For the lovers of maths!

Calculus Level pending

Consider,

S = 2 [ 1 + x + ∑ k = 1 ∞ ( x ( 4 k ) ( 4 k ) ! + x ( 4 k + 1 ) ( 4 k + 1 ) ! ) ] S = 2 \left [1+x+\sum_{k=1}^{\infty}{ \left(\dfrac{x^{(4k)}}{(4k)!}+\dfrac{x^{(4k+1)}}{(4k+1)!} \right)} \right ]

If S = 0 S=0 , then find all possible numbers of solution for − 6 < x < 0 -6 < x < 0 .


The answer is 2.

Akash Shukla by Akash Shukla , 26, India

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1 solution

Jeko Lonardo
May 29, 2016

First, expand the equation using Taylor series

S= 2(1+x+ x 4 4 ! \frac {x^4}{4!} + x 5 5 ! \frac {x^{5}}{5!} + x 8 8 ! \frac {x^{8}}{8!} + x 9 9 ! \frac {x^{9}}{9!} +...)+( x 2 2 ! \frac {x^{2}}{2!} + x 3 3 ! \frac {x^{3}}{3!} + x 6 6 ! \frac {x^{6}}{6!} + x 7 7 ! \frac {x^{7}}{7!} +...)-( x 2 2 ! \frac {x^{2}}{2!} + x 3 3 ! \frac {x^{3}}{3!} + x 6 6 ! \frac {x^{6}}{6!} + x 7 7 ! \frac {x^{7}}{7!} +...)

S= ( x 0 0 ! \frac {x^0}{0!} + x 1 1 ! \frac {x^1}{1!} + x 4 4 ! \frac {x^4}{4!} + x 5 5 ! \frac {x^{5}}{5!} + x 8 8 ! \frac {x^{8}}{8!} + x 9 9 ! \frac {x^{9}}{9!} +...)+( x 2 2 ! \frac {x^{2}}{2!} + x 3 3 ! \frac {x^{3}}{3!} + x 6 6 ! \frac {x^{6}}{6!} + x 7 7 ! \frac {x^{7}}{7!} +...)+( x 0 0 ! \frac {x^0}{0!} + x 1 1 ! \frac {x^1}{1!} + x 4 4 ! \frac {x^4}{4!} + x 5 5 ! \frac {x^{5}}{5!} + x 8 8 ! \frac {x^{8}}{8!} + x 9 9 ! \frac {x^{9}}{9!} +...)-( x 2 2 ! \frac {x^{2}}{2!} + x 3 3 ! \frac {x^{3}}{3!} + x 6 6 ! \frac {x^{6}}{6!} + x 7 7 ! \frac {x^{7}}{7!} +...)

S= ( x 0 0 ! \frac {x^0}{0!} + x 1 1 ! \frac {x^1}{1!} + x 2 2 ! \frac {x^2}{2!} + x 3 3 ! \frac {x^{3}}{3!} + x 4 4 ! \frac {x^{4}}{4!} + x 5 5 ! \frac {x^{5}}{5!} + x 6 6 ! \frac {x^{6}}{6!} + x 7 7 ! \frac {x^{7}}{7!} + x 8 8 ! \frac {x^{8}}{8!} + x 9 9 ! \frac {x^{9}}{9!} +...)+( x 1 1 ! \frac {x^1}{1!} - x 3 3 ! \frac {x^3}{3!} + x 5 5 ! \frac {x^5}{5!} - x 7 7 ! \frac {x^{7}}{7!} + x 9 9 ! \frac {x^{9}}{9!} -...)+( x 0 0 ! \frac {x^{0}}{0!} - x 2 2 ! \frac {x^{2}}{2!} + x 4 4 ! \frac {x^{4}}{4!} - x 6 6 ! \frac {x^{6}}{6!} + x 8 8 ! \frac {x^{8}}{8!} -...)

S= e x e^x +sin(x)+cos(x)

Input x= -2 π \pi → \rightarrow S= 1.0018...

Input x= - π \pi → \rightarrow S=-0.956...

Input x= 0 → \rightarrow S= 2

Then, for S=0, there are 2 values for x, one between -2 π \pi and - π \pi and the other between - π \pi and 0

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution, but in last steps, you can find the range of 'x', just for additional maths.

e x + s i n ⁡ x + c o s ⁡ x = 0 e^x+sin⁡x+cos⁡x=0 , e x = − √ 2 [ s i n ⁡ ( π 4 + x ) ] e^x=-√2[sin⁡(\frac{π}{4}+x)]

e x e^x can never be negative or zero ,so s i n ⁡ ( π 4 + x ) sin⁡(\frac{π}{4}+x) must be negative.

∴ − π < π 4 ⁡ + x < 0 -π< \frac{π}{4}⁡+x<0 ,

∴ − 5 π 4 < x < − π 4 \frac{-5π}{4}<x<\frac{-π}{4} ,

At x= − 5 π 4 \frac{-5π}{4} ,L.H.S= e − 5 π 4 e^\frac{-5π}{4} = 0.xxx and R.H.S = 0. So. L.H.S >R.H.S.

At x= − 3 π 4 \frac{-3π}{4} ,L.H.S= e − 3 π 4 > e − 5 π 4 e^\frac{-3π}{4}>e^\frac{-5π}{4} and R.H.S = 1.41 > L.H.S. So their graph will intersect. Now again, at x = − π 4 x=\frac{-π}{4} , L.H.S = e − π 4 > e − 3 π 4 e^\frac{-π}{4}>e^\frac{-3π}{4} and R.H.S = 0 , so their graph will again intersect. Hence we get two solutions.

Akash Shukla - 5 years ago

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