How do you find a one-decimal place approximation for $-root4 200$ ?

Question

How do you find a one-decimal place approximation for $-root4 200$ ?

1 Answer

Impact of this question

1918 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-23T22:07:15

Use some reasonings about $3^4$ , $4^4$ and $sqrt(200)$ to choose approximation $3.8$ , then use one step of Newton's method to refine the approximation, finding that $3.8$ is good to one decimal place already.

Explanation:

$3^4 = 81$ and $4^4 = 256$ , so $root(4)(200)$ lies somewhere between $3$ and $4$ .

More specifically, $200 ~~ 196 = 14^2$ so $root(4)(200) ~~ sqrt(14)$ and $sqrt(14) ~~ 4-1/4 = 3.75$ , so $root(4)(200) ~~ 3.8$

To find the $4$ th root of a number $n$ , we could find its square root and then the square root of that, but instead let's choose a reasonable first approximation $a_0$ , then use the following formula to iterate:

$a_(i+1) = a_i + (n - a_i^4)/(4a_i^3)$

Let $a_0 = 3.8$

Then:

$a_1 = a_0 + (200-3.8^4)/(4*3.8^3)$

$=3.8 + (200-208.5136)/219.488$

$=3.8 - 8.5136/219.488 ~~ 3.76$

So $3.8$ is a good approximation of $root(4)(200)$ to one decimal place and $-3.8$ is a one decimal place approximation for $-root(4)(200)$

Science

Math

Humanities

... and beyond

How do you find a one-decimal place approximation for $-root4 200$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find a one-decimal place approximation for -root4 200-root4 200?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find a one-decimal place approximation for $-root4 200$ ?