Area Frenzy

Calculus Level 4

Let ∣ x ∣ < 1 |x| < 1 and ( 0 < a < 1 ) (0 < a < 1) .

If f ( x ) = ∑ n = 1 ∞ n 6 x n f(x) = \sum_{n = 1}^{\infty} n^6 x^n and g ( x ) = ∑ n = 1 ∞ n 5 x n g(x) = \sum_{n = 1}^{\infty} n^5 x^n and the region R R bounded by f f and g g on [ 0 , 1 − a ] [0,1 - a] has area A R = 11 a 6 − 94 a 5 − 9609 a 4 + 3170 a 3 + 330 a 2 − 488 a + 119 a 6 − 2 ln ⁡ ( a ) A_{R} = \dfrac{11a^6 - 94a^5 - 9609a^4 + 3170a^3 + 330a^2 - 488a + 119}{a^6} - 2\ln(a) , find the real value of a a and express the answer as the value of A R A_{R} to six decimal places.


The answer is 75708175.605170.

Rocco Dalto by Rocco Dalto , 67, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rocco Dalto
May 11, 2018

Let ∣ x ∣ < 1 |x| < 1 and ( 0 < a < 1 ) (0 < a < 1) .

∑ n = 1 ∞ n x n = ∑ j = 1 ∞ ∑ n = j ∞ x n = 1 1 − x ∑ j = 1 ∞ x j = 1 1 − x ( x ) 1 1 − x = x ( 1 − x ) 2 \sum_{n = 1}^{\infty} n x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2}

( x x − 1 ) ( 1 x ) ( x x − 1 ) = x ( x − 1 ) 2 . (\dfrac{x}{x - 1})(\dfrac{1}{x})(\dfrac{x}{x - 1}) = \dfrac{x}{(x - 1)^2}.

∑ n = 1 ∞ n 2 x n = ∑ j = 1 ∞ ( ∑ n = j ∞ n x n ) = \sum_{n = 1}^{\infty} n^2 x^n = \sum_{j = 1}^{\infty} (\sum_{n = j}^{\infty} n x^n) = 1 1 − x ∑ j = 1 ∞ ( j x j ) + x ( 1 − x ) 2 ∑ j = 1 ∞ x j = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} (j x^j) + \dfrac{x}{(1 - x)^2}\sum_{j = 1}^{\infty} x^j =

( 1 1 − x ) ( x ( 1 − x ) 2 ) + ( x ( 1 − x ) 2 ) ( x 1 − x ) = (\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x}) = x 2 + x ( 1 − x ) 3 \dfrac{x^2 + x}{(1 - x)^3} .

∑ n = 1 ∞ n 3 x n = ∑ j = 1 ∞ ∑ n = j ∞ n 2 x n = \sum_{n = 1}^{\infty} n^3 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^2 x^{n} = 1 1 − x ∑ j = 1 ∞ j 2 x j + 2 x ( 1 − x ) 2 ∑ j = 1 ∞ j x j + x 2 + x ( 1 − x ) 3 ∑ j = 1 ∞ x j = ( 1 1 − x ) ( x 2 + x ( 1 − x ) 3 ) + \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{2x}{(1 - x)^2}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^2 + x}{(1 - x)^3}\sum_{j = 1}^{\infty} x^j = (\dfrac{1}{1 - x})(\dfrac{x^2 + x}{(1 - x)^3}) + ( 2 x ( 1 − x ) 2 ) ( x ( 1 − x ) 2 ) + ( x 2 + x ( 1 − x ) 3 ) ( x 1 − x ) = (\dfrac{2x}{(1 - x)^2})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^2 + x}{(1 - x)^3})(\dfrac{x}{1 - x}) = x 3 + 4 x 2 + x ( 1 − x ) 4 \dfrac{x^3 + 4x^2 + x}{(1 - x)^4} .

∑ n = 1 ∞ n 4 x n = ∑ j = 1 ∞ ∑ n = j ∞ n 3 x n = \sum_{n = 1}^{\infty} n^4 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^3 x^{n} = 1 1 − x ∑ j = 1 ∞ j 3 x j + 3 x ( 1 − x ) 2 ∑ j = 1 ∞ j 2 x j + 3 ( x 2 + x ) ( 1 − x ) 3 ∑ j = 1 ∞ j x j + x 3 + 4 x 2 + x ( 1 − x ) 4 ∑ j = 1 ∞ x j = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^3 x^j + \dfrac{3x}{(1 - x)^2}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{3(x^2 + x)}{(1 - x)^3}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}\sum_{j = 1}^{\infty} x^j =

