tayor series

**charikaar** · February 10th 2010, 04:52 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, 05:11 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, 06:26 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, 06:44 AM

Thank you very much for your replies.

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

Math Help - tayor series

LinkBack

Thread Tools

Display

tayor series

Similar Math Help Forum Discussions

Exploiting geometric series with power series to find Taylors series

question about power series, taylor series, maclaurin series

tayor expansion

calculus II areas about axis, pressure, infinite series, MacLaurin series

Tayor Series

Search Tags