How do you find the power series for #f'(x)# and #int f(t)dt# from [0,x] given the function #f(x)=Sigma 10^nx^n# from #n=[0,oo)#?

Answer 1

# f'(x) = sum_(n=0)^oo \ n10^n x^(n-1) #

# int_0^x \ f(t) \ dt=sum_(n=0)^oo \ 10^n / (n+1) \ x^(n+1) #

We have:

# f(x) = sum_(n=0)^oo \ 10^nx^n #

And so:

# f'(x) = d/dx sum_(n=0)^oo \ 10^nx^n # # " " = sum_(n=0)^oo \ d/dx10^nx^n # # " " = sum_(n=0)^oo \ 10^nd/dxx^n # # " " = sum_(n=0)^oo \ 10^n nx^(n-1) # # " " = sum_(n=0)^oo \ n10^n x^(n-1) #

And:

# int_0^x \ f(t) \ dt= int_0^x \ sum_(n=0)^oo \ 10^nt^n \ dt# # " "= sum_(n=0)^oo \ int_0^x \ 10^nt^n \ dt# # " "= sum_(n=0)^oo \ 10^n \ int_0^x \ t^n \ dt# # " "= sum_(n=0)^oo \ 10^n \ [ t^(n+1)/(n+1) ]_0^x #
# " "= sum_(n=0)^oo \ 10^n \ { x^(n+1)/(n+1) - 0 } # # " "= sum_(n=0)^oo \ 10^n \ x^(n+1)/(n+1) # # " "= sum_(n=0)^oo \ 10^n / (n+1) \ x^(n+1) #
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Answer 2

To find the power series for (f'(x)), differentiate the given power series term by term. To find the power series for (\int f(t) , dt) from ([0,x]), integrate the given power series term by term with respect to (t) from (0) to (x).

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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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