Examples of Calculus

For mellin transform value we are going to follow the below steps:
Step 1: In the first step we use Mellin transform definition for given value,
First we write the given requirement about mellin value,
^(s) = _n=1∑^+∞ λ_kµ_k^-s,
Here we have given polylogarithm function
 ^(s) = _n=1∑^+∞ d(k)/k^-s ξ²(s),
Where ‘w’ and ‘k’ are equal values which belong to natural numbers.
And here λ_k =d(k), µ_k = k   and g(x) = e^-x all values are use in Mellin definition .
 D*(s) = Г(s)ξ²(s) , where s is belongs to (1,+∞),
Step 2: In the second step we are going to explain the Mellin series.
Now we,
 D*(s)^[_n=0∑^+∞ 1/(s-1)² + y/(s-1)] + [1/4s]-_n=0∑^+∞ ξ² (-2k - 1)/(2k +1)! 1/s+ 2k +1 where s is belongs to (-∞ to +∞),
Step 3: Now we separate the resulting function in Mellin form.
D(x) 1/x (-log x + y) + ¼ - _k=0∑^+∞ ξ² ((-2k -1)/(2k +1)!)x^2k+1,
At last we get Mellin transform value.

For Mellin transform value we are going to follow the below steps:
Step 1: In the first step we use Mellin transform definition for given value.
First we write the given requirement about mellin value,
L_w(x) = _n=1∑^+∞ e^-nx/n^w,
Here we have given poly logarithm function,
 Li_w (z) = _n=1∑^+∞ zⁿ n^-w,
Where ‘w’ and ‘k’ is equal value which are belongs to natural number.
And here λ_n = 1/n^w  , µ_k = n   and g(x) = e^-x all values are use in Mellin definition .
^(s) = ξ(s + k), L_k*(s) = ξ(s + k)Г(s) , where s is belongs to (1, +∞),
Step 2: In the second step we are going to explain the Mellin series.
Now we,
 L_k*(s)^ _n=0∑^+∞ (-1)ⁿ ξ(k-n)/n! 1/(s+n) + (-1)^k-1/(k-1)![1/(s+k -1)² + H_k-1/s+k-1] where s is belongs to (-∞ to +∞)
Step 3: Now we separate the resulting function
L_k(x) = (-1)^k-1/(k-1)! x^k-1 [-log x + H_k-1]+ _n=0∑^+∞ (-1)ⁿ ξ(k-n)/n! xⁿ,
L_1/2 (x) = √(л/x) + _n=0∑^+∞ (-1)ⁿ ξ(1/2 - n)/n! xⁿ .
(Here we take k= ½ then after we are resulting the value at k=1/2) by these expression we got L_1/2 value.
At last we get Mellin transform value.

Here we are going to follow the below steps.
Step 1: In the first step we use to given function in mellin form.
l(x) = log Г(x+1) – λx = _n=1∑^+∞ [x/n – log (1+ x/n)] where s is belongs to -2 to +∞,  
Here by this expression we got the  λ_n = 1 , µ_n = 1/n ,
 g(x)= x-log(1+x),
Step 2: In this step we write the expression in Mellin transform from.
^(s) = _n=1∑^+∞ λ_k µ_k^-s = _n=1∑^+∞ n^s = ξ(-s),
l*(s) = ^(s)g*(s) = -ξ(-s) л/s sin лs,
In this way we get Mellin transform value.

For solving the mellin transform function we are going to follow the below steps:
Step 1: In the first step we explain function by using given value.
^(s) = _k=1∑^+∞ λ_k µ_k^-s = _k=1∑^+∞ k ^-1+s  =ξ(1-s),
Step 2: In this step we are going to use the Mellin transform properties.
H*(s) = ^(s) g*(s) = -ξ(1-s) л/sin лs  s is belong to -1 to 0  [after using Mellin transform we got mellin transfrom].
ξ(s) = 1/s-1 +λ +….., ξ(1-s) = -1/s +λ (Here we got ξ(s) value by mellin expression),
h*(s)^ [1/s² – λ/s] – _k=1∑^+∞ (-1)^k ξ(1-k)/(s-k)   where s is belongs to -1 to +∞,
H_n  = log n + у + _k≥1∑ (-1)^k B_k / k 1/(n^k),
= log n +  у + 1/2n -1/2n +1/120 n⁴ ……,
After solving we got H_n series in Mellin transform series.

