Macluarine expansion, get the coefficient of x^2

[YehiaMedhat]

The question is: The coeffiecient of $x^2$ in Maclurine expansion of $cosh(\sqrt{3x})$ is ...
So, trying to solve the question I have to diffrentiate, don't I? so, the first deriviative is $\frac {3sinh(\sqrt{3x})}{2\sqrt{3x}}$, and as it's Maclurine expansion how would I insert the value of x=0 to the deriviative? Is there's a mistake I have done?

 

[topsquark]

The question is: The coeffiecient of $x^2$ in Maclurine expansion of $cosh(\sqrt{3x})$ is ...
So, trying to solve the question I have to diffrentiate, don't I? so, the first deriviative is $\frac {3sinh(\sqrt{3x})}{2\sqrt{3x}}$, and as it's Maclurine expansion how would I insert the value of x=0 to the deriviative? Is there's a mistake I have done?

You didn't find the derivative, you found the integral!

What is the derivative of $cosh(\sqrt{3x})$?

-Dan

 

[YehiaMedhat]

Aren't these the deriviatives, then by the chain rule you deriviate the inside function, is it wrong?

 

[topsquark]

View attachment 34779Aren't these the deriviatives, then by the chain rule you deriviate the inside function, is it wrong?

Ew! Yes, you had it right. My mistake.

All I can suggest is to take $\displaystyle \lim_{x \to 0} \dfrac{d}{dx} cosh(\sqrt{3x})$. It works, but I'm not sure why this is giving a problem.

-Dan

 

[YehiaMedhat]

Ew! Yes, you had it right. My mistake.

All I can suggest is to take $\displaystyle \lim_{x \to 0} \dfrac{d}{dx} cosh(\sqrt{3x})$. It works, but I'm not sure why this is giving a problem.

-Dan

Ok, don't worry if I tell little about my amazingly silly mistakes it'll blow your mind??
But wondering about how could I approach the answer. Should I find the expansion for standard cosh(), and then substitute for $\sqrt{3x}$, but, wait a sec, actually this gives the answer b, but it's not correct according to the model answer attached.

 

[BigBeachBanana]

Ok, don't worry if I tell little about my amazingly silly mistakes it'll blow your mind??
But wondering about how could I approach the answer. Should I find the expansion for standard cosh(), and then substitute for $\sqrt{3x}$, but, wait a sec, actually this gives the answer b, but it's not correct according to the model answer attached.

Let $z = \sqrt{3x}$.
Find the expansion for $\cosh(z)$ and plug in $\sqrt{3x}$ afterwards.
Please share your work so we can see where it went wrong (or right).

 

[Dr.Peterson]

But wondering about how could I approach the answer. Should I find the expansion for standard cosh(), and then substitute for $\sqrt{3x}$, but, wait a sec, actually this gives the answer b, but it's not correct according to the model answer attached.

But you didn't show us the choices, or the answer you are comparing to. Might it be wrong? We don't know yet, because you're withholding information.

Please quote the entire problem, and your work, and, when available, the answer you are given; it saves a lot of trouble for the people trying to help!

 

[YehiaMedhat]

But you didn't show us the choices, or the answer you are comparing to. Might it be wrong? We don't know yet, because you're withholding information.

Please quote the entire problem, and your work, and, when available, the answer you are given; it saves a lot of trouble for the people trying to help!

I did say that i make amazingly silly ones. The answer came after plugging in the $\sqrt{3x}$ was $\frac {3}{2!}$.
The supposed correct answer is answer $\frac{9}{2!}$

 

[Dr.Peterson]

I did say that i make amazingly silly ones. The answer came after plugging in the $\sqrt{3x}$ was $\frac {3}{2!}$.
The supposed correct answer is answer $\frac{9}{2!}$

Please show how you got that answer. What is the series you plugged it into? What did that produce? Which coefficient did you look at? And did they show anything other than the answer itself? (Both are wrong, unless I've made a mistake myself.)

And, again, it may still be important to see the list of choices, of which yours was (b). For that matter, can you show us an image of the entire actual problem?

 

[BigBeachBanana]

I did say that i make amazingly silly ones. The answer came after plugging in the $\sqrt{3x}$ was $\frac {3}{2!}$.
The supposed correct answer is answer $\frac{9}{2!}$

I agree with @Dr.Peterson. The answer is not $\frac{9}{2!}$.

 

[YehiaMedhat]

I got it, sorry!!

 

[Dr.Peterson]

I got it, sorry!!

And the other things I requested?

Which answer do you now get, and how? And when you mentioned a "model answer", did that mean that they show work somewhere, or just what they say here?

I see that they say the correct answer is (c), which is what I get. Do you agree yet?

 

[YehiaMedhat]

I agree, ?
I mostly cancel show my full work because the notation of math Jax is still too long to write a few steps, especially on a touch keyboard, so I try to summarize it verbally, nevertheless I can use it.

 

[Dr.Peterson]

I agree, ?

That's good.

I mostly cancel show my full work because the notation of math Jax is still too long to write a few steps, especially on a touch keyboard, so I try to summarize it verbally, nevertheless I can use it.

I hope you recognize that you can use an image of handwriting to show your work, or just type using basic text notation rather than Latex.

 

[YehiaMedhat]

Ok, thanks @Dr.Peterson