Confounding geometric series convergence problem

[hoppfull]

Use the geometric series to compute the Taylor series for

. . . . .\(\displaystyle f(x)\, =\, \dfrac{1}{2\, -\, x}\)

Where does this series converge? Hint:

. . . . .\(\displaystyle \dfrac{1}{2\, -\, x}\, =\, \dfrac{1}{2}\, \dfrac{1}{1\, -\, \frac{x}{2}}\)

---

I know the following series converges while |x| < 1:

. . . . .\(\displaystyle \displaystyle \dfrac{1}{1\, -\, x}\, =\, \sum_{k=0}^{\infty}\, x^k\)

Which must mean that

. . . . .\(\displaystyle \displaystyle \dfrac{1}{2}\, \dfrac{1}{1\, -\, \frac{x}{2}}\, =\, \dfrac{1}{2}\, \sum_{k=0}^{\infty}\, \left(\dfrac{x}{2}\right)^k\)

converges while

. . . . .\(\displaystyle \displaystyle \bigg|\dfrac{x}{2}\bigg|\, <\, 1, \Longleftrightarrow\, |x|\, <\, 2\)

What if I had approached this problem in a different way?

. . . . .\(\displaystyle \displaystyle \dfrac{1}{2\, -\, x}\, =\, \dfrac{1}{1\, -\, (x\, -\, 1)}\)

Which means that

. . . . .\(\displaystyle \displaystyle \dfrac{1}{1\, -\, (x\, -\, 1)}\, =\, \sum_{k=0}^{\infty}\, (x\, -\, 1)^k\)

converges while

. . . . .\(\displaystyle \displaystyle |x\, -\, 1|\, <\, 1\, \Longleftrightarrow\, 0\, <\, x\, <\, 2\)

What am I missing here? Both answers can't be correct. Is neither correct?

 

[Deleted member 4993]

Use the geometric series to compute the Taylor series for

. . . . .\(\displaystyle f(x)\, =\, \dfrac{1}{2\, -\, x}\)

Where does this series converge? Hint:

. . . . .\(\displaystyle \dfrac{1}{2\, -\, x}\, =\, \dfrac{1}{2}\, \dfrac{1}{1\, -\, \frac{x}{2}}\)

---

I know the following series converges while |x| < 1:

. . . . .\(\displaystyle \displaystyle \dfrac{1}{1\, -\, x}\, =\, \sum_{k=0}^{\infty}\, x^k\)

Which must mean that

. . . . .\(\displaystyle \displaystyle \dfrac{1}{2}\, \dfrac{1}{1\, -\, \frac{x}{2}}\, =\, \dfrac{1}{2}\, \sum_{k=0}^{\infty}\, \left(\dfrac{x}{2}\right)^k\)

converges while

. . . . .\(\displaystyle \displaystyle \bigg|\dfrac{x}{2}\bigg|\, <\, 1, \Longleftrightarrow\, |x|\, <\, 2\)

What if I had approached this problem in a different way?

. . . . .\(\displaystyle \displaystyle \dfrac{1}{2\, -\, x}\, =\, \dfrac{1}{1\, -\, (x\, -\, 1)}\)

Which means that

. . . . .\(\displaystyle \displaystyle \dfrac{1}{1\, -\, (x\, -\, 1)}\, =\, \sum_{k=0}^{\infty}\, (x\, -\, 1)^k\)

converges while

. . . . .\(\displaystyle \displaystyle |x\, -\, 1|\, <\, 1\, \Longleftrightarrow\, 0\, <\, x\, <\, 2\)

What am I missing here? Both answers can't be correct. Is neither correct?

What are your thoughts?

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[hoppfull]

What are your thoughts?

