Tesla3
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Hi,
Could anybody help me with the following question? I have attached the question in my book and my solution so far. But I am stuck and can't figure out how to continue as the integral gets too complicated to integrate. Is there something I am doing wrong or missing? I would be really grateful if anyone could help.
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ghostwalker
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(Original post by Tesla3)
Hi,
Could anybody help me with the following question? I have attached the question in my book and my solution so far. But I am stuck and can't figure out how to continue as the integral gets too complicated to integrate. Is there something I am doing wrong or missing? I would be really grateful if anyone could help.
It "says" approximately, which suggest some sort of power series.

Spoiler:
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binomials
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Tesla3
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(Original post by ghostwalker)
It "says" approximately, which suggest some sort of power series.

Spoiler:
Show

binomials
Could you please elaborate? like Maclurin Series?
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ghostwalker
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(Original post by Tesla3)
Could you please elaborate? like Maclurin Series?
It's in the spoiler of my first post!
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Tesla3
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(Original post by ghostwalker)
It's in the spoiler of my first post!
Can you show how? I am a bit confused......
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ghostwalker
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(Original post by Tesla3)
Can you show how? I am a bit confused......
Check your textbook. What's the expansion of (1+x)^n? Then just replace the x and n with the appropriate values. You have two binomials to expand. Have a go.
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Tesla3
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(Original post by ghostwalker)
Check your textbook. What's the expansion of (1+x)^n? Then just replace the x and n with the appropriate values. You have two binomials to expand. Have a go.
I get that the denominator can be expanded using the binomial theorem but what about the numerator? It has three terms e^2 , 1 and 2ecos(theta).
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Tesla3
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After using the binomial theorem I get this........
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ghostwalker
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(Original post by Tesla3)
After using the binomial theorem I get this........
Well expanding the square in the denominator hardly qualifies as requiring the full might of the binomial expansion.

Let's rewrite the integrand to make it clearer.

 \displaystyle [1+e\cos\theta ]^{-2}[1+(2e\cos\theta +e^2)]^{\frac{1}{2}}

Now try.
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Tesla3
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(Original post by ghostwalker)
Well expanding the square in the denominator hardly qualifies as requiring the full might of the binomial expansion.

Let's rewrite the integrand to make it clearer.

 \displaystyle [1+e\cos\theta ]^{-2}[1+(2e\cos\theta +e^2)]^{\frac{1}{2}}

Now try.
wait let me try.....
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Tesla3
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After expansion and multiplying I get something like this. Also I neglect any e terms with power equivalent to 2 or higher. Then I simplify and integrate to get the desired result. Is it okay? ghostwalker
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ghostwalker
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(Original post by Tesla3)
After expansion and multiplying I get something like this. Also I neglect any e terms with power equivalent to 2 or higher. Then I simplify and integrate to get the desired result. Is it okay? ghostwalker
Looks fine.

You made a minor slip in the second expansion - should have e^2/2 - but it doesn't effect the final outcome.
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Tesla3
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(Original post by ghostwalker)
Looks fine.

You made a minor slip in the second expansion - should have e^2/2 - but it doesn't effect the final outcome.
Thanks
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