hi,
i m stuck at the ...part for both an & bn..
how to proceed thence?
thanks
atupid
i m stuck at the ...part for both an & bn..
how to proceed thence?
thanks
atupid
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I think you will find that ...
\(B_{n}=-\frac{2cos(n\pi)}{n} \ for \ n=1,2,3,...\)
So
\(B_{1}=+\frac{2}{1}\)
\(B_{2}=-\frac{2}{2}\)
\(B_{3}=+\frac{2}{3}\)
\(B_{4}=-\frac{2}{4}\)
etc.
Well, let's break it down:
\(cos(n\cdot\pi)=\)
\(=1\ \ \ for\ n=0\)
\(=-1\ for\ n=1\)
\(=1\ \ \ for\ n=2\)
\(=-1\ for\ n=3\)
etc...
Now let's take a look at:
\((-1)^n=\)
\(=1\ \ \ for\ n=0\)
\(=-1\ for\ n=1\)
\(=1\ \ \ for\ n=2\)
\(=-1\ for\ n=3\)
Do you see that they are actually the same series?
The (-1)^(n+1) part simply ensures that with n odd, the sign of the harmonic amplitude is positive and with n even, the sign is negative. It's just a neat way of writing the solution without expressly including the cosine function.the power in question is n+1 not merely n
how did the +1 came about?
i m wondering what could be going wrong with my simple asking?
all i wanted to know is how did the power of n+1 come about, not how to compute it.
The (-1)^(n+1) part simply ensures that with n odd, the sign of the harmonic amplitude is positive and with n even, the sign is negative. It's just a neat way of writing the solution without expressly including the cosine function.
First of all (-1)^n+1 could be interpreted aslet me try one more time...
some of other answers have (-1)^n
others have (-1)^n+1
why some have "+1" while others no?
regards,
stupid