Well, my thoughts are why does one approach yield this answer?
\(\displaystyle \displaystyle \bigg|\dfrac{x}{2}\bigg|\, <\, 1 \Longleftrightarrow\, -2 <\, x\, <\, 2\)
And another approach (which seems equally valid to me) yield this answer?
\(\displaystyle \displaystyle |x\, -\, 1|\, <\, 1\, \Longleftrightarrow\, 0\, <\, x\, <\, 2\)

 

[Ishuda]

Use the geometric series to compute the Taylor series for

. . . . .\(\displaystyle f(x)\, =\, \dfrac{1}{2\, -\, x}\)

Where does this series converge? Hint:

. . . . .\(\displaystyle \dfrac{1}{2\, -\, x}\, =\, \dfrac{1}{2}\, \dfrac{1}{1\, -\, \frac{x}{2}}\)

---

I know the following series converges while |x| < 1:

. . . . .\(\displaystyle \displaystyle \dfrac{1}{1\, -\, x}\, =\, \sum_{k=0}^{\infty}\, x^k\)

Which must mean that

. . . . .\(\displaystyle \displaystyle \dfrac{1}{2}\, \dfrac{1}{1\, -\, \frac{x}{2}}\, =\, \dfrac{1}{2}\, \sum_{k=0}^{\infty}\, \left(\dfrac{x}{2}\right)^k\)

converges while

. . . . .\(\displaystyle \displaystyle \bigg|\dfrac{x}{2}\bigg|\, <\, 1, \Longleftrightarrow\, |x|\, <\, 2\)

What if I had approached this problem in a different way?

. . . . .\(\displaystyle \displaystyle \dfrac{1}{2\, -\, x}\, =\, \dfrac{1}{1\, -\, (x\, -\, 1)}\)

Which means that

. . . . .\(\displaystyle \displaystyle \dfrac{1}{1\, -\, (x\, -\, 1)}\, =\, \sum_{k=0}^{\infty}\, (x\, -\, 1)^k\)

converges while

. . . . .\(\displaystyle \displaystyle |x\, -\, 1|\, <\, 1\, \Longleftrightarrow\, 0\, <\, x\, <\, 2\)

What am I missing here? Both answers can't be correct. Is neither correct?

Both are correct. They are both representations of the original function valid in different regions. For example suppose we want the reciprocal 999. Since 999 is (1000-1) we can use your equation above of
.\(\displaystyle \displaystyle \dfrac{1}{1\, -\, x}\, =\, \sum_{k=0}^{\infty}\, x^k\)
but not directly. We do the same sort of thing you did and rewrite the equation. That is
. .\(\displaystyle \displaystyle \dfrac{1}{999}\, =\,\dfrac{1}{1000}\,\dfrac{1}{1\,-\dfrac{1}{1000}}\, =\, \dfrac{1}{1000}\,\sum_{k=0}^{\infty}\,\left( \dfrac{1}{1000}\right)^k\)

Thus we can actually get a convergence for \(\displaystyle |x|\, >\, 1\) with a simple rearrangement

 

[hoppfull]

Both are correct. They are both representations of the original function valid in different regions. For example suppose we want the reciprocal 999. Since 999 is (1000-1) we can use your equation above of
.\(\displaystyle \displaystyle \dfrac{1}{1\, -\, x}\, =\, \sum_{k=0}^{\infty}\, x^k\)
but not directly. We do the same sort of thing you did and rewrite the equation. That is
. .\(\displaystyle \displaystyle \dfrac{1}{999}\, =\,\dfrac{1}{1000}\,\dfrac{1}{1\,-\dfrac{1}{1000}}\, =\, \dfrac{1}{1000}\,\sum_{k=0}^{\infty}\,\left( \dfrac{1}{1000}\right)^k\)

Thus we can actually get a convergence for \(\displaystyle |x|\, >\, 1\) with a simple rearrangement

This is very confusing. I can see how the two approaches are equivalent. However I can't see why \(\displaystyle \sum_{k=0}^{\infty}\left(x - 1\right)^k\) would yield a less complete answer than \(\displaystyle \dfrac{1}{2} \sum_{k=0}^{\infty}\left(\dfrac{x}{2}\right)^k\). It actually seems like the first approach is the safer approach since we're not breaking anything out before we analyze it.