Obtain the MacLaurin series for 1/(2-x) by making an appropriate substitution into the MacLaurin series for 1/(1-x).
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The MacLaurin series for 1/(1-x) = Σ x^k
I substitue (x-1) in for x, because 1/(2-x) = 1/(1-(x-1))
Making the same substitution in the MacLaurin series gives Σ (x-1)^k
If I manually calculate the MacLaurin series for 1/(2-x), I get Σ x^k/2^(k+1)
Those two don't match. What did I do wrong? Or does that substitution method not work?
1/(1-x) = Σ0∞xk
write 1/(2-x) as (1/2)(1/(1-x/2))
then
1/(2-x)
= (1/2)(1/(1-x/2))
= (1/2)Σ0∞x/2k
=Σ0∞xk+1
as you have corrected determined.
Sorry, the last two lines should read:
=(1/2)Σ0∞(x/2)k
=Σ0∞xk/2k+1
Thanks so much!
I can't read your symbols, but I can make out what you are doing and it results in the right answer.
Sorry, I was accidentally on unicode.
You would be able to read the symbols with unicode encoding. If you are on Windows XP, you can use the menu to go
view/character encoding/utf-8
However, utf-8 may not be available automatically on all computers.
I will take more care with the encoding next time. Thanks.
The substitution method you tried is not incorrect, but it seems there might be a misunderstanding in the way you performed the substitution.
To obtain the MacLaurin series for 1/(2-x) by making an appropriate substitution into the MacLaurin series for 1/(1-x), you need to replace the variable x in the series for 1/(1-x) with (2-x), not just (x-1).
Here's the correct approach:
1. Start with the MacLaurin series for 1/(1-x), which is Σ x^k.
2. Substitute (2-x) in place of x. This gives us Σ (2-x)^k.
3. Expand this expression using the binomial theorem: Σ (2-x)^k = Σ C(k,0)*2^k*(-x)^0 + C(k,1)*2^(k-1)*(-x)^1 + C(k,2)*2^(k-2)*(-x)^2 + ...
(where C(k,n) represents the binomial coefficient)
4. Simplify further. Note that C(k,0) = 1, so the first term in the series simplifies to 2^k. The rest of the terms can be rearranged to collect powers of x, resulting in Σ (2^k * (-1)^n * C(k,n) * x^n).
5. Finally, we can rewrite the series as Σ (2^k * C(k,n) * x^n).
So, the MacLaurin series for 1/(2-x) is Σ (2^k * C(k,n) * x^n), where n ranges from 0 to infinity.
I hope this clarifies the correct approach for obtaining the MacLaurin series for 1/(2-x) using substitution.