= ( 1 1 − x ) ( x 3 + 4 x 2 + x ( 1 − x ) 4 ) + = (\dfrac{1}{1 - x})( \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}) + ( 3 x ( 1 − x ) 2 ) ( x 2 + x ( 1 − x ) 3 ) + ( 3 ( x 2 + x ) ( 1 − x ) 3 ) ( x ( 1 − x ) 2 ) + (\dfrac{3x}{(1 - x)^2})(\dfrac{x^2 + x}{(1 - x)^3}) + (\dfrac{3(x^2 + x)}{(1 - x)^3})(\dfrac{x}{(1 - x)^2}) + x 3 + 4 x 2 + x ( 1 − x ) 4 ( x 1 − x ) = x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}(\dfrac{x}{1 - x}) = \dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5} .

g ( x ) = ∑ n = 1 ∞ n 5 x n = ∑ j = 1 ∞ ∑ n = j ∞ n 4 x n = g(x) = \sum_{n = 1}^{\infty} n^5 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^4 x^{n} =

1 1 − x ∑ j = 1 ∞ j 4 x j + 4 x ( 1 − x ) 2 ∑ j = 1 ∞ j 3 x j + 6 ( x 2 + x ) ( 1 − x ) 3 ∑ j = 1 ∞ j 2 x j + 4 x 3 + 4 x 2 + x ( 1 − x ) 4 ∑ j = 1 ∞ j x j \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^4 x^j + \dfrac{4x}{(1 - x)^2}\sum_{j = 1}^{\infty} j^3 x^j + \dfrac{6(x^2 + x)}{(1 - x)^3}\sum_{j = 1}^{\infty} j^2 x^j + 4\dfrac{x^3 + 4x^2 + x}{(1 - x)^4}\sum_{j = 1}^{\infty} j x^j

+ x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 ∑ j = 1 ∞ x j = + \dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5}\sum_{j = 1}^{\infty} x^j =

1 1 − x ( x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 ) \dfrac{1}{1 - x}(\dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5}) + 4 x ( 1 − x ) 2 ( x 3 + 4 x 2 + x ) ( 1 − x ) 4 \dfrac{4x}{(1 - x)^2}(\dfrac{x^3 + 4x^2 + x)}{(1 - x)^4} + 6 ( x 2 + x ) ( 1 − x ) 3 ( x 2 + x ( 1 − x ) 3 ) + 4 ( x 3 + 4 x 2 + x ) ( 1 − x ) 4 ( x ( 1 − x ) 2 ) + ( x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 ) ( x 1 − x ) \dfrac{6(x^2 + x)}{(1 - x)^3}(\dfrac{x^2 + x}{(1 - x)^3}) + \dfrac{4(x^3 + 4x^2 + x)}{(1 - x)^4}(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5})(\dfrac{x}{1 - x})

= x 5 + 26 x 4 + 66 x 3 + 26 x 2 + x ( 1 − x ) 6 = \dfrac{x^5 + 26x^4 + 66x^3 + 26x^2 + x}{(1 - x)^6}

f ( x ) = ∑ n = 1 ∞ n 6 x n = ∑ j = 1 ∞ ∑ n = j ∞ n 5 x n = f(x) = \sum_{n = 1}^{\infty} n^6 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^5 x^{n} = 1 1 − x ∑ j = 1 ∞ j 5 x j + 5 x ( 1 − x ) 2 ∑ j = 1 ∞ j 4 x j + 10 ( x 2 + x ) ( 1 − x ) 3 ∑ j = 1 ∞ j 3 x j + 10 ( x 3 + 4 x 2 + x ) ( 1 − x ) 4 ∑ j = 1 ∞ j 2 x j \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^5 x^j + \dfrac{5x}{(1 - x)^2}\sum_{j = 1}^{\infty} j^4 x^j + \dfrac{10(x^2 + x)}{(1 - x)^3}\sum_{j = 1}^{\infty} j^3 x^j + \dfrac{10(x^3 + 4x^2 + x)}{(1 - x)^4}\sum_{j = 1}^{\infty} j^2 x^j

+ 5 ( x 4 + 11 x 3 + 11 x 2 + x ) ( 1 − x ) 5 ∑ j = 1 ∞ j x j + + \dfrac{5(x^4 + 11x^3 + 11x^2 + x)}{(1 - x)^5}\sum_{j = 1}^{\infty} j x^j + ( x 5 + 26 x 4 + 66 x 3 + 26 x 2 + x ( 1 − x ) 6 ) ∑ j = 1 ∞ x j = (\dfrac{x^5 + 26x^4 + 66x^3 + 26x^2 + x}{(1 - x)^6})\sum_{j = 1}^{\infty} x^j =

