Is expanding via Taylor series true only for values less than 1 ??

Question

How do I prove this result??

prove $1+x/2-x^3/8<(1+x)^{1/2}$ , for $x>0$

And while we're at it, I had a deeper doubt, Is expanding via Taylor series true only for , $x<1$ ?

Because, we easily write the error term as the smallest power of x that is not in the expansion i.e if we expand upto $n^\text{th}$ power, we take error term as $(x-a)^{n+1}*f^{(n+1)thderivative}(a)$$/(n+1)!$ , which would only be true if $x<1$ otherwise the most significant term should be the last term in the series which would tend to infinity for $x>1$.

This doubt arises because I have used Maclaurin series several times and there we always take $x$ tending to zero

thanks!

Answer 1

Yes, you are right in this case. The value of a Taylor series (or power series in general) may not be defined at some region since the value will go wild. A concept called Radius of convergence is there to determine how far can a power series converge.

See for example, the solution of this to find out how to calculate the radius of convergence.

Answer 2

For the first result, the formula in the question appears to be in error. I think you are supposed to use the Taylor expansion of $(1+x)^{1/2},$ which is $$ (1+x)^{1/2} = 1 + \frac12 x - \frac18 x^2 + - \cdots $$ (continuing with alternating $+$ and $-$ signs). Note that it's $\frac18 x^2$ rather than $\frac18 x^3,$ suggesting that there's a typo in the question. (Did you check carefully after copying the formulas?)

For $0 < x \leq 1$ this is a convergent alternating series and you can apply the Alternating Series Estimation Theorem. (For a theorem statement and example, see Alternating Series Estimation Theorem and this series.)

For $x < 0$ the result you're supposed to prove is false.

For $x > 1$ the result is true but the Taylor series around $0$ does not converge. There are other methods you could apply, such as squaring the quantities on both sides when $1 + \frac12 x - \frac18 x^2$ is positive.

It depends on exactly what you were supposed to prove--that is, for which $x$ you were supposed to prove the inequality.

---

As for Taylor series for $x > 1,$ consider a polynomial such as $p(x) = 1 + x + 3x^2 - 4x^3.$ The formula for this polynomial is literally its own Taylor series, with zero coefficients for the $x^4$ term and all higher terms (because the $4$th derivative of this polynomial is zero). The Taylor series is good for any real number $x.$

Now consider the Taylor series around $0$ for $$f(x) = \frac1{1+x} = 1 - x + x^2 - x^3 +- \cdots .$$ This converges for $\lvert x\rvert < 1$ and diverges for $\lvert x\rvert > 1$.

But now look at $$g(x) = \frac1{1+2x} = 1 - 2x + 4x^2 - 8x^3 +- \cdots .$$ This diverges for $\lvert x\rvert > \frac12$.

On the other hand, $$g(x) = \frac1{1+\frac12x} = 1 - \frac12x + \frac14x^2 - \frac18x^3 +- \cdots $$ converges for $\lvert x\rvert < 2$.

So in the case of a Taylor series around $0,$ the limit on $\lvert x\rvert$ does not have to be $1.$ It could be any number.

For any particular Taylor series centered at $a$, if there is a number $r$ for which the series converges whenever $\lvert x - a\rvert < r$ and diverges whenever $\lvert x - a\rvert > r$, we say that $r$ is the radius of convergence of the series. If the series converges for all real $x$, we say the radius of convergence is infinite.

Answer 3

Convergence of a Taylor series depends on how fast the coefficients decrease. For example

$$e^x=1+x+\frac{x^2}2+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$

they decrease so fast that the series converges for any $x$.

In general, the series converges up to the closest singularity of the function. For instance,

$$\frac1{1-x}=1+x+x^2+x^3+\cdots$$ cannot converge past $x=1$ because the function goes to infinity.

But obviously, by a simple rescaling

$$\frac1{1-2x}=1+2x+4x^2+8x^3+\cdots$$ converges up to $x=\dfrac12$, while

$$\frac3{3-x}=1+\frac x3+\frac{x^2}9+\frac{x^3}{27}+\cdots$$ converges up to $x=3$.

Answer 4

The inequality $$1+\frac{x}{2}-\frac{x^3}{8}<\sqrt{x+1}$$ is false, in general.

It is true for $x>\alpha$ where $\alpha\approx 0.743014$ is the largest root of $$1+\frac{x}{2}-\frac{x^3}{8}-\sqrt{x+1}=0$$ (see picture below)