What is the taylor series for #x(e^(2x))#?

Answer 1

#x+2x^2+2^2/(2!)x^3+2^3/(3!)x^4+2^4/(4!)x^5+\cdots#
#=x+2x^2+2x^3+4/3 x^4+2/3 x^5+4/15 x^6+4/45 x^7+8/315 x^8+\cdots#

This converges for all #x# and equals #xe^(2x)# for all #x#.

You can first use the well-known series for #e^(x)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+\cdots#
Replace #x# with #2x#:
#e^(2x)=1+2x+(2x)^2/(2!)+(2x)^3/(3!)+(2x)^4/(4!)+\cdots#
Then multiply everything by #x# and simplify:
#xe^(2x)=x+2x^2+2^2/(2!)x^3+2^3/(3!)x^4+2^4/(4!)x^5+\cdots#
#=x+2x^2+2x^3+4/3 x^4+16/24 x^5+32/120 x^6+64/720 x^7+128/5040 x^8+\cdots#
#=x+2x^2+2x^3+4/3 x^4+2/3 x^5+4/15 x^6+4/45 x^7+8/315 x^8+\cdots#
You can also do this by using the formula #f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+(f''''(0))/(4!)x^4+\cdots#, where #f(x)=xe^(2x)#.
Calculating some derivatives here gives #f'(x)=e^(2x)+2xe^(2x)#, #f''(x)=2e^(2x)+2e^(2x)+4xe^(2x)=4e^(2x)+4xe^(2x)#, #f'''(x)=8e^(2x)+4e^(2x)+8xe^(2x)=12e^(2x)+8xe^(2x)#, #f''''(x)=32e^(2x)+16xe^(2x)#, etc...
Hence, #f(0)=0#, #f'(0)=1#, #f''(0)=4#, #f'''(0)=12#, #f''''(0)=32#, etc..., resulting in
#x+4/(2!)x^2+12/(3!)x^3+32/(4!)x^4+\cdots#
#=x+2x^2+2x^3+4/3 x^4+\cdots#
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Answer 2

The Taylor series for ( x \cdot e^{2x} ) can be found by applying the Taylor series expansion for ( e^{2x} ) and then multiplying it term by term by the Taylor series expansion for ( x ).

The Taylor series expansion for ( e^{2x} ) is:

[ e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} ]

The Taylor series expansion for ( x ) is simply:

[ x = x ]

Multiplying these two series term by term gives:

[ x \cdot e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n \cdot x}{n!} ]

This can be further simplified to:

[ x \cdot e^{2x} = \sum_{n=0}^{\infty} \frac{2^n \cdot x^{n+1}}{n!} ]

So, the Taylor series for ( x \cdot e^{2x} ) is:

[ \boxed{\sum_{n=0}^{\infty} \frac{2^n \cdot x^{n+1}}{n!}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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