1 1 − x ( x 5 + 26 x 4 + 66 x 3 + 26 x 2 + x ( 1 − x ) 6 ) + \dfrac{1}{1 - x}(\dfrac{x^5 + 26x^4 + 66x^3 + 26x^2 + x}{(1 - x)^6}) + 5 x ( 1 − x ) 2 ( x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 ) + 10 ( x 2 + x ) ( 1 − x ) 3 ( x 3 + 4 x 2 + x ( 1 − x ) 4 ) \dfrac{5x}{(1 - x)^2}(\dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5}) + \dfrac{10(x^2 + x)}{(1 - x)^3}(\dfrac{x^3 + 4x^2 + x}{(1 - x)^4})

+ 10 ( x 3 + 4 x 2 + x ) ( 1 − x ) 4 ( x 2 + x ( 1 − x ) 3 ) + 5 ( x 4 + 11 x 3 + 11 x 2 + x ) ( 1 − x ) 5 ( x ( 1 − x ) 2 ) + ( x 5 + 26 x 4 + 66 x 3 + 26 x 2 + x ( 1 − x ) 6 ) ( x 1 − x ) + \dfrac{10(x^3 + 4x^2 + x)}{(1 - x)^4}(\dfrac{x^2 + x}{(1 - x)^3}) + \dfrac{5(x^4 + 11x^3 + 11x^2 + x)}{(1 - x)^5}(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^5 + 26x^4 + 66x^3 + 26x^2 + x}{(1 - x)^6})(\dfrac{x}{1 - x}) .

= x 6 + 57 x 5 + 302 x 4 + 302 x 3 + 57 x 2 + x ( 1 − x ) 7 = \dfrac{x^6 + 57x^5 + 302x^4 + 302x^3 + 57x^2 + x}{(1 - x)^7} .

Let u = 1 − x ⟹ d u = − d x ⟹ u = 1 -x \implies du = -dx \implies

∫ 0 1 − a g ( x ) d x = ∫ 1 a u 5 − 31 u 4 + 180 u 3 − 390 u 2 + 360 u − 120 u 6 d u = \int_{0}^{1 - a} g(x) dx = \int_{1}^{a} \dfrac{u^5 - 31u^4 + 180u^3 - 390u^2 + 360u - 120}{u^6} du =

∫ 1 a 1 u − 31 u − 2 + 180 u − 3 − 390 u − 4 + 360 u − 5 − 120 u − 6 d u = \int_{1}^{a} \dfrac{1}{u} - 31u^{-2} + 180u^{-3} - 390u^{-4} + 360u^{-5} - 120u^{-6} du =

( ln ⁡ ( u ) + 31 u − 90 u 2 + 130 u 3 − 90 u 4 + 24 u 5 ) ∣ 1 a = − 5 a 5 + 31 a 4 − 90 a 3 + 130 a 2 − 90 a + 24 a 5 + ln ⁡ ( a ) (\ln(u) + \dfrac{31}{u} - \dfrac{90}{u^2} + \dfrac{130}{u^3} - \dfrac{90}{u^4} + \dfrac{24}{u^5})|_{1}^{a} = \dfrac{-5a^5 + 31a^4 - 90a^3 + 130a^2 - 90a + 24}{a^5} + \ln(a)

and,

∫ 0 1 − a f ( x ) d x = ∫ 0 1 − a x 6 + 57 x 5 + 302 x 4 + 302 x 3 + 57 x 2 + x ( 1 − x ) 7 d x = \int_{0}^{1 - a} f(x) dx = \int_{0}^{1 - a} \dfrac{x^6 + 57x^5 + 302x^4 + 302x^3 + 57x^2 + x}{(1 - x)^7} dx = − ∫ 1 a u 6 − 63 u 5 + 602 u 4 − 2100 u 3 + 3360 u 2 − 2520 u + 720 u 7 d u = -\int_{1}^{a} \dfrac{u^6 - 63u^5 + 602u^4 - 2100u^3 + 3360u^2 - 2520u + 720}{u^7}du = − ∫ 1 a 1 u − 63 u − 2 + 602 u − 3 − 2100 u − 4 + 3360 u − 5 − 2520 u − 6 + 720 u − 7 d u = -\int_{1}^{a} \dfrac{1}{u} - 63u^{-2} + 602u^{-3} -2100u^{-4} + 3360u^{-5} - 2520u^{-6} + 720u^{-7} du = − ( ln ⁡ ( u ) + 63 u − 301 u 2 + 700 u 3 − 840 u 4 + 504 u 5 − 120 u 6 ) ∣ 1 a = -(\ln(u) + \dfrac{63}{u} - \dfrac{301}{u^2} + \dfrac{700}{u^3} - \dfrac{840}{u^4} + \dfrac{504}{u^5} - \dfrac{120}{u^6})|_{1}^{a} = − ( ln ⁡ ( a ) + 63 a − 301 a 2 + 700 a 3 − 840 a 4 + 504 a 5 − 120 a 6 − 6 = -(\ln(a) + \dfrac{63}{a} - \dfrac{301}{a^2} + \dfrac{700}{a^3} - \dfrac{840}{a^4} + \dfrac{504}{a^5} - \dfrac{120}{a^6} - 6 = 6 a 6 − 63 a 5 + 301 a 4 − 700 a 3 + 840 a 2 − 504 a + 120 a 6 − ln ⁡ ( a ) \dfrac{6a^6 - 63a^5 + 301a^4 - 700a^3 + 840a^2 - 504a + 120}{a^6} - \ln(a)

