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warwick

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Lagrange asserted that any function can be expanded in a power series. Could you give an example to show that his assertion is incorrect?
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How do I find a function that can't be expressed in a power series?
 
You'll have to define "Power Series". Is there a difference between:

1) All subsequent derivatives are zero, and

2) The derivative fails to exist?
 
Try expressing \(\displaystyle \sqrt{x}\) as a power series.

If it were written as a power series, then it would have the form

\(\displaystyle \sqrt{x}=a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdot\cdot\cdot\)

because \(\displaystyle \sqrt{x}=0\) when x=0.

Now, square both sides and see a contradiction. This does not have a power series.

This is because of the two values we have associated with a square root.

For instance, \(\displaystyle 1\cdot 1=1\) and \(\displaystyle (-1)(-1)=1\)
 
galactus said:
Try expressing \(\displaystyle \sqrt{x}\) as a power series.

If it were written as a power series, then it would have the form

\(\displaystyle \sqrt{x}=a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdot\cdot\cdot\)

because \(\displaystyle \sqrt{x}=0\) when x=0.

Now, square both sides and see a contradiction. This does not have a power series.

This is because of the two values we have associated with a square root.

For instance, \(\displaystyle 1\cdot 1=1\) and \(\displaystyle (-1)(-1)=1\)
Click to expand...

How many terms in the expansion do I go out? One or two? I'm not seeing the contradiction.
 
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You didn't answer my query. It will direct you to the answer to your question.
 
No, and why would it? It is in the fundamental development of what you are studying so that such a problem would be reasonable to test your knowledge. Galactus gave you an excellent example.
 
What about?

\(\displaystyle \sqrt{x}=a_{1}x\)

x = a12x2

If you take the derivative twice, you get

1 = 2a12x

0 = 2a12
 
tkhunny said:
No, and why would it? It is in the fundamental development of what you are studying so that such a problem would be reasonable to test your knowledge. Galactus gave you an excellent example.
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How many terms out do go on the right? I went out one, two, and three but didn't see a contradiction.

I took the derivative above. Does that relate to your query?
 
Please find \(\displaystyle \frac{d}{dx}\sqrt{x}\) = ??

Show your result.
 
tkhunny said:
Please find \(\displaystyle \frac{d}{dx}\sqrt{x}\) = ??

Show your result.
Click to expand...

I know what the derivative is. I took the derivative AFTER I squared both sides.

(1/2)x-1/2

Do I just need to find a function that's not differentiable over some interval?
 
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Hold on there. No one questioned your knowledge of the derivative.

It was hoped that the demonstration might encourage you to think about the solution to your problem. The point is that the derivative fails to exist at x = 0.
 
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