nϕ

Cubist is correct .... rechecked OP .................... (going to the corner before Jomo jumps on me.

Back to task at hand>

B * e^(-iΦ) = 1 + a*e^(iΦ) + {a*e^(iΦ)}^2 + {a*e^(iΦ)}^3 ......

B * e^(-iΦ) * {1 - a*e^(iϕ)} = 1 - [a*e^(iΦ)]^(n) = = 1 - a^n * e^(inΦ)] incorrect - follow Cubist's method of "conjugate"

Continue......