For solving mellin transform we need to follow the below steps:
Step 1: In the first step we explain given exponential function
f(x) = e^-x = 1- x+ x²/2! –x³/3! + …….[After explanation exponential function],
This expression can be write like = _x→0∑^m_j=0 (-1)^j / j! x^j + o(x^m+1),
Step 2: In this step we used the Mellin transform definition,
f*(s) = Г(s) ,  where x is belong to all integer limit (0 , +∞),
f*(s) is meromorphically containable to (-M -1 , +∞),
f*(s) ^ _x→0∑^m_j (-1)^j/j! 1/(s+j) where s is belong limit (-M , -1 , +∞)
Step 3: In this step we finalize the resulting expression
Finally we get,
Г(s) ^_j=0∑^∞ (-1)^j /j! x 1/s+j  , where s is belong to all integer value.
In this way we got final value of Mellin transform is  
Г(s) ^_j=0∑^∞ (-1)^j /j! x 1/s+j. 

For solving this problem we need to follow the below steps:
Step 1: In the first step, we write the definition of Mellin transform function,
M [∑_k λ_k f(µ_kx); s] = (∑_k λ_k µ^-s)f*(s),
Step 2: In this step we find the value of λ_k , µ_k , f(x) = e^-x,
λ_k  =1, µ_k  =k , f(x) = e^-x (here we got the value of λ_k, µ_k by mellin transform properties.
g*(s) = (1/1^s + 1/2^s + 1/3^s ….) [This Mellin transform expression],
M[e^-x ;s]  
Now we get final value of mellin transform value which is g*(s) = (1/1^s + 1/2^s + 1/3^s …….).

For solving differential equation first we will write the separate form.
First we write the given differential equation then we integrate equation.
 ∫2x dx = ∫(y² +1) dy,
First we integrate ‘2x’ with respect to ‘x’ and then ‘y² +1’ is integrated with respect to ‘y’
[Integration of y² is 1/3 y³ and Integration of ‘2’x is 2x²/2]
         2x²/2= 1/3 y³ + y +c,
           x² = y³/3 +y +c  [where c is integration constant ],
In this way we got the separate differential equation, which is x² = y³/3 +y +c.

First we will calculate velocity vector.
To calculate the velocity vector we have to differentiate the position vector x(t) and y(t), the process of calculation of velocity vector is shown below:
Therefore velocity vector v(t)= d(t⁵)/dt, d(t³)/dt
=> v(t) = d(t⁵)/dt, d(t³)/dt = (5t⁴), (3t²),
Therefore the velocity vector is equals to v(t) = (5t⁴),(3t²),  
Now we will calculate the acceleration vector.
To calculate the acceleration vector we have to differentiate the velocity vector, therefore we get- 
a(t) = d(5t⁴)/dt, d(3t²)/dt = 20t³, 6t
Therefore the acceleration vector is equals to a (t) = 20t³, 6t.

First we will calculate velocity vector.
 
To calculate the velocity vector we have to differentiate the position vector x(t) and y(t), the process of calculation of velocity vector is shown below-
 
Therefore velocity vector v(t) = d(t³ + 1) / dt, d(t²)/dt
 
=> v(t) = d(t³ + 1)/dt, d(t²)/dt = (3t²), (2t)
 
Therefore the velocity vector is equals to v(t) = (3t²), (2t)  
 
Now we will calculate the acceleration vector.
 
To calculate the acceleration vector we have to differentiate the velocity vector, therefore we get-
 
a(t) =  d(3t²)/dt, d(2t)/dt = 6t, 2
 
Therefore the acceleration vector is equal to a(t) = 6t, 2.

First we will calculate velocity vector.
 
To calculate the velocity vector we have to differentiate the position vector x(t) and y(t), the process of calculation of velocity vector is shown below-
 
Therefore velocity vector v(t) = d(t⁴ + 3 t)/dt, d(t³ – t²)/dt
 
=> v(t)= d(t⁴ + 3t)/dt, d(t³ – t²)/dt = (4t³ + 3) , (3t² + t)
 
When t = 2 then velocity vector will be equals to-
 
=> v (2) = (4(2) 3 + 3),(3(2)²+ 2) = 35, 14
 
Therefore the velocity vector is equals to 35, 14 when t = 2.
 
