How do you use differentiation to find a power series representation for #1/(6+x)^2#?

Answer 1
Basically: 1. Integrate to get a form of #1/(1-x)#. 2. Modify the equation to achieve getting precisely #1/(1-X)# where X is some variant of x, whether it's #x/6# (here), #-y/2#, #theta/pi#, etc. 3. Write out the power series with #x# = some substituted value, like #x/6#, here. 4. Reverse what you did to re-acquire the original function. i.e. re-multiply by what you divided by (-1 and 1/6, here), then re-differentiate. Whatever returns your original function.
Notice how the power series #1/(1-x)# can be written as the power series: #1 + x + x^2 + x^3 + ...# #= sumx^n#
Similarly, use -x instead of x. Every odd power is negative, and every even power is positive by virtue of squaring to some order of magnitude (e.g. #(x^2)^n#).
#= sum(-1)^nx^n = 1 - x + x^2 - x^3 + x^4 - ...#
Integrate #1/(6+x)^2# to get #-1/(6+x)#. Divide by -(1/6) to get #1/(1 + x/6)#.
x/6 is your new x. Plug it in, use this alternating series from a few lines up, factor in the 1/6 to get back to #1/(6+x)#, and incorporate the negative to get back to #-1/(6+x)#.
#=> -(1/6)[(-1)^0(x/6)^0 + (-1)^1(x/6)^1 + (-1)^2(x/6)^2 + (-1)^3(x/6)^3 + ...]# #=> -1/6 + x/36 - x^2/216 + x^3/1296 - ...#
Then, re-differentiate the result to get back to #1/(6+x)^2#.
#=> 1/36 - x/108 + (x^2)/432 - ...#
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Answer 2

To find the power series representation for ( \frac{1}{{(6+x)}^2} ), we can use differentiation. Here's the step-by-step process:

  1. Start with a known power series representation: [ \frac{1}{{1-x}} = \sum_{n=0}^{\infty} x^n ]

  2. Differentiate both sides with respect to ( x ): [ \frac{d}{{dx}} \left( \frac{1}{{1-x}} \right) = \frac{d}{{dx}} \left( \sum_{n=0}^{\infty} x^n \right) ]

  3. Use the power rule for differentiation and recognize that the derivative of a power series is just another power series: [ \frac{1}{{(1-x)^2}} = \sum_{n=1}^{\infty} nx^{n-1} ]

  4. Now, to find the power series representation for ( \frac{1}{{(6+x)}^2} ), substitute ( x = \frac{-x}{6} ) into the series above: [ \frac{1}{{(6+x)}^2} = \sum_{n=1}^{\infty} n \left( \frac{-x}{6} \right)^{n-1} ]

  5. To adjust the index of summation, let ( m = n-1 ), then rewrite the series: [ \frac{1}{{(6+x)}^2} = \sum_{m=0}^{\infty} (m+1) \left( \frac{-x}{6} \right)^m ]

  6. Simplify the series: [ \frac{1}{{(6+x)}^2} = \sum_{m=0}^{\infty} (-1)^m \frac{(m+1)x^m}{6^m} ]

Therefore, the power series representation for ( \frac{1}{{(6+x)}^2} ) is: [ \sum_{m=0}^{\infty} (-1)^m \frac{(m+1)x^m}{6^m} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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