Advertisements
Advertisements
Question
Find `root(3)(10001)` approximately (two decimal places
Solution
`root(3)(10001) = (10001)^(1/3)`
= `(1000 + 1)^(1/3)`
= `{1000(1 + 1/1000)}^(1/3)`
= `(1000)^(1/3) [1 + 1/10^3]^(1/3)`
= `10{1 + 1/3(1/10^3) + (1/3((-2)/3))/3 (1/10^3)^2 ...}`
= `10{1 + 1/3000 - 2/18000000 ...}`
= 10[1 + 0.000333 ...]
= 10(1.000333)
= 10.0033
APPEARS IN
RELATED QUESTIONS
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid
`1/(5 + x)`
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid
`2/(3 + 4x)^2`
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid
`(5 + x^2)^(2/3)`
Prove that `sqrt((1 - x)/(1 + x))` is approximately euqal to `1 - x + x^2/2` when x is very small
Write the first 6 terms of the exponential series
`"e"^(-2x)`
Write the first 4 terms of the logarithmic series
log(1 + 4x) Find the intervals on which the expansions are valid.
Write the first 4 terms of the logarithmic series
`log((1 + 3x)/(1 -3x))` Find the intervals on which the expansions are valid.
Write the first 4 terms of the logarithmic series
`log((1 - 2x)/(1 + 2x))` Find the intervals on which the expansions are valid.
If p − q is small compared to either p or q, then show `root("n")("p"/"q")` ∼ `(("n" + 1)"p" + ("n" - 1)"q")/(("n"- 1)"p" +("n" + 1)"q")`. Hence find `root(8)(15/16)`
Find the coefficient of x4 in the expansion `(3 - 4x + x^2)/"e"^(2x)`
Find the value of `sum_("n" = 1)^oo 1/(2"n" - 1) (1/(9^("n" - 1)) + 1/(9^(2"n"- 1)))`
Choose the correct alternative:
The coefficient of x6 in (2 + 2x)10 is
Choose the correct alternative:
If a is the arithmetic mean and g is the geometric mean of two numbers, then
Choose the correct alternative:
The sum up to n terms of the series `1/(sqrt(1) +sqrt(3)) + 1/(sqrt(3) + sqrt(5)) + 1/(sqrt(5) + sqrt(7)) + ...` is
Choose the correct alternative:
The coefficient of x5 in the series e-2x is