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Taylor's series question [Solved!]

My question

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

1. Taylor Series

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

X

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

Re: Taylor's series question

Hello Snowboy

Your problem is very similar to the example on the page you came from:

1. Taylor Series

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

Re: Taylor's series question

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?

X

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?

Re: Taylor's series question

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.

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.

Re: Taylor's series question

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?

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?

Re: Taylor's series question

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:

Forum graph - svgphp-00.511.522.50.511.52-0.5xy

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`.

Re: Taylor's series question

Thanks a lot!

X

Thanks a lot!

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