tayor series

**charikaar** · February 10th 2010, 03:52 AM

Let f(x) = log(1 + x).

Let

f : R--->R be infinitely differentiable. Prove or disprove the following two statements.
(i) ‘The Taylor series of f always converges for at least one point.’

(ii) ‘The Taylor series of f always converges to the function for at least two points.’

I don't understand this question so can you please help?

thanks

**Prove It** · February 10th 2010, 04:11 AM

Originally Posted by charikaar

Let f(x) = log(1 + x).

Let

f : R--->R be infinitely differentiable. Prove or disprove the following two statements.
(i) ‘The Taylor series of f always converges for at least one point.’

(ii) ‘The Taylor series of f always converges to the function for at least two points.’

I don't understand this question so can you please help?

thanks

Try finding the Taylor series first...

Assume that $f(x)$ can be expressed as a polynomial.

Then $\ln{(1 + x)} = a + bx + cx^2 + dx^3 + \dots$ .

By letting $x = 0$ we see that $a = 0$ .

Take the derivative of both sides

$(1 + x)^{-1} = b + 2cx + 3dx^2 + 4ex^3 + \dots$ .

By letting $x = 0$ we see that $b = 1$ .

Take the derivative of both sides

$-(1 + x)^{-2} = 2c + 3!dx + 4\cdot 3 ex^2 + 5\cdot 4 fx^3 + \dots$ .

By letting $x = 0$ we see that $2c = -1$ and so $c = -\frac{1}{2}$ .

Take the derivative of both sides

$2(1 + x)^{-3} = 3!d + 4! ex + 5 \cdot 4 \cdot 3 fx^2 + 6\cdot 5 \cdot 4 gx^3 + \dots$ .

By letting $x = 0$ we see that $3!d = 2$ , so $d = \frac{2}{3!} = \frac{1}{3}$ .

Take the derivative of both sides

$-3!(1 + x)^{-4} = 4!e + 5!fx + \dots$

By letting $x = 0$ we see that $4!e = -3!$ and so $d = -\frac{1}{4}$ .

You should be able to see that if we were to keep going we would have

$\ln{(1 + x)} = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \dots$

$= \sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}x^n}{n}$ .

Can you find for which values of $x$ that this series will converge for?

Hint: A geometric series converges for $|r| < 1$ ...

**HallsofIvy** · February 10th 2010, 05:26 AM

Originally Posted by charikaar

Let f(x) = log(1 + x).

Let

f : R--->R be infinitely differentiable. Prove or disprove the following two statements.
(i) ‘The Taylor series of f always converges for at least one point.’

(ii) ‘The Taylor series of f always converges to the function for at least two points.’

I don't understand this question so can you please help?

thanks

What, exactly, is the question? You first say "Let f(x) = log(1 + x)" and then say "Let f : R--->R be infinitely differentiable". Which is it? log(1+ x) is not defined for $x\le -1$ . If that is the function in question, then it is easy to show, perhaps by actually calculating its Taylor's series about some x= a, as Prove It does, that its Taylor's series converges for all x in it radius of convergence.

If the problem is to determine whether the Taylor's series of any infinitely differentiable function converges to the function value in at least one x value, then look at the general formula for the Taylor's series, at x= a,
$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$
and see what happens if x= a.

If the problem is to determine whether the Taylor's series of any infinitely differentiable function converges to the function value in at least two x values, then that is much harder.

Look at the function
$f(x)= e^{-\frac{1}{x^2}}$ is $x\ne 0$
f(0)= 0

It's not too difficult to show that all derivatives exist at 0: its derivatives, of all orders, for x not 0, are $e^{-\frac{1}{x^2}}$ over a polynomial in x. As x goes to 0, the exponential "dominates" and so this goes to 0. Taking the nth derivative to be 0 at x= 0 gives us an infinitely differentiable function. But its derivative is 0 at x= 0 for all orders of derivative so its Taylor's series at x= 0 is just
$\sum_{n= 0}^\infty \frac{0}{n!}x^n= 0+ 0+ 0+ \cdot\cdot\cdot$ .

That clearly converges, to 0, for all x but $f(x)\ne 0$ except at x= 0.

**charikaar** · February 10th 2010, 05:44 AM

Thank you very much for your replies.

Please ignore Let f(x) = log(1 + x).

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