my question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p

my question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p

Relevant page
<a href="/series-expansion/1-taylor-series.php">1. Taylor Series</a>
What I've done so far
I think I need to differentiate it several times.
Here it is:
`f(x) = x^(1/2)`
`f'(x) = 1/2 x^(-1/2) = 1/(2 x^(1/2))`
`f''(x) = - 1/4 x^(-3/2) = -1/(4 x^(3/2))`
`f'''(x) = 3/8 x^(-5/2) = 3/(8 x^(5/2))`
But I'm stuck then

Now that you've differentiated, you need to substitute the necessary values, just like I did in that example.

Or is you problem with the square root approximation?

Regards

X

Hello Snowboy
Your problem is very similar to the example on the page you came from:
<a href="/series-expansion/1-taylor-series.php">1. Taylor Series</a>
Now that you've differentiated, you need to substitute the necessary values, just like I did in that example.
Or is you problem with the square root approximation?
Regards

So `a=1`, right?
`f(1) = 1`
`f'(1) = 1/2`
`f''(1) = -1/4`
`f'''(1) = 3/8`
Plugging it into the formula:
`1 + 1/2 (x-1)` ` -1/4(x-1)^2` ` + 3/8(x-1)^3-...`
How many steps do we have to do?

So you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.

X

The question says you need 4 d.p. accuracy.
So you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.

I can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.

Maybe I should do another step to make sure.

`f''''(x) = -15/16 x^(-7/2)`

`f''''(1) = -15/16`

So the 5th term is `-15/16(x-1)^4`

Plugging in `1.1` gives `-0.00009375`

Adding this gives `1.047875-0.00009375 = 1.04778125`

To 4 d.p. it's still `1.0478`.

But checking on my calculator, I get `sqrt(1.1) = 1.04880884817`

To 4 d.p., that's `1.0488`. Is my answer wrong?

X

It says "at 1.1".
Is this OK?
`sqrt(1.1) = 1 + 1/2(0.1) - 1/4(0.1)^2 + 3/8(0.1)^3`
` = 1 + 0.05 - 0.0025 + 0.000375` ` = 1.047875`
I can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.
Maybe I should do another step to make sure.
`f''''(x) = -15/16 x^(-7/2)`
`f''''(1) = -15/16`
So the 5th term is `-15/16(x-1)^4`
Plugging in `1.1` gives `-0.00009375`
Adding this gives `1.047875-0.00009375 = 1.04778125`
To 4 d.p. it's still `1.0478`.
But checking on my calculator, I get `sqrt(1.1) = 1.04880884817`
To 4 d.p., that's `1.0488`. Is my answer wrong?

There's s problem with your function: 1` ` + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3

Your thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.

Our Taylor's Series approximation is only "good" right near the number we expand about (in your case, `x=1`).

To see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:

The green one is `y=sqrt(x)` and the magenta curve is `y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3`.

You can see it's not a very good approximation as we get away from `x=1`.

X

Your thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.
Our Taylor's Series approximation is only "good" right near the number we expand about (in your case, `x=1`).
To see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:
[graph]310,250;-0.3,2.5;-0.5,2,0.5,0.5;sqrt(x),1` ` + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3[/graph]
The green one is `y=sqrt(x)` and the magenta curve is `y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3`.
You can see it's not a very good approximation as we get away from `x=1`.