Expand ln(1+x) and ln(1-x) and hence deduce that
ln((y+1)/(y-1))=[2/y]+[2/(3(y^3))]+[2/(5(y^5))]+[2/(7(y^7))]. State the range of values of y for which the expansion is valid. Hence calculate ln(101) correct to 4 significant figures given that ln(99)=4.59507.
I managed to expand ln(1+x) and ln(1-x). Can someone help me to continue the problem?
To expand ln(1+x), we can use the Taylor series expansion:
ln(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...
To expand ln(1-x), we can use the Taylor series expansion as well:
ln(1-x) = -x - (x^2)/2 - (x^3)/3 - (x^4)/4 - ...
Now, let's add the two expansions together:
ln(1+x) + ln(1-x) = (x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...) + (-x - (x^2)/2 - (x^3)/3 - (x^4)/4 - ...)
= 2(x - (x^3)/3 + (x^5)/5 - ...)
Simplifying further, we get:
ln(1+x) + ln(1-x) = 2(x - (x^3)/3 + (x^5)/5 - ...)
Now, let's substitute y = x^2 into this equation:
ln(1+y^(1/2)) + ln(1-y^(1/2)) = 2(y^(1/2) - (y^(3/2))/3 + (y^(5/2))/5 - ...)
To express y^(n/2) as (y^n)/(y^2), we can write:
y^(n/2) = (y^n)/(y^2)
So, our equation becomes:
ln(1+y^(1/2)) + ln(1-y^(1/2)) = 2(y^(1/2) - (y^(3/2))/3 + (y^(5/2))/5 - ...)
Which simplifies to:
ln(1+y^(1/2)) + ln(1-y^(1/2)) = 2[(y^2)/(y^2) * (y^(1/2)) - (y^3)/(3 * y^2) * (y^(1/2))^3 + (y^5)/(5 * y^2) * (y^(1/2))^5 - ...]
ln(1+y^(1/2)) + ln(1-y^(1/2)) = 2[(y^2)/(y^2) * y^(1/2) - (y^3)/(3 * y^2) * y^(3/2) + (y^5)/(5 * y^2) * y^(5/2) - ...]
ln(1+y^(1/2)) + ln(1-y^(1/2)) = 2(y + (y^2)/3 + (y^3)/5 + (y^4)/7 + ...)
Notice that the right-hand side is the expansion we are looking for:
2(y + (y^2)/3 + (y^3)/5 + (y^4)/7 + ...)
Now, let's address the range of values of y for which this expansion is valid. The expansion ln((y+1)/(y-1)) = 2/y + 2/(3(y^3)) + 2/(5(y^5)) + 2/(7(y^7)) is valid when the absolute value of y is less than 1, i.e., |y| < 1.
To calculate ln(101) using this expansion, we need to find a value of y that satisfies |y| < 1. Since ln(99) = 4.59507, we can set y = (99-1)/(99+1) = 98/100 = 0.98.
Substituting y = 0.98 into the expansion, we get:
ln(101) ≈ 2(0.98 + (0.98^2)/3 + (0.98^3)/5 + (0.98^4)/7 + ...)
Now, evaluate the right-hand side of the equation using the expansion and round the result to 4 significant figures to find the approximate value of ln(101).