Doubt with time in the exponent (jerk, Jounce...)

Question

I'm doing something wrong, but I don't know what it is.

Suppose I have a position equation $x = 3^{t/2}$ so in $t=0$ is in $x=1$ and in $t=2$ is in $x=3$.

Because the first derivative is $v=(1/2)(\log(3))3^{t/2}$ and the second $a=(1/4)(\log(3)^2)3^{t/2}$ and the third $J=(1/8)(\log(3)^3)3^{t/2}$, and so on, I tried to calculate the position in $t=2$ using this equation:

${\displaystyle {\vec {x}}={\vec {x}}_{0}+{\vec {v}}_{0}t+{\frac {1}{2}}{\vec {a}}_{0}t^{2}+{\frac {1}{6}}{\vec {\jmath }}_{0}t^{3}+{\frac {1}{24}}{\vec {s}}t^{4}\cdots}$

An infinite sum. I have an initial velocity $v=log(3)/2$ and initial position $x=1$ in $t=0$, so

${\displaystyle {\vec {x}}= 1+ (log(3)/2)*t + {\frac {1}{2}}{\vec {a}}_{0}t^{2}+{\frac {1}{6}}{\vec {\jmath }}_{0}t^{3}+{\frac {1}{24}}{\vec {s}}t^{4}...}$

Substituting the derivatives:

$$\displaystyle {x}=1 + (log(3)/2)*t + \sum_{n=2}^{\infty} \frac {\log(3)^n \cdot3^{t/2}}{n! 2^n}t^n $$

For $t=2$ it is supposed we must obtain $x=3$, so I substitute

$$\displaystyle {{x}}=1 + log(3) + \sum_{n=2}^{\infty} \frac {\log(3)^n \cdot3}{n!} $$

(Note that $1/2^n$ disappear with $t^n$ )

and Wolfram Alpha says this sum is $x=1+ log(3)+6-3\log(3)$ that is greater than $3$

What is wrong, here?

Answer 1

When you say you have no initial velocity, you are incorrect.

You calculated your velocity to be $v=\log (3)/2 \cdot 3^{t/2}$, so at $t=0$ your velocity is $\log (3)/2$.

Also, in your final substitution, you're missing a factor of $\frac{1}{2^n}$.

Answer 2

I finally find the right answer that is not, of course, a factor is missing, but it is true, that the initial velocity was missing and I edited the question.

Nobody in Physics or Mathematics Stack Exchange has been able to solve it, but put a negative vote...wonderful, although the right answer is very interesting.

The error was because when you do not have a constant acceleration, you have to put the initial one, and if you do not have a constant Jerk, yo must put the initial one...and so on.. and so on... and the initial ones with $t=0$ are

${\displaystyle {\vec {x}}_{0}=1}$

${\displaystyle {\vec {v}}_{0}=\frac {log(3)}{2}}$

${\displaystyle {\vec {a}}_{0}=\frac {log(3)^2}{4}}$

${\displaystyle {\vec {j}}_{0}=\frac {log(3)^3}{8}}$

${\displaystyle {\vec {s}}_{0}=\frac {log(3)^4}{16}}$

The equation:

$${\displaystyle {\vec {x}}={\vec {x}}_{0}+{\vec {v}}_{0}t+{\frac {1}{2}}{\vec {a}}_{0}t^{2}+{\frac {1}{6}}{\vec {\jmath }}_{0}t^{3}+{\frac {1}{24}}{\vec {s}}_{0}t^{4}\cdots}$$

substituing

$\displaystyle {x}=1 + \frac {log(3)}{2}*t + \sum_{n=2}^{\infty} \frac {\log(3)^n }{n! 2^n}t^n$

and for $t=2$ then

$\displaystyle {x}=1 + log(3) +2 - log(3) = 3$ As we expected.