Now we will calculate the acceleration vector.
 
To calculate the acceleration vector we have to differentiate the velocity vector, therefore we get-
 
a(t) =  d(4t³ + 3)/dt, d(3t² + t)/dt = 12t², 6t + 1
 
When t = 3 then acceleration vector a(t) will be equals to:
=> a(3) = 108, 19
 
Therefore the acceleration vector is equals to 108, 19 when t = 3.

First we will calculate the velocity vector.
 
To calculate the velocity vector we will need to differentiate the position vectors, the process of calculation of velocity vector is shown below-
 
velocity vector v(t) = r'(t) = [d (4 sin (t)/dt, d(9 cos t)/dt]
 
=> [d (4sin(t)/dt, d(9cost)/dt] = [4 cos (t), - 9sin(t)]
 
Therefore velocity vector for our problem will be equals to v(t) = [ 4 cos ( t ) , - 9 sin (t)]
 
Now we will calculate the acceleration vector.
 
Acceleration vector can be calculated by differentiating the velocity vector v(t), the process is shown below-
 
a(t) = r'' (t) = [d (4 cos(t)) /dt, d(- 9 sin(t))/dt]
 
=>  a(t) = r'' (t) = [d (4 cos(t))/dt, d(- 9sin(t))/dt] = [- 4 sin(t), - 9cos(t)].
 
Therefore acceleration vector  a(t) is equals to  [- 4 sin (t), -9 cos(t)].

Velocity vector can be calculated by differentiating the position vector.
 
 
Therefore velocity vector will be equals to v(t) = [d(sin (3t))/dt,
 d(cos (5t))/dt].
 
=> v(t) = [3 cos(3t), -5sin(5t)],
 
Therefore velocity vector is equals to v(t) = [3cos(3t), -5sin(5t)].
 
Now we will calculate acceleration vector.
 
Acceleration vector can be calculated by differentiating the velocity vector  or double differentiating the position vector, above process is shown below-
 
Acceleration, a(t) = [d(3 cos(3t))/dt, d(-5 sin(5t))/dt = [-9 sin (3t), -25cos(5t)].
 
Therefore acceleration vector of the moving body is equals to a(t) = [- 9sin(3t), -5cos(5t)].

To calculate the velocity vector we will differentiate the given coordinates, therefore we get-
 
[d(sin t)/dt, d(t²/2)/dt], therefore we will get (cos t, t).
 
Therefore velocity vector v(t) = (cos t, t)
 
Now we will find the acceleration vector.
 
To find acceleration vector we will need to differentiate the velocity vector, therefore we get-
 
[d(cos t)/dt, d(t)/dt] = (–sin t, 1)
 
Therefore acceleration vector will be equals to a(t) = (-sint , 1).

Velocity vector can be calculated as-
 
=> v(t) = (dx/dt, dy/dt) = [d(t³ + 4t²)/dt, d(t⁴ – t³)] = [3t² + 8t, 4t³ – 3t²].
 
Now velocity vector at t = 1 is given by-
 
v(1) = (11, 1)
 
Now finding acceleration vector-
 
a(t) = (d²x/dt, d²y/dt) = [d(3t² + 8t)/dt, d(4t³ – 3t²)] = (6t + 8, 12t² – 6t)
 
When t = 2 then acceleration vector will be-
 
a(2) = (20, 36).

Here our parametric equations will be x = t and y = t³
 ,
Now, velocity vector can be calculated as v(t) = r ' ( t ) = i + 3t² j + 3k,
 
and Acceleration vector can be calculated as a(t) = r'' ( t ) = 6t j.
 
Now when t = 1, then velocity vector is given as-
 
v(t) = r' (1) = i + 3j + 3k,
 
Acceleration vector curves can be given as-
 
a(1) = r''(1) = 6 j.
 
Hence velocity vector is equals to (i + 3j + 3k) and acceleration vector is equals to 6 j.