⟹ \implies

∫ 0 1 − a f ( x ) − g ( x ) d x = 11 a 6 − 94 a 5 + 391 a 4 − 830 a 3 + 930 a 2 − 528 a + 120 a 6 − 2 ln ⁡ ( a ) = \int_{0}^{1 - a} f(x) - g(x) dx = \dfrac{11a^6 - 94a^5 + 391a^4 - 830a^3 + 930a^2 - 528a + 120}{a^6} - 2\ln(a) = 11 a 6 − 94 a 5 − 9609 a 4 + 3170 a 3 + 330 a 2 − 488 a + 119 a 6 − 2 ln ⁡ ( a ) \dfrac{11a^6 - 94a^5 - 9609a^4 + 3170a^3 + 330a^2 - 488a + 119}{a^6} - 2\ln(a)

⟹ 10000 a 4 − 4000 a 3 + 600 a 2 − 40 a + 1 a 6 = 0 ⟹ \implies \dfrac{10000a^4 - 4000a^3 + 600a^2 - 40a + 1}{a^6} = 0 \implies ( 10 a − 1 ) 4 = 0 ⟹ a = 1 10 ⟹ ∫ 0 9 10 f ( x ) − g ( x ) d x = 75708175.605170 (10a - 1)^4 = 0 \implies a = \boxed{\dfrac{1}{10}} \implies \int_{0}^{\frac{9}{10}} f(x) - g(x) dx = \boxed{75708175.605170} to six decimal places.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

If you want the final equation to simplify to 10000 a 4 − 4000 a 3 + 600 a 2 − 40 a + 1 a 6 = 0 , \frac{10000a^4 - 4000a^3 + 600a^2 - 40a + 1}{a^6} = 0, the coefficient of a 4 a^4 in 11 a 6 − 94 a 5 + 9609 a 4 + 3170 a 3 + 330 a 2 − 488 a + 119 a 6 − 2 log ⁡ a \dfrac{11a^6 - 94a^5 + 9609a^4 + 3170a^3 + 330a^2 - 488a + 119}{a^6} - 2 \log a should be − 9609 -9609 .

Jon Haussmann - 3 years ago

Log in to reply

I had -9609 in my notes. It was a typing error. Thanks! I changed it. I appreciate that someone notified me. Sorry about that.

Below is a slightly altered version I posted of this problem:

Let ∣ x ∣ < 1 |x| < 1 and ( 0 < a < 1 ) (0 < a < 1) .

If f ( x ) = ∑ n = 1 ∞ ( − 1 ) n + 1 n 6 x n f(x) = \sum_{n = 1}^{\infty} (-1)^{n + 1} n^6 x^n and g ( x ) = ∑ n = 1 ∞ ( − 1 ) n + 1 n 5 x n g(x) = \sum_{n = 1}^{\infty} (-1)^{n + 1} n^5 x^n and the region R R bounded by f f and g g on [ a − 1 , 0 ] [a - 1,0] has area A R = 11 a 6 − 94 a 5 − 9609 a 4 + 3170 a 3 + 330 a 2 − 488 a + 119 a 6 − 2 ln ⁡ ( a ) A_{R} = \dfrac{11a^6 - 94a^5 - 9609a^4 + 3170a^3 + 330a^2 - 488a + 119}{a^6} - 2\ln(a) , find the real value of a a and express the answer as the value of A R A_{R} to six decimal places.

Rocco Dalto